PH752:

Superconductivity and Magnetism

Lecture Notes

Dr. Jorge Quintanilla

SEPnet and Hubbard Theory Consortium, University of Kent

and STFC Rutherford Appleton Laboratory

i

Lecture I. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

1 About this course

We will have 28 lectures, roughly split as 14 on Magnetism and 14 on Superconductivity.

I will make handouts available after each lecture. The handouts can be found at

http://blogs.kent.ac.uk/strongcorrelations/teaching/superconductivity-and-magnetism/ Be sure to check thatpage regularly.

Check also the reading list available online. There are two recommended texts that you should be veryfamiliar with by the end of this course:

• Stephen Blundell, Magnetism in Condensed Matter (OUP 2001).

• James F. Annett, Superconductivity, Superuids and Condensates (OUP 2004).

References to these two books will be abbreviated SB and JFA, respectively, in these lecture notes. Anexpanded reding list with additional references will be made available as the course progresses.

There will be 6 problem sheets, with a few practice problems each. The problem sheets will be handed outor made available online 2 weeks before they are due. Every problem sheet will have an assesed component.There will also be two class tests: one on Magnetism on week 14 (one week after the Winter break) and oneon Superconductivity on week 19. Finally, in the third term we will have an exam. The assesment patterncan be found on the web.

In addition to the lectures, we will have a number of workshops: two just before the rst class tests, twojust before the second one, and a number of optional ones after the lectures and class tests are over. Inprinciple these have been scheduled in the latter part of the second term but we will agree whatever is mostconvenient for everyone nearer the date.

My room is Ingram 230 however I do spend a lot of time doing research the ISIS Facility, STFC RutherfordAppleton Laboratory, Harwell Science Campus, Didcot, Oxfordshire, OX11 0QX. The best form of contactin any case is email: j.quintanilla@kent.ac.uk. You may also want to try Skype: j.quintanilla.

2 What is special about magnetism and superconductivity?

2.1 History

Magnetism and superconductivity are two of the most fascinating phenomena known to humankind.

Magnetism, as manifested in the ability of some materials to attract iron and each other and to sense thedirection of the magnetic North, has been known for millenia. The earliest known magnetic artifact datesfrom the 4th centruy BC. It is a magnetic compass from China:1

1 National High Magnetic Field Laboratory, Early Chinese Compass, http://www.magnet.fsu.edu/education/tutorials/museum/chinesecompass.html(accessed 29 July 2011).

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(This was not used to navigate, but to nd the spiritually most desirable alignment for new houses!) Eventhe word magnetism itself is very old. It comes from the Greek magnes lithos, the Magnesian stone,referring to a geographical region which was known to be rich in magnetite.2

The magnetic compass uses ferromagnetism. Other magnetic phenomena include paramagnetism and anti-ferromagnetism. Some of them have been discovered recently - in fact new forms of magnetism are beingdiscovered even nowadays.

In contrast to magnetism, Superconductivity is a much less ancient subject. Kammerlingh Onnes foundsuperconductivity serendipitously in 1911 (this year is the 100th anniversary!) when he was studying theelectrical properties of metals at very low temperatures. He found that the resistivity of Mercury (Hg)completely vanished when it was cooled below a critical value of about 4.2K. Here is an old photo of thegentleman alongside his historic plot of resistivity vs temperature:3

2.2 Applications

Both magnetism and superconductivity have important applications.

Magnetism is ubiquitous in our technology. Sometimes the applications are quite humble, e.g. we use it tokeep doors closed. More important is its application in magnetic storage. Computer hard disks store bitsin the form of microscopic magnetisation domains on a ferromagnetic surface. The information is then readusing a device that converts the magnetic eld into an electric signal (a magnetic read head).

The read heads improved considerably with the discovery of giant magnetoresistance (GMR). In GMR, anelectron current is passed through a stack of ultra-thin ferromagnetic layers. The magnetic elds created bydierent layers point in dierent directions. This scatters the electrons as they move from one layer to thenext, increasing resistance. Under the inuence of an externally-applied magnetic eld, the dierent layersalign their contributions, leading to a large drop in resistance:4

2 Douglas Harper, Magnet, in The Online Etymology Dictionary, http://www.etymonline.com/index.php?term=magnet(accessed 29 July 2011).

3 Images from Rudolf de Bruyn Ouboter, Heike Kamerlingh Onnes's Discovery of Superconductivity, Scientic American(March 1997) and from Dirk van Delft and Peter Kes, The discovery of superconductivity, Physics Today (September 2010).

4 The gure is from Giant Magnetoresistance, in Physics Central, American Physical Society,http://www.physicscentral.com/explore/action/magnetoresistance-1.cfm (accessed 28 September 2011).

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These drops in resistence can be used to read the bits recorded on the magnetic disc underneath a readhead. Thanks to GMR read heads, the density of magnetic storage kept increasing exponentially with timeinto the 21st century. This is what allowed, among other tings, the miniaturisation of the iPods. Fert andGrunberg, the discoverers of GMR, shared the Nobel prize in 2007.

The increase in magnetic information storage density with time is even faster than the famous Moore's lawfor transistors in microchips. It is called Kryder's law. It is the great technological triumph of magnetism.Here is a graph:5

Superconductivity also has important applications, although it is not as ubiquitous as magnetism because,while many useful magnetic properties occur at room temperature, all known superconductors need to bekept at liquid nitrogen or lower temperatures in order to remain superocnducting. Nevertheless supercon-ductivity is routinely used wherever it is important to generate large magnetic elds, such as in medicalMagnetic Resonance Imaging. High-temperature superconductors are still poorly understood so mostapplications employ low-temperature, or conventional superconductors. Finding a room-temperature su-perconductor remains one of the holy grails of physics.

2.3 More is dierent

What is special, from a more fundamental point of view, about magnetism and superconductivity? First ofall, they are both properties of matter whose understanding necessitatates quantum mechanics. Secondly,they are both emergent phenomena.

5 The gure is from Giant Magnetoresistance: Research, in Physics Central, American Physical Society,http://www.physicscentral.com/explore/action/mr-research.cfm (accessed 28 September 2011).

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Let me explain what I mean by emergent. The laws of physics that govern the behaviour of dierentmaterials under dierent conditions are always the same. Yet some materials are magnetic, others aresuperconducting, and many others do none of this. What is more, a superconducting or magnetic materialwill lose its properties if we change its thermodynamic conditions - for example, by raising the temperature.It is clear that just knowing the laws of physics is not enough to understand phenomena such as magnetismand superconductivity. We need to understand the principles of self-organisation whereby the∼ 1023 particlesthat make up the sample cooperate to allow new behaviour to emerge from the physical laws. The Nobelprize-winning American physicist, Phil W. Anderson, summed this magic up with the motto

More is dierent

3 Magnetism: statement of the problem

We will do Magnetism rst because it is conceptually simpler and the theoretical tools required are lessadvanced, so by looking at the subjects in that order the learning curve will be less steep. (On the other handas you will see superconductivity is a simpler subject in the sense that it deals with a single phenomenon.In contrast, in the rst part of the course we will be looking at paramagnetism, ferromagnetism, anti-ferromagnetism...)

3.1 The essential featue of magnetism: spontaneous magnetic induction

Our starting point are Maxwell's equations:

∇ ·E = ρ/ε0 (1)

∇ ·B = 0 (2)

∇×E = −∂B∂t

(3)

∇×B = µ0J + ε0µ0∂E

∂t(4)

These equations determine the electric eld E and magnetic induction B given the charge density ρ andcurrent density J at each point in space r and moment in time t. The universal constants ε0 and µ0 arethe electric permitivity and magnetic permeability of free space, respectively. What the equations tell us,respectively, is the following:

• Gauss' law for electric elds: electric elds ow out of electric charges.

We prove this by integratin g both sides of the equation over a region of volume V , then using Gauss'theorem to show that the integral on the LHS is the same as the ux of the vector eld E through theoutside surface S of V . Thus the ux of E in or out of the region equals the total amount of chargecontained in the region, divided by ε0.

• Gauss' law for magnetic elds: there are no magnetic charges (also called magnetic monopoles).

The proof is similar to that for the electric eld.

• Faraday's law: time-varying magnetic induction leads to circulating electric elds. (A direct conse-quence of this is that a time-dependent magnetic eld going through a loop of wire will induce a currentin the wire.)

To understand we calculate the ux of the quantities on the LHS and RHS through an open surfaceS. We then invoke Gauss' theorem according to which the intergal on the LHS is the same as a lineintegral around the curve C bounding the surfacde S (i.e. its perimeter).

• Ampere's law: currents induce circulating magnetic eld (assuming that the electric eld is static).

Again we can use Stokes' theorem to see this.

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We know that magnets have certain properties like their ability to attract each other, etc. What is the reasonfor these properties? The essential feature of magnetic materials is this: magnetic materials are capableof generating magnetic induction without any energy input. By this we mean that normally togenerate a magnetic induction one would have to set up a current going around a wire, however we ndthat in the presence of any magnetic material we can detect and additional magnetic induction which is anintrinsic property of the magnet. The study of magnetism is the study of how this magnetic induction comesabout, and what form it takes.

3.2 How do we know this? Measuring magnetism

How do we know that magnets have their own, intrinsic magnetic induction? We know it because we canmeasure this induction. The device used to do this is called a magnetometer. There are various design ofmagnetometer, but the majority are based on Faraday's law. One way to attain this is to build an electriccircuit with a closed loop in it. We then place the sample in the middle of the loop.

Then we either withdraw the sample from the loop (extraction magnetometer) or move our magnetic samplerepeatedly in and outirrespectful of the mag tirrespectful of the maghrough the loop (vibration magnetome-ter). All the time we measure the electric current through the loop with an Amp-meter. There will becurrent if the circulation of the electric eld is nite. But this is given by

˛CE · du = − ∂

∂t

ˆSB · ds, (5)

which is the integral form of Faraday's law. If the sample does not have a magnetic induction associatedwith it then the RHS is zero and we will see no current. If, on the other hand, there is a nite magneticinduction generated by the sample, then as we move the sample the ux of induction through the surface Swill change with time:

This will make the RHS of the above equation nite and induce a circulating electric eld. That will createa current around the loop. Thus the detection of the current shows that there is a nite magnetic inductionaround the sample. In this way we nd that some magnetic materials do indeed generate their own magneticinduction.

This is not the only way of measuring the magnetic induction associated with magnetic materials - andin fact it wouldn't work for all magnetic materials. For an antiferromagnet, for example, this would notwork, beacuse it is only capable of detecting a macroscopic magnetic inductance existing outside the sample.For antiferromagnets, there is an alternating arrangement of magnetic inductance that cancels outside thesample and can only be detected inside the sample and on a microscopic (10−10m) scale. There are other,more sophisticated methods that we will see later in the course.

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Lecture II. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In the previous lecture we introduced the subjects of magnetism and superconductivity and then stated theproblem of magnetism. We remembered Maxwell's equations for electromagnetism and saw that magneticinduction is due to currents. We then showed how one can measure magnetic induction. We stated thatcertain materials are capable of generating their own magnetic induction. In this lecture we are going tointroduced the dening property of magnetic materials: magnetisation.

1 Magnetisation

For static elds the electric eld and magnetic induction completely decouple and Maxwell's equationssimplify considerably. We can then describe magnetism with only two equations:

∇ ·B = 0 (1)

∇×B = µ0J (2)

The second of these equations, Ampere's law, tells us that loops of magnetic induction form around currents:

˛CB · du = µ0

ˆSJ · ds (3)

In the absence of a net electron ow, this means that all magnetic inductance is created by closed loops ofelectric current. For example, a wire bent around in a loop will generate a magnetic eld:

That means that the intrinsic magnetic induction of a magnetic sample must be due to small loop-likecurrents inside the sample:

1

We will now introduce two elds M and H dening two distinct components of the magnetic induction:

B = µ0 (M + H) (4)

The rst component, H, is called themagnetic eld and is the component of B due to charge owing freelythrough a material:

This eld can spill out as it does when we use an electric circuit to generate an inductance:

The second component, M, is called the magnetization and is the component of the magnetic inductiondue to charge owing in microscopic closed loops inside the sample.

We say that M arises from bound charges, which can only circulate around a microscopic region inside thesample (e.g. around an atom) while H arises from free charges, which can ow along the whole sample. Wedecompose the current similarly to how we have decomposed the magnetic induction:

J = Jbound + Jfree. (5)

M and H obey similar equations to that obeyed by B involving each of the two components of the current:

˛CH · du =

ˆSJfree · ds (6)

˛CM · du =

ˆSJbound · ds (7)

Now suppose we have set up an electric circuit to generate a given magnetic induction that we wish to applyto our magnetic sample. In the absence of the sample, there is no magnetization so we have a magneticinduction

Bapplied = µ0Happlied. (8)

When the sample is introduced, M goes from zero to a nite value (whatever intrinsic magnetization thesample has) and H also changes because, according to Gauss' law,

∇ ·B⇒ ∇ ·H = −∇M. (9)

This means that in general when we apply a given magnetic induction to a sample we don't know what thevalue of the magnetic eld H inside the sample is. However if the magnetization is not very large, M H,one can approximate

H ≈ Happlied. (10)

We will always assume this to be the case except when we state otherwise. The details of what happneswhen this is not true can be found in SB Sec. 1.1.4.

2

Magnetic materials are those for whichM 6= 0. (11)

The actual form that the magnetization takes depends on the type of magnetism. The study of the dierentforms of magnetization and how they come about is the subject of study of magnetism.In the previous lecturewe saw that magnetic induction B results from nite currents J. We split the current into two components,one due to free charges that can roam around a sample and another one due to charges that are bound tomove in very small microscopic loops inside a sample. The two types of current give rise to the magneticeld H and the magnetization M, respectively: B = µ0 (H + M) . Magnetism is the study of M. In thislecture we will begin to look at how a nite magnetization may arise.

2 Magnetic moments

So far the denition of M (and therfore of H) has been rather loose - how do we distinguish between boundand free charges. Here we are going to give a somewhat more rigorous denition in terms of magneticmoments. We shall dene the magnetization M as the volume density of magnetic moments.

What is a magnetic moment? A magnetic moment is a very small object made up of an electric current Igoing around in a small closed loop C:

The magnetic moment gives rise to a nite magnetic induction B (r) in the space around it. For distances rwhich are than the linear dimensions of the current loop (this is what we mean by the current loop beingsmall), this is given by 6

B =µ0

4πr3[3 (m · r) r−m] (12)

where we have dened the vector quantity m which measures the strength and direction of the magneticmoment. m is the product of the current I and a vector giving the area and orientation of the minimalsurface enclosed by the loop C:

m = I S (13)

The key point is that B (r) at a given r depends only on m, not on other features such as the shape of theloop of current (e.g. an ellipse or a circle). This is because the moment is small.

Now let us consider a sample that has a nite density of innitesimally-small magnetic moments, dm. Themagnetization is the volume density of such moments, obtained by dividing dm by the corresponding volumeelement:

M =dm

d3r. (14)

The magnetic induction at a point r due to such density of magnetic moments is obained by adding up allthe contributions from all the individual moments:

B (r) =

ˆdB =

ˆµ0

4πr3[3 (dm · r) r− dm] . (15)

We mow substitute Md3r = dm and reduce this to an integral of the magnetization:

B (r) =

ˆµ0

4πr3[3 (M (r) · r) r−M (r)] d3r. (16)

As we can see the relationship between the magnetic induction generated by the sample and its magnetizationis not at all trivial. That is why it is convenient to focus on the magnetization rather than the magneticinduction when discussing magnets.

6 This is a well-known result from the theory of electromagnetism. It follows from Maxwell's equations, which we recalled inthe previous lecture. There are many textbooks that prove particular cases of this result. See, for example, Feynman Lectureson Physics, Vol. II, Sec. 14-5 (note that this text does not use S.I. units; to convert the equations to the S.I. we must substitute1/4πε0c

2 → µ0/4π).

3

3 Relationship of magnetic moments to angular momentum

Let us now see how a magnetic moment is aected by an applied eld.

First we will establish the relationship between magnetic moment and angular momentum. Consider aparticle of charge q and mass mq orbiting the origin of coordinates. Let us assume that the orbit is circularand has radius r. The orbiting particle creates a current along the ciruclar orbit. This current is

I =qv

2πr(17)

where v is the velocity of the particle. The magnetic moment due to this orbiting particle is this times thearea enclosed by the orbit, πr2 :

m =qv

2πr× πr2 =

qvr

2. (18)

On the other hand the angular momentum of our particle is

L = r×mqv⇒ L = rmqv. (19)

Thus there is a relationship between magnetic moment and angular momentum:

m =q

2mqL. (20)

Since both m and L point in the same direction we can make this a vectorial relation:

m = γqL, (21)

where the quantity

γq =q

2mq(22)

is called the gyromagnetic ratio of our particle. Note that it depends on the charge and mass of theparticle alone, not on the size of the orbit.

4 Larmor precession

The relationship found above between a magnetic moment and angular momentum allows us to work outhow a magnetic moment m reacts to an externally-applied magnetic induction B. We know from classicalmechanics that the force on a charged particle in the presence of an electromagnetic eld is

F = q (E + v ×B) . (23)

In the absence of an electric eld, we are left only with the second term: the Lorentz force

F = q (v ×B) . (24)

This will act on the charge or charges making up a magnetic moment, causing a torque that makes it re-orientitself. The total torque on the particles in is given by

G = r× F.

= q r× (v ×B) = q1

2(r× v)×B

=q

2mqL×B = γq L×B, (25)

4

where in the second line we have assumed only the total torque averaged over an orbit need be consideredand in the third line we have used that the angular momentum is L = r ×mqv. This must be equated tothe rate of change of the angular momentum:

d

dtL = G (26)

⇒ d

dtL = γq L×B. (27)

We now use the gyromagnetic ratio to turn this into an equation for the time-evolution of the magneticmoment:

d

dtm = γqm×B. (28)

The key thing to notice here is that the rate of change of the magnetic moment is perpendicular to themagnetic moment itself - therefore a magnetic induction can never lead to an increase or decrease of the sizeof the magnetic moments - they merely precess in periodic orbits.

To see this explicitly let us solve the above equatio of motion. First, let us choose our system of coordinatesto that B is parallel to the z axis:

B = (0, 0, B) . (29)

The vector product then gives

m×B =

∣∣∣∣∣∣i j kmx my mz

0 0 B

∣∣∣∣∣∣ = (myB,−mxB, 0) (30)

Substituting this back into (28) we get

d

dtmx = γqBmy (31)

d

dtmy = −γqBmx (32)

d

dtmz = 0 (33)

The third of these equations tells us that the component of m parallel to the magnetic induction does notchange. The other two coupled equations are solved by

mx = mρ sin (ωLt+ φ) (34)

my = mρ cos (ωLt+ φ) , (35)

where mρ and φ are integration constants, if we take for the Larmor frequency

ωL = γqB. (36)

(This can be proven by substitution and we leave it as an exercise.)

This tells us that in the presence of a magnetic induction a magnetic moment will presrve its tilt with respectto the magnetic induction and precess around it at a constant frequency which is proportional value of theinduction and to the gyromagnetic ratio:

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This has a very important consequence: an externally-applied magnetic induction B cannot be used tomagnetize a sample. The spins will start precessing around the applied eld, all of them at the samefrequency, and therefore the total amount of magnetization will change direction but not size (it is not justthat each magnetic moment will remain the same size, but also that the way they add up will also remain thesame). This seems at odds with our observations of magnetic phenomena! In reality when we applyan induction to almost any system the total size of the magnetization changes, either for or against the eld(paramagnetism and diamagnetism, respectively). Also, in many magnetic systems such as ferromagnetsand antiferromagnets there is a spontaneous magnetization even in the absence of an applied eld whichis due to each magnetic moment reacting to the induction created by all the other moments. But Larmorprecession suggests that none of this can take place. Indeed the Lorentz force (24) is always perpendicularto a particle's displacement, therefore it cannot do any work on a system.

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Lecture III. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In the previous lectures we introduced the concepts of magnetization and magnetic moments. Magnetizationis a macroscopic concept: the contribution that a material makes to magnetic induction in and around it.Magnetic moment, on the other hand, is a microscopic concept: it is the microscopic origin of magnetization.So far however we have only discussed magnetic moments from the point of view of classical mechanics. How-ever the electron is a quantum particle. In this lecture we will introduce a quantum-mechanical descriptionof the magnetic moment of an electron.

1 The Bohr magneton

Something we have ignored until now is that angular momentum, and therefore also magnetic moments, arequantised. Consider the action associated with the motion in a circular orbit of a particle giving rise to amagnetic moment:

S =

˛Cmqv · dr (1)

According to the Bohr-Somerfeld quantisation rules that action must be an integer number of times thequantum of action, which is Planck's constant:

S = νh, ν = 0, 1, 2, . . . (2)

But the action is nothing butS = mqv2πr = 2πL, (3)

where L = rmqv is the magnitude of the angular momentum, so this is just the quantisation of angularmomentum,

L = ν~. (4)

This, in turn, quantises magnetic moment, µ:7

µ = γqL = γq~ν. (5)

[Here we have used the gyromagnetic ratio γq to relate the magnetic moment µ to the angular momentumL. See the previous Lecture 2, Eqs. 21,22.] Magnetic moment is thus quantised in units of

γq~ =q~

2mq. (6)

For an electron we have mq = me, q = −e and this is called the Bohr magneton:

µB = − e~2me

. (7)

Just as ~ is the quantum of angular momentum, µB is the quantum of magnetic moment:

µ = µBν. (8)

The negative sign of the Bohr magneton means that, for electrons, the magnetic moment points in theopposite direction to the angular momentum (this is due to the negative charge of electrons: when they turnaround clockwise, the corresponding current turns anti-clockwise).

As we will see the quantisation of magnetic moment plays an essential role in magnetism.

7 Note we use the notation µ for magnetic moment (c.f. previous lectures where we used m).

1

2 Orbital angular momentum

Let us now be a bit more rigorous.

In Quantum Mechanics the electron has two components to its angular momentum: orbital angular momen-tum and spin angular momentum. Both of them contribute to the electron's magnetic moment.

Let us treat rst the orbital angular momentum. This is the quantity that corresponds, in the classicallimit, to the classical angular momentum that we considered before. The corresponding quantum-mechanicaloperator is therefore obtained straight-forwardly from the corresponding classical expression:

L = r× p→ L = r× p (9)

= r× ~i∇r (10)

= −i~(y∂

∂z− z ∂

∂y, z

∂

∂x− x ∂

∂z, x

∂

∂y− y ∂

∂x

)≡(Lx, Ly, Lz

). (11)

From the last expression one can show that the three components Lx, Ly, Lz obey the angular momentumcommutation relations: [

Lx, Ly

]= i~Lz ,

[Ly, Lz

]= i~Lx ,

[Lz, Lx

]= i~Ly. (12)

Thus dierent components of the angular momentum do not commute. In other words, when one of thecoordinates of the spin vector has a well-dened value, the other two are undened.

These commutation relations are very unusual. Indeed they can be written as

L× L = i~L (13)

which means the vector L is perpendicular to itself!

On the other hand the magnitude squared of the angular momentum, L2 = L2x + L2

y + L2z, does commute

with each of the individual components:[L2, Lx

]=[L2, Ly

]=[L2, Lz

]= 0. (14)

Thus we can nd a set of states that are simultaneous eigenstates of L2 and Li for a given i, e.g. i = z. Thatset takes the form8

|l,ml〉 l = 0, 1, 2, . . . ,∞ml = −l,−l + 1, . . . , l − 1, l

(15)

and the eigenvalues are l (l + 1) ~2 and ml~, respectively. In other words

L2|l,ml〉 = l (l + 1) ~2|l,ml〉, (16)

Lz|l,ml〉 = ml~|l,ml〉. (17)

8 Bra-ket notation: to denote the eigenstates of the angular momentum operator we have used `ket' notation. This meansthat instead of denoting a state by its wave function we just write its quantum nubers, e.g. `l,ml', between `|' and `〉', like this:|l,ml〉.What's wrong with just giving the wave function, ψl,ml (r)? There is nothing wrong, of course, but it does mean that what

we write is only valid in a particular basis. Thus, if r represents position, ψl,ml (r) is the wave function in the position basis.In the momentum basis, for example, the same quantum-mehcanical state would have a dierent wave function, say φl,ml (p).It is useful to have a way of designating the state that does not require specifying which basis we are working on. This is whatthe ket notation |l,ml〉 aords us.From the ket of the state we are interested in, we can easily construct wave functions in dierent bases. This requires using

a `bra' to make a `bra-ket', like this: ψl,ml (r) = 〈r|l,ml〉 and φl,ml (p) = 〈p|l,ml〉. |r〉 and |p〉 denote position and momentumeigenstates, with well-dened values of the position and momentum (given by r and p, respectively). The bra-kets denote theprojections of the state |l,ml〉 onto each of these two states, giving the amplitudes of probability that the partcile will be foundat position r or with momentum p, respectively.

2

Thus the size of the angular momentum is

L =√l (l + 1)~ (18)

and its projection along the z direction isLz = ml~. (19)

Note that although L and Lz are xed, Lx and Ly are completely undetermined. We can represent thisgraphically as follows (for l = 3, as an example):

3 Orbital magnetic moments

Now we can use the gyromagnetic ratio of the electron, which we derived in the last lecture, to work out thecontribution of an electron's quantised angular momentum to its magnetic moment, µL:

µL = γ−eL (20)

= −µBL/~. (21)

From this follows that the magnetic moment's size and projection along a given direction are quantised:

µL =√l (l + 1)µB where l = 0, 1, 2, . . . ,∞; (22)

µL,z = −µBml where ml = −l,−l + 1, . . . , l − 1, l. (23)

Also, if µL,z has a well-dened value then µL,x and µL,y are completely undetermined (likewise in a statein which µL,x is well-dened µL,y and µL,z are undened, and so on). So we can see that from a quantum-mechanical point of view the magnetic moment becomes quite a strange object compared to our previous,classical treatment.

4 Spin angular momentum

In addition to the orbital angular momentum L, quantum particles also have another form of angularmomentum called `spin' and denoted by S. Unlike L, spin does not come from quantising a classicalexpression [c.f. Eq. (9)]. Spin is an intrinsically quantum-mechanical quantity with no classical analogue.

The components of the spin obey analogous commutation relations to those of orbital angular momentum[Eqs. (12-14)]. We can therefore nd simultaneous eigenstates of S2 and Sz. The corresponding quantumnumbers are s (the analogue of l) and ms (playing the role of ml).

The intrinsically quantum-mechanical nature of S has a number of important consequences:

1. For a given type of particle s will always have the same value.

2. s can take either integer or half-integer values, depending on wether the particle is a boson or a fermion.For electrons s = 1/2 (this makes electrons fermions).

Moreover we shall now see that the gyromagnetic ratio is dierent for spin than for orbital angular momentum.

3

5 Spin magnetic moments

The gyromagnetic ratio for spin angular momentum is dierent than for orbital angular momentum - itdiers a factor g called the g-factor:

µS = gSγ−eS, (24)

= −µBgSS/~. (25)

For an electron, the g-factor can be calculated using relativistic Quantum Mechanics and turns out to beg = 2. Quantum electrodynamics corrects this to g = 2.0023... but we will ignore that. Thus the magneticmoment of an electron's spin is given by

µS = gS√s (s+ 1)µB with s = 1/2 and gS ≈ 2 (26)

≈√

3µB; (27)

µS,z = −gSµBms with ms = −1/2 or + 1/2 (28)

≈ ±µB. (29)

Note that the magnetic moment coming from the spin angular momentum of an electron is bound to theelectron. Thus, in the language we used in Lecture 1, it behaves like a magnetic moment arising from boundcharges even if the electron itself is unbound. Because of this the spin angular momentum is particuarlyimportant for itinerant magnetism. This will be discussed later on in the course.

6 Nuclear magnetism

Since the nucleus also contains orbiting charges, namely the protons, there is also a magnetic moment of thenucleus that results from the quantised angular momentum of the protons. As in the case of the electrons, theprotons have spin 1/2 and therefore both orbital and spin angular momenta. Nuclear magnetic moments canthus be described in a very similar way to electronic moments, with one important dierence: the quantumof magnetic moment for a proton is not the Bohr magneton µB, but the nuclear magneton µN .

µN can be obtained from µB by changing the electron charge −e for the proton charge +e and the electronmass me for the proton mass mp:

µN =e~

2mp. (30)

Since mp ∼ 103me, we have µN ∼ 10−3µB, implying that the nuclear contribution to magnetic phenomenacan often be neglected. We will do this from now on unless explicitly mentioned otherwise.

4

Lecture IV. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In the previous lecture we introduced the concepts of the orbital and spin angular momentum of an electronfrom a quantum-mechanical point of view. In this lecture we will nish the process of constructing the`building blocks' of magnetism (namely, the magnetic moments) by learning how to combine these momentsto make the total moment of the electron, and those of dierent electrons to make that of an atom or ion.This will lay the ground work for the discussion of magnetic states of matter resulting from the collectivebehaviour of many individual magnetic moments starting from the next lecture.

1 Combining angular momenta

The magnetic moment of an atom or ion in a solid is made up of contributions from dierent electrons. Also,even the magnetic moment of a single electron is made up of orbital and spin magnetic moments. To ndthe combined magnetic moment of the atom we will need to combine dierent types of angular momentum.

Let us consider the sum of two angular momenta:

J = J1 + J2. (1)

For example, we could be dealing with the orbital and spin angular momenta of the electron:

J = L + S. (2)

Alternatively, we could be talking about the angular momenta of two electrons on the same atom or thetotal orbital andgular momentum of all the electrons in one atom and the total spin angulat momentum ofthe same electrons.

In any case since the two angular momentum operators act on dierent degrees of freedom, they commute.Therefore we can nd a complete basis set of simultaneous eigenstates of J2

1, J1,z, J22, and J2,z, which we will

label|j1,m1, j2,m2〉. (3)

These states are built by the direct product of the bases for the two individual angular momenta,

|j1,m1, j2,m2〉 = |j1,m1〉 ⊗ |j2,m2〉. (4)

This means that the state is made up of two states living in dierent spaces: J1 only acts on the rst subspaceand J2 on the second one. For given j1, j2 the projection quantum numbers are given by

m1 = −j1,−j1 + 1, . . . , j1 − 1, j1 and (5)

m2 = −j2,−j2 + 1, . . . , j2 − 1, j2, (6)

giving (2j1 + 1) (2j2 + 1) states in total.

One can prove that J is also an angular momentum, obeying angular momentum commutation rules:

J× J = i~J. (7)

Because of this, there must be another complete basis set formed by simultaneous eigenstates of J2 and Jz.Moreover Jz and J2 commute with J2

1 and J22 (but not with J1,z or J2,z) so we can actually nd a complete

basis set of simultaneous eigenstates of J2, Jz, J21, J

22, which we label

|j1, j2, j,m〉. (8)

The two bases are related through

|j1, j2,m1,m2〉 =∑j,m

〈j1, j2, j,m|j1, j2,m1,m2〉|j1, j2, j,m〉.

1

The projections 〈j1, j2, j,m|j1, j2,m1,m2〉 are the Clebsch-Gordan coecients. We will not derive them here[for a derivation of their values see any good Quantum Mechanics textbook, e.g. Bransden & Joachain] butwe will note that they are non-zero only for certain combinations of j,m,m1 and m2. This allows us to workout, for given j1, j2, which values are allowed for the new quantun numbers j and m: we starts with the theallowed values of m1 and m2, which we know (we jkust gave them above), and each of these combinationswill give us a value of j and m. The combinations for which the corresponding coecient is zero are notallowed. We nd that

j = |j1 − j2| , |j1 − j2|+ 1, . . . , j1 + j2 − 1, j1 + j2 and (9)

m = m1 +m2 = −j,−j + 1, . . . , j − 1, j, (10)

giving again a total of (2j1 + 1) (2j2 + 1) states.

The rst of these equations tell us that the two angular momenta can combine in oppposition or re-inforceeach other, the most extreme cases corresponding to j = |j1 − j2| and j = j1 + j2, respectively. Once thetotal angular momentum of length J = ~

√j (j + 1) has been formed, its component along the quantisation

direction can take values Jz = ~m, with m = −j,−j + 1, . . . , j as usual for an angular momentum.

2 Combining magnetic moments

Our interest in the combination of angular momenta is that the combined angular momentum will give riseto its own magnetic moment through a gyromagnetic ratio γ:

µ = γJ. (11)

The gyromagnetic ratio γ of the combined moment is related to those of the individual moments, γ1 and γ2:

µ1 = γ1J1 (12)

µ2 = γ2J2 (13)

To put γ in terms of γ1 and γ2 we write µ = µ1 + µ2 using the above formulae:

γJ = γ1J1 + γ2J2. (14)

We now multiply by J, obtainingγJ2 = γ1J · J1 + γ2J · J2, (15)

and use

J21 =

(J− J2

)2= J2 + J2

2 − 2J · J2 ⇒ J · J2 =1

2

(J2 + J2

2 − J21

)(16)

J22 =

(J− J1

)2= J2 + J2

1 − 2J · J1 ⇒ J · J1 =1

2

(J2 + J2

1 − J22

)(17)

to re-arrange the expression as

γJ2 = γ11

2

(J2 + J2

1 − J22

)+ γ2

1

2

(J2 + J2

2 − J21

). (18)

Let us now make each term on the LHS and RHS of the above equation act on the same basis state|j1, j2, j,m〉:

γJ2|j1, j2, j,m〉 = γJ2|j1, j2, j,m〉 (19)

γ11

2

(J2 + J2

1 − J22

)|j1, j2, j,m〉 = γ1

1

2

(J2 + J2

1 − J22

)|j1, j2, j,m〉 (20)

γ21

2

(J2 + J2

2 − J21

)|j1, j2, j,m〉 = γ2

1

2

(J2 + J2

2 − J21

)|j1, j2, j,m〉 (21)

HereJ = ~

√j (j + 1) , J1 = ~

√j1 (j1 + 1) , J2 = ~

√j2 (j2 + 1) (22)

2

are the size of each of the three angular momenta, respectively. Putting all this back into (18) we nd

γ = γ1j (j + 1) + j1 (j1 + 1)− j2 (j2 + 1)

2j (j + 1)+ γ2

j (j + 1) + j2 (j2 + 1)− j1 (j1 + 1)

2j (j + 1). (23)

Evidently if γ1 = γ2 then the new gyromagnetic ratio coincides with the old one, γ = γ1 = γ2 (even if thequantum numbers of the two combined angular momenta were dierent).

It is customary to express the gyromagnetic ratio in terms of the Bohr magneton by introducing the Landég-factor

γ ≡ −gJµB/~⇒ µ = −gJµBJ/~. (24)

Then the above expression becomes a formula giving the g-factors:

gJ = g1j (j + 1) + j1 (j1 + 1)− j2 (j2 + 1)

2j (j + 1)+ g2

j (j + 1) + j2 (j2 + 1)− j1 (j1 + 1)

2j (j + 1). (25)

For the total magnetic moment of an electron, resulting form its spin and orbital angular momenta, we havegL = 1 and gS = 2 (as we saw in the previous lecture) so Eq. (25) gives

gJ =3

2+

3/4− l (l + 1)

2j (j + 1). (26)

More generally if we are adding a large spin, resulting from combining several spins, and a large orbitalmoment, with quantum numbers stot and ltot, we get

gJ =3

2+stot (stot + 1)− ltot (ltot + 1)

2j (j + 1). (27)

3 Fine structure

Until now, whether we use the bases |j1, j2,m1,m2〉j1,j2,m1,m2or |j1, j2, j,m〉j1,j2,j,m to describe a com-

pound angular momentum has been treated as immaterial. However, if what we are trying to describe arestationary states (i.e. eigenstates not just of the above operators, but also of the Hamiltonian describing thedynamics of our particles) then we must choose a basis that corresponds to a set of operators that commutewith the Hamiltonian.

Consider a Hamiltonian featuring a term of the form

V =λ

~2J1 · J2, (28)

i.e. a quadratic coupling betweem the two spins. This can be, for example:

1. the spin-orbit relativistic correction that couples the orbital and spin angular momenta of an electron,an atom or an ion;

2. the exchange interaction between two adjacent magnetic moments (we will see where that can comefrom later on).

Then |j1,m1, j2,m2〉 cannot be an eigenstate of the Hamiltonian because V does not commute with J1,zor J2,z. On the other hand eigenstates of the Hamiltonian proportional to |j1, j2, j,m〉 can be found since

J2 =(J1 + J2

)2= J2

1 + J22 + 2J1 · J2 and therefore

V =λ

2~2(J2 − J2

1 − J22

), (29)

which commutes with J2, J21, J

22 and also with Jz (since J2, J2

1, and J22 all commute with it).

3

Now we can evaluate perturbatively the change in energy, ∆E, due to the above perturbation. To lowestorder, this corresponds to simply evaluating the expectation value of the perturbation, V , in the unperturbedground state, which can be taken to be a simultaneous eigenstate of J,J1 and J2. We get

∆E ≈⟨V⟩

=λ

2[j (j + 1)− j1 (j1 + 1)− j2 (j2 + 1)] . (30)

We now think about what happens when a term of the form (28) is switched on. Initially our energylevels were specied by j1 and j2 and had degeneracy equal to (2j1 + 1) (2j2 + 1). Then as we turn on thecoupling term the levels split into dierent ones corresponding to dierent values of j, as given by (30). Theirdegeneracy thus gets lowered to just 2j + 1.

Example: If we ignore relativistic corrections, an electron in a p orbital (l = 1, s = 1/2) has available(2l + 1) (2s+ 1) = 3× 2 = 6 states, namely

ml = 1 , ms = 1/2

ml = 1 , ms = −1/2

ml = 0 , ms = 1/2

ml = 0 , ms = −1/2

ml = −1 , ms = 1/2

ml = −1 , m = −1/2

These 6 states are all degenerate. Alternatively, we may describe them in terms of the total angularmomentum quantum numbers j and mj . j goes from |l − s| = 1/2 to l+ s = 3/2 and mj from −j to j:

j = 3/2 , mj = 3/2

j = 3/2 , mj = 1/2

j = 3/2 , mj = −1/2

j = 3/2 , mj = −3/2

j = 1/2 , mj = 1/2

j = 1/2 , mj = −1/2

Now we turn on perturbatively a spin-orbit coupling term λL · S. With this term on, ml and ms areno longer quantum numbers, but j and mj still are. The corresponding perturbation energy is

∆E ≈⟨λL · S

⟩=

λ

2[j (j + 1)− l (l + 1)− s (s+ 1)] =

λ

2

[j (j + 1)− 11

4

]=

λ

2for j = 1 +

1

2=

3

2(degeneracy 4),

−λ for j =

∣∣∣∣1− 1

2

∣∣∣∣ =1

2(degeneracy 2).

Thus the 6-fold degenerate energy level splits, under the inuence of spin-orbit coupling, into a 4-folddegenrate level and a 2-fold degenerate one:

This degeneracy can be lifted further by application of a magnetic eld (we will see that later on).

4

4 Magnetic moment of an atom: Hund's rules

Let us assume for a moment that electrons in atoms do not interact with each other, in other words, thatall the electron sees is the potential created by the atomic nucleus. We will also, for the time being, ignorespin-orbit coupling. Then each electron in the atom goes on a particular energy shell with quantum numbersn = 1, 2, 3, . . . and l = 0, 1, 2, . . . , n− 1. Each of these energy levels can in principle accommodate 2 (2l + 1)electrons, corresponding to the dierent values of the additional quantum numbers j = |l − s| , . . . , l + sand mj = −j, . . . , j. Once all the electrons have been placed, their angular momenta can be added up asdiscussed in the previous lecture, giving rise to the magnetic moment of the atom.

Now, if all the states in a given energy level are occupied, then the contribution of that energy level tothe total angular momentum, and therefore the magnetic moment, is zero, because for every electron withJz = ~mj there is another one with Jz = −~mj , so they compensate. If an atom has all its shells completelylled like this, then it is not very interesting from the point of view of magnetism.

For example, a Helium atom has two electrons which ll the n = 1, l = 0 shell. Their quantum numbers are

n = 1, l = 0, s = 1/2,ml = 0,ms = 1/2 (i.e. j = 1/2,mj = 1/2)

n = 1, l = 0, s = 1/2,ml = 0,ms = −1/2 (i.e. j = 1/2,mj = −1/2)

The total angular momenta areml +ms = 1/2 and − 1/2,

which add up to zero.

On the other hand, if an atom or ion has a partially-lled shell then the magnetic moments do notnecessarily compensate and potentially the atom can have a magnetic moment that may contribute tomagnetism. The problem is that when the shell is not full there are potentially many dierent ways ofdistributing the available electrons among the available orbitals. Each way has the same energy but it leadsto a dierent nal value for the magnetic moment. To sort this out we need to remove our initial assumptions:let electrons interact with each other so that dierent ways of distributing electrons among orbitals will leadto dierent energies. Let us also allow spin-orbit coupling to lift some of the degeneracy of the individualenergy levels.

The accurate calculation of the energy of a few-electron system forming an atom is actually a very dicultproblem. Here we give a set of rules-of-thumb called `Hund's rules' which do a farily good job at describinghow the electrons are distributed among the dierent orbitals. They allow us to determine the spin, orbital,and total angular momentum quantum numbers of an atom or ion resulting from all their individual electrons:stot, ltot and jtot, respectively. This in turn enables us to work out the g-factor and therefore the magneticmoment.

Here are the rules for a shell with angular momentum quantum number l containing N electrons. They mustbe applied in the order in which they are given here:

• Zeroth rule: Pauli's exclusion principle.- there can be no two electrons with the same value of ml andms.

• Hund's rst rule.- Make the total spin quantum number stot as big as possible.

By the rules for addition of angular momenta, which we saw above, this will force us to align thespins of all the electrons. That will in turn make them occupy dierent orbitals (by Pauli's exclusionprinciple). In this way their mutual Coulomb replusion will be minimized as we avoid payimng thepairing energy price of having two electrons in the same orbital.

• Hund's second rule.- Make the total orbital angular momentum quantum number ltot as large aspossible, while remaining consistent with rule number 1.

Again, this will lower Coulomb repulsion because of the form the angular momentum wave functionshappen to have.

5

• Hund's third rule.- Make the total total angular momentum quantum number jtot as large as possibleif N ≥ 2l + 1 (shell more than half-lled) and as small as possible if N ≤ 2l + 1 (shell less thanhalf-lled).

This minimizes the spin-orbit coupling term λL·S = λ2~2[j2tot (jtot + 1)− l2tot (ltot + 1)− s2tot (stot + 1)

],

taking into account that the coupling constant λ tends to be > 0 for less than half-lled shells and < 0for more than half-lled shells.

These rules are best understood through an example:

Example: application of Hund's rules to the Ho3+ ion. The atom Ho has the electronic structure [Xe]4f116s2.Removing 3 electrons will leave us with [Xe]4f10. The f -shell has l = 3 and therefore enough room foronly 2 (2l + 1) = 14 electrons, so it is more than half-lled by our 10 electrons. These are the statesthat are available:

l s ml ms

3 1/2 3 1/2-1/2

2 1/2-1/2

1 1/2-1/2

0 1/2-1/2

-1 1/2-1/2

-2 1/2-1/2

-3 1/2-1/2

Let us apply Hund's rst rule: distribute electrons among the spin states so as to generate the highestpossible value of stot. First we place the rst 7 electrons all with their spins pointing up:

l s ml ms

3 1/2 3 1/2 •-1/2

2 1/2 •-1/2

1 1/2 •-1/2

0 1/2 •-1/2

-1 1/2 •-1/2

-2 1/2 •-1/2

-3 1/2 •-1/2

We still have to put 3 more electrons, and those will lower our spin by 3/2 whatever we do. To choosewhere to put them we need, therefore, to consult Hund's second rule: in order to maximise ltot, weput them on the available states with the highest value of ml:

6

l s ml ms

3 1/2 3 1/2 •-1/2 •

2 1/2 •-1/2 •

1 1/2 •-1/2 •

0 1/2 •-1/2

-1 1/2 •-1/2

-2 1/2 •-1/2

-3 1/2 •-1/2

We now have completely distributed all the electrons among the available states. The spin and orbitalangular momenta quantum numbers have worked out to be stot = 7/2−3/2 = 2 and ltot = 3+2+1 = 6.Now to work out how stot and ltot combine to give the total angular momentum quantum number Ineed to apply the third rule: since the shell is more than half-lled, I choose the maximum possiblevalue for jtot = stot + ltot = 8. The magnetic moment will be

µ = −gJµB√jtot (jtot + 1) (31)

with the g-factor given by

gJ =3

2+stot (stot + 1)− ltot (ltot + 1)

2jtot (jtot + 1). (32)

This yields

µ = −5

4µB√

8 (8 + 1) ≈ −10.7µB.

The spin conguration of an atom or ion is denoted ACB, where A = 2stot+ 1, B = jtot and C = ltot (withltot = 0, 1, 2, 3, 4, 5, 6, . . . denoted S, P,D, F,G,H, I, . . .)

For example,

for Ho we have 2stot + 1 = 2× 2 + 1 = 5, jtot = 8 and ltot = 6 ≡ I so we write 5I8.

5 Crystal elds

Interactions between electrons in dierent magnetic ions or atoms are crucial to magnetism, as they areresponsible for the collective behaviour that we anticipated in Lecture 1. They often involve virtual quatum-mechanical tunnelling of electrons from one lattice site to another (exchange interactions). This will be seenin future lectures. For the time being we are interested only in the eect electrostatic interactions withelectrons in the immediate environment of a given atom or ion. These electrostatic interactions would notbe present in an isolated atom or ion but manifest themselves when the atom/ion is placed inside a crystalwhere it is surrounded by other atoms or ions. They take the form of an electric eld that the electrons see:the `crystal eld'.

Hund's rules work fairly well for elements with unlled 4f shells (lanthanides). For those with unlled 3dshells (the rst row of transition metals) they do not work. In general, they work better the deeper inside theatom the unlled shell is located. In that case, the electrons in the unlled shell are quite isolated from thecrystal eld. This is good because Hund's rules assume an atom or ion which is suspended in empty space, ina spherically-symmetric environament. In reality, the magnetic atom or ion is often encaged by a numberof non-magnetic atoms or ions that surround it. This makes the atom or ion's environment non-spherically

7

symmetric due to the crystal eld. If the unlled shell is deep inside the atom, the eect of the crystal eldis notwhen the so great.

The crystal eld prevents the electrons from orbiting freely around the nucleus, which may result in an ltot =0 state rather than the higher value suggested by Hund's second rule. This is called orbital `quenching'.Evidently in this case we have L = 0⇒ J = S.

In addition to orbital quenching, the crystal eld splits the energy levels as dierent congurations havedierent level of overlap with the adjacent atoms. For example, an l = 2 unlled shell in an atom thatis surrounded by negatively-charged ions placed in an octahedral conguration (with the magnetic ion inthe centre of the octahedron, and a non-magnetic ion in each vertex) will see its 2 × (2× 2 + 1) = 10-folddegenerate level split into a 2 × 3 = 6-fold degenerate level called t2g and a 2 × 2 = 4-fold degenerate onecalled eg. Hund's rst rule still applies, but the electrons are distributed only among the 3 t2gorbitals oramong the 5 t2g and eg orbitals depending on whether the pairing energy is smaller than or larger than thet2g − eg splitting, respetively. [See Fig. 3.4 in the text by Blundell for a picture of this.]

8

Lecture V. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In Lecture 1 we introduced the notion of magnetism as a collective phenomenon due to the combinedbehaviour of many quantum-mechanical entities (the other major example being superconductivity, whichwe will treat later on in the course) and we stated the problem as follows: there are materials that createa nite magnetic induction in and around them, and the science of magnetism tries to explain its origin.Then, in Lecture 2, we formulated the problem in terms of a vector quantity called the magnetization,M, which is nite within a magnetic sample and zero without. We also introduced the notion of magneticmoments, µ, which are the microscopic entities that combine to give a macroscopic magnetization. InLecture 3 we established the quantum nature of the magnetic moments of electrons and in Lecture 4 wesaw how those combine to give the total magnetic moments of atoms and ions in solids. We are now ina position where we understand what the building blocks of manetism are and we can start to tackle theproblem of combining them to yield the macroscopic phenomena of interest. In these two lectures we willintroduce the basic tools of statistical mechanics that we will be using to do that and we will apply themto describe the simplest magnetic phenomenon: paramagnetism.

1 From magnetic moments to magnetization

We saw in Lecture 1 that the magnetization of a solid (whose measurement and calculation one could sayis the object of Magnetism) is just the density of magnetic moments:

M =dµ

d3r.

(Lecture 1, Eq. 14)

Therefore if we knew the positions Rj of all the magnetic sites in the system, designated by the indexj = 1, 2, . . . ,N , and at each of those sites we knew the magntiude and direction of the correspondingmagnetic moment, µj , then we would be able to compute

dµ =∑j

Rj ∈ d3r

µj (1)

and therefore M. If all dipoles are identical, and they are distributed uniformly around the sample withdensity ρ, then

M = ρµ . (2)

We have seen in previous lectures how a magnetic moment can arise for an electron or a whole atom. Wehave not said anything about what those magnetic moments are doing, though. They could, for example, beacting in unison, all pointing in the same direction; or they could be oriented randomly. In the latter casethe above sum might give dµ = 0 and therefore zero magnetization, M = 0, even though each individualmoment µj may be quite large. Which is it?

2 Combining magnetic moments

Statistical Mechanics comes to the rescue here: it tells us how, once we know the microscopic physics ofindividual magnetic moments, their behaviours combine to re-inforce or counter-act each other for a givenstate of thermal equilibrium at a given temperature, T .

1

For our purposes, the important statement that Statistical Mechanics makes is the following. Suppose thatwe know all the stationary states of the system,

|n〉 = |0〉, |1〉, |2〉, |3〉, . . . (3)

and their energies,En = E0, E1, E2, E3, . . . (4)

These means that |n〉 and En are the eigenvectors and eigenvalues of the system's Hamiltonian, H:

H|n〉 = En|n〉, n = 0, 1, 2, . . . (5)

Once we know the eigenstates, we can compute the expectation value of any operator, including the operator

giving the components of the jth magnetic moment. Thus in the ground state it will be given by

〈0|µj |0〉. (6)

More generally, in an arbitrary stationary state |n〉 it will be given by

〈n|µRj|n〉. (7)

In this way we could, in principle, compute the magnitude and direction of each magnetic moment in eachpossible state of the system.

Now, in reality the system is not normally in a stationary state (either the ground state or any excited state).Instead it is in thermal equilibrium at some temperature T , meaning that it is in a mixture of dierentstationary states. Statistical mechanics tells us that the probability of state |n〉 in this mixture is

pn =e−En/kBT

Z(8)

where kB is Boltzmann's constant. The normlization constant Z is determined by requiring that the proba-bility that the system is in some state is 1:

∞∑n=0

pn = 1. (9)

We thus obtain the partition function of the system,

Z =∞∑n=0

e−En/kBT . (10)

The above formula giving pn can be used to compute the average value of any given magnetic moment:⟨µj⟩

=∑n

pn〈n|µj |n〉 (11)

=∑n

e−En/kBT

Z〈n|µj |n〉 (12)

=1

Z

e−E0/kBT 〈0|µj |0〉+ e−E1/kBT 〈1|µj |1〉+ e−E2/kBT 〈2|µj |2〉+ · · ·

(13)

Having introduced the tools of Statistical Mechanics necessary to describe magnetic phenomena, we willnow use them to tackle paramagnetism. These are the simplest magnetic phenomena because they donot involve in any essential way interactions between magnetic moments (i.e. they are only collective in thesense that they result from the sum of many independent magnetic behaviours, not in the deeper sense thatwe will encounter later on).

Our rst take on the problem of paramagnetism will in fact neglect interactions between magnetic momentscompletely and assume that we can treat each magnetic moment independently. This means that theHamiltonian is

H = H1 +H2 + . . .+HN (14)

where each of the Hj describes the behaviour of a single magnetic moment µj . Then the state of the wholesystem can be specied by the individual states of the magnetic moments, considered in isolation, i.e. in theabove formuale we can intepret |n〉 as a state of one in particular of the magnetic moments and En as itsenergy.

2

3 Interaction of a magnetic moment with an applied eld

Before we plunge into the many-body problem, we need to learn one last thing about magnetic moments:their interaction with an applied mangetic eld. In Lecture 2 we already saw that a magnetic momentwill precess in a magnetic eld with the Larmor frequency ωL = γqB. Let us compute the energy that themagnetic moment posesses by virtue of being in the presence of the magnetic eld it is precessing about.

The torque that is responsible for Larmor precession is

G = γq L×B

(Lecture 2, Eq. 25)

which givesG = µ×B. (15)

Let us call θ the angle between B and µ. If the magnetic induction and the magnetic moment are parallel,θ = 0, we have G = 0. At nite angle θ we have G = µB sin θ. Therefore when the magnetic moment andthe magnetic induction are rotated adiabatically from θ = 0 to a nite angle θ the energy changes by anamount ∆E equal to the work done against the torque:

∆E =

ˆ θ

0Gdθ =

ˆ θ

0µB sin θ dθ = µB (1− cos θ) . (16)

Now, using µB cos θ = µ ·B we conclude that the energy change of a magnetic moment µ when a magneticinduction B is turned on is, up to a constant,

∆E = −µ ·B . (17)

Now from (Lecture 4, Eq. 24) we haveµ = gJ |µB| J/~. (18)

Moreover the component of the total angular momentum J along a given quantization direction is

Jz = ~m, m = −J,−J + 1, . . . , J − 1, J. (19)

All this leads toµz = gJ |µB|m, (20)

so we have (choosing the quantization direction to be parallel to the magnetic eld)

∆E = −gJ |µB| mB, wherem = −J,−J + 1, . . . , J − 1, J

(21)

Thus a magnetic eld splits degenerate angular momentum states, reecting that the energy of a magneticmoment depends on the relative orientation to the applied eld.

Let us look at an example from Lecture 4: the stationary states of an electron in a p orbital. In the absenceof spin-orbit coupling and without an applied magnetic eld, these states are six-fold degenrate. In Lecture4 we saw that spin-orbit coupling partially lifts this degenracy by splitting levels diering in the value of thetotal angular momentum, J (page 4 of the corresonding handout):

3

In a magnetic eld, the degenracy is much less: even without spin-orbit coupling, dierent values of m arenon-degenerate, so the only degeneracy we have left is that between states with the same value of m butdierent values of j:

Since in reality the latter degeneracy was already lifted by spin-orbit coupling, then that cannot be neglectedwe have no degeneracy left at all:

4

Lecture VI. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

1 Paramagnetism

Let us now consider a system made up of independent magnetic moments: a solid with magnetic ions, i.e.those with unlled shells that have a net magnetic moment (as discussed in previous lectures). Let us assumethatthere arer no interactions between the magnetic moments (we will see what is the eect of interactions inlater lectures). Consider rst the situation with no applied magnetic eld. Then assuming all the magneticmoments are identical and have angular momentum quantum number jtot. Every stationary state of themagnetic moment with magnetic quantum number mj = −jtot, ..., jtot is degenerate. Thus each one willcompensate with another one with the opposite value of that quantum number,−mj (this is true whether ornot spin-orbit coupling is present). Evidently these two states have values of µz which are of opposite signand equal in magntidue, so the results is that (choosing the equal energy fo all the states as the origin ofenergy)

〈µz〉 =

jtot∑m=−jtot

1

ZgJ |µB|m (1)

=1

ZgJ |µB|

jtot∑m=−jtot

m (2)

= 0 (3)

Now let us put in an applied magnetic induction:

〈µz〉 =

jtot∑m=−jtot

1

Ze−−gJ |µB |mB

kBT gJ |µB|m (4)

=1

ZgJ |µB|

jtot∑m=−jtot

meβgJ |µB |Bm (5)

[β ≡ 1/kBT ]. (6)

where we have introduced the very useful notation of Statistical Mechanics for the inverse temperature. Now

Z =

jtot∑m=−jtot

eβgJ |µB |Bm (7)

⇒ ∂Z

∂H= βgJ |µB|

jtot∑m=−jtot

meβgJ |µB |Bm. (8)

This allows us to write

〈µz〉 =1

βZ

∂Z

∂B(9)

id est

〈µz〉 = kBT∂

∂BlnZ. (10)

So all we have to compute is the partition function Z. Now from Eq. (1) we have

Z =

jtot∑m=−jtot

rm with r = eβgJ |µB |B (11)

= r−jtot

2jtot∑m=0

rm (12)

1

Remember the formula for a geometric series:9

M∑m=0

rm =rM+1 − 1

r − 1(13)

Applying this we get

Z = r−jtotr2jtot+1 − 1

r − 1(14)

=r2jtot+1 − 1

rjtot+1 − rjtot(15)

[r−

2jtot+1

2

r−2jtot+1

2

× ] =r

2jtot+1

2 − r−2jtot+1

2

r1/2 − r−1/2(16)

Now we put inr = eβgJ |µB |B = ex (17)

with

x ≡ gJ |µB|BkBT

(18)

and we get

Z =sinh

[(2jtot + 1) x2

]sinh

[x2

] . (19)

Thus for 〈µz〉 we obtain

〈µz〉 = kBT∂

∂BlnZ (20)

= kBTβgJ |µB|∂

∂xln

sinh[(2jtot + 1) x2

]sinh

[x2

] (21)

= gJ |µB|

2jtot + 1

2coth

(2jtot + 1)x

2− 1

2coth

x

2

(22)

The magnetization M = ρ 〈µz〉 is thus

M = ρgJ |µB|

2jtot + 1

2coth

(2jtot + 1)x

2− 1

2coth

x

2

(23)

9 Proof: Let (. . .) ≡ 1 + r + r2 + . . . + rM . Then r (. . .) = r + r2 + . . . + rM + rM+1= (. . .) + rM+1 − 1⇒r (. . .) = (. . .) +

rM+1 − 1⇒(. . .) = rM+1−1r−1

, Q.E.D.

2

Lecture VII. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In the last lecture we ended with an expression for the magnetization of a paramagnet, modelled as amacroscopic set of independent magnetic dipoles:

M = ρgJ |µB|

2jtot + 1

2coth

(2jtot + 1)x

2− 1

2coth

x

2

(Lecture 6, Eq. 23)

The variable x was given by

x ≡ gJ |µB|BkBT

(Lecture 6, Eq. 18)

Here ρ is the density of magnetic dipoles, jtot is the total angular momentum quantum number of each dipole,and gJ its gyromagnetic ratio. B is the externally-applied magnetic induction and T is the temperature. Weare now going to investigate what the above result means.

Since

coth ξ =eξ + e−ξ

eξ − e−ξ→ 1 as ξ →∞, (1)

for large x the expression between curly braces becomes

2jtot + 1

2coth

(2jtot + 1)x

2− 1

2coth

x

2≈ jtot, (2)

implying that the magnetization saturates in large elds:

M ≈ Ms ≡ ρgJ |µB| jtot for gJ |µB|B kBT. (3)

This is the saturation magnetization and it corresponds to all the magnetic moments being alignedwith the eld so their component in the eld direction has the maximum value, µz = gJ |µB| jtot (thismaximum value corresponds to the maximum allowed value of the magnetic quantum number, mjtot =−jtot,−jtot + 1, . . . , jtot − 1, jtot).

In contrast,

for small ξ, coth ξ ≈ 1 + ξ + 1− ξ1 + ξ − (1− ξ)

=1

ξ, (4)

giving2jtot + 1

2coth

(2jtot + 1)x

2− 1

2coth

x

2≈ 2jtot + 1

2

2

(2jtot + 1)x− 1

2

2

x= 0, (5)

i.e.M → 0 as gJ |µB|B/kBT → 0. (6)

This happens when the magnetic energy −µzB = −gJ |µB| jtotB that tends to align the moments in theeld direction is negligible compared to the energy of thermal uctuations, kBT , which tend to scramble thedirection of all the magnetic moments so they average to zero.

The above results suggest expressing our results in terms of the new variable

y ≡ gJ |µB| jtotBkBT

(7)

which is the ratio between the magnetic energy of a fully-aligned moment and the typical energy of a thermaluctuation. Evidently y = xjtot. The magnetization then takes the form

M = MsBjtot

(gJ |µB| jtotB

kBT

)(8)

1

where Bjtot (y) is the Brillouin function,

Bjtot (y) =

2jtot + 1

2jtotcoth

(2jtot + 1) y

2jtot− 1

2jtotcoth

y

2jtot

(9)

As we can see it is easier to overcome thermal uctuations and fully polarise a spin 1/2 magnetic momentthan one with a large spin. This is because the low spin has to choose between pointing along the eld (lowenergy) and against it (high energy):

j=1/2

There have to be very strong thermal ucatuations before the spin decides that it is worth pointint againstthe eld from time to time. In contrast, a large spin has many possible congurations that are only pointingslightly away from the applied eld:

2

j=2

Thermal uctuations can easily tilt the spin slightly in that way at little cost in terms of magnetic energy.

3

Lecture VIII. Jorge Quintanilla

Magnetism and Superconductivity (PH752)

In the rst half of this lecture we will complete our study of paramagnets by computing their magneticsusceptibility. We will see how measurements of magnetic susceptibility can be combined with measurementsof saturated magnetization (previous lecture) to obtain microscopic information about the magnetic momentsmaking up a magnetic material and to verify the quantum-mechanical picture of magnetic moments developedin Lectures 1-4. We will then say some brief words about diamagnetism and start to think, in the secondhalf of the lecture, about interactions between magnetic moments - which we have ignored until now.

1 Magnetic susceptibility

To conclude our initial study of paramagnetism, we are going to characterise the magnetic behaviour of ourparamagnet in terms of an important quantity called the magnetic susceptibility. This is dened by

χ =∂M

∂H. (1)

Note that the derivative is with respect to the applied magnetic eld, rather than the magnetic induction:B = µ0H. In particular we are interested in the low-eld susceptibility

χ0 =∂M

∂H

∣∣∣∣H→0

.

We have to expand to an order higher than we did before:

coth ξ =ex + e−x

ex − e−x≈

1 + ξ + 12ξ

2 + 1− ξ + 12ξ

2

1 + ξ + 12ξ

2 −(1− ξ + 1

2ξ2) =

2 + ξ2

2ξ=

1

ξ+ξ

3, (2)

to obtain

Bjtot (y) ≈ 2jtot + 1

2jtot

[2jtot

(2jtot + 1) y+

(2jtot + 1) y

2jtot × 3

]− 1

2jtot

[2jtoty

+y

2jtot × 3

](3)

=

(2jtot + 1

2jtot

)2 y

3−(

1

2jtot

)2 y

3=jtot + 1

jtot

y

3(4)

Thus

M ≈Msjtot + 1

jtot× gJ |µB| jtotµ0H

3kBTfor

gJ |µB| jtotµ0HkBT

1 (5)

and hence

χ0 = Msjtot + 1

jtot× gJ |µB| jtotµ0

3kBT(6)

= µ0ρg2Jµ

2Bjtot (jtot + 1)

3kBT(7)

We now can recognizejtot (jtot + 1) = J2/~2 (8)

and thereforeg2Jµ

2Bjtot (jtot + 1) = [(gJ |µB| /~) J ]2 . (9)

But gJ |µB| /~ is the gyromagnetic ratio for our magnetic moment (Lecture 4, Eq. 3), so the expressionbetween square brackets is nothing but the magnitude of the magnetic moment,

µ =√⟨µ2⟩

= gJ |µB|√jtot (jtot + 1), (10)

1

Hence

χ0 = µ0ρµ2

3kBT, (11)

which is the celebrated Curie law:

Note that the Curie law is independent of the value of µ, i.e. it works just as well for highly quantum-mechanical jtot = 1/2 magnetic moments as for almost-classical high-jtot moments.

2 Determining the magnetic moments' parameters

Plotting χT vs T gives a constant and allows the

...identication of the value of µ:

Therefore one can always use the Curie law to determine µ and therefore the combination of parametersg2Jjtot (jtot + 1). Although at low temperatures there may be deviations due to interactions between mag-netic moments (whcih we have neglected here), at suciently high temperatures, when kBT becomes greaterthan the characteristic energy of those interactions, the Curie behaviour is recovered.

Additionally, one could use a measurement of the saturation magnetization, Eq. (3), to determine gJjtot.Again, interaction eects may be important at moderate elds, but they become negligible at large enoughelds and do not alter the value of the saturation magnetization.

Thus from these two measurements we can deduce both the g-factor gJ and the angular momentum quantumnumber jtot of our magnetic moments.

It is experiments of this type that enable us to check that Hund's rules do work, and also to establish wherethey don't. More generally, they allow us to conrm, even in systems with quite strong interactions, the

2

validity of our microscopic picture of individual, quantum-mechanical magnetic moments as the buildingblocks of magnetism.

The quantum-mechanical nature of magnetic dipole moments has a spectacular manifestation in the abovemeasurements. For a classical magnetic moment, we would expect saturation to occur when the full magneticmoment µ is aligned along the eld: µz = µz,max = µ (z is the direction of the magnetic eld). Quantum-mechanically, on the other hand,

µz,max/µ = jtot/√jtot (jtot + 1). (12)

For large jtot, we have√jtot (jtot + 1) ≈ jtot and the classical behaviour is recovered. But for magnetic

moments with small jtot there can be a signicant dierence. In the extreme quantum-mechanical limit ofthe smallest possible angular momentum, jtot = 1/2, we have jtot/

√jtot (jtot + 1) = 1/

√3 ≈ 0.58 i.e. the

maximum polarization a magnetic moment can achieve is little more than half the actual size of the spin!This can be appreciated in the gure on page 2.

3 Diamagnetism

So far we have always focused on systems where the atoms have unlled shells which can possess nitemagnetic moments in zero eld. It turns out that even systems with lled shells have a magnetic response,as the atoms can acquire a magnetic moment when they polarise in an applied magnetic eld. However,unlike paramagnetism, diamagnetism is quite unrelated to the more intersting interaction-induced magneticphenomena that we will study later on in the course. We will therefore not say more about diamagnetismfor the time being. (We will return to it when we tackle superconductivity, an interaction-induced phase ofmetals one of whose dening characteristic is perfect diamagnetism. In that sense supeconductivity can beregarded as a very exotic type of itinerant magnetic state.)

4 Magnetic interactions: the energy scales

In Lectures 1-4 we built the Lego bricks out of which magnets are made, namely the magnetic moments ofindividual electrons and atoms. In Lectures 5-7 we showed how the collective behaviour of many individualmagnetic moments may lead to macroscopic magnetic behaviour - in particular, to paramagnetism, whichis the ability of some materials to develop a magnetization in repsonse to an applied magnetic eld. Toshow this, we treated the many magnetic moments in a paramagnet as non-interacting entities, each oneresponding to the applied eld and to thermal uctuations independently. However, there are importantmagnetic phenomena that do not conform to this description. In ferromagnetic materials, for example,a magnetization develops even in the absence of an externally-applied eld. This is contrary to what weshowed for our model of indepedent magnetic dipoles, where each dipole points in a dierent direction so themagnetization averages to zero unless a eld is applied. Note that ferromagnetism was the rst magneticstate to be discovered and indeed that it was through ferromagnetism that magnetic forces in general cameto be known thousands of years ago (remeber from Lecture 1 that magnet comes from magnet lithos, thestone from Magnesia, a lodestone and therefore a ferromagnet). An inescapable consequence of this is thatthe knowledge we have built up since the start of this course is still a couple of millenia behind!

Of course we do have now some understanding of where the magnetization of a ferromagnet comes from: itis an emergent phenomenon resulting from interactions between magnetic moments. These interactionstend to make the moments line up. This means that it becomes energetically favourable for the magneticdipole moments to collectively make up their minds on a single direction that they all point in, so they canall minimize that interaction energy.

The above paragraph gives the correct qualitative picture, which we will make quantitative in the followinglectures, of how ferromagnetism (and also other magnetized ground states) emerges. However, it still leavesus facing the problem of the mechanism: where did the interaction between magnetic moments come fromin the rst place? In the reminder of this lecture we will take our rst steps towards tackling this problem,by identifying where the energy dierence which lines up magnetic moments could come from.

3

5 Magnetostatics

The rst, obvious candidate are the interactions between magnetic moments and the eld created by othermagnetic moments. This is, in fact, what makes two macroscopic magnets (e.g. two pieces of lodestone)attract each other. As we saw in Lecture 2 an individual magnetic dipole µ creates an induction around itgiven by the formula

B =µ0

4πr3[3 (µ · r) r− µ]

(Lecture 2, Eq. 12)

Let us now consider a second magnetic dipole, µ′, and ask what energy E it has due to being in the presenceof the magnetic induction due to the rst dipole. Substituting the above formula for B in the result we havealready derived for the energy of a dipole in a generic magnetic induction,

E = −µ′ ·B. (13)

(Lecture 5, Eq. 17),

we haveE =

µ04πr3

[µ′ · µ− 3 (µ · r)

(µ′ · r

)], (14)

where r is the vector that goes from dipole µ to dipole µ′. If we assume that the magnitudes of the magneticdipole moments are

µ, µ′ ∼ |µB| =e~

2me= 9.274 10−24Am2 (15)

and that the distance between the to dipoles is

r ∼ 5oA, (16)

we have, using µ0 = 4π10−7NA−2,

E ∼ µ04πr3

µ2B =4π10−7NA−2

4π (5 10−10m)3(9.274 10−24Am2

)2 ≈ 10−25J. (17)

Now let us equate this to a thermal energy

kBT = E ⇒ T =E

kB=∼ 10−25J

1.4 10−23JK−1∼ 0.01K. (18)

This is clearly far too small to explain ferromagnetic transition temperatures of order 1000K or more encout-nered in many materials. For example, the Curie temperature of Fe is 1043K, while its magnetic moments(determined from high-temperature susceptibility data in the paramagnetic state above TC , as describedwhen we dealt with paramagnets) have a size of 2.2µB, corresponding to a temperature scale of ∼ 0.05K.So whatever the interactions between magnetic moments leading to ferromagnetism in Fe, they have to leadto interaction energies ∼ 5 orders of magntidue greater! Clearly another source for these energies has to besought.

(That said, in some materials the dipolar interactions are important. In spin ices, which are pyrocholore-structure magnets with Dy or Ho ions, the magnetic dipole per ion is ∼ 10µB, which makes µ2 100 timeslarger meaning that dipolar interactions become important at temperatures ∼ 1K or more, where much ofthe interesting physics of these materials such as emergent magnetic monopoles happens.)

6 Electrostatics

What interaction energy is ∼ 1000K in magnets? Consider the Coulomb interaction between two electronsin adjacent atoms. The energy of one electron in the eld of another electron a distance r apart is

E =1

4πε0

e2

r. (19)

4

Substituting the same distance as above, i.e. r ∼ 5oA, the charge of the electron, e ∼ −1.6 10−19C, and the

electric permitivity of free space, ε0 = 8.85 10−12Fm−1, we obtain

E ∼ 1

10−10Fm−1(10−19C)2

10−10m= 10−18J (20)

aso the corresponding temperature is

T ∼ E

kB∼ 10−18

10−23= 105K. (21)

Now we are talking! The Coulomb repulsion between electrons in dierent atoms is a much greater energyscale than any magnetic interactions, and it is certainly large enough to explain phase transitions with criticaltemperatures ∼ 1000K or more.

The problem is, how does an electrostatic repulsion become an interaction between magnetic moments? Thekey is that the Coulomb interaction inuences the way the electrons move and this is correlated with thespin of the electrons through the Pauli exclusion principle. This is called the exchange interaction. In thenext lecture we will see how this works through a very simplied calculation which nevertheless captures thephysics of what is going on.

(Note: a similar problem presents itself for high-temperature superconductivity: the energy scale of thephonon-mediated electron-electron interaction that is responsible for superconductivity in conventional su-perconductors is far too small to explain the superconducting critical temperatures ∼ 100K or more of thehigh-Tc's. As in magnetism, the electrostatic interaction between electrons seems to be strong enoug. On theother hand for the high-Tc superconductors the mechanism whereby the Coulomb replusion between elec-trons is converted into a pairing interaction that can lead to superconductivity is still unclear. In contrast,for ferromagnets we do understand quite well how Coulomb repulsion can become an interaction tending toalign magnetic moments. This is what we will do in the next lecture.)

5

Jorge Quintanilla

Magnetism and Superconductivity (PH752)

Problem Sheet 1

Information needed to tackle these problems can be found in the handouts for Lectures 1-6X8. You will need toconsult also a periodic table of the elements with electronic structure information and some physical constants(both can be found for example behind the front and back covers, respectively, of Setephen Blundell, Magnetismin Condensed Matter, OUP 2001). Give all results in SI units.

Problems marked with an asterisk (*) will be assessed.

Problem 1: Consider a magnetic moment of strength equal to one Bohr magneton. If the moment where due

to electric charges orbiting a nucleus at a distance of 2oA, what would be the corresponding current?

If this was due to a single electron, how many r.p.m.'s (revolutions per minute) would the electron bemaking around the nucleus? [Ignore the spin of the electron.]

Problem 2: Consider a magnetic moment, of strength equal to one Bohr magneton, sitting at the origin ofcoordinates and pointing along the z direction. Compute the direction and magnitude (in SI units)

of the corresponding magnetic induction at distances of 5oA and 10

oA along the z axis, the x axis and

the (1, 0, 1) diagonal (1oA = 0.1nm). On the basis of these six values and of symmetry considerations,

sketch the vector eld B around the magnetic moment.

Problem 3: Enumerate the quantum numbers l, s, j,mj of all the states available to an electron in an f -shell (l = 3 orbitals). For each state, give the values of the electron's orbital, spin, and total angularmomenta as well as the projection of the latter along the quantisation axis. Sketch the partial liftingof degeneracy due to spin-orbit coupling.

Problem 4: (*) Use Hund's rules to derive the magnetic quantum numbers s, l, j of a Pm3+ ion, and hencethe standard `term symbol' describing this ion's mangetic state (describe in detail the application ofeach of the Hund rules). How much degeneracy does this state have? Derive the g-factor and use it tocalculate the size of the corresponding magnetic moment. Sketch how the degeneracy is lifted underan applied eld of increasing strength.

Assume a magnetic eld equivalent to 1T is applied to a salt containing Pm3+ ions. Calculate, usingthe results obtained above, the value of the energy split between the state that is most aligned with theapplied eld and the next one downxxxxup in energy. Use this value to estimate the temperature at whichthe magnetization of the salt will be, to a good approximation, completely saturated by the applied

eld. If there are 2 magnetic moments per unit cell, and the unit cell dimensions are 5oA × 5

oA × 5

oA,

what is the staurated value of the magnetization? How does it compare to the strength of the appliedeld?

Finally, compute the magnetic susceptibility of the salt at 1K.

Problem 6X5: Obtain an expression for the partition function of a magnetic moment with jtot = 1 in anapplied magnetic eld. Derive expressions for the magnetization and the magnetic susceptibility atzero eld. Verify that the expressions you obtain are particular cases of the more general formulaevalid for arbitrary jtot which we obtained in the lectures.

Problem 7X6: Calculate the magnetic moment of Mn3+ rst using Hund's rules and then assuming orbitalquenching (i.e. ltot = 0). Ignore the crystal eld in both cases. Write the term symbols for bothstates.

Experimentally for a particular salt containg this ion it is found that the susceptibility is χ0 = 23.2×µ0ρµ2BkBT

. Is the orbital angular momentum quenched in this system? Would the result be the same if thecrystal eld were larger or smaller? In what way?

1

Jorge Quintanilla

Magnetism and Superconductivity (PH752)

Problem Sheet 1

SOLUTIONS

Problem 1 solution:

[Solution]

Problem 2 solution:

[Solution]

Problem 3 solution:

[Solution]

Problem 4 solution:

The atom Pm has the electronic structure [Xe]4f56s2 (from the periodic table). Removing 3 electronswill leave us with [Xe]4f4 [1 point]. The f -shell has l = 3 and therefore enough room for only2 (2l + 1) = 14 electrons, so it is less than half-lled by our 4 electrons (note I am using l, s to refer tothe orbital and spin angular momentum quantum numbers of individual electrons; I will use stot, ltotand jtot to denote the spin, orbital, and total angular momentum quantum numbers of the ion as awhole). These are the states that are available for each individual electron:

l s ml ms

3 1/2 3 1/2-1/2

2 1/2-1/2

1 1/2-1/2

0 1/2-1/2

-1 1/2-1/2

-2 1/2-1/2

-3 1/2-1/2

Let us apply Hund's rst rule: distribute electrons among the spin states so as to generate the highestpossible value of stot. This is achieved by placing all 4 electrons in the same spin state, ms = 1/2,with their spins pointing up (we can place all electrons in the same spin state because the shell is lessthan half-lled).

This still leaves some freedom as to where the electrons go, so we need to invoke Hund's second rule:in order to maximise ltot, we put ll rst the ms = 1/2 states with the highest value of ml:

1

l s ml ms

3 1/2 3 1/2 •-1/2

2 1/2 •-1/2

1 1/2 •-1/2

0 1/2 •-1/2

-1 1/2-1/2

-2 1/2-1/2

-3 1/2-1/2

We now have completely distributed all the four electrons among the available states. The spin andorbital angular momenta quantum numbers have worked out to be

stot = 4× 1/2 = 2 [4 points]

andltot = 3 + 2 + 1 + 0 = 6 .[4 points]

Now to work out how stot and ltot combine to give the total angular momentum quantum number Ineed to apply Hund's third rule: since the shell is less than half-lled, I choose the minimum possiblevalue for

jtot = |stot − ltot| = 4 .[4 points]

The term symbol is ACB, where A = 2stot + 1, B = jtot and C = ltot (with ltot = 0, 1, 2, 3, 4, 5, 6, . . .denoted S, P,D, F,G,H, I, . . .) For Pm3+ we have 2stot+1 = 2×2+1 = 5, jtot = 4 and ltot = 6 ≡ I,so the term symbol is

term symbol = 5I4 .[2 points]

Since jtot = 4, we have

degeneracy = 2jtot + 1 = 2× 4 + 1 = 9 , [10 points]

which corresponds to the 9 distinct states

mj = −4,−3,−2,−1, 0, 1, 2, 3, 4.

The g-factor is given by

gJ =3

2+stot (stot + 1)− ltot (ltot + 1)

2jtot (jtot + 1)

=3

2+

2 (2 + 1)− 6 (6 + 1)

2× 4 (4 + 1)

3

2+

6− 42

8× 5=

3

2− 36

40=

3

2− 9

10=

12

20

=3

5. [7.5 points]

The magnetic moment isµ = γJ,

2

where γ = gJ |µB| /~ is the gyromagnetic ratio. Therefore its size will be

µ =√µ2 = γ ×

√J2 = (gJ |µB| /~)×

√jtot (jtot + 1) ~2

= gJ |µB|√jtot (jtot + 1)

This yields

µ =3

5|µB|

√6 (6 + 1) ≈ 3.89 |µB| ≈ 36.1 10−24 JT−1 . [7.5 points]

Under an applied magnetic induction B the Pm3+ ion will experience an energy change

∆E = −µ ·B = −µzB,

where we have chosen the z-axis to be pointing along the applied eld direction. Here

µz = γJz = (gJ |µB| /~)× ~mjtot= gJ |µB|mjtot

with

mjtot= −jtot,−jtot + 1, . . . , jtot − 1, jtot= −4,−3,−2,−1, 0, 1, 2, 3, 4,

the 9-fold degeneracy into 9 distinct energies that change linearly with the magnetic eld with dierentslopes:

∆E = −4gJ |µB| ×B∆E = −3gJ |µB| ×B∆E = −2gJ |µB| ×B∆E = −gJ |µB| ×B∆E = 0

∆E = gJ |µB| ×B∆E = 2gJ |µB| ×B∆E = 3gJ |µB| ×B∆E = 4gJ |µB| ×B

This is shown by the following plot (x ≡ gJ |µB|B, y ≡ ∆E):

[10 points]

3

The dierent lines correspond to mjtot= −4,−3,−2, . . . , 3, 4 (from top to bottom).

The state most aligned with the eld has energy

Emost aligned = −4gJ |µB| ×B

The next one up in energy hasEnext one up = −3g |µB| ×B

The dierence is therefore∣∣∣Emost aligned − Enext one up∣∣∣ = gJ |µB| ×B

=3

5|µB| ×B

=3

5× 9.274 10−24JT−1 × 1T

= 5.56 10−24J [10 points]

Suppose we have applied a eld of 1T. This will tend to align the moments with the eld. Atwhich temperature T will the moments be completely aligned? That will happen when the thermaluctuations, that have energy kBT , are no longer strong enough to bring a magnetic moment from themost aligned state to the next one up in energy, i.e. when

kBT .∣∣∣Emost aligned − Enext one up∣∣∣

⇒ T =∣∣∣Emost aligned − Enext one up∣∣∣ /kB

≈ 5.56 10−24J/(1.38 10−23JK−1

)≈ 0.4K . [10 points]

Below this temperature, to a good approximation, all magnetic moments are aligned with the eld.

The density of magnetic moments is

ρ = 2/(5 10−10m× 5 10−10m× 5 10−10m)

= 1.6 1028m−3 [3 points]

therefore the saturated magnetization is

Ms = ρµz,max = ρ× 4gJ |µB|

= 1.6 1028m−3 × 4× 3

5× 9.27 10−24JT−1

= 3.84 1028 |µB|

= 3.56 105 JT−1m−3 . [7 points]

The magnetic induction isB = µ0 (M + H) .

Outside the sample B = Bext, H = Hext (the applied eld) and M = 0 so

Bext = µ0Hext.

Thus the externally-applied eld is

Hext =1

µ0Bext =

1

4π 10−7Hm−11T

≈ 7.96 105 JT−1m−3

≈ 2.24×Ms ⇒ Ms ≈ 0.46×Hext . [10 points]

4

The (zero-eld) magnetic susceptibility is [Lect. VIII, Eq. (11)]

χ0 = µ0ρµ2

3kBT.

At T = 1K this gives

χ0 ≈ 4π 10−7Hm−1 × 1.6 1028m−3 ×(2.49 10−23 JT−1

)23× 1.3807 10−23JK−1 × 1K

≈ 0.3 . [10 points]

Problem 5 solution:

[Solution]

Problem 6 solution:

[Solution]

5

Lecture IX. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In the past three lectures we discussed paramagnetism - the tendency of some magnetic materials toalign their magnetic moments with an applied magnetic eld, leading to an additional magnetic induction.Paramagnetism is a collective phenomenon, since many magnetic moments have to act in unison, and we thushad to describe it using the tools of Statistical Mechanics (i.e. the partition function). On the other hand,our discussion of paramagentism treated each magnetic moment in the material as an independent entity.Such theory falls short when we try to use it to describe magnetic materials that present a spontaneousmagnetization even in the absence of an applied eld, such as ferromagnets or antiferromagnets. Indeedour theory of paramagnets predicted (correctly) zero magnetization in the absence of an applied eld, whichis not what happens in ferromagnets or antiferromagnets.

At the end of the last lecture we put forward the idea that in order for the magnetic moments in a materialto spontaneously self-organise without an external stimulus it is necessary to take into account interactionsbetween dierent magnetic moments. We discussed the possible origin of such interactions. We discarded apurelymagnetostatic origin on the grounds that the energy scales that come out of such assumption are fartoo small - the eects should not be noticable above temperatures 1K, whereas ferromagnetic transitiontemperatures can be ∼ 1000K or more. We then pointed out that electrostatic interactions betweenelectrons are much stronger and indeed perfectly capable, at least as far as energy scales are concerned, tolead to magnetism. However, we did not discuss the mechanism, i.e. how can repulsion between electronsgive rise to an interaction tending to align magnetic moments with respect to each other in a particular way?That is the question that we address in what follows.

1 The exchange interaction: qualitative picture

Consider, for simplicity, an ion that has an unlled shell with a single electron. Moreover let us assume thatthe shell has only s orbitals so the magnetic moment of the ion comes exclusively from the electron's spin.

Now let us bring near it other identical ions, to form a crystal. Our unpaired electron will be repelled by theelectrons in the other ions and will therefore tend to localise more tightly on its own ion. In this way, theelectron minimizes the Coulomb interaction energy between our electron and all the other electrons:

U =1

4πε0

∑j

e2

|r− rj |(1)

where r and rj are the position vectors of our electron and of the jth electron, respectively.

If that were all, we would have a paramagneticMott insulator with each unpaired electron tightly localisedaround its ion. Indeed, since the Coulomb interaction energy does not depend on where the spins of ourelectrons are pointing, the magnetic moments of such system would be completley disordered in the absenceof an applied magnetic eld:

However, our electron also has kinetic energy,

T =p2

2me, where p = −i~∇2

r. (2)

To minimize this energy the electron would want to be in a plane wave state,

〈r|p〉 ∝ eip.r/~. (3)

1

However the probability density of a plane wave extends uniformly across the lattice:

-1

-0.5

0

0.5

1

1.5

2

-20 -15 -10 -5 0 5 10 15 20

wa

ve

fu

nctio

n

position (a.u.)

Real partImaginary part

|...|2

Evidently this is very bad from the point of view of the Coulomb interaction as the electron overlaps withall the other electrons in the crystal!

The optimal situation is intermediate between the electron being completely localised and being completelyspread out.

Indeed Heisenberg's uncertainty principle tells us that there is a relationship between the spread ofvalues taken by an electron's position along a given coordinate axis, ∆x, and by its momentum along thatsame axis, ∆p:

∆x∆p ≥ ~2. (4)

If the electron is highly localised, ∆x is small, implying large ∆p which means that the electron's wave packetis made up of many plane waves, some of them with quite high momentum, impying a higher kinetic energy.Thus the electron will be neither completely localised on its ion, nor extended throughout the whole lattice,but will be centred on its ion but spread onto a few of the neighbouring ions:

The system remains a Mott insulator, but there is some additional interaction between dierent ions. (Themagnetism of metals will be discussed later on in the course.)

Now, here comes the crucial part: when the electron is sitting tightly on its ion its spin can point either upor down indepedently of what the other electrons in the other ions are doing; however, hopping onto anotherion is more or less likely depending on what the direction of our electron's spin is relative to that of thesurrounding electrons. For instance:

• The electron may hop onto an orbital that already contains one electron, so by Pauli's exclusionprinciple (Hund's rule #0) its spin must be the opposite to that of that electron.

• The electron may hop onto a shell where there are free orbitals and all electrons have their spins aligned,in which case (by Hund's rule #1) it will prefer to have its own spin aligned with the preferreddirection.

Either way the ability of the electron to hop onto an adjacent ion, and therefore to lower its kinetic energyby spreading out its wave function, depends on the direction of its spin. The electron will therefore achievelower energies when its spin direction maximises this ability. In this way the direction of its spin has becomecorrelated with that of the spins of the adjacent unpaired electrons. The resulting interaction betweenadjacent magnetic moments is called the exchange interaction. The two cases mentioned above correspondto anti-ferromagnetic exchange (which tends to make adjacent spins point in opposite directions) andferro-magnetic exchange (tending to make all spins parallel to each other).

2

2 The exchange interaction: a very simple model

Let us now make this a bit more quantitative using a simple calculation. We will model the behaviour ofour electron using a very small Hilbert space consisting of three states:

|L ↑〉, |L ↓〉, |R ↓〉 (5)

These states correspond to the electron being, respectively, in its original ion (which we have artbitrarilylabelled L for Left) with its spin pointing up, in the same ion with its spin pointing down, and in theadjacent ion with its spin also pointing down. The |R ↑〉 state is not allowed because we assume that theadjacent ion already contains an electron with its spin pointing up - so by Pauli's exclusion principle ourelectron cannot hop onto R unless its spin points down.

Let us now construct a matrix describing the Hamiltonian of our electron for this simple model.

Since there are no applied elds, if we ignore the inuence of the neighbouring ion the energies that theelectron has when it points up or down are equivalent:

〈L ↑ |H|L ↑〉 = 〈L ↓ |H|L ↓〉. (6)

We can choose this to be the origin of energy i.e. the electron staying on its site with the spin up or downhas energy = 0:

〈L ↑ |H|L ↑〉 = 〈L ↓ |H|L ↓〉 = 0. (7)

On the other hand if the electron is on the other ion, R, then its electrostaic (Coulomb) energy increases bya certain amount U due to the stronger repulsion with the other electron when they are both on the samesite:

〈R ↓ |H|R ↓〉 = U. (8)

This extra energy cost is called the Hubbard U after the British physicist John Hubbard, who introducedthe concept in his eponymous model of 1963.

Finally, the Hamiltonian must be able to make our electron hop spontaneously from L to R, or vice versa:

〈L ↓ |H|R ↓〉 = 〈R ↓ |H|L ↓〉 = −t. (9)

Thus this simple model contains the three essential ingredients of the exchange interaction:

3

1. Coulomb repulsion U tending to localise the electron on its own ion.

2. Kinetic energy −t tending to delocalise the electron onto the neighbouring ion.

3. A local constraint on the electron's spin when it is on the neighbouring ion (in this case, coming fromthe Pauli exclusion principle).

I emphasize that this is a very simplistic model. In particular, it makes no real sense to discuss the dynamicsof one electron while considering the spin and position coordinates of the other electron xed - evidently ifour electron can hop from L to R, so can the electron on the other ion hop from R to L! However the mainresult we will obtain is quantitatively correct, which is why we are using it.

In the above basis, and using the above results, the Hamiltonian matrix is

H =

0 0 00 0 −t0 −t U

. (10)

The diagonal matrix elements represent the energies associated with being in a particular state:

H1,1 = 〈L ↑ |H|L ↑〉 = 0 (11)

H2,2 = 〈L ↓ |H|L ↓〉 = 0 (12)

H3,3 = 〈R ↓ |H|R ↓〉 = U (13)

The o-diagonal matrix elements represent the quantum-mechanical probability amplitude to spontaneouslygo from one quantum-mechanical state to another one.

H2,1 = H1,2 = 〈L ↓ |H|L ↑〉 = 0 (14)

H3,1 = H1,3 = 〈R ↓ |H|L ↑〉 = 0 (15)

H3,2 = H2,3 = 〈R ↓ |H|L ↓〉 = −t (16)

Note we have used that the Hamiltonian matrix has to be Hermitian and therefore, given that all its elementshappen to be real, symmetric. We have also assumed that the Hamiltonian does not feature any terms capableof ipping the elctron's spin.

To nd the energies E of stationary states |ψ〉 of this Hamiltonian we have to solve the time-independentSchrödinger equation:

H|ψ〉 = E|ψ〉. (17)

We do that in the next lecture.

4

Lecture X. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

Continuing from the previous lecture, the eigenvalues E can be obtained by solving the characteristic poly-nomial:

|H− E1| = 0, (1)

where

1 =

11

1

(2)

is the unit matrix. This gives ∣∣∣∣∣∣ −E −E −t

−t U − E

∣∣∣∣∣∣ = 0 (3)

⇒ E2 (U − E) + t2E = 0 (4)

⇒ E =

t

U2t −

√(U2t

)2+ 1

,

0, or

t

U2t +

√(U2t

)2+ 1

(5)

We see that the energy depends on1

λ≡ U

2t, (6)

a dimensionless constant measuring the relative strength of the hopping t and the electrostatic repulsion U .Our description of our electron hopping only to the adjacent atom is valid if this quantity is large, i.e. ifthe repulsion U hemming in the electron is much larger than the hopping amplitude (otherwise, we have ametal, not an insulator; the magnetism of metals will be discussed later on in the course).

Expanding the expressions between curly braces to linear order in λ we obtain

E =

t

1λ −

√1λ2

+ 1≈ −tλ2 = − t2

U +O(t3

U2

),

0, or

(similarly) ≈ U + t2

U +O(t3

U2

),

where we have used the Taylor expansion√1

λ2+ 1 =

1

λ+λ

2+ o

(λ2). (7)

Let us now see what is the structure of the ground state |ψ〉. This is obtained by plugging the correpsondingenergy back into the time-independent Schrödinger equation:

0 0 00 0 −t0 −t U

〈L ↑ |ψ〉〈L ↓ |ψ〉〈R ↓ |ψ〉

= − t2

U

〈L ↑ |ψ〉〈L ↓ |ψ〉〈R ↓ |ψ〉

. (8)

Note that we have expressed the state |ψ〉 as a column vector in terms of its projections on the various basisstates |L ↑〉, |L ↓〉, |R ↑〉. This is a system of three equations with the three unkowns 〈L ↑ |ψ〉, 〈L ↓ |ψ〉, 〈R ↑

1

|ψ〉 which when solved yields

(〈L ↑ |ψ〉, 〈L ↓ |ψ〉, 〈R ↓ |ψ〉) =

0, 1,

√(U

2t

)2

+ 1− U

2t

(9)

=

(0, 1,

√1

λ2+ 1− 1

λ

)≈(

0, 1,λ

2

)(10)

=

(0, 1,

t

U

)(11)

(note that we have not normalised this state). Thus the ground state does correspond to basically the |L ↓〉state with a bit of |R ↓〉 mixed in, as we expected.

In contrast, the rst excited state, with energy E = 0, is localised on the L site with the spin pointing up:

(〈L ↑ |ψ〉, 〈L ↓ |ψ〉, 〈R ↓ |ψ〉) = (1, 0, 0) . (12)

Our main result is thus that the ground state involves the electron being localised on L, with its spin pointingdown, and a little bit of spillover onto R, and the corresponding ground state energy is lowered by an amount

J ≈ −t2/U (13)

with respect to the state where the particle is competely localised on L, with the spin up, which is the rstexcited state. This explains why the particle will choose to have its spin pointing down.

1 Superexchange

A distinction is sometimes made between direct exchange and superexchange. Direct exchange is whatwe have described above. Superexchange occurs when the magnetic ions are separated by an intervening,non-magnetic ion. Suppose for example, following Blundell, that we have a transition metal oxide featuringtwo transition metal ions (M) with, for simplicity, one unpaired electron each occupying identical d orbitals.Imagine that the two ions are separated by a non-magnetic oxygen ion featuring two electrons in a single porbital:

The electrons in the d orbitals cannot hop directly from M to M because they are too far away. Neithercan they hop onto the O2− ion because the only p orbital that points in the direction of the M ions isfull. However, the two electrons in that p orbital can lower their kinetic energies by hopping onto the twoM ions. That makes them couple antiferromagnetically to the two electrons in the two M ions. Since thetwo electrons in the p orbital must also have their spins pointing in opposite directions (by Hund #0), anantiferromagnetic correlation is induced between the two M ions.

2

2 Magnetic hamiltonians: the Heisenberg model

Evidently the situation in a real magnet is a lot more complicated than what we have described above.However, for magnetic insulators it is often possible to write a Hamiltonian that depends only on the angularmomentum operators on each ion:

H (J1,J2,J3, . . .) . (14)

This is based on the fact that the unpaired electrons are mostly localised on each magnetic ion (see above).

There are some things that can be said quite generally about the specic form of such Hamiltonian. Let usstart by carrying out a Taylor expansion in terms of all those variables:10

H (J1,J2,J3, . . .) = constant +∑i

H(1)i · Ji +

∑i,j

Ji ·H(2)i,j · Jj + . . . (17)

We can simplify the above Hamiltonian considerably by a few general considerations:

• The rst term on the RHS can be ignored by choosing the origin of energy appropriately

⇒ H =∑i

H(1)i · Ji +

∑i,j

Ji ·H(2)i,j · Jj + . . . (18)

• In the absence of an externally-applied magnetic eld the system must be symmetric (i.e. invariant)under the ipping of all the angular momenta:

Jj → −Jj , (19)

meaning that the linear term, and inded all odd-order terms, must vanish: H(1) = H(3) = H(5) =. . . = 0

⇒ H =∑i,j

Ji ·H(2)i,j · Jj +O

(J4)

(20)

• Another symmetry that we can use for many crystals is the invariance of the system under identical

rotations of all the spins. Then the tensor H(2)must be a scalar: H(2)i.j = H(2)

i,j

⇒ H =∑i,j

H(2)i,j Ji · Jj +O

(J4)

(21)

• Moreover, we can assume, for an insulator,that the size of the magnetic moment on each site is xed(so the magnetic interactions only change the moments' directions, not their size). This will be givenby its angular momentum quantum number, jtot. Assuming for simplicity that all sites have the samevalue of jtot we have Ji · Ji = J2 = ~2jtot (jtot + 1) [see Lecture IV, page 2]. The diagonal elementsof the quadratic term thus take the form∑

i,i

H(2)i,i J

2i =

∑i,i

H(2)i,i ~

2jtot (jtot + 1) for i = j. (22)

This does not depend on the direction the spins are pointing in and therefore it is another constantthat can be eliminated from the Hamiltonian by a further oset of the origin of energy

⇒ H =∑i 6=jH(2)i,j Ji · Jj +O

(J4). (23)

10 Here we have used a shorthand notation whereby H(1)i is a vector, H(2)

i,j is a third-rank tensor, and so on:∑i

H(1)i · Si =

∑i

∑α=x,y,z

H(1)i,αS

αi (15)

∑i,j

Si ·H(2)i,j · Sj =

∑i,j

∑α,β=x,y,z

Sαi H(2)i,j,α,βS

βj (16)

. . .

3

• To conclude we shall assume that the higher-order terms can be neglected to obtain the (very famous!)Heisenberg model

H = −∑i 6=j

Ji,j Ji · Jj , (24)

where we have introduced the so-called exchange constants11

Ji,j ≡ −~2H(2)i,j . (25)

Note that we have introduced a sign convention whereby J > 0 for ferromagnetic coupling (favouringparallel spins) and J < 0 for antiferromagnetic coupling (favouring anti-parallel spins). We have alsoreplaced each angular momentum operator Ji with its diemnsionless version

Ji ≡ Ji/~

so that the exchange constant Ji,j has the dimensions of an energy.

If the orbital angular momentum is quenched, L = 0⇒ J = S [Lecture 1, page 8], and we assume that Ji,jis negligible for all but nearest-neighbour sites, the Heisenberg model reads

H = −J∑〈i,j〉

Si · Sj , (26)

where 〈i, j〉 indicates that i, j must be nearest neighbours. Then on the basis of the earlier discussion weexpect that for nearest-neighbour exchange coupling J ∼ −t2/U [Eq. 2], with the meaning of t and of Uthat we discussed above.

More generally, the exchange coupling J is related to the microscopic parameters governing the dynamics ofindividual electrons in a more complicated way.

In the next lecture we will start to tackle the problem of solving Hamiltonians of the type given above usingmean eld theories.

11 The exchange coupling between sites i and j, Ji,j , is not to be confused with the spins on those sites, Ji and Jj .

4

Lecture XI. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In the last lecture we argued for a Hamiltonian describing magnetic insulators. It features unpaired electronsand exchange constants arising from the interactions between the spins of those electrons. The exchangeinteraction comes from an interplay between the Coulomb repulsion between unpaired electrons on dierentions, tending to localise them, and their kinetic energy, tending to delocalise them. Here we are goingto explore the consequences of such Hamiltonian. This requires a theory taking into account interactionsbetween magnetic moments. We will use a very successful, though approximate approach called mean-eldtheory.

1 Curie-Weiss mean eld theory of magnetic insulators

Our starting point is the Heisenberg Hamiltonian which we derived in the last lecture:

H = −∑j

∑i 6=j

Ji,j Ji · Jj .

(Lecture X, Eq. 24)

Here we have introduced the dimensionless angular momentum

Jj ≡ Jj/~ (1)

so the exchange constants Ji,jhave dimensions of energy. Ji,j gives the interaction strength between amagnetic moment that is at site i and another one that is at site j:

Let us suppose that Ji,j > 0. If Ji and Jj are parallel (or, at any rate, the component of one along the axisdened by the other is positive - as in the gure) then Ji · Jj is positive and thus the magnetic moments ati and j contribute to lowering of the energy (note the negative sign in the front - this is a convention). Incontrast, if Ji and Jj have an anti-parallel component then the corresponding term in the Hamiltonian willincrease the energy and the pair will make a positive contribution to the energy. Thus Ji,j > 0 means the thetwo magnetic moments have a ferromagnetic interaction, i.e. one tending to align the magnetic moments

to each other. Conversely, if Ji,j < 0 then the ith and jth sites have an anti-ferromagnetic interactioni.e. one that tends to make the spins point in opposite directions.

The above Hamiltonian describes magnetic moments interacting with each other but not subject to anyexternal inuences. As in the case of paramagnets (Lectures V-VIII), we can also apply externally a magneticinduction:

H = −∑j

∑i 6=j

Ji,j Ji · Jj −∑j

µj ·B. (2)

1

The applied magnetic induction couples to the magnetic moment of each individual site j, µj . Now

µj = gJ |µB|Jj/~ (3)

= gJ |µB| Jj , (4)

so

H = −∑j

∑i 6=j

Ji,jg2Jµ

2B

µi · µj −∑j

µj ·B (5)

Unfortunately, the above problem is incredibly dicult - in fact, exact solutions exist only for one-dimensionalsystems (chains of spins) and some particularly simple examples in two dimensions (the two-dimensional Isingmodel, whose solution is due to Lars Onsager). In three dimensions, for accurate knowledge of the behaviourof such systems formed by many interacting magnetic moments we rely on computer simulations (e.g MonteCarlo) and, of course, on experiments. However there is an approximation which not only works quite wellin many instances but moreover aords a mathematically simple and physically insightful description of thisand many other many-body problems: the mean eld approximation.

We start by noticing that the above Hamiltonian can be re-written in the form

H = −∑j

∑i 6=j

Ji,jg2Jµ

2B

µi + B

· µj . (6)

We now note that this looks the same as the Hamiltonian of a paramagnet made up of non-interacting

magnetic moments if we say that the magnetic moment at the jth site experiences an applied eld of theform

∑i 6=j(Ji,j/g

2Jµ

2B

)µi + B. In eect, putting in

Beff,j =∑i 6=j

Ji,jg2Jµ

2B

µi + B (7)

we have

H = −∑j

Beff,j · µj . (8)

which looks, formally at least, exactly like the Hamiltonian of a paramagnet.

Of course µi is itself a uctuating magnetic moment - in other words, unlike the case of a true non-interactingparamagnet, we cannot really write H in the form

H = H1 +H2 + . . .+HN ,

(Lecture V, Eq. 14)

where each Hj depends on the jth magnetic moment only, because the molecular eld∑

i 6=j λi,jµidepends also on all the other magnetic moments µi. This is where the approximation comes in: we willreplace the molecular eld with a mean eld obtained by approximating the congurations of all themoments that interact with µj by their thermal averages:

Beff,j ≈∑i 6=j

Ji,jg2Jµ

2B

〈µi〉+ B . (9)

This captures the notion that the environment of one magnetic moment is just like the environment of anyother: once the moments settle on a particular form of collective behaviour, they will all follow it. This isa physically-reasonable approximation that works very well in many cases. The mean eld approximation,developed initially by Curie and Weiss in the context of magnetism, is the bedrock on which much of ourknowledge of emergence and many-body physics rests. Further applications of mean eld theory include theBardeen-Cooper-Schrieer theory of superconductivity (which was one of the great triumphs of Twientieth-century physics and we will see later on).

2

We can now write the Hamiltonian in the separable form recalled above with the Hamiltonian for eachindividual site j given by

Hj = −Beff,j · µj , (10)

orHj = −Be,jµz,j (11)

where µz,j is the comopnent of µj along the direction of Beff,j .

Now recall that for a paramagent

M = MsBjtot

(gJ |µB| jtotB

kBT

)(Lect. VII, Eq. 8)

Using M = ρµj,z andMs ≡ ρgJ |µB| jtot

(Lect. VII, Eq. 3)

and substituting the eective eld Be,j instead of B we obtain

〈µj,z〉 = µsatz Bjtot

(µsatz Be,jkBT

)(12)

whereµsatz ≡ gJ |µB| jtot. (13)

µsatz is the maximum value that the magnetic moment on site j, µz,j , can take (the saturated magneticmoment along the eld direction; note that it depends on the angular momentum quantum number jtotand therefore, in general, on the site index j, as dierent ions may have dierently-sized magnetic moments;we will assume below that all ions are equivalent i.e. jtot is the same everywhere and, as in the above two

equations, we will not show the dependence of µsatz on jtot explicitly).

The above result (12) is just a recycling of our earlier formulae for a paramagnet but with the appliedmagnetic induction B replaced with the mean eld Beff,j . The dierence between our mean eld theory andour earlier theory of a non-interacting paramangnet is that the eective mean Be,j depends on the value ofthe magnetic moment itself. Thus equation (12) is to be solved self-consistently with (9), which also relates〈µz,j〉 to Be,j . This is very complicated in general so we will assume now a few particularly simple cases.

2 Validity of the mean eld approximation

Let us say a little bit about the validity of the mean eld approximation. Any system kept at constanttemperature T will always tend to minimise its free energy,

F = 〈H〉 − TS. (14)

The rst term is the energy of the system. The second term is the temperature times the entropy, with anegative sign. Thus the thermal equilibrium state of any system will strike a balance between minimizingthe energy 〈H〉 and maximising the entropy S.

Consider the mean eld theory of a ferromagnetic system. The magnetic moments have collectively settledon a single direction which they are pointing along. If they were all perfectly `locked' into that position, therewould be no uctuations, so the entropy would be zero (only one microstate compatible with the observedmacrostate). In mean eld theory uctuations are allowed, but only at the level of a single magnetic moment,i.e. each magnetic moment uctuates, as in a paramagnet, but it does so in the presence of an eective meaneld that does not. Mean eld theory starts to break down when there are collective uctuations that involvemore than one individual magnetic moment ipping or re-orienting in sync. Such collective excitations

3

cost some energy, so at very low temperatures (where the rst term in the above equation is dominant) theydo not take place. However at any nite temperature they can potentially proliferate, as they lead to anincrease in entropy.

To ascertain whether collective excitations will proliferate, leading to a failure of the mean eld theory,imagine a system with ferromagnetic interactions in its thermal equilibrium state. For simplicity let ussay that interactions take place only between nearest neighbours, with exchange nocstant J . Let us nowhypothesize a collective uctuation, in the form of a small region, of size l, where a group of magneticmoments have spontaneously ipped to point in a dierent way:

Locally, both inside and outside the region where the uctuation has happened, each magnetic moment seesother magnetic moments pointing the same way. However, on the surface separating the uctuation from thebulk, there are magnetic moments that are pointing in opposite directions and which therefore raise the totalenergy of the system. The increase in total energy is ∼ J× l2 where l2 (times a constant factor) is the area ofthe surface separating the bulk from the uctuation. Therefore as the uctuation grows its energy increases.For low enough temperatures, this will tend to make the uctuations small and short-lived, validating themean-eld approximation.

Now consider the case of a one-dimensional system, i.e. a chain of magnetic dipole moments where eachmagnetic moment interacts with the one in front and with the one behind it. Let us ip all the spins in aregion of the chain to create a uctuation, as we did above:

As we can see, there are a couple of bonds along which two adjacent spins are pointing in opposite directions.This costs an energy ∼ J (2J , to be precise). Now let us make the uctuation grow:

4

Unlike the 3D case, where the energy increased as the uctuation grew, in the 1D case there are still only twounstatised bonds, so the energy cost of the uctuation is still ∼ J . Thus the energy cost of the uctuationis independent of its size, l. Therefore, for any given T > 0, uctuations will tend to grow, so as to increasethe entropy, and once they form their size will be un-bounded.

More generally, the energy cost associated with the boundary of a uctuation of linear size l grows like∼ Jld−1, where d is the dimensionality of the system. For large d, this energy cost is high and thereforeonly single-moment uctuations like the ones considered in mean eld theory are important. Indeed meaneld theories become exact in the academic, but conceptually important limit d → ∞ (it is conceptuallyimportant because it gives us an anchoring point where we can prove that mean eld theory must work).For small d, the larger uctuations become very cheap as a means to increase entropy so they proliferate andmean eld theory becomes inadequate. The main consequence of this is that the mean eld approximationoften works in 3D, though not always, never works in 1D, and rarely works in 2D.

5

Lecture XII. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In the previous lecture we saw that the problem with a system of many interacting magnetic moments isthat each moment sees, in addition to any externally-applied magnetic elds, an additional eld made up ofcontributions from the other magnetic moments. In general this molecular eld uctuates and the problemis very dicult to solve. The mean eld approximation consists of neglecting those uctuations, i.e. assuming,for the purpose of the computation of the molecular eld, that the magnetic moments that contribute toit can be replaced with their thermal averages. We are then left with the problem of determining thoseaverages, which reduces to one we have already solved: a non-interacting paramagnet, except that this timeinstead of the applied eld we have a mean eld made up of the applied eld plus the frozen moleculareld. Now we are going to solve the equations for the simplest case of a ferromagnetic state.

1 Ferromagnetism

Let us assume that the interactions between magnetic moments are all ferromagnetic:

Ji,j > 0, (1)

tending to align the magnetic moments in the same direction. It is therefore reasonable to suppose that allmagnetic moments 〈µi〉 will have the same magnitude and direction everywhere:

〈µi〉 = 〈µz〉uz for all i. (2)

This will of course be the direction of the applied magnetic eld. Then our two self-consistent equations(Lect. XI, Eqs.9,12) become

〈µz〉 = µsatz B

(µsatz BekBT

)(3)

Be = λj 〈µz〉+B (4)

where B (y) is the Brillouin function (Lecture VII, Eq. 9; note that we have ommitted the dependence onjtot - again, we are assuming jtotis the same everywhere) and

λj ≡1

g2Jµ2B

∑i 6=j

Ji,j (5)

is a dimensinless coupling constant quantifying the strength of interaciton of the spin at j with all the otherspins in the sample. In what follows we will assume that the exchange constant Ji,j does not depend onanything but the relative position of the two sites i and j,

Ji,j = J (Ri −Rj) , (6)

and therefore λ will be site-independent:λj = λ. (7)

The above equations can be recast as

〈µz〉µsatz

= B

(µsatz BekBT

)(8)

µsatz BekBT

=µsatz

2

kBTλ〈µz〉µsatz

+µsatz B

kBT(9)

1

Re-arranging the second equation we obtain(µsatz

2

kBTλ

)−1(µsatz BekBT

− µsatz B

kBT

)=〈µz〉µsatz

(10)

Dening the dimensionless parameter

y ≡µsatz BekBT

(11)

we obtain〈µz〉µsatz

= Bjtot (y) (12)

and

〈µz〉µsatz

=1

λ

(kBT

µsatz2 y −

B

µsatz

)= T y − B . (13)

To solve this system of equations we plot the quantity 〈µz〉 /µsatz vs the parameter y using each of theabove two formulae. Where the curves cross we have found a solution. The three dimensionless parameterscontrolling the problem are

jtot (14)

T ≡ kBT

λµsatz2 =

1

λ

kBT

(gJµBjtot)2 =

kBT

j2tot∑

i 6=j Ji,j(15)

B ≡ B

λµsatz

=gJµBB

jtot∑

i 6=j Ji,j(16)

The rst of these three parameters is the angular momentum quantum number of the magnetic moments;the second compares the temperature to the energy of interaction between magnetic moments; and the thirdcompares the applied magnetic induction to the molecular eld. The two cruves whose intersection givesour mean eld solutions are the Brillouin function, which we have already encountered in the theory ofparamagnetism and whose shape is given by jtot, and a straight line whose slope T and whose intercept ofthe vertical axis, at y = 0, is −B. The second curve is a straight line. In what follows we will talk aboutsmall temperatures, large temperatures, small elds and large elds meaning small and large valuesof T and B, respectively, i.e. small values of T and B compared to the relevant energy scales in the problem.

Example: Magnetization of ferromagnet whose magnetic moments have angular momentum quanrtum num-

ber jtot = 1 at temperature T = (1/4) k−1B λµsatz

2and in an applied magnetic induction B =

(1/5)µsatz λ (i.e. T = 1/4, B = 1/5).

We have

2

giving three solutions, two of which point against the direction of the eld and which are therefore un-physical. The solution where 〈µz〉 points in the eld direction is the physical one. Once we have locatedthat solution approximately we can zoom in graphically on the solution or determine it numerically by,for example, the bisection method. A few bisections give, for the above parameters, y ≈ 4.76(6), forwhich we nd 〈µz〉 ≈ 0.992µsatz (i.e. for this value of the eld the magnetization is almost completelysaturated).

Now let us see the kind of solutions we get more generally.

1.1 Magnetisation

First, note that for nite B we always get at least one solution:

This is to be expected: as in the case of a non-interacting paramagnet, when we apply a eld we immmediatelypolarize the magnetic moments. If the interactions between magnetic moments are also tending to makethem point in a given direction, the eect is only re-inforced.

In fact we can nd up to three solutions, as in the previous example:

Evidently the solution with the highest magnetization in the direction of the eld will have the lowest energy,as we already mentioned in the example.

3

Now look at what happens when we decrease the eld. In the case of the paramagnet, remember that forzero eld the magnetic moment (and therefore the magnetization) was zero. Here it depends:

• For large T the straight line is very steep and as the magnetic eld is turned o the magnetization weobtain tends to zero:

The three straight lines in the above plot correspond to three dierent values of the eld, B = 2/5, 1/5, 0(lower, middle and upper lines, respectively). As we can see as we lower the eld the magnetizationdecreases and for zero eld it is equal to zero. Thus for large T the system behaves as a paramagnet.

• For low T then even at B = 0 there is still a nite magnetization:

In this plot we have decreased eld from B = 0.6 (lower straight line) through B = 0.4 and 0.2 toB = 0. We go from having one solution at positive y to having three solutions (the two new ones beingunphysical, as discussed above).

The B = 0 curve in the above example deserves additional comment. Here there are still three solutions ofour equations. The one with 〈µz〉 = 0 is still unphysical (since it corresponds to the B → 0 limit of oneof the two unphysical solutions going against the applied eld direction). On the other hand the positiveand negative solutions are equivalent in the absence of an applied magnetic eld. As soon as we add aninnitesimal eld, the one in the direction of the eld becomes slightly preferable. However int he absenceof an applied eld the system will choose one of the two solutions at random: spontaneous symmetrybreaking.

4

Within this theory, the explanation is that the molecular eld remains nite even in zero applied eld. Thisis the way the mean-eld theory captures the tendency of magnetic moments to self-organize as a rsult oftheir mutual interactions.

Another way to look at this is as a function of temperature, for xed magnetic eld. At zero eld, since thetemperature controls the slope of the straight line, we see that there is a critical temperature where theslope is such that there will be a nite magnetization:

This critical temperature will depend on T . It is the temperature at which the slope of the straight line andof the Brillouin function at y = 0 coincide. The former is just equal to T , i.e.

T =1

λ

kBT

µsatz2 =

1

λ

kBT

(gJµBjtot)2 . (17)

The latter can be easily obtained from the low-y approximation to the Brillouin function that we obtainedin Lect. VIII to calculate the susceptibility of a paramagnet:

B (y) ≈ jtot + 1

jtot

y

3(18)

⇒ ∂B(y)∂y

∣∣∣y=0

=jtot + 1

3jtot. (19)

Equating these two slopes at T = Tc we obtain

T =jtot + 1

3jtot(20)

⇒ kBTC =1

3λ (gJµB)2 jtot (jtot + 1) , (21)

i.e. kBTC =1

3λµ2 , (22)

where µ is the size of the magnetic moments:

µ =√⟨µ2⟩

= gJ |µB|√jtot (jtot + 1)

(Lect. VIII, Eq. 10)

The above equation gives the critical temperature, or Curie temperature, for ferromagnetism to developin the absence of an applied eld. It tells us that ferromagnetism will develop at a higher temperature if

5

either the interactions between magnetic moments are strong (large λ) or the magnetic moments themselvesare large (large µ).

Putting in the formulae for λ and µ explicitly we see that TC indeed only depends on jtot and on Ji,j :

kBTC =1

3jtot (jtot + 1)×

∑i 6=j

Ji,j . (23)

A particularly simple case is obtained when

Ji,j =

J if i, j are nearest neighbours,0 otherwise.

(24)

Then

kBTC =2zJ

3jtot (jtot + 1) , (25)

where z is the number of nearest neighbours each lattice site has.

As always the magnetization is found form the magnetic moment by

M = ρµ . (26)

(Lect. V, Eq. 2).

As the temperature is lowered, we go from zero magnetization at T > TC to non-zero at T < TC . Thetransition takes place continuously, i.e. M statrs by becoming innitesimally small as we go from T = T+

C

to T−C , instead of jumping from zero to a nite value:

This is called a second-order phase transition or instability. Once the system is below TC it becomesunstable towards develping a spontenous magnetization, breaking the symmetry betweenM and −M : sym-metry breaking. Magnetization is the order parameter of this instability: its nite value indicates thatthe symmetry has broken, leading to a more ordered state (lower entropy). Such concepts were invented todeal with superconductivity and magnetism and then found much wider applicaiton, e.g. in particle physics(the Higgs mechanism).

Under a magnetic eld, the temperature dependence of magnetization is dierent. It looks as follows (redcurve):

6

We see that M is always nite though at high enough T it tends asymptotically to zero. At very low T , onthe other hand, it tends to Ms just as in the zero-eld case. This can be obtained using the same geometricconstruction used above.

1.2 Zero-eld susceptibility in the disordered state

As in the case of the paramagnet, let us look at the low-eld behaviour of the magnetization to derive thezero-eld magnetic susceptiblity,

χ0 =∂M

∂H

∣∣∣∣H→0

= µ0∂M

∂B

∣∣∣∣B→0

= µ0ρ∂ 〈µz〉∂B

∣∣∣∣B→0

.

Remember that for a paramagnet

χpara0 =

ρµ0µ2

3kB

1

T. (27)

(Lecture VIII, Eq. 11)

We wish to see how this expression is modied when interactions between magnetic moments are taken intoaccount.

We will focus on the case where T > TC , i.e. we are in the disordered (paramagnetic) state. Now we useEq. 12. In view of the denition of y Eq. 11 we see that at low eld and in the disordered state,

y =µsatz BekBT

=µsatz (λ 〈µz〉+B)

kBT(28)

is small because both B and 〈µz〉 are small (B is small because we are calculating the susceptibility in theB → 0 limit and 〈µz〉 is small because for T > TC the only net magnetization is that induced by the smallapplied eld). Therefore for low B and above TC we can do a low-y expansion (18) in (12):

〈µz〉µsatz

= B (y) ≈ jtot + 1

jtot

y

3. (29)

Putting the last two equatiosn together we have

〈µz〉 = µsatz

jtot + 1

jtot

1

3

µsatz (B + λ 〈µz〉)kBT

(30)

=

(µsatz

)23kBT

jtot + 1

jtot(B + λ 〈µz〉) (31)

=g2Jµ

2Bjtot (jtot + 1)

3kBT(B + λ 〈µz〉) (32)

=µ2

3kBT(B + λ 〈µz〉) (33)

7

where we have used µ = gJµBJ/~ with J2 = ~2jtot (jtot + 1) . From this

〈µz〉

1− λ µ2

3kBT

=

µ2

3kBTB (34)

⇒ 〈µz〉 =13µ

2

kBT − 13λµ

2B (35)

Now, in the previous lecture we derived kBTC = 13λµ

2, therefore

〈µz〉 =13µ

2

kBT − kBTCB (36)

whence

χferro0 =ρµ0µ

2

3kB

1

T − TC(for T > TC) (37)

which, interestingly, is the same as the zero-eld magnetic susceptibility of a paramagnet but with the 1/Ttemperature dependence replaced with 1/ (T − TC):

The divergence of the susceptibility is one of the main features of a continuous phase transition (also called,as we mentioned above, a second-order phase transition or an instability). As we lower the temperature, andwe approach the critical temperature at which the system will develop its own, spontaneous magnetization,inducing a magnetization by application of an external eld becomes easier and easier. Eventually, rightabove TC , the system becomes innitely susceptible to developing a magnetization: and innitesimally smallapplied eld will do. The system has become unstable towards a ferromagnetic state.

All instabilities are accompanied by diverging susceptibilities, though it is not always the same susceptibility.In a ferromagnet it is the magentic susceptiblity, as we have just shown, but in a superconductor, for example,the susceptiblity that diverges at the superconducting critical temperature is the pairing susceptibility, whichmeasures how susceptible electrons are to pair up in the presence of a pairing eld created by anothersuperconductor.

8

Lecture XIII. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

Having described ferromagnetism, let us now turn to the theory of another ordered state of magnetic matter:antiferromagnetism.

Antiferromagnetism is a form of magnetic order that can take place when the exchange interactions be-tween magnetic moments tend to make them point in opposite directions. In the context of the HeisenbergHamiltonian we have been dealing with,

H = −∑j

∑i 6=j

Ji,jg2Jµ

2B

µi · µj −∑j

µj ·B,

(Lect. XI, Eq. 5)

this means thatJi,j < 0. (1)

As we will see this can lead to an instability into a form of order where the magnetic moments point alternatelyup and down in such way that the total magnetization averages to zero (unlike the case of a ferromagnet).However, let us rst discuss the magnetic susceptiblity of a system with antiferromagnetic interactions inthe disordered state, i.e. above the critical temperature.

1 Magnetic susceptibility of ferromagnets and antiferromagnets in the paramagnetic state

The discussion of the zero-eld magnetic susceptibility of an antiferromagnet above its critical temperatureis entirely analogous to that of a ferromagnet. We will nevertheless go through the entire argument againhere, just keeping it a bit more general.

Recall that within the mean-eld approximation, the thermal average of the z component of the magnetic

moment at the jth magnetic ion is given by

〈µj,z〉 = µsatz Bjtot

(µsatz Be,jkBT

).

(Lect. XI, Eq. 12)

Here the eective eld at the jth ion is given by

Beff,j ≈∑i 6=j

Ji,jg2Jµ

2B

〈µi〉+ B

(Lect. XI, Eq. 9)and the z-axis is the direction it takes locally at that ion.

Now let us assume that we start from a situation where the applied magnetic indcution is B = 0 and also weare at high enough temperature that the system is disordered (paramagnetic) so

⟨µj⟩

= 0 for all j. Then thesecond of the above two equations gives Beff,j = 0 (we can check that this is self-consistent: if we substituteBeff,j = 0 in the rst equation we get back

⟨µj,z

⟩= 0, as we assumed).

Now suppose we apply a small magnetic induction, B 6= 0. This will make Beff,j nite, which will then leadto a nite 〈µj,z〉 . The direction of the induced magnetic moment and the associated mean eld will be thesame everywhere, so the situation is the same as the paramagnetic state of a ferromagnet (i.e. a ferromagnetabove its Curie temperature). The same arguments we used in the last lecture apply so we get

〈µz〉 =13µ

2

kBT − 13λµ

2B,

1

(Lecture XII, Eq. 35)where

λ ≡ 1

g2Jµ2B

∑i 6=j

Ji,j .

(Lecture XII, Eqs. 5,7)

In Lecture XII we then went on to identify 13λµ

2 with kBTC , the energy associated with the Curie tempera-ture, and deduced

χferro0 =ρµ0µ

2

3kB

1

T − TC, (for T > TC)

(Lect. XII, Eq. 37)which generalised the expression valid for a paramagnet (λ = 0),

χpara0 =

ρµ0µ2

3kB

1

T.

(Lecture VIII, Eq. 11)

On the other hand, for negative exchange constants we may have λ < 0 so we need to generalise the aboveexpression by dening the quantity θ in the following way:

kBθ ≡1

3λµ2, (2)

i.e. kBθ =1

3jtot (jtot + 1)×

∑i 6=j

Ji,j . (3)

This new temperature scale allows us to write the zero-eld magnetic susceptibility in the following, verygeneral way:

χ0 =ρµ0µ

2

3kB

1

T − θ, (paramagnetic state) (4)

which is valid, within mean eld theory, whatever the sign of the interactions, as long as we are at highenough temperatures to be within the paramagnetic state.

For a ferromanget (λ > 0), θ coincides with the Curie temperature, TC , and thus (4) coincides with (Lect. XII,Eq. 37). For a paramagnet (λ = 0) we have θ = 0 and thus (4) reproduces (Lect. VIII, Eq. 11), also above.Finally, for a system with antiferromagnetic interactions (λ < 0) θ is negative and it does not, in general,have the meaning of a critical temperature.

The following plot shows χ0 as a function of T/θ for θ > 0 (right), θ = 0 (centre) and θ < 0 (left):

2

Taking the θ = 0 curve as a reference, which corresponds to a paramagnet with no interactions betweenmagnetic moments (Lecture VIII), we see that ferromagnetic interactions tend to make the system more sus-ceptible to magnetise in the presence of an applied magnetic induction, while antiferromagnetic interactionshave the opposite eect: they react against the tendency of the magnetic eld to align all the moments inthe same direction and thus decrease the magnetic susceptibility, compared to the non-interacting case.

In the ferromagnetic case, the susceptiblity eventually diverges when T = θ = TC as the system goes throughthe ferromagnetic instability that we discussed in Lecture XII. In the antiferromagnetic case, the magneticsusceptibility never diverges. However there can be, as in the case of a ferromagnet, an instability towardsan ordered state - it is just not an ordered state characterised by a nite magnetization. The description ofsuch anti-ferromagnetically ordered state will be the subject of the main lecture.

3

Lecture XIV. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In this lecture we continue to develop the mean eld theory of antiferromagnetism. In the previous lec-ture we computed the magnetic susceptiblity of an antiferromagnet in the disordered (i.e. paramagnetic)state. We saw that unlike a ferromagnet, where the susceptiblity is enhanced by interactions between mag-netic moments, compared to the non-interacting case, in an antiferromagnet the zero-eld susceptibility issuppressed compared to the non-interacting case. Now we will work out the mean eld theory of the anti-ferromagnetic instability that leads, in this case, to an ordered state below a critical temperature called theNeel temperature, TN (after Louis Neel).

1 Mean eld theory of the antiferromagnetic state

Unlike the ferromagnetic state, which always takes the same form (all magnetic moments aligned in the sameway), ferromagnetism can take many forms depending on the exact type of anti-ferromagnetic interactionsin the system. We will assume here the simplest possible type of interaction, where each magnetic momentinteracts antiferromagnetically with its immediate neighbours, and that's it:

Ji,j =

−J < 0 if i and j are nearest neighbours,

0 otherwise.(1)

We will also assume, unlike the case of ferromagnetism, that there is no externally-applied eld (this intro-duces considerable complications in the case of the anti-ferromagnetic state - see Blundell).

Let us now think of what kind of magnetic order we should nd at low temperatures in this case. To illustrateit, consider the simple case of a square lattice:

We are saying that the magnetic moment on a certain site j interacts only with its z nearest neihgbours (inthis square lattice example, z = 4):

1

Since the exchange constant −J is always negative, it will tend to align the four neighbouring magnetic

moments in the opposite direction to that adopted by the jth one:

This is the reason why the susceptibility is suppressed (see the last lecture): since each magnetic momentwants to point against the direction chosen by the surrounding ones, there is a penalty, coming from inter-actions, for aligning all the magnetic moments in the same direction.

What kind of magnetic order can such interaction lead to? It is easy to guess if we forget now about themagnetic moment that we coloured in red and think of the four moments coloured in blue. Each of themhas four neighbours (including the red magnetic moment) and each of those neighbours will tend to be anti-aligned to the corresponding blue spin - i.e. they will tend to adopt the same orientation as the original redspin:

Clearly as this propagates through the lattice we end up with two square sublattices, one made up of redmagnetic moments and another one composed of blue ones:

2

We will call the two sublattices the A sublattice and the B sublattice. This suggests that we look forsolutions of the form

〈µz,j〉 =

〈µz,A〉 if j is in the "A" sublattice;

〈µz,B〉 = −〈µz,A〉 if j is in the "B" sublattice;(2)

to the mean-eld self-consistency equations

Beff,j ≈∑i 6=j

Ji,jg2Jµ

2B

〈µi〉+ B

(Lect. XI, Eq. 9)

〈µj,z〉 = µsatz Bjtot

(µsatz Be,jkBT

)

(Lect. XI, Eq. 12)

Substituting the above type of solution, with the interaction given by Eq. (1), above, and applied magneticinduction B = 0, the rst of the above self-consistency equations reduces to12

Beff,ABeff,B

=∑〈i,j〉

−Jg2Jµ

2B

×〈µz,B〉〈µz,A〉

= ±

∑〈i,j〉

J

g2Jµ2B

×〈µz,A〉〈µz,A〉

(3)

⇒ Beff,B = −Beff,A and Beff,A = λ 〈µz,A〉 , (4)

where

λ ≡∑〈i,j〉

J

g2Jµ2B

. (5)

The second self-consistency equation in turn becomes

〈µz,A〉〈µz,B〉

=

〈µz,A〉− 〈µz,A〉

= µsatz Bjtot

(µsatz

kBT×

Beff,A−Beff,B

)(6)

= µsatz Bjtot

(±µ

satz

kBT×Beff

)(7)

= ±µsatz Bjtot

(µsatz

kBT×Beff

)(8)

12 We introduce the notation∑〈i,j〉 . . . meaning sum over all values of i and j subject to the constraint that the ith and jth

ions are nearest neighbours.

3

i.e.〈µz,A〉µsatz

= Bjtot

(µsatz

kBTBeff

). (9)

Denining, as before,

y ≡µsatz BeffkBT

(Lect. XII, Eq. 11)we have

〈µz,A〉µsatz

= T y with T =kB

λ(µsatz

)2 (10)

and〈µz,A〉µsatz

= Bjtot (y) . (11)

These self-consistency equations are entirely analogous to the ones we obtained for the ferromagnetic statein Lecture XII, with the magnetic moment in the A sublattice playing the role of the uniform magneticmoment. We solve the equation graphically in the same way as before

and nd the same temperature-dependence of the magnetic moment in the A sublattice that we found backthen for the uniform magnetic moment:

The critical temperature at which the magnetic moment becomes nite, TN , is obtained in the same way asthe Curie temperature [this is left as an exercise] and is given by

kBTN =1

3λµ2 = −2

3Jzjtot (jtot + 1) , (12)

4

where as usual µ =√⟨µ2⟩

= gJ |µB|√jtot (jtot + 1) is the r.m.s. expectation value of the magnetic moment

strength and z is the number of nearest nieghbours (z = 4 is the square lattice example).

Note, however, that unlike ferromagnetism the magnetization in the antiferromagnetic state is

M = ρ〈µz,A〉+ 〈µz,B〉

2= 0 (13)

in the anti-ferromagnetic state. The staggered magnetization,

Mstaggered ≡ ρ〈µz,A〉 − 〈µz,B〉

2, (14)

on the other hand does become nite as we go through TN and can be taken as the order parameter of theantiferromagnetic instability. It is the susceptibility to develop such staggered magnetization that divergesas the system undergoes the antiferromagnetic instability at TN (rather than the magnetic susceptibilitywhich, as we saw in the preivous lecture, never diverges for an antiferromagnet).

2 Broken symmetry

The dierent nature of ferromagnetism and antiferromagnetism can be better appreciated in terms of theconcept of broken symmetry.13

Let us go back to the general Heisenberg Hamiltonian,

H = −∑j

∑i 6=j

Ji,jg2Jµ

2B

µi · µj −∑j

µj ·B.

(Lect. XI, Eq. 5)

In the absence of an externally-applied magnetic induction (B=0) the above Hamiltonian becomes

H = −∑j

∑i 6=j

Ji,jg2Jµ

2B

µi · µj , (15)

which is symmetric under the reversal of all the magnetic moments in the system. That is to say that underthe operation

µj → −µj for all j (16)

the Hamiltonian stays exactly the same. This is because the dot product

µi · µj (17)

does not change when I reverse both µi and µj .

Such symmetry under the inversion of all the magnetic moments can also de regarded as a symmetry undertime reversal, since in order to ip a magnetic moment all we need to do is reverse the currents giving rise toit, which could be achieved by turning time backwards so as to reverse the trajectories of all the charges:14

Timereversal

13 The wonder and deep implications of broken symmetry are beautifully brought forth in Nobel laureate Phil W. Anderson'sclassic essay More is dierent, Science 177, 393-396 (1972).14 Of course, spin does not emerge from circulating currents in any trivial way, but nevertheless it behaves in the same way

as orbital angular momentum under time reversal.

5

In the paramagnetic state, the time-reversal operation (inverting all the magnetic moments) would leave allobservables unchanged. In particular the magnetization, which is zero in the paramagnetic state (remem-ber that there is no applied eld in this discussion), would remain zero after time-reversal. In contrast,in the ferromagnetic state, where the magnetic moments have spontaneously chosen to point in one par-ticular direction, time reversal would have the eect of inverting the magnetization, so it would have amacroscopically-observable eect. We thus say that ferromagnetism breaks time-reversal symmetry.

Time-reversal symmetry breaking is a notable phenomenon: how does the system decide which value ofthe magnetisation to adopt, given that the Hamiltonian (which describes fully the dynamics of the system)cannot distinguish between values related by time-reversal symmetry? The key to this conundrum is thedivergence of the susceptibility to magnetise as the Curie temperature is approached from above. Because ofthis, any small perturbation, e.g. a tiny applied magnetic eld (e.g. the Earth's magnetic eld), will makea macroscopic region of the magnet point in one direction rather than another.

In antiferromagnets, time-reversal symmetry is also broken, since ipping all the magnetic moments in thesample would have the eect of swapping the A and B sublattices, changing the sign of the staggeredmagnetization. However, in addition, another symmetry is broken, namely the symmetry under latticetranslations. This says that if all the sites on the lattice are displaced by exactly one lattice constant in thesame direction, so that each occupies the site of one of its nearest neighbours, the Hamiltonian, again, staysthe same (this is obviously only valid for ininite systems with no boundaries - but for all practical purposeswithin this discussion any macroscopic chunk of matterial can be ocnsidered innite). This symmetry holdstrue in the paramagnetic state, and also for a ferromagnet (since all sites develop the same magnetizationand therefore are still equivalent). However, when we carry out the operation in the antiferromagnetic statewe nd that it has the eect of interchanging the A and B sublattices, which is the same as ipping all themagnetic moments, again changing the sign of the order parameter (that is, the staggered magnetization):

Thus the antiferromagnetic state breaks time-reversal symmetry and, in addition, also the symmetry underlattice translations.

6

Lecture XV. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In the past couple of lectures we have discussed a dierent type of ordered magnetic states: anti-ferromagnetism.However, given that our description of such magnetic order has been approximate (based on mean-eld the-ory) one wonders: how do we know that such order exists?

The case of ferromagnetism is pretty clear, as it leads to a spontaneous magnetization that can be detectedusing amagnetometer (see Lecture 1). However anti-ferromagnetism, having no net magnetization, cannotbe detected in this way.

In this lecture we will discuss two of the main techniques used to detect the microscopic arrangement ofmagnetic moments in a material: neutron scattering and muon spin rotation.

1 Magnetic neutron scattering

Neutrons have no charge, therefore they are very penetrating and can be used to probe the bulk of a magneticmaterial without being absorbed or reected near the surface. At the same time, they do have a spin (withangular momentum quantum number s = 1/2) so they interact magnetically with the magnetic dipolemoments inside the sample. This is what makes them valuable in the study of magnetism.

They are not just useful for magnetism. In addition, they also bounce o the nuclei in the sample due to thestrong-force interaction with the other nuclei in them, so they can also be used to study the crystal structure.

Neutrons can be produced in two ways: through nuclear ssion (in a sision reactor) and through spalla-tion (using a proton accelerator).

In nuclear ssion a heavy nucleus is hit by by a neutron from an earlier ssion event and splits into twolighter nuclei, ejecting some excess neutrons in the process. Some of these neutrons go on to split othernuclei while the rest escape the reactor core and can be used to generate energy (by heating some water thatevaporates and moves a turbine) or, even better (!), to do neutron scattering!

In a spallation neutron source, a proton beam from a synchrotron or cyclotron is red at a target made ofsome heavy element (e.g. Tungsten). The nuclei get hot and evaporate some of their neutrons.

In either case, the neutrons have to go through a moderator: a chunk of material that is kept at somesuitable temperature where the neutrons can thermalise. When they leave the moderator, the neutrons havea known, thermal (or quasi-thermal) energy distribution, corresponding to the moderator's temperature, T .They have become thermal neutrons. The typical enegies of thermal neutrons are given by

kBT ∼p2

2mn, (1)

where p and mn are the linear momentum and mass of the neutron, respectively. The momentum is relatedto the neutron wavelength λ by means of the de Broglie wave-particle duality equation,

p =h

λ. (2)

Combining these two equations yields the thermal de Broglie wavelength of the neutron,

λ ∼ h√2mnkBT

. (3)

Substituting the mass of the neutron,

mn ≈ mp ≈ 1.7 10−27Kg, (4)

we get

λ ∼ 1 nm√T/K

. (5)

1

Thus for moderator temperatures T ∼ 100K we get neutron wavelengths λ ∼ 1 nm/10 = 1 Angstrom, whichof the order of the atomic spacings of solids. At these wavelengths interference eects result from featuresthat vary within the sample on the atomic scale. This is what makes neutron scattering useful for the studyof magnetism.

The choice of moderator will depend on what temperatures (i.e. what wavelengths) we wish to achieve,e.g. to study biological systems or long-wavelength helical magnetism (see Lecture XVI) we may wish forsomewhat longer wavelengths than to study, say, Neel order. Some common substances to use are water,solid hydrogen and solid methane.

Once the neutrons have been thermalised, they are guided towards our sample of magnetic material. Whilegoing through the sample, the neutrons will get deected a certain angle 2θ from their incoming trajectory.

Each neutron is detected after being scattered and its exact energy is recorded. Recording the energy is ofcourse the same as recording the wavelength, since, using again two of the relations employed above,

E =p2

2mn(6)

=h2

2mnλ2. (7)

The energy of the neutron is determined as follows:

1. In a pulsed neutron source, the neutrons are produced in bunches, each lasting a very short time (ofthe order of a nanosecond). Since we know when each pulse of neutrons hits the sample, we can workout the energy of a neutron from the additional time t it takes it to hit the detector (time-of-ight).If the distance between the detector and the sample is l, then the neutron's velocity is v = l/t and thusits momentum is p = mnv = mn (l/t) . The energy is of course E = p2/2mn = mnl

2/2t2.

Most spallation neutron sources are based on a proton synchrotron, where protons are acceleratedin compact bunches, and are thus pulsed sources.

2. In a continuous source, there is a constant ow of neutrons so the time-of-ight technique cannot beused. This is the case of most reactor-based neutron sources. In that case there are two potions (sic!):

(a) We can place a chopper between the source and the sample. The chopper is a rapidly-rotatingcylinder with two apertures which eectively chops the neutron beam into pulses.

3. Alternatively, and more commonly, we can use a monochromator, which is a crystal which scattersneutrons of dierent energies in dierent directions (using Bragg diraction, which we explain below).By orienting the monochromator crystal in dierent directions we can select neutrons of dierentwavelengths to reach our sample:

Here are the two arrangements:

2

chopper

Thermal neutrons

(continuous)

Thermal neutrons(pulsed)

Sample

Scatteredneutrons

Detec-tors

mono-chromator

Thermal neutrons

(continuous) neutron shield

mono-chromatic neutrons

(continuous)

Sample

Scatte

red

neutrons

Detec-

tors

While reactors generate much higher uxes of neutrons than spallation sources, in a pulsed source data fromall neutrons can be collected, while in a continuous one we have to throw away most of the neutrons. Theresult is that in the end both technologies have their advantages and disadvantages.

The scattering angle θ depends on the crystalline and magnetic structure of the system. It is very useful toregard a regular crystal lattice as a collection of sets of crystallographic planes. Each set is compoased ofmany parallel, equi-spaced planes containing an identical array of atoms. For a given crystal lattice, there aremany such sets of crystallographics planes, each of them with a given periodicity (the spacing between the

3

planes) and direction (their orientation). For example, here we draw two of the many sets of crystallographicplanes that make up a square lattice:

When neutrons scatter o a crystal, trajectories corresponding to dierent scattering angles interfere. Thiswill lead to more scatering intensity for some values of θ than others. In particular, there will be intensescatering if the angle is such that trajectories correpsonding to reections on parallel crystallographics planesinteract construcitvely. Fro this to happen two conditions are necessary:

1. The angle θ must correspond to specular reection on a set of crystallographic planes existing in thecrystal.

2. The lengths of the optical paths corresponding to reection on subsequent planes must dier by aninteger number of wavelengths, so as to ensure constructive interference.

The latter condition leads to the Bragg diraction law:

2d sin θ = nλ, (8)

where d is the spacing between the relevant set of crystallographic planes and n = 0, 1, 2, 3, . . .

d

θ

θθ

θ

π/2−

θ

π/2−θ

Throughout this discussion we are assuming elastic scattering, i.e. that the neutron does not exchangeenergy with the system. Most neutron scattering events are like this, so it is in general a good approximation

4

to assume that this is the case always. (In inelastic experiments, on the other hand, neutrons that haveactually exchanged some energy are deliberately selected. We will discuss this later.) Under this assumptionthe magnitudes of the neutron momentum before and after it went through the sample are identical. Interms of the neutron's wave vector k, which is related to the momentum via

p = ~k, (9)

this translates to|k| =

∣∣k′∣∣ . (10)

For elastic scattering, the scattering vectorQ = k′ − k, (11)

that is, the change in the neutron's wave vector on going through the sample, depends exclusively on therelative direction of k and k′, i.e. on the scattering angle θ. Its magnitude is given by

Q/2

k= sin θ ⇒ Q = 2k sin θ , (12)

as can be seen from the following geometric construction:

2θ

k k'

Q

k

Evidently the scattering vector Q corresponding to a Bragg relfection will be perpendicular to the corre-sponding set of crystallographic planes.

Regarding its magnitude, it follows from the Bragg diraction law (8). First we notice that k is related tothe wavelength of the neutrons via de de Broglie relation, Eq. (2):

~k =h

λ⇒ k =

2π

λ, (13)

thus

Q =4π

λsin θ. (14)

This allows us to re-write the Bragg law in the form 2d λ4πQ = nλ i.e.

Q = n2π

d. (15)

5

The scattering vectors Q that are perpendicular to crystallogrpahic planes and whose magnitudes are torelated to the planes' spacing by the above formula form the crystal's reciprocal lattice.

A neutron diraction experiment will show bright Bragg spots at the detectors corresponding to reciprocallattice vectors. When the system orders magnetically, the scattering of neutrons will become stronger.

The crucial reason why neutron scattering is so useful in the study of magnetism is that anti-ferromagneticcan alter the reciprocal lattice. Take the square lattice we looked at before. Assume we cool down the crystaland it goes through a Neel instability, so it develops anti-ferromagnetic order:

Since the neutrons couple to the magnetic moments, they are sensitive to the broken translational symmetry.This has been broken by the AF order, with the result that, for example, the set of crystallographic planesthat we colored above in green has its periodicity halved (since spin-up and spin-down planes are nolonger equivalent):

The corresponding reciprocal lattice vectors are twice as small:

QT>TN = n2π

d→ QT<TN = n

2π

2d=

1

2QT>TN . (16)

Giving values to n = 0, 1, 2, . . . we see that all the reciprocal lattice vectors that we had originally are stillthere, but now there are smaller ones that weren't present in the paramagnetic crystal:

At T > TN

6

At T < TN

As the temperature is lowered further, the magnetic peaks become more intense as the magnetic momentsbecome stronger:

7

This is how antiferromagnetism was discovered and the main way to study it nowadays.

2 Muon spin rotation

Muons are sub-atomic particles. Like the electron, they are leptons, and they carry the same charge chargeas the electron, −e (negative muon, µ−) or the same charge of a positron, +e (positive muon, µ+). Also likethe electron and positron, muons have s = 1/2. Thus lke the neutron the muon can interact with magneticmoments - though the way we exploit that is quite dierent, as we shall see.

Unlike the electron, however, the muon has a farily large mass,mµ ≈ 200me, and a nite lifetime, τµ ≈ 2.2µs.

Muons consitute some of the most common forms of cosmic ray. We are regularly bombarded with muons.They are generated when high-energy particles hit nuclei in the ionisphere. Interestingly, at the velocitieswith which they reach the planet's surface they wouldn't be able to make the journey form the ionospherein the brief time spanned by their lifetimes, were it not for time dilation (resulting from their relativisticspeeds) which means that from their pespective time passes mucho more slowly than from the point of viewof our stationary frame of reference.

The muons we use in condensed matter research are not of the cosmic variety. They are generated here,on Earth, using proton beams. The muons are created by hitting a nucleus (any nucleus will do, but C orBe are commonly used) with a proton beam. When the proton hits the nucelus, it produces a pion whichquickly disintegrates, emitting a muon and the corresponding neutrino:

8

Here there's an extra bit of particle physics that helps us a lot: it turns out that the pion's disintegrationinvolves the electro-weak interaciton, which violates parity conservation. As a result, the muon is producedwith its spin pointing anti-parallel to its momentum (the neutrino is also produced with its spin anti-parallelto its momentum; this way both spins add up to the pion's s = 0). This means that the muon beam weproduce has a well-dened spin polarisaiton.

Usually for condensed matter experiments positive muons are employed. The are implanted in a sample andthen we wait for them to spontaneously disintegrate. Typically the muons will come into the sample in shortpulses (~100ns or so, which is much shorter than the muon lifetime, see above). When they go into thesample, they will very quickly come to rest in side it. When the muons disintegrate, they emit a positronand two neutrinos. Again the electro-weak interaction comes to help us here, because the positron is emittedwith higher probability in the direction in which the muon's spin is pointing at the time of disintegration:

It is the positrons that we detect, but because of this coupling between their direction of emission and themuon's spin direction, we gain information about where the muon's spin was pointing inside the sample.

Because the spins of all the mouns were pointing in the same direction to begin with, if there are absolutely nomagnetic elds (either internal to the sample or externally-applied) then the moments would just continue topoint in that direction indenitely and all the positrons will be detected in the same direction, too. However,if there is an applied magnetic eld or, more interestingly, the sample has an intrinsic, internal magneticinduction (e.g. an eective exchange eld Beff responsible for ferromagnetic or anti-ferromagnetic order)then the muon'sm magnetic moment will precess about this local eld:

9

This precssion will happen at the Larmor frequency of the muon. The derivation is entirely as what we didfor electrons early in this course, so we quote the result here:

ωLarmor = gµγµBlocal (17)

The gyromagnetic ratio and g-factor of the muon are

γµ =|e|

2mµ(18)

gµ ≈ 2M (19)

giving

ωLarmor ≈ 2π 135.5MHz ×BlocalT

. (20)

This precession can be seen as a rotation in the direction in which we detect positrons for about 10µs afterthe muon pulse has been implanted in the sample (after that time, most of the muons in the pulse havedied out, so we have to wait until the next pulse comes along before we can collect more data). Very often,what is measured is the asymmetry between the positron detections in two detectors pointing in dierentdirections. This asymmetry oscillates as the mouns precess around the internal elds in the sample.

The following two animations illustrate much better than any words one could write how a muon rotationexperiment works:15

http://neutron.magnet.fsu.edu/images/uSR.gif

http://neutron.magnet.fsu.edu/images/uSR2.gif

Like neutron scattering, muon spin rotation alolows us to detect the magnetic moments that turn up belowTN in anti-ferromagnets. The key is that because the muon sits at a particular site within the crystallattice, it can see the magnetic moment of the nearest magnetic ion, without averaging the magnetizationfrom dierent ions on dierent magnetic sites. Muon spin rotation therefore allows us to determine thetemperature-dependence of the magnetic moment of each particular sub-lattice:

15 These videos come from Muon spin relaxation, Quantum Materials Group, Florida State University,http://neutron.magnet.fsu.edu/muon_relax.html (retrieved 23 November 2011).

10

3 Neutron and muon facilities

Muon spin rotation and, specially, neutron scattering underpin all modern research in magnetism and thetechnologies that are based on it (such as magnetic data storage). Although individual experiments arecarried out by small groups of researchers, the neutron and muon sources themselves are huge enterprises(each of them has many dierent instruments on which dierent teams can be working with dierent samplesat the same time). The following video describing the ISIS facility at the Science and Technology FacilitiesCouncil's Rutherford Appleton Laboratory, in Oxfordshire, gives an idea of the scale:16

http://www.isis.stfc.ac.uk/science/a-video-tour-of-the-isis-facility4469.html

16 ISIS Facility, http://www.isis.stfc.ac.uk (retrieved 10 November 2011).

11

Lecture XVI. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

In the past four Lectures, XI-XIV, we showed how the interactions between magnetic moments (whoseorigin we had discussed in Lectures IX-X) can lead to more correlated states than the non-interactingparamagnets discussed earlier in the course. Mean eld theory allows us to exploit those earlier resultsobtained for non-interacting paramagnets by taking account of interactions by means of an eective meaneld that approximately describes the eect of the other magnetic moments on a given one. We discussedhow interactions modify the susceptibility of a paramagnet and also the ferromagnetic and antiferromagneticinstabilities into ordered states below the critical Curie and Neel temperatures.

At the end of the last lecture we discussed, briey, ferromagnetism and the anti-ferromagnetism of a bipartitelattice in terms of the powerful concept of broken symmetry. However, these are not the only correlatedstates of magnets, so in this lecture we will describe a few more.

1 Competing orders

Remember that the Heisenberg Hamiltonian can be written in the form

H = −∑j

Beff,j · µj .

Lect. XI, Eq. 8)

In the mean eld approximation the eective eld Beff,j is given by

Beff,j ≈∑i 6=j

Ji,jg2Jµ

2B

〈µi〉+ B .

Lect. XI, Eq. 9)

This equation relates the eective eld to the expectation value of the magnetic moment, which can, in turn,be computed from the above Hamiltonian, giving

〈µj,z〉 = µsatz Bjtot

(µsatz Be,jkBT

).

Lect. XI, Eq. 12)

Note that this gives the component of⟨µj⟩along the direction of the local eective eld at the jth site,

Beff,j . In the ground state (at T = 0) we have

Bjtot

(µsatz Be,jkBT

)→ ±1 for Be,j ≷ 0, (1)

so the magnetic moments saturate i.e. they are perfectly aligned (as far as is quantum-mechanically allowed17)with the direction of the local eective eld. Yet this equation does not give the direction of the local eldBeff,j itself. In fact, there can in general be many solutions to the self-consistency equations correspondingto dierent directions of the eective elds (e.g. for the bi-partite anti-ferromagnet of Lecture XIV, a ferro-magnetic arrangement with 〈µz,A〉 = 〈µz,B〉 would also have been a solution, as long as we chose the localeective eld to point always in the same direction; but that solution is not physical).

17 Even the an angular momentum reaches its maximum alignment with a given direction, there is always uncertainty left inthe value of the comopnents perpendicualr to that direction - see Lectures III and VIII.

1

To determine which one of several low-temperature solutions to the self-consistency equations is the correctone we need to compute the ground-state energy. Within the mean eld approximation, this is

〈H〉 = −∑j

⟨Beff,j · µj

⟩(2)

= −∑j

Beff,j ·⟨µj⟩

(3)

≈ −∑j

∑i 6=j

Ji,jg2Jµ

2B

〈µi〉+ B

· ⟨µj⟩ (4)

= −∑j

∑i 6=j

Ji,jg2Jµ

2B

〈µi〉 ·⟨µj⟩−B ·

∑j

⟨µj⟩

(5)

Given several competing solutions to the self-consistency equations, the above formula tells us which one hasthe lowest energy.

The RHS of the above equation has two terms: an interaction term featuring the dot products of pairsof magnetic moments; and a Zeeman term giving the energy due to the externally-applied magnetic eld.The interaction term will depend on the relative orientations of the magnetic moments and therefore willallow us to decide between co-linear and non co-linera magnetism.

2 Ferrimagnetism

While our discussion of ferromagnetism (Lectures XI-XII) was fairly general, as was that of the correlatedparamagnetic state of ferromagnets and anti-ferromagnets above their critical temperatures (Lecture XIII),our treatment of the Neel instability and of anti-ferromagnetic order below the corresponding critical tem-perature, TN (Lecture XIV) relied on particularly simplied assumptions about the crystal lattice and aboutthe magnetic interactions. In particular we assumed a bipartite lattice i.e. one that can be notionallydivided into two equivalent lattices made up of totally identical sites. Let us discuss the state that emergeswhen this assumption is relaxed.

We consider a magnet with nearest-neighbour antiferromagnetic interactions but where, unlike the case thatwe already discussed, rather than a bi-partite lattice with all identical sites we have a bi-partite lattice wherethe sites in the A sublattice are of one type and those in the B sublattice are of another - for example, themagnetic moment residing at an A site may have one angular momentum quantum number jtot,A whilethat at a B site may have a dierent size given by a dierent angular momentum quantum number jtot,B.We will still expect the magnetic moments to form a bipartite anti-ferromagnetic arrangement, but this timethe spins that are pointing one way will have one value of their magnetic moment, while those pointing theother way will haev a dierent value:

2

This has a very important consequence: this anti-ferromagnetic state has a net magentization, like aferromagnet.

Indeed, even from the more fundamental point of view of symmetry-breaking, this state is more akin to a fer-romagnet than to an anti-ferromagnet, because it only breaks time-reversal symmetry, and not translationalsymmetry.

To emphasize the fact that the moments are anti-ferromagnetically aligned, however, we call such statesferrimagnetic rather than ferromangetic. Ferrimagnetism, unlike ferromagnetism, can be driven by anti-ferromagnetic interactions.

Why does this state not break translational symmetry? Even at high temepratures in the absence of amagnetic eld, where the system is paramagnetic and the magnetic moments will all average to zero, thesites are inequivalent, so the spatial periodicity in the situation with no magnetic order is the same as in themagnetically-ordered state - to reproduce the whole crystal, we have to make copies of the region within thedotted lines:

This is the same unit we would need to repeat in order to reproduce the magnetic structure drawn above.

3 Non co-linear magnetism

Our discussions of paramagnetism (Lectures VI-VIII), ferromagnetism (Lecture XII), anti-ferromagnetism(Lectures XIII-XIV), and ferrimagnetism (above) all relied on the assumption that the magnetic momentson distinct magnetic ions are co-linear, i.e. always pointing either in the parallel or anti-parallel directionswith respect to other magnetic moments.

Co-linearity is, indeed, a natural assumption as, on the one hand, the applied magnetic eld will tend tomake moments parallel to each other and, on the other hand, moment-moment interactions of the form

− Ji,jµi · µj (6)

will tend to make the moments either parallel (Ji,j > 0) or anti-parallel (Ji,j < 0). However, there are manysituations where such co-linearity does not necessarily follow. We will give two examples below.

3.1 Spin-op phase

Firstly let us consider an anti-ferromagnet in the presence of an externally-applied magnetic induction.Then the two tendencies we mentioned above (for each magnetic moment to align parallel to the eld andfor magnetic moments to align anti-parallel to each other) compete. This may lead the system to choose aphase where the orientation is intermediate between anti-ferromagnetic (each magnetic moment anti-alignedto its neighbours) and paramagnetic (aligned with the externally-applied eld).

To achieve this, the system can exploit the Heisenberg Hamiltonian's symmetry under global rotations ofall the spins (Lecture X). This allows it, for small applied elds, to orient the antiferromagnetically-aligned

3

moments in a direction perpendicular to the applied eld, so as to maintain the anti-ferromagnetic order whileavoiding having any spins pointing against the eld, and allows, as the applied eld grows, for the magneticmoments to tilt increasingly in the direction of the eld until eventually, for high elds, a ferromagneticarrangement in the eld direction develops:

Apply small magnetic field

Uniform spin rotation

Apply stronger magnetic field

The component of magnetization parallel to the eld develops gradually and is present, albeit small, as soonas an external eld is applied (however tiny).

Such phase is termed a spin-op phase because of what happens if there is, additionally, some magneticanisotropy (i.e. some intrinsic tendency, due to spin-orbit coupling, for each magnetic moment to pointalong a given easy axis, independently of moment-moment interactions). In that case the symmetry underglobal rotations of the Heisenberg Hamiltonian no longer holds. The anti-ferromagnetic moments can nolonger rotate freely, and therefore, for weak elds, they stay in their un-altered anti-ferromagnetic cong-uration, even if it means some magnetic moments are pointing against the applied eld. When the eld

4

is strong enough, they suddenly op onto a more ferromagnetic conguration. This is called a spin-optransition. The spin-op phase then evolves continuously, as the eld is increased further, towards thefully-ferromagnetic conguration:

Apply larger magnetic field

Apply small magnetic field

Even stronger magnetic field

If the anisotropy is very large, so that magnetic moments can really only point either parallel or anti-parallelto the eld, there is no spin-op but there is instead a spin-ip transition going directly from the anti-ferromagnetic to the ferr-magnetic conguration:

5

Apply small magnetic field

Apply stronger magnetic field

3.2 Helical anti-ferromagnetism

Another feature of many real anti-ferromagnets that our discussion of the anti-ferromagnetic state has ignoredare interactions beyond nearest-neighbours. In particular, anti-ferromagnetic interactions betweennext-nearest-neighbours may induce helical anti-ferromagnetic order. This often happens, for example,in a ferromagnet with layered crystal structure when in addition to the ferromagnetic interactions betweenthe magnetic moments within a layer and between immediately adjacent layers there is an anti-ferromagneticinteraction between moments in next-nearest-neighbour layers. Each layer will order ferromagnetically and bemostly aligned with the two nearest-neighbour layers, but with a slight tilt so as to lower the antiferromagneticinteraction energy with the next nearest-neighbour ones. As we go from layer to layer, tha small additionalangle builds up so that after some distance 2π/q the magnetization of the layers has completed one fullrevolution. Thus this is very long-wavelength anti-ferromagnetic order:

6

q is called the pitch of the helix.

q may not be related to the lattice constant in any obvious way i.e. the helical order may be incommensu-rate with the crystal lattice. An example of this behaviour is Dy.

4 Frustration

An important class of systems with anti-ferromagnetic interactions are those where such interactions arefrustrated. In a frustrated anti-ferromagnet it is not possible to nd a unique conguration of the magneticmoments that minimises the energy. Instead, there is a manifold of degenerate ground states.

For example, consider magnetic moments on a triangular lattice. If there is an anti-ferromagnetic interactionbetween every pair of moments, then once two of the moments are aligned anti-parallel with one another thethird one cannot be anti-parallel to both of its nearest-neighbours:18

Indeed, whether the third moment points parallel to one of the other two or anti-parallel makes no dierenceto the total energy, leading to macroscopic ground-state degeneracy: a large number of states areequally likely at T = 0. This means that there is nite entropy in the ground state. (In general, whatwill happen then is that other terms in the Hamiltonian, such as longer-ranged components of the interaction,will break the degeneracy, but sometimes we have to go to even lower temperatures to see that.)

Frustrated magnets are a very active area of current research.

18 Image from Ludovic D. C . Jaubert, Topological Constraints and Defects in Spin Ice (PhD thesis, University of Lyon,2009).

7

Jorge Quintanilla

Magnetism and Superconductivity (PH752)

Problem Sheet 2

Information needed to tackle these problems can be found in the handouts for Lectures 9-16. You will need toconsult also a periodic table of the elements with electronic structure information and some physical constants(both can be found for example behind the front and back covers, respectively, of Setephen Blundell, Magnetismin Condensed Matter, OUP 2001). Give all results in SI units.

Problems marked with an asterisk (*) will be assessed.

Problem 1: (*) Consider an isotropic ferromagnet (i.e. one where the Heisenberg model can be applied)composed of Ho3+ ions arranged in a simple cubic lattice:19

All other atoms and ions in the material are non-magnetic. The Curie temperature of this material isTC = 500K.

Assuming that the magnetism is dominated by exchange interactions between nearest-neighbour mag-netic ions, what is the value of the corresponding exchange constant?

Compute the zero-eld magnetic susceptibility at T = 1000K. Does it increase or decrease when wecool down to 600K? By how much, and why?

Use the mean-eld self-consistency equations to derive the size of the magnetic moments and the corre-sponding local eective magnetic inductionBe (in Tesla) at the ion site at T = 700K, 300K, and 100K.Estimate the precession frequency of muons implanted in the sample at those temperatures (assumingthat the muon sees the same eective eld as the magnetic ions).

Hint: You will need the parameters for a Ho3+ ion, which we worked out in Lect. IV.

Problem 2: A magnetic material is made up of two types of magnetic ions arranged in a NaCl structure:20

As shown in the gure, this crystal structure has two types of site. Let us call them A sites and B sites.Suppose that A sites have magnetic moments of strength µ and B sites have magnetic moments ofhalf that strength, µ/2. Let us also assume that the magnetic interactions take place between nearestneighbours only and are antiferromagnetic.

Sketch the ground-state magnetic structure and name what type of magnetism it corresponds to. Whatsymmetries are broken by that state?

Derive (explaining how you arrive at the result) the zero-eld magnetic susceptibility in the paramag-netic state (i.e. at temperatures above the critical temperature).

Compare the formula you have obtained for the suscetibility with the result valid for ferromagnetism,anti-ferromagnetism and paramagnetism (Lecture XIII, Eq. 4) and comment on the similarities and/ordierences.

19 Image from Wikipedia, Cubic crystal system (retrieved 21 November 2011).20 Image from P. G. Nelson, Introduction to Inorganic Chemistry, http://www.hull.ac.uk/chemistry/intro_inorganic/ (re-

trieved 21 November 2011).

1

Problem 3: Consider the Heisenberg model with nearest-neighbour exchange interactions (Ji,j = J if i andj are nearest neighbours, 0 otherwise). Now add some large anisotropy i.e. assume that each site hasan easy axis such that its magnetic moment can only point in one of the two directions along that axis(up or down). This is called the Ising model.

Describe the two ground states of the Ising model for J > 0 and J < 0 in the absence of externally-applied elds.

For the case when the exchange constant is set to a xed value J < 0, calculate the size of an externally-applied eld that will make the system prefer the other ground state. At this value of the eld thereis a so-called spin-ip transition. Describe what that consists of.

Problem 4: Consider a magnetic material made up of identical magnetic ions, of magnetic dipole momentstrength µ, arranged on a tetragonal crystal lattice:21

This is best viewed as a stack of two-dimensional layers, each layer having a square-lattice geometry.All other atoms and ions in the material are non-magnetic. Let us say that the magnetism of ourmaterial is dominated by exchange interactions between nearest-neighbours (i.e. between momentswithin the same layer) and between next-nearest-neighbours (i.e. between moments lying on top ofeach other on dierent layers), with exchange constants J and J ′, respectively.

Assuming that J > 0 and J ′ < 0, what will be the ground-state magnetic conguration? How will thisbe reected in a neutron diraction experiment?

Derive, in analogy with the thoery of bipartite anti-ferromagnetism of Lecture XIV, the self-consistencyequations describing this state at nite temperature. Dene an appropriate order parameter.

Derive a formula for the critical temperature. How does it depend on J and J ′?

21 Image from Wikipedia, Tetragonal crystal system, http://en.wikipedia.org/wiki/Tetragonal (retrieved 21 November 2011).

2

Lecture XVII. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

Today we conclude our study of magnetism.

Let us re-visit the ground state of a ferromagnet and enquire what are the low-energy excited states (i.e. therst that will activate when we make the temperature still small, but nite).

Let us go back to the Heisenberg model, in the absence of an applied eld:

H = −∑i 6=j

J (Ri −Rj) σi · σj .

(Lect. X, Eq. 24)

Hereσi ≡ Ji/~ =

(Si + Li

)/~ (1)

is the total angular momentum of the ith magnetic ion, measured in units of ~ (what we called Ji in LectureX). We have also showed that the exchange coupling between ions i and j depends only on the relativepositions of the two ions, Ri and Rj , respectively:

Ji,j = J (Ri −Rj) . (2)

We will assume all sites to have the same angular momentum quantum number, jtot.

The ferromagnetic ground state |0〉 will have all the magnetic moments pointing along the same direction.Let us choose that direction to be the z axis. Then we can denote

|0〉 = |(site 1)jtot ,

(site 2)jtot ,

(site 3)jtot , . . .〉 (3)

soσi,z|0〉 = jtot|0〉 for all sites i. (4)

We are going to check that this state is an eigenstate of the Hamiltonian. This is not trivial - we hypothesizedthe above ground state as part of a mean eld theory but there was no guarantee that it was the exact groundstate of the Hamiltonian. We will see that it is.

It is useful to introduce the raising and lowering operators

σi,± ≡ σi,x ± iσi,y. (5)

One can use the commutation relations for angular momentum to show that22

σi,±| . . .(site i)mjtot

. . .〉 = C(jtot,mjtot

)| . . .

(site i)mjtot

± 1 . . .〉, (6)

where C(jtot,mjtot

)is a normalisation constant:

C(jtot,mjtot

)=

√(jtot ∓mjtot

)(jtot + 1±mjtot

). (7)

In other words, σi,± raises (lowers) the magnetic quantum number mjtotby one or, equivalently, they tilts

the spin towards (away from) the z axis so as to make its component along that axis change by ~:22 We can nd the proof in ny good introductory Quantum Mechanics textbook, e.g. B.H. Bransden and C.H. Joachain,

Introduction to Quanutm Mechanics (Longman, 1994), Section 6.5.

1

z-axis

If the spin at i is already pointing as parallel to the z-axis as possible, then σi,+ annihilates the state:

σi,±| . . .(site i)jtot . . .〉 = 0 . (8)

In terms of these operators, the Hamiltonian is

H = −∑i 6=j

Ji,j (σi,xσj,x + σi,yσj,y + σi,zσj,z) (9)

= −∑i 6=j

Ji,j (σi,−σj,+ + σi,zσj,z) . (10)

When we apply the Hamiltonian to the ferromagnetic ground state we get

H|(site 1)jtot ,

(site 2)jtot ,

(site 3)jtot , . . .〉 = −

∑i 6=j

Ji,j(0 + j2tot

)|0〉 (11)

= E0|0〉. (12)

i.e. |0〉 is, indeed, an eigenstate of the Hamiltonian with eigenvalue E0. In other words |0〉 is a stationarystate of the system,

H|0〉 = E0|0〉 , (13)

with energy

E0 = −∑i 6=j

Ji,jj2tot . (14)

This proves that |0〉 is an eigenstate of H. In fact it is the lowest-energy eigenstate (we have not proved thathere, but basically because this state is getting as low an energy as you can get from every bond, any chnagewe made ot that state would actually increase the energy). So indeed the kind of ground state given by theT → 0 limit of our Curie mean eld theory is exact.

(For antiferromagnetism, it is a dierent story: the ground state of the Neel mean eld theory turns out notto be an eigenstate of the Hamiltonian, and it is therefore only an approximation. The true ground state isa lot more complicated, riddled with quantum ucutations.)

Let us now study the low-energy excited states. The simplest thing we could do to the ground state would

be to ip one of its spins, say the spin at the lth site:

σl,−|0〉 = σl,−|(site 1)jtot ,

(site 2)jtot , . . .〉 (15)

=√

2jtot|(site 1)jtot ,

(site 2)jtot , . . . ,

(l)

jtot − 1, . . .〉 (16)

≡√

2jtot|l〉. (17)

2

Note that we have introduced the shorthand notation |l〉 for the state obtained from |0〉 by lowering the

angular momentum of the lth magnetic moment by a single quantum of angular momentum, ~, leaving theother magnetic ions unperturbed. It is easy to show, however, that |l〉 is not an eigenstate of H. Indeed

H|l〉 = −∑i 6=j

Ji,j (σi,−σj,+ + σi,zσj,z) |l〉 (18)

with

σi,−σj,+|l〉 = σi,−σj,+| . . .(l)

jtot − 1 . . .〉 (19)

= δj,lσi,−√

2jtot|0〉 (20)

= δj,l√

2jtotσi,−|0〉 (21)

= δj,l2jtot|i〉 (22)

and

σi,zσj,z|l〉 =

j2tot|l〉 if l 6= i, j

jtot (jtot − 1) |l〉 if l = i or l = j.(23)

Thus

H|l〉 = −∑i:i 6=l

Ji,l 2jtot|i〉 −∑

i,j:i 6=j,i6=l,j 6=lJi,j j

2tot|l〉

−∑j:l 6=j

Jl,j jtot (jtot − 1) |l〉 −∑i:i 6=l

Ji,l jtot (jtot − 1) |l〉 (24)

= −∑i:i 6=l

Ji,l 2jtot|i〉 −∑i,j:i 6=j

Ji,j j2tot|l〉

+∑j:l 6=j

(Jl,j + Jj,l) jtot|l〉 (25)

= −∑i:i 6=l

Ji,l 2jtot (|i〉 − |l〉)−∑i,j:i 6=j

Ji,j j2tot|l〉 (26)

= −∑i

Ji,l 2jtot (|i〉 − |l〉) + E0|l〉. (27)

(In the last line we used that i = l ⇒ |i〉 − |l〉 = 0.) Thus the elementary excitations above the groundstate are not simply re-orientations of individual quantum-mechanical spins. This already tells us that themean eld theory will be more limited at nite temperature since it assumed a description in terms of aparamagnet in an eective eld. In a paramgnet, the excitations at nite temperature are changes in theorientation of individual spins. In a true ferromagnet, the excitations are more complicated.

Let us now show that the following linear superposition of states with turned spins is an eigenstate:

|k〉 ≡ 1√N

∑l

eik·Rl |l〉 . (28)

This is a plane wave with wave vector k (⇒ wavelength λ = 2π/k) i.e. a state with well-dened mo-mentum, p = ~k, and which is therefore completely delocalised throught the material. This is consistentwith Heisenberg's uncertainty principle, ∆p∆x & ~. However, unlike a plane wave state of a free electron,the object which delocalises throught the sample here is not a particle, but the site at which the magneticmoment is slightly rotated away from the z axis.

3

Let us prove that |k〉 is an eigenstate of H. First we apply the Hamiltonian to |k〉:

H|k〉 =1√N

∑l

eik·RlH|l〉 (29)

=1√N

∑l

eik·Rl

∑i

Ji,l 2jtot (|i〉 − |l〉) + E0|l〉

(30)

= −2jtot1√N

∑l

eik·Rl∑i

J (Ri −Rl) |i〉 (31)

+2jtot1√N

∑l

eik·Rl∑i

J (Ri −Rl) |l〉 (32)

+1√N

∑l

eik·RlE0|l〉. (33)

Now, for the rst of the above three terms we have

− 2jtot1√N

∑l

eik·Rl∑i

J (Ri −Rl) |i〉 = −2jtot1√N

∑Rl,Ri

eik·RlJ (Ri −Rl) |Ri〉 (34)

[Ri −Rl ≡ R,Ri ≡ R′

]= −2jtot

1√N

∑R,R′

eik·(R′−R)J (R) |R′〉 (35)

= −2jtot∑R

eik·RJ (R)1√N

∑R′

eik·R′ |R′〉 (36)

= −2jtot∑R

eik·RJ (R) |k〉; (37)

for the seoncd term we have

2jtot1√N

∑l

eik·Rl∑i

J (Ri −Rl) |l〉 = 2jtot1√N

∑Ri,Rl

eik·RlJ (Ri −Rl) |Rl〉 (38)

[Ri −Rl ≡ R,Rl ≡ R′

]= 2jtot

1√N

∑R,R′

eik·R′J (R) |R′〉 (39)

= 2jtot∑R

J (R)1√N

∑R′

eik·R′ |R′〉 (40)

= 2jtot∑R

J (R) |k〉; (41)

and for the third term we have1√N

∑l

eik·RlE0|l〉 = E0|k〉. (42)

Putting all this together we arrive at

H|k〉 = −2jtot∑R

eik·RJ (R) |k〉+ 2jtot∑R

J (R) |k〉+ E0|k〉 (43)

=

[E0 + 2jtot

∑R

J (R)(

1− eik·R)]|k〉 (44)

This shows that |k〉 is a stationary state of the system,

H|k〉 = Ek|k〉 , (45)

with energy

Ek = E0 + 2jtot∑R

J (R)(

1− eik·R). (46)

4

The states |k〉 are spin waves and correspond to the lowest-energy excited states of the system. Ek is thespin-wave dispersion relation giving the energy Ek of the spin wave as a function of its wave vector, k.

A spin wave is a delocalised spin-1 excitation, since it corresponds to lowering the angular momentum of thesystem by ~ (this is what σi,− does; the spin wave is built by spreading this lowering of a spin by ~ equallythroughout the crystal). Therefore insofar as spin waves can be described as independent quasi-particlesthey have spin quantum number s = 1, which is an integer, and therefore they behave as bosons i.e. theyobery Bose-Einstein statistics (free electrons, have s = 1/2, therefore behaving as fermions and obeyingFermi-Dirac statistics). Such quantised spin wave quasiparticles are known as magnons.

(The independent spin-wave approximation is all right at low temperatures, when there are only a few ofthem spread throughout the system, but it is not so ne at higher temepratures when they start to collidewith each other.)

Let us now consider the above expression and expand for low wave vectors:

1− eik·R = 1−[1 + ik ·R− 1

2(k ·R)2 + . . .

](47)

= −ik ·R +1

2(k ·R)2 + . . . (48)

This leads to

Ek = E0 + 2jtot∑R

J (R)

[−ik ·R +

1

2(k ·R)2 + . . .

](49)

= E0 + jtot∑R

J (R) (k ·R)2 + . . . , (50)

which has a quadratic dependence on the wave vector of the spin wave, k :

(The plot shows the excitation energy with respect to the ground state, Ek − E0.)

This is a very generic and very important result. It tells us that, for isotropic ferromagnets, there is nogap to spin-wave excitations at low energies. There is a very deep reason for this: in the limit of longwavelengths, k → 0, the spin wave corresponds to lowering the z comoponent of the angular momentum(i.e. rotating the magnetic moments away from the preferred direction) uniformly throughout the sample.Since the total size of the angular momentum,

√J2 = ~

√σ2 = ~

√jtot (jtot + 1), cannot change while we

do this, this corresponds to a uniform rotation of all the magnetic moments in the sample. But as weknow the Heisenberg model is symmetric under such uniform rotations, i.e. the Hamiltonian of the systemis invariant under such continuous deformations. As a result, one can uniformly roate all the magnetic

5

moments of the sample at no energy cost. As k becomes nite, then the rotation is not the same everywhere,and it is no longer a symmetry of the Hamiltonian: there is a twist of nite wavelength λ = 2π/k...

λ/2

...and that twist, of course, does cost energy.

Quite generally, the ungapped, low-energy excitations that appear in a system whose ground state breaksa continuous symmetry of the Hamiltonian are called Goldston modes or Goldstone bosons. This is avery general idea that goes well beyond magnetism.

Magnons are very important in ferromagnets. For a start, because they are ungapped, they appear as soonas the temperature is non-zero. Since each magnon lowers the component of the magnetization along thedirection of the ferromagnetic order, the magnons lead to a lowering of the magnetization, below the valuepredicted by mean-eld theory, at any non-zero temperature. This is called Bloch's T 3/2 law, because ofthe dependence of the suppression of magnetization on temperature (for small temperatures).

The inuence of these Goldstone bosons goes even further for systems that are eectively two- or one-dimensional (i.e. made up of un-coupled planes or chains of magnetic moments). Becuase of the largedensity of states available for magnons at low energies in one or two dimensions, their proliferation at nitetemperature is so great that they completely suppress the magnetisation, killing the ferromagnetism. Thisis the Mermin-Wagner-Berezinskii theorem.

Because the Mermin-Wagner-Berezinskii theorem depends only on the existence of Goldstone bosons, it isvery general. In its more general formulation it says that you can have no spontenous symmetry breakingat nite temperature in systems with a continuous symmetry (i.e. with Glodston bosons) in less than threedimensions.

In addition to their inuence on thermodynamic properties, magnons can actually be observed directly usinginelastic neutron scattering (INS). INS works by selecting the energy of the neutrons before and afterthey scatter from the sample, which allows us to count the number of neutrons that have exchanged a givenamount of energy ~ω and momentum ~k with the sample. These are quite rare: most neutrons scatterelastically (this is the main assumption of neutron diraction experiments like the ones we discussed before).Some of the inelastically-scattered neutrons will have ipped their spin, transmiting a quantum of angularmomentum, ~, to the sample. This creates one magnon. By looking at many neutrons, and recording, forthose whose spin has ipped on interacting with the sample, their individual values of ~ω and k, we can seeat what combinations of those quantities scattering is maximum. Taking Ek = ~ω, those combinations givethe spin-wave dispersion relation.

6

Lecture XVIII. Jorge Quintanilla,

Magnetism and Superconductivity (PH752)

Let us now turn to superconductivity.

23

23 Images from Rudolf de Bruyn Ouboter, Heike Kamerlingh Onnes's Discovery of Superconductivity, Scientic American(March 1997) and from Dirk van Delft and Peter Kes, The discovery of superconductivity, Physics Today (September 2010).

1

2

Jorge Quintanilla

Magnetism and Superconductivity (PH752)

Problem Sheet 3

Information needed to tackle these problems can be found in the handouts for Lecture 17.

Problems marked with an asterisk (*) will be assessed.

Problem 1: Calculate the spin-wave dispersion relation Ek for the ferromagnetic Heisenberg model withjtot = 1/2.

Assume a 1D chain geometry:

Take the exchange constant to be equal to J between nearest-neighbours, zero elsewhere.

Compute the quadratic approximation to the dispersion relation valid at low values of the wavevectork.

Derive the mass of the magnons, mmag, by comparing to the formula

Ek = constant +~2k2

2mmag. (1)

Problem 2: (*) Calculate the spin-wave dispersion relation Ek for the ferromagnetic Heisenberg model withjtot = 1/2.

Assume a one-dimensional square lattice geometry:

Take the interactions to be of strength J between nearest neighbours (whether in the direction parallelor perpendicular to the ladder), and zero elsewhere.

1

Jorge Quintanilla

Magnetism and Superconductivity (PH752)

Problem Sheet 4

Information needed to tackle these problems can be found in the handouts for Lectures 9-16. You will need toconsult also a periodic table of the elements with electronic structure information and some physical constants(both can be found for example behind the front and back covers, respectively, of Setephen Blundell, Magnetismin Condensed Matter, OUP 2001). Give all results in SI units.

Problems marked with an asterisk (*) will be assessed.

Problem 1: A superconductor ring is installed on top of a solenoid of the same diameter: [DRAWING]The ring is kept in the superconducting state at all times. Initially there is zero current going throughthe solenoid. When we pass a current, is there any magnetic ux going through the ring? What ishappening inside the supercondcutor to produce that result? Now imagine that the ring were muchbroader than the solenoid: [DRAWING] Will any net magnetic ux go through the ring when westart passing current theorugh the solenoid? Will the magnetic induction on the plane of the ring bethe same everywhere? Sketch the magnetic eld lines around the ring.

1

Lecture XIX. Introduction: electrons in solids (1 lecture)

Introduction

Role of magnetism in current technologies: Krymer's law (vs Moore's law) (==> Ipods and Iphones):http://www.scienticamerican.com/article.cfm?id=kryders-law (request from library and use pictures)

Since then many magnetic and superconducting states with ever more exotic electromagnetic properties havebeen discovered, and the study of superconductivity and magnetism remains one of the most challengingareas of research.

Although some materials are capable of displaying magnetic or superconducting propeties while others arenot, it is important to emphasize that neither magnetism nor superconductivity are intrinsic properties of agiven substance. Instead, they occur only when the temperature of the sample is lowered below its criticaltemperature, Tc. For example, in Kammerlingh Onnes' experiment the sample of Hg went from having afairly conventional resistance, R ≈ 0.1Ω, at a temperature of T ≈ 4.3K to vanishing resistance, R < 10−5Ω,at T ≈ 4.1K (see Fig. ??). This change from ordinary metallic behaviour to superconducting behaviourtakes place quite sharply at Tc ≈ 4.2K. Similarly, magnetic states such as ferromagnetism also have criticaltemperatures. For example, iron is ferromagnetic only below its Curie temperature, Tc = 770K. Since thisis well above ordinary room temperatures, we are used to thinking of Fe as an intrinsically ferromagneticmaterial. However, if we heat Fe above its critical temperature it will suddenly loose its ferromagnetism,even though other properties such as crystal structure or electrical resistivity will be changing only smoothlywith temperature within this range.

How can a material undergo, without any change in composition or in the interactions between its con-stituents, radical changes in its properties such as those that take place when it enters a magnetic or super-conducting state? It all boils down to the balance between energy and disorder. At a given temperature,any system in equilibrium must minimize its Helmholtz free energy,

F = U − TS (1)

Here U is the internal energy of the system, while S is its entropy - a measure of how disordered the systemis. Evidently at high temperature the best way to lower F is to increase S, while at low temperatures keepingU low becomes more important. Thus magnetism and superconductivity are more ordered states than theirhigh-temperature non-magnetic and non-superconducting counterparts. The ordered states develop in orderto lower the internal energy of the system.

Just as the solid state is a state of matter, distinct from, for example, the liquid state, superconductivityand magnetism are also states of matter. What this means is that a sample will enter its superconductingor magnetic state as soon as it is placed in the right thermodynamic conditions, independently of the paththrough parameter space that was taken to place it there (as long as the sample is always in thermalequilibrium24).

When a more ordered state of matter is reached upon cooling there is often an associated broken symmetry.To understand this let us consider the example of crystallisation (the change from liquid to solid - Fig. 1). Theequations describing the microscopic motion of water molecules have a number of symetries. For example,they are homogeneous (the same weverywhere) and isotropic (the same in all directions). Yet when watercrystallises, forming ice, the molecules arrange themselves in an inhomogeneous, anisotropic state. This hasmacroscopic manifestations such as the beautiful shape of a snowake (Fig. 1). The snowake is symmetricunder rotations by π/3 and its multiples around its central axis but not, for example, under π/2 rotations.This is quite distinct from liquid water, which is symmetric under all rotations, and it reects the crystallinearrangement of the molecules on the microscopic scale. Since the collective state of the macroscopically-largenumber of molecules forming the snowake has lower symmetry than the microscopic equations of motion thatdescribe their motion we say that the symmetry is 'broken'.25 Similarly, the magnetic and superconducting

24 What this retriction amounts to, in practice, is that any change in the sample's conditions must have taken place veryslowly.25 The wonder and deep implications of broken symmetry were discussed by Nobel laureate Phil W. Anderson in his classic

article More is dierent, Science 177, 393-396 (1972).

1

Fig. 1: A water droplet (left) and a snowake (right).

states that a crystal may enter upon cooling further, below Tc, break additional symmetries. What thosesymmetries are we will get to in due course.

In principle, we can obtain the free energy of any system, and therefore all its thermodynamic properties,by computing its partition function,

Z =∑

n=1,2,3,...

e− EnkBT , (2)

where kB = 1.3807 10−23JK−1 (Boltzmann's constant). This requires knowing the possible microstates ofthe system, labelled by n = 1, 2, 3, . . . and their energies En. In other words we need to solve the mechanicsof the system under consideration.

For any crystal, the mechanics are given by the following Hamiltonian:

H = ... (3)

The above Hamiltonian is the theory of everything as far as condensed matter physics is concerned.26 [Twoapproaches: brute force (but no one knows how - you have to lower your demands on what you want to getand then you obtain DFT, which requires uncontrolled approximations) and models (the route we will followhere, which is used for the interesting cases when DFT does not work and also is more useful in a coursebecause it leads to understanding).]

In a metal, the crystal lattice is made up of positively-charged ions, with the remaining charge belonging tofree electrons that are responsible for the conduction of electricity.27 Superconductivity takes place whenthose free electrons re-arrange themselves into a collective state where such conduction occurs without anyresistance.

In insulators, on the other hand, every electron is bound to a specic atom, ion or molecule. Conductionof electricity is therefore strongly suppressed as the temperature is lowered. Insulators can never becomesuperconductors.28

Both metals and insulators possess magnetic moments due to the orbital motion of free electrons throughthe crystal or bound electrons around their atoms. In addition, the electron has an intrinisic angularmomentum, its spin, which is often the main contribution.

26 For conciseness, we have ommitted relativistic eects and assumed that there are no applied elds. However these are easyto put back in without altering the throust of our discussion.27 The positive ions are xed in deep potential wells at their lattice positions. The exception to this rule are super-ionic

conductors (not to be confused with superconductors) where both the free electrons and the ions are mobile. However, this israre, and throughout this course we will assume it doesn't happen.28 Interestingly, some of the superconductors with the highest known values of Tc behave very much like insulators until they

enter the superocnducting state. These systems are poorly understood.

2

It is clear from the above that both magnetism and superconductivity are instances of electronic orderi.e. it is the electrons inside the material which go from a disordered to and ordered state at Tc. This isquite dierent from crystallisation, the example of broken symmetry that we discussed above, where it is thenuclei that go into a more ordered state.

The crucial consequence of the electronic nature of magnetism and superconductivity is that it makes quan-tum mechanics essential to describe these states. Consider rst the case of a bound electron. The electronis conned to a space the size of an atom or molecule. The uncertainty in its position is therefore ∆x . 1nm.Now consider the case of the free electrons in a metal. These electrons have very low uncertainty in theirlocation, as they can roam throghout the sample: ∆x ∼ L (where L is the linear dimension of the sample).This gives a very low uncertainty of the momentum. At temperature T , the typical thermal energy of oneof these electrons will be ∼ kBT . At high temperatures, the unbound electrons will increase their energy byincreasing their kinetic energy, p2/2me ∼ kBT. From this we deduce a typical momentum for one of theseelectrons: p ∼

√2mekBT . Now, in quantum mechanics a particle is also a wave and its wavelength is given

by the famous de Broglie expression

λ =h

p. (4)

This leads to

λ ∼ h√2mekBT

(5)

which is the thermal de Broglie wavelength of the electron.

29

29 Here by disordered we mean non-magnetic and non-superconducting. The fact that the sample is crystalline alreadyimplies some form of order as discussed above. However in general the crystal structure does not change at Tc.

3

Reading list

The main texts for this course:

On superconductivity:J. F. Annett; Superconductivity, Superuids and Condensates (OUP, 2004)

On magnetism:S. Blundell; Magnetism in Condensed Matter (OUP, 2001)

These two are published as relatively inexpensive paperbacks so I would advise that all students taking thiscourse buy a copy of each if they can.

By the end of this course students should have read the above two titles from the rst page to the last.

More advanced texts that may be useful:

P G de Gennes; Superconductivity of Metals and Alloys (Westview Press, March 31, 1999)

This is a reprint of the classic text on superconductivity - still the most lucid exposition half a century afterit was rst published.

J B Ketterson and S N Song; Superconductivity (CUP, 1999)

This includes detailed discussions of more recent topics that have emerged after de Gennes' wrote his classictext such as superuidity in 3He, unconventional pairing in superconductors, etc. It is quite pedagogical,with the claculations carried out in explicit detail.

R M White, Quantum Theory of Magnetism; (Springer; softcover reprint of hardcover 3rd ed. 2007 edition- November 23, 2010)

This is a more advanced text on magnetism, emphasizing theoretical methods and also with a very goodsection on neutron scattering.

General solid state physics texts:

Aschroft and Mermin; Solid State Physics (Thomson Press India, paperback edition, 2003)

Charles Kittel; Introduction to Solid State Physics

These give a good grounding on solid state concepts useful to follow the course.

Other useful texts:

Michael Tinkham; Introduction to Superocnductivity (2nd Edition, 1996)

S. Elliot: The Physics and Chemistry of Solids (1998)

D.C. Mattis The theory of magnetism made simple (2004)

Tilley and Tilley; Superuidity and Superconductivity, (1990)

4

Index

broken symmetry, 1

critical temperature, 1

free energy, 1

Hamiltonian, 2

Magnetism, 1

partition function, 2

quantum mechanics, 3

states of matter, 1Superconductivity, 2

Maxwell's equations [SB B.12-15]

∇E = ρ/ε0

∇ ·B = 0

∇× E = −∂B∂t

∇×B = µ0J + ε0µ0∂E∂t

Tight-binding band structure30

Consider the Hamiltonian of an electron orbiting an ion located at R:

HR =p2

2me+

1

4πε0

ZRe2

|r−R|. (6)

The rst term is the kinetic energy and the second term is the Coulomb potential due to the attractionbetween the electron and the ion. The solutions to the corresponding time-independent Schrödinger equationgive the atomic orbitals and energy levels:

HR|ψRm〉 = Em|ψR

m〉. (7)

Note that the energy levels Em are independent of the position of the atom, R.

Now consider the Hamiltonian of the same electron but in the presence of a whole Bravais lattice of ions.Let R be the set of all Bravais lattice vectors (the positions of the ions). The Hamiltonian is

H =p2

2me+∑R

1

4πε0

Ze2

|r−R|(8)

≡ p2

2me+∑R

VR

= HR +∑

R′ 6=R

VR′ . (9)

where in the second line we have introduced the notation

VR =1

4πε0

Ze2

|R− r|(10)

for the Coulomb interaction between the electron and the nuceus located at position R. We wish to nd anapproximate solution to the corresponding time-independent Schödinger equation,

H|ψ〉 = E|ψ〉. (11)

The Schrödinger equation is HR +∑

R′ 6=R

VR′

|ψ〉 = E|ψ〉 (12)

and we will treat the term∑

R′ 6=R VR′ as a perturbation.

Let us expand the state of the electron, |ψ〉, in terms of the atomic orbitals|ψRm〉m,R

30 We follow quite closely the treatment in Ashcroft and Mermin, Solid State Physics.

1

|ψ〉 =∑m,R

|ψRm〉〈ψR

m|ψ〉. (13)

This expression assumes that|ψRm〉m,R

is a complete set:∑

m,R |ψRm〉〈ψR

m| = 1. The brackets 〈ψRm|ψ〉 are

the coecients of the expansion. We now require that the state of the electron obey Bloch's theorem:31

〈r + R|ψ〉 = eik.R〈r|ψ〉 for any position vector r and lattice vector R. (14)

We can attain this condition by choosing

〈ψRm|ψ〉 = eik.R〈ψ0

m|ψ〉, (15)

Under this assumption the expansion of the state of the electron becomes

|ψ〉 =∑m,R

eik.R〈ψ0m|ψ〉|ψR

m〉 (16)

=∑m,R

eik.Rψm|ψRm〉,

where we have introduced the shorthand notation

ψm ≡ 〈ψ0m|ψ〉

for the amplitudes to be determined.32

We now substitute (16) into (12) and project both sides of the equation on 〈ψ0n|. The LHS takes the formHR +

∑R′ 6=R

VR′

∑m,R

eik.Rψm|ψRm〉 =

∑m

ψmEm∑R′′

eik.R′′〈ψ0

n|ψR′′m 〉+

∑R′′

eik.R′′〈ψ0

n|∑R′ 6=0

VR′ |ψR′′m 〉

while the RHS becomes ∑m

ψmE∑R′′

eik.R′′〈ψ0

n|ψR′′m 〉.

Thus what we have∑m

ψm(Em − E)∑R′′

eik.R′′〈ψ0

n|ψR′′m 〉+

∑R′′

eik.R′′〈ψ0

n|∑R′ 6=0

VR′ |ψR′′m 〉 = 0,

31 For a proof see the derivation of Eq. (8.6) in Ashcroft and Mermin.32

Proof of Eq. (15): Projecting (16) on r+R we obtain

〈r+R|ψ〉 = 〈r+R|∑m,R′

eik.R′〈ψ0m|ψ〉|ψR′

m 〉

=∑m,R′

eik.R′〈ψ0m|ψ〉〈r+R|ψR′

m 〉

=∑m,R′

eik.R′〈ψ0m|ψ〉〈r|ψR′−R

m 〉

[R′′ ≡ R′ −R

]=

∑m,R′′

eik.(R′′+R)〈ψ0

m|ψ〉〈r|ψR′′m 〉

= eik.R〈r|∑m,R′′

eik.R′′〈ψ0m|ψ〉|ψR′′

m 〉

= eik.R〈r|ψ〉, Q.E.D.

We have used the fact that 〈r|ψ0m〉 = 〈r+R|ψR

m〉 i.e. the amplitude at r of the mth atomic orbital in the atom atthe origin is the same as the amplitude at r+R of the same atomic orbital but for the atom at R.

2

which is a matrix equation for the vector

ψ =

ψ0

ψ1...

,

namelyA.ψ = 0.

Here

An,m ≡∑R′′

eik.R′′

(Em − E) 〈ψ0n|ψR′′

m 〉

+∑R′′

eik.R′′〈ψ0

n|∑R′ 6=0

VR′ |ψR′′m 〉.

Take the simplest case of a 1D system when 〈ψRn |ψR′

m 〉 = δn,mδR,R′and 〈ψ0n|∑

R′ 6=0 VR′ |ψR′′m 〉 = 0 except

when n = m and R′′ = ±a, when it equals −tm. Then

A =

E0 − E − 2t0 cos (ka) 0 0 · · ·

0 E1 − E − 2t1 cos (ka) 0 · · ·0 0 E2 − E − 2t2 cos (ka) · · ·...

......

. . .

so we have

A =

E0 − 2t0 cos (ka) 0 0 · · ·

0 E1 − 2t1 cos (ka) 0 · · ·0 0 E2 − 2t2 cos (ka) · · ·...

......

. . .

ψ0

ψ1

ψ2...

= E

ψ0

ψ1

ψ2...

,

chracteriand eigenvalue problem whose solutions are given by the charcteristic ploynomial equation∏m

[Em − 2tm cos (ka)− E] = 0⇒

E = Em − 2tm cos (ka) . (17)

Near the bottom of the lowest band we have

E ≈ E0 + tma2k2 ≡ ~2k2

2meff(18)

leading to

meff =~2

2tma2.

To have this equal to an electron mass we would need

tma2 =

~2

2me.

The discretised form of the kinetic energy − ~22me

∂2ψ∂x2

is

− ~2

2me

ψ(x+ a)− 2ψ (x) + ψ (x− a)

a2

so the Schrodinger equation with a potential V (x) is

− ~2

2me

ψ(x+ a)− 2ψ (x) + ψ (x− a)

a2+ V (x)ψ (x) = Eψ (x)

3

i.e.

~2

2mea2

2 + v0 −1 0 · · ·−1 2 + v1 −1 0 · · ·0 −1 2 + v2 −1 0 · · ·... 0 −1 2 + v3 −1 0 · · ·

... 0 −1 2 + v4 −1. . .

.... . .

. . .. . .

. . .

ψ0

ψ1

ψ2

ψ3

ψ4...

= E

ψ0

ψ1

ψ2

ψ3

ψ4...

where vj = V (ja) /

(~2

2mea2

). Thus solving this problem reduces to nding the eigenvalues of the above

matrix i.e. solving the characteristic polynomial equation∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

2 + v0 − ε −1 0 · · ·−1 2 + v1 − ε −1 0 · · ·0 −1 2 + v2 − ε −1 0 · · ·... 0 −1 2 + v3 − ε −1 0 · · ·

... 0 −1 2 + v4 − ε −1. . .

.... . .

. . .. . .

. . .

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0.

Additional material

Broken symmetry

Fitzgerald: The rich are dierent from us.

Hemingway: Yes, they have more money.

Quoted in P. W. Anderson,More is dierent, Science 177, 393-396 (1972).

Interestingly, the increased order in the superconducting and magnetic states manifests itself in the form ofbroken symmetries. This is a very fundamental concept in Physics that is best understood by analysinga more familiar phase transition rst: crystallisation.

Consider a system of N classical, point-like particles, described by the Hamiltonian

H =∑

i=1,2,...,N

p2i

2mi+

∑i,j=1,2,...,N

V (|ri − rj |) . (19)

Here mi is the mass of the ith particle, pi is its momentum, ri its position, and V (|ri − rj |) is a potential

describing the interaction between the particles at positions ri and rj . Note that the interaction potentialonly depends on distance |ri − rj | between the two particles. Similarly, ... This Hamiltonian has a numberof symmetries. For example, if we translate the system by a vector R,

ri → ri + R for all i, (20)

the Hamiltonian stays the same. Similarly, if we rotate the system by a given angle the Hamiltonian alsodoes not vary. ...

that is very well illustrated by comparing the structure of a water droplet to that of a snowake (Fig. 1).Both are made up of identical consituents, namely large quantities of H2O molecules. Yet the snowake ismade up of crystalline water, which has much lower symmetry than liquid water. This is quite clearly visiblein the shape of the snowake, which is symmetric under rotations by exactly 60o, or one of its multiples.Liquid water, on the other hand, is symmetric under all rotations (i.e. it stays the same after we rotate it by

1

an arbitrary angle). So the snowake is much less symmetric than the droplet - we say that in the snowakerotational symmetry is broken. Indeed liquid water has all the symmetries of free space, while crystallinewater (ice) lacks many translational and rotational symmetries. What is remarkable about broken symmetryis that the state of the system has lower symmetry than the Hamiltonian that describes it.33

As we will see magnetism and superconductivity provide particularly rich examples of symmetry-breaking.

Can one predict a broken-symmetry state from a microscopic model of a given system? In principle, themethods of statistical mechanics allow us to compute any observable for a system in thermal equilibrium atsome given temperature T . This presumes that we can solve the mechanics of the system, i.e. that we canenumerate all its possible microstates n = 1, 2, 3, . . . and nd the corresponding energies E1, E2, E3, . . . Thenwe can compute the partition function

Z =∑

n=1,2,3,...

e− EnkBT . (21)

Here kB = 1.3807 10−23JK−1 is Boltzmann's constant. Once we know this, the probability of any microstateis given by

pn =1

Ze− EnkBT . (22)

This in turn allows us to compute the thermal average of any microscopic observable On by averaging overmicrostates:

O ≡ 〈On〉 =∑n

pnOn. (23)

Alternatively, we can use Z to derive the free energy using

F = −kBT lnZ (24)

and employ thermodynamic relations to obtain other observables by dierentiating the free energy.

For example, the internal energy is the thermal everage of the energy En over all the microstates n:

U = 〈E〉 =∑n

pnEn. (25)

This could be used to compute U directly or alternatively one could compute the free energy F and thenuse the thermodynamic relation

U = −kBT 2 ∂

∂T

(F

kBT

). (26)

This can be expressed more compactly using

U =∂

∂β(βF ) (27)

where we have introduced the inverse thermal energy

β ≡ 1

kBT. (28)

Exercise: Prove that (27) is equivalent to (26) using the chain rule. Then prove that this thermodynamicrelation is equivalent to the statistical-mechanical expression (25).Hints: it is much better to put all expressions in terms of β rather than T rst. Then substitute (21)into (24) and that into (27). Substitute (22) into (25) and compare the results.

33 The deep implications of broken symmetry were discussed by Nobel laureate Phil W. Anderson in his classic article Moreis dierent, Science 177, 393-396 (1972).

2

What makes magnetism and superconductivity special as broken-symmetry states is that to compute thespectrum Enn=1,2,... it is indispensable to employ quantum mechanics. In magnetism, time-reversalsymmetyr is borken, while in superconductivity is the symmetry with respect to an overall change of thewave function. Classical physics gives no mechanism for the breaking of either of these symmetries (in thecase of superconductivbity, the pahse of the wave function is not even a classical concept).

The problem isH = ...

but of course we cannot solve this problem so we work with models.

Problem: Using dimensional analysis, show what the superocnducting and magnetic Tc can depend on.

Now let us put more esh into the above discussion by considering the particular case of a ferromagnetictransition. To keep things simple, let us assume that the system is an insulator so we do not have the addedcomplications of electrical conductivity and possible superocnducting behaviour. [...]

This is made up of ∼ 1023 molecules of water that move about in a complicated fashion following (toa rst approximation) the laws of classical mechanics. Nevertheless on the macroscopic leand QuantumMechanics is the deepestngth scales corresponding to the above picture the liquid making up the dropletseems stationary, uniform and isotropic. This is not suprising i.e. it has the same symmetries quite uniformrepresents the water droplet level Insideaccording to the droplet, we nd a liquid that is homogeneous andisotropic: although the chaotic trajectories of the water molecules inside it

...

are entered What happened at Tc was not a change in the sample's constituents, a Coulomb-interactingmixture of ∼ 1023 particles (positive ions vibrating around their crystal lattice positions34 and conductionelectrons moving around them), but of the collective state of those constitutents.

ions, the electrons : both above and below Tc, Onnes' sample of Hg was always made up of the same ∼ 1023

Mercury ions, vibrating around their equilibrium positions on the crystal lattice , with the same conductionelectrons moving through it. The interactions between the electrons, between the electrons and the ions, andbetween the ions were also exactly the same above and below the critical temperature.

Indeed the whole physics of most superocnductors can be described by the relatively simple Hamiltonian

H =∑i

P 2i

2Mi+∑j

p2j2me

(29)

+1

4πε0

∑i,i′

ZiZi′e2∣∣∣Ri − Ri′∣∣∣ +

∑j,j′

e2∣∣rj − rj′∣∣ −∑i,jZie

2∣∣∣Ri − rj∣∣∣

which is, of course, independent of temperature. 35

-

To understand this consider the de-Broglie wavelength of an electron with momentum p,

λ =h

p. (30)

Here h = 6.6262 10−34 Js is Planck's constant (the quantum of action). At a given temperature T , the kineticenergy of the electron will be

p2

2me. kBT, (31)

34 Mercury crystallises at Tm = 234.32K [http://en.wikipedia.org/wiki/Mercury_%28element%29 (accessed 19 August 2011)].35 Here the indices j, j′ designate electrons, i, i′ designate ions, −e and me are the charge and mass of an electron, respectively,

Zie and Mi are is the charge and mass of an ion, and the position and momenta of electrons and ions are given by the operatorsrj , Ri, pj , Pi, respectively.

3

where me = 9.1095 10−31Kg is the rest mass of an electron (for free electrons we would have written ∼;the . sign take into account that the electron may be bound inside an atom or interacting with otherelectrons). Combining the last two expressions we obtain the thermal de Broglie wavelength of theelectron,

λ & λT =h√

2mekBT. (32)

At room temperature, T ≈ 300K, we have λT = 1.3 10−8m ∼ 10nm. Given that

Let us rst rst consider an isolated atom. As we know from undergraduate quantum mechanics, the atomhas a postiviely-charged nucleus surrounded by a cloud of electrons that occupy a series of discrete energylevels. We will ignore for the time being the interactions between the electrons. Then the Hamiltonian ofeach electron can be written individually. Assuming that we can take the nucleus to be xed (using the factthat its mass is much larger than that of the electron) the Hamiltonian in question is

H =p2

2me− 1

4πε0

Ze2

|r|, (33)

where p is the momentum of the electron, me is its mass, e its charge, and Z the atomic number (so Zeis the charge of the nucleus). The rst term on the RHS is the kinetic energy and the second term isthe potential energy due to the interaction with the nucleus. The factor 1/4πε0 in front of the Coulombinteraction features the permitivity of free space ε0 = 8.85419 10−12 Fm−1 (we will use SI units throughout)., and describe exclusively the interaction between each electron and the nucleus. At the non-relativistic levelthis is given by Schrödinger's equation

p2

2mr− 1

4