Notes on Magnetism for Solid State Physics 1

Version: Semester 2 AY 2012-2013

Leek Meng Lee email: mlleek@ntu.edu.eg

Contents

1 Basic Concepts and Quantities 2

2 Overview and Classification 3

3 Bohr-van Leeuwen Theorem 4

4 Diamagnetism 54.1 General: Atom in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54.2 Diamagnetism of core electrons: Larmor or Langevin Diamagnetism . . . . . . . . . . . . 64.3 Diamagnetism of conduction electrons: Landau Diamagnetism . . . . . . . . . . . . . . . 7

5 Paramagnetism 95.1 Atomic Paramagnetism: Curie or Langevin Paramagnetism . . . . . . . . . . . . . . . . . 95.2 Paramagnetism of Conduction Electrons: Pauli Paramagnetism . . . . . . . . . . . . . . . 115.3 Van Vleck Paramagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.4 Isentropic Demagnetization or Adiabatic Demagnetization . . . . . . . . . . . . . . . . . . 14

6 Collective Magnetism 156.1 Direct Exchange Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.2 Ferromagnetic Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6.2.1 Weiss Mean Field theory of Ferromagnetism (Simplest theory of ferromagnetism) . 196.2.2 Magnons (Quantized Spin Waves) . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2.3 Ferromagnetic Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

[Why do domains form?] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25[Types of Domain Walls] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26[Estimation of a 180o Blochs Walls energy:] . . . . . . . . . . . . . . . . . . . . . 27[Magnetization and Hysteresis] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.3 Antiferromagnetic and Ferrimagnetic Order . . . . . . . . . . . . . . . . . . . . . . . . . . 296.3.1 Weiss Mean Field Theory of Antiferromagnetism . . . . . . . . . . . . . . . . . . . 29

7 Tutorial Problems 31

8 Tutorial Solutions 33

1

1 Basic Concepts and Quantities

Here, we briefly review some concepts from electromagnetism.

1. Magnetization ~M :

~M = limV0

1

V

over Vi

~mi (1)

where ~mi is the magnetic moment of the ith atom and V is a small volume. Thus magnetizationis also the magnetic dipole moment per unit volume.

2. ~H-field:

The magnetization current is ~JM = ~ ~M and the total ~B-field is~ ~B = 0

(~J + ~JM

)(2)

where ~J is the ordinary current. Define ~H-field as

~H =1

0~B ~M = ~ ~H = ~J (3)

3. Magnetic susceptibility :

Assume the constitutive equation ~M = ~H then

~B = 0( ~H + ~M) = 0(1 + ) ~H (4)

4. Energy of magnetic moment:

Figure 1: A current loop and the area element vector.

d~m = Id ~A (5)in standard notation

= d~ = Id ~A (6)~ = I

d ~A (7)

E = ~ ~B from electromagnetism (8)

The quantum Hamiltonian will be inspired from that expression.

5. Bohr magneton B:

Consider the hydrogen in the n = 1 ground state. In Bohr-Sommerfeld quantization, mevr = h.

magnetic moment: = I A = r2I (9)= r2

et

(10)

= r2e

2r/v(11)

2

Figure 2: The Hydrogen atom.

= evr2

(12)

| use Bohr-Sommerfeld quantization rule (13)= eh

2me(14)

| define the Bohr magneton B = eh2me

(15)

= B (16)

2 Overview and Classification

Magnetic Phenomena can be classified into 3 main groups:

1. Diamagnetism

susceptibility dia < 0 external field induces magnetic dipoles which orientate themselves antiparallel to the fieldfollowing Lenzs law.

Thus it is a property of all materials. However it is a weak effect that is usually overwhelmed by other magnetic effects. Superconductors are perfect diamagnets, i.e. the external field is completely negated in thesuperconductor, or dia = 1.

2. Paramagnetism

susceptibility para > 0 permanent magnetic dipoles

Localized moments due to partially filled inner electron shells (Curie, Langevin Param-agnetism).

Itinerant moments due to free conduction electrons (Pauli Paramagnetism).

Collective or Ordered Magnetism susceptibility is a complicated function = (T,H,history of material)

This type of magnetism is due to (quantum mechanical) exchange interaction betweenthe permanent magnetic dipoles.

There is a critical temperature TC and below TC , there is spontaneous magnetization.

3 main types: Ferromagnetism, Ferrimagnetism and Antiferromagnetism.

3

3 Bohr-van Leeuwen Theorem

The Bohr-van Leeuwen theorem is

In a classical system, there is no thermal equilibrium magnetization.

Proof:Assume a 3D solid with N identical ions with magnetic moment ~m each. The magnetization is

~M =N

V~m (17)

where ~m is done using the classical statistical average,

~m =dx1 dx3Ndp1 dp3N ~m eHdx1 dx3Ndp1 dp3NeH (18)

where = 1kBT (kB is the Boltzmann constant and T is the thermodynamic temperature) and from

E = ~ ~B we can write

~ = E ~B

= ~m = H ~B

(19)

where H is the classical Hamiltonian. Then we can write

~m =dx1 dx3Ndp1 dp3N

(H

~B

)eH

dx1 dx3Ndp1 dp3NeH (20)

| define the classical partition function Z = 1N !h3N

dx1 dx3Ndp1 dp3NeH (21)

| note Z ~B

=1

N !h3N

dx1 dx3Ndp1 dp3N

(H

~B

)eH so we can write (22)

=1

Z

Z

~B(23)

The classical Hamiltonian with magnetic field ~B = ~ ~A is in the form

H =1

2m

3Ni=1

(~pi + e ~Ai

)2+ other terms without ~A (24)

Z =1

N !h3N

dx1 dx3Ndp1 dp3Ne

2m

3Ni=1(~pi+e

~Ai)2+ (25)

| change variables ~ui = ~pi + e ~Ai and note the integration limits for ~pi goes from to +(26)=

1

N !h3N

dx1 dx3Ndu1 du3Ne

2m

3Ni=1 ~u

2i+ (27)

which gives the result that Z is independent of ~B. In fact, Z is the same as that without the magneticfield. Thus Z

~B= 0 and ~m = 0. PROVED.

We provide a cartoon picture of the theorem. See figure 3.

4

Figure 3: The cartoon picture of Bohr-van Leeuwen theorem. The closed orbits and skipping orbitscontribute opposite magnetic moments and so they cancel out.

4 Diamagnetism

4.1 General: Atom in a Magnetic Field

Assume there are Z (core) electrons in the atom in a magnetic field ~B = ~ ~A, the quantum Hamiltonianis (where V is the atomic potential)

H =

Zi=1

1

2me

(~pi + e ~Ai

)2+ V (28)

=Zi=1

1

2me

(~p 2i + e(~pi ~Ai + ~Ai ~pi) + e2 ~Ai ~Ai

)+ V (29)

| choose Coulomb gauge ~ ~A = 0 (30)| writing ~A(~r) = 1

2( ~B ~r) will satisfy both ~ ~A = 0 and ~B = ~ ~A (31)

=Zi=1

1

2me

(~p2i + e(~pi ~Ai + ~Ai ~pi)e2

(1

2( ~B ~ri)

)2)+ V (32)

| note that in position representation, using ~i ~Ai = 0, (33)| we can write ~pi ~Ai + ~Ai ~pi ih~i ( ~Ai) + ~Ai (ih~i) = 2ih ~Ai ~i 2 ~Ai ~pi(34)| then substitute ~Ai = 1

2( ~B ~ri) (35)

=

Zi=1

1

2me

(~p2i + e(

~B ~ri) ~pi + e2

4( ~B ~ri)2

)+ V (36)

| write ( ~B ~ri) ~pi = (~ri ~pi) ~B which is the triple product identity (37)| and note that ~ri ~pi = h~Li where ~Li is the dimensionless orbital angular momentum (38)

=Zi=1

1

2me

(~p2i + eh

~Li ~B + e2

4( ~B ~ri)2

)+ V (39)

| include spin of the electrons interacting with the magnetic field Hspin = Bg0 ~S ~B (40)| where g0 2 and B = eh

2meis the Bohr magneton (41)

5

=Zi=1

~p2i2me

+ V +

Zi=1

B

(~Li + g0~Si

) ~B

paramagnetism

+

Zi=1

e2

8me( ~B ~ri)2

diamagnetism

(42)

We said diamagnetism is a phenomena in all materials but the Hamiltonian seems to indicate thatparamagnetism and diamagnetism are universal phenomena. Actually, the paramagnetism term vanishesfor certain states of the atom. We can see that by first rewriting the paramagnetic Hamiltonian. (weconsider only 1 term in the sum and drop the index i for simplicity)

Hpara = B(~L+ g0~S) ~B (43)= B ~B ~J 1~J2

(~J ~L+ g0 ~J ~S

)(44)

| note ( ~J ~L)2 = ~J2 + ~L2 2 ~J ~L where ~J = ~L+ ~S and ~J ~L commutes with ~L ~J (45)| a similar expression is used for ~J ~S (46)= B ~B ~J 1

2 ~J2

(~J2 + ~L2 ( ~J ~L)2 + g0( ~J2 + ~S2 ( ~J ~S)2)

)(47)

| use ~J = ~L+ ~S (48)

=B ~B ~J2 ~J2

(~J2 + ~L2 ~S2 + g0( ~J2 + ~S2 ~L2)

)(49)

| recall that the square of (dimensionless) angular momentum operators (50)| are: ~J2 = J(J + 1), ~L2 = L(L+ 1) and ~S2 = S(S + 1) (51)| this is written in atomic physics notation (52)= B ~B ~J

(1 + g02

+g0 12

(S(S + 1) L(L+ 1)

J(J + 1)

))(53)

| recall that g0 2 (54)= B ~B ~J

(1 +

J(J + 1) L(L+ 1) + S(S + 1)2J(J + 1)

)(55)

= geffB ~B ~J (56)

where geff is called the Lande g-factor. And so in a state |J = 0 (which is determined by Hundsrules in atomic physics), there is no paramagnetism and thus diamagnetism is truly universal.

4.2 Diamagnetism of core electrons: Larmor or Langevin Diamagnetism

We will consider the state |J = 0 since in other states, the presence of paramagnetism will overwhelm dia-magnetism. So now Hdia is the perturbation Hamiltonian. Recall from first order Rayleigh Schrodingertime independent perturbation theory, the energy correction is,

E(1)n = n(0)|Hint|n(0) (57)

E(1)0 =

0

Zi=1

e2

8me

(~B ~ri

)2 0

(58)

| where we wrote |J = 0 = |0 (59)| assume the magnetic field is in the z-axis ~B = (0, 0, B) (60)

=

Zi=1

e2

8me0 |(Bxi yiB)z (Bxi yiB)z| 0 (61)

=Zi=1

e2B2

8me0|x2i + y2i |0 (62)

6

| assume the state is spherically symmetric which is reasonable for J = 0 (63)| then the expectation values x2i = y2i =

1

3r2i (64)

=e2B2

12me

Zi=1

0|r2i |0 (65)

Recall the magnetic moment is ~ = E ~B

, so the moment of this single atom is

= E(1)0

B= e

2B

6me

Zi=1

0|r2i |0 (66)

Assume a 3D solid of volume V with N identical such atoms, the magnetization is (recall ~M = NV ~m),

M =N

V = N

V

e2B

6me

Zi=1

0|r2i |0 (67)

Recall ~M = ~H and ~H = 10~B ~M . Since M

=Zi=1

1

2me

(p2ix + p

2iz + (piy + eBx)

2)

(72)

| we can rewrite it to make it look more familiar, define xi0 = piyeB

and c =eB

me(73)

=Zi=1

(p2ix2me

+1

2mec(xi xi0)2 + p

2iz

2me

)(74)

The (time independent) Schrodinger equation is H = E and for position representation pix ih xi ,piy ih yi and piz ih zi . Considering only 1 electron (thus dropping index i),[

h2

2me

2

x2+1

2me

2c

(x ih

eB

y

)2 h

2

2me

2

z2

] = E (75)

write in the form = eikzzeikyy(x) | (76)[ h

2

2me

2

x2+1

2me

2c

(x+

hkyeB

)2+h2k2z2me

](x) = E(x) (77)

[ h

2

2me

2

x2+1

2me

2c

(x+

hkyeB

)2](x) =

(E h

2k2z2me

)(x) (78)

This is simply a (shifted potential) harmonic oscillator problem. Hence the energy eigenvalues are

E h2k2z2me

=

(n+

1

2

)hc (79)

E =

(n+

1

2

)hc +

h2k2z2me

(80)

Thus the electron is quantized in the x-y plane perpendicular to the field and it is a free particle inthe z-direction which is the direction of the field.

Now we will only outline the statistical mechanics calculation leading to the susceptibility. The(quantum) partition function for a canonical ensemble is calculated using

Z =

dkz

n

e EkBT =

dkz

n

e 1kBT

((n+ 1

2)hc+

h2k2z2me

)(81)

Then the Helmholtz free energy F is given by

F = kBT lnZ (82)The magnetization is obtained from

M = FB

= F0H

(83)

Finally, the magnetic susceptibility is obtained from

=

(M

H

)H=0

(84)

The result for Landau diamagnetism is

Landau = N2V

02B

F (85)

where F is the Fermi energy.

8

5 Paramagnetism

5.1 Atomic Paramagnetism: Curie or Langevin Paramagnetism

Consider an atom with electrons in states giving a net angular momentum J (integer or half integer). 1

The paramagnetic Hamiltonian is

Hpara = geffB ~B ~J (86)Take ~B = Bz so ~B ~J = BJz and denoting atomic states as |J,mJ, the energy eigenvalues are

Hpara|J,mJ = geffBBJz|J,mJ = geffBBmJ |J,mJ (87)Note that there is no h as Jz is defined to be dimensionless.The calculation leading to the susceptibility is done using statistical mechanics. The (quantum)

partition function for a canonical ensemble is

Z =J

mJ=Je EkBT (88)

=J

mJ=Je geffBBmJ

kBT (89)

| note that we can rename the summation to absorb the minus sign (90)| and let x = geffBB

kBT(91)

=

JmJ=J

exmJ (92)

= exJ + ex(J1) + . . . + ex(J1) + exJ (93)

= exJ(1 + ex + e2x + . . .+ ex(2J1) + ex2J

)(94)

| which is a finite geometric series with first term = 1, common factor = ex (95)

| so use SN =Nk=0

rk =1 rN+11 r and here there are 2J + 1 terms so N + 1 = 2J + 1 (96)

= exJ1 ex(2J+1)

1 ex (97)

=exJ ex(J+1)

1 ex (98)

=exJex/2 ex(J+1)ex/2

ex/2 ex/2 (99)

=e

x2(2J+1) ex2 (2J+1)e

x2 ex2 (100)

| use sinh z = 12(ez ez) (101)

=sinh

(x2 (2J + 1)

)sinh

(x2

) (102)Magnetization is calculated from

M = FB

= B

(kBT lnZ) (103)1This is determined using Hunds Rules in atomic physics.

9

= kBT1

Z

Z

B(104)

| but recall x = geffBBkBT

= xB

=geffBkBT

(105)

= geffB1

Z

Z

x(106)

= geffB1

Z

sinh(x2

)sinh

(x2 (2J + 1)

) ( 2J+12 cosh (x2 (2J + 1))sinh

(x2

) sinh (x2 (2J + 1)) cosh (x2 )2 sinh2

(x2

))(107)

= geffBJ

(2J + 1

2Jcoth

(2J + 1

2JxJ

) 12J

coth

(xJ

2J

))(108)

| define the Brillouin function BJ(y) = 2J + 12J

coth

(2J + 1

2Jy

) 12J

coth( y2J

)(109)

= geffBJBJ(xJ) (110)

The susceptibility can be calculated using =(MH

)H=0

= 0(MB

)B=0

but this will make the makethe algebra very messy. We will calculate 2 specific cases only.

Case 1: Smallest angular momentum value of J = 12 .

BJ(xJ)|J=1/2 =2(12

)+ 1

2(12

) coth(2(12

)+ 1

2(12

)) 12(12

) coth(x(12

)2(12

))

(111)

= 2 coth x coth(x2

)(112)

| use the addition identity coth x = coth2(x2

)+ 1

2 coth(x2

) (113)=

1

coth(x2

) (114)= tanh

(x2

)(115)

M = geffBJ tanh(x2

)(116)

= 0

(M

B

)B=0

= 0

(M

x

x

B

)B = 0 or x = 0

(117)

= 0

(geffB

12

2

2 (x2

) geffBkBT

)x=0

(118)

=g2eff0

2B

4kBT(119)

Case 2: x

=1

Jx+(2J + 1)2x

12J 1Jx

x12J

(123)

=4J2 + 4J

12Jx (124)

=J + 1

3x (125)

M = geffBJJ + 1

3x (126)

= 0

(M

x

x

B

)B = 0 or x = 0

(127)

= 0

(geffB

J(J + 1)

3

geffBkBT

)x=0

(128)

=0g

2eff

2BJ(J + 1)

3kBT (129)

This relation of 1T is known as the Curies law of paramagnetism.

5.2 Paramagnetism of Conduction Electrons: Pauli Paramagnetism

For conduction electrons, they contribute a paramagnetic moment because each electron has spin 12 . Forsimplicity, the derivation will be done assuming T = 0.

We split the density of states of conduction electrons into spin up and spin down components.

(E) = (E) + (E) (130)

The magnetic field is assumed to be in the positive z-direction ~B = Bz and ms = +12 (parallelto ~B) and ms = 12 (antiparallel to ~B).

When there is no magnetic field, there is an equal number of -electrons as -electrons and (E) =(E) and there is zero net magnetization.

Figure 4: The density of states for free electrons separated into spin up and spin down parts. Thereare equal numbers of spin up and spin down electrons on average. The electrons are filled to F , theFermi energy.

On application of the magnetic field ~B = Bz, and considering only spin, the Hamiltonian of theproblem is the paramagnetic Hamiltonian Hpara = g0B ~S ~B.

Spin up electrons: Hpara| = 2B ~S ~Bms = +12 = 2BBSz ms = +12 = BB ms = +12

Spin down electrons: Hpara| = 2B ~S ~Bms = 12 = 2BBSz ms = 12 = BB ms = 1211

Thus spin up electrons gain BB energy and spin down electrons lose BB energy. The density ofstates are modified,

(E) (E BB) and (E) (E + BB) (131)

Figure 5 shows the situation clearly.

Figure 5: The density of states for free electrons separated into spin up and spin down parts underthe influence of a magnetic field in the spin up direction. Spin up and spin down electrons gains BBand loses BB respectively. Then the system reaches new thermal equilibrium at F .

There is now excess spin down electrons and this causes a net magnetization.

M =BV

(N N) = B(n n) (132)

The number densities of up and down electrons are denoted by n and n respectively. We need toevaluate the number densities.

n = +

dE(E BB)fFD(E) (133)

| where fFD is the Fermi-Dirac distribution (134)=

+BB

dE(E BB)fFD(E) (135)

| make a change of variables E = E BB (136)=

0

dE(E)fFD(E + BB) (137)

| since BB

| recall that (E) = (E) and (E) + (E) = (E) (143)

= 2BB 0

dE(E)fFD(E)

E(144)

We can calculate the susceptibility now.

Pauli = 0

(M

B

)B=0

(145)

= 02B 0

dE(E)fFD(E)

E(146)

| at low temperature, we approximate fFD(E) (F E) which is the step function(147)

| thus fFD(E)E

E

(F E) = (F E) and evaluate the integral (148)= 0

2B(F ) (149)

| recall in free electron theory (F ) = 32

N

V

1

F(150)

=3N

2V

02B

F(151)

Notice that there is a relationship between Landau and Pauli:

Landau = 13Pauli (152)

Thus the total susceptibility of a metal in a magnetic field is

metal = Landau + Pauli + Larmor + Curie (153)

= 13Pauli + Pauli + Larmor + Curie (154)

=2

3Pauli + Larmor + Curie (155)

It turns out that the dominant contribution to metal depends very much on the material.

5.3 Van Vleck Paramagnetism

Recall in the discussion of diamagnetism, in the state |J = 0 = |0, the paramagnetic Hamiltonian givesno contribution in first order perturbation, i.e.

For Hint =

Zi=1

B(~Li + g0~Si) ~B + e2

8me( ~B ~ri)2 (156)

=Zi=1

geffB ~B ~Ji + e2

8me( ~B ~ri)2 (157)

E(1)0 = 0|Hint|0 (158)

=Zi=1

geffB0| ~B ~Ji|0+ e2

8me0|( ~B ~ri)2|0 (159)

| and the first term is zero (160)

=e2B2

12me

Zi=1

0|r2i |0 (161)

13

Van Vleck paramagnetism is simply the second order contribution from Hpara =Z

i=1 geffB~B ~Ji.

Thus recall the second order perturbation theory energy correction expression:

E(2)n =

m(6=n)

|n|Hint|m|2En Em (162)

| in this case, the state for correction is |n = |0 and Hint = Hpara (163)

E(2)0 =

m(6=0)

0 Zi=1 geffB ~B ~Jim2E0 Em (164)

Note that the state |J = 0 = |0 is usually the lowest energy state (atomic physics), thus EmE0 > 0and E

(2)0 < 0.

Again take ~B = (0, 0, B) so ~B ~Ji = BJiz.

E(2)0 =

m(6=0)

Zi=1

g2eff2BB

2 |0|Jiz |m|2E0 Em (165)

The moment of this single atom is

= E(2)0

B=

m(6=0)

Zi=1

g2eff2BB

|0|Jiz |m|2Em E0 (166)

The magnetization of N identical atoms in volume V is, M = NV . The susceptibility is,

0MB

=N

V

0B (167)

=N

V0

m(6=0)

Zi=1

g2eff2B

|0|Jiz |m|2Em E0 (168)

This positive susceptibility is called Van Vleck paramagnetism and it is very small. It is temperatureindependent because this is a quantum mechanical derivation and not a statistical mechanical derivation.It turns out, from a more detailed derivation, that the above expression for is essentially correct andis temperature independent.

5.4 Isentropic Demagnetization or Adiabatic Demagnetization

This is a technique to cool paramagnetic materials. Recall the expression of magnetization in atomicparamagnetism,

M = geffBJ(J + 1)

3x = g2eff

2B

J(J + 1)

3

B

kBT(169)

Thus increasing the B-field increases the magnetization and increasing T decreases M , which iscompatible with common sense.

The required concepts in entropy are:

Boltzmann definition of entropy: S = kB ln where is the number of (micro)arrangements. At low temperatures, all the magnetic moments align with the magnetic field and there is only 1arrangement, = 1, therefore S = 0.

At high temperatures or zero magnetic field, the system is disordered and in state J , there aredim(mJ) = 2J + 1 possibilities for each atom. For N identical atoms, = (2J + 1)

N givingS = kB ln(2J + 1)

N = NkB ln(2J + 1).

14

The 2 steps of isentropic cooling are:

1. Isothermal magnetization:

Field is applied to magnetize the material. The temperature of the sample is kept constant by ahelium bath that removes heat from the sample. As seen from above, the entropy decreases onmagnetization.

2. Adiabatic Demagnetization:

The sample is now isolated (adiabatic) and the magnetic field is slowly reduced to zero (quasi-static). This demagnetization increases the entropy of magnetic moments. But because it is anadiabatic process (Q = 0), the entropy change is zero, thus there is a compensating decrease inentropy elsewhere. This decrease in entropy is in the phonons and hence the material cools.

Figure 6: The entropy-temperature S T diagram showing the 2 steps of isentropic cooling.

6 Collective Magnetism

Notice that we could understand the phenomena of diamagnetism and paramagnetism without assumingany explicit interaction among the magnetic moments (or among the electrons that gave rise to thesemagnetic moments).

For collective magnetism, there is interaction between the magnetic moments and this give rise tocooperative or collective phenomena (ferromagnetism, ferrimagnetism and antiferromagnetism). Theseinteractions are called under the title of exchange interactions. The signature of cooperative phe-nomena is a spontaneous ordering of magnetic moments below a critical temperature. In ferro- andferrimagnetism, this is called the Curie temperature TC and in antiferromagnetism, this is called theNeel temperature TN .

We make a broad classification first:

Insulators magnetic moments are from partially filled electron shells

can be ferro-, ferri- or antiferromagnetic

well described by Heisenberg model (see later)

Metals Band magnetism (Beyond the scope here.)

15

conduction electrons are responsible for magnetism exchange interaction causes a spin-dependent band shift (for T < TC) and thus a partic-ular spin orientation is preferred

simplest model is the Hubbard model Localized magnetism

magnetic moments are not from conduction electrons exchange interaction is between localized electrons that contributes magnetic momentsand itinerant conduction electrons

6.1 Direct Exchange Interaction

The physics behind direct exchange interaction is extremely simple: Coulomb interaction + Paulisexclusion principle.

We will illustrate this using the standard Heitler-London treatment in molecular physics. We willalso derive the Heisenberg model.

Figure 7: Hydrogen molecule consisting of 2 hydrogen atoms.

Consider the model of a hydrogen molecule (or 2 hydrogen atoms). The Hamiltonians of the 2separate atoms (in position representation) are,

First atom: Ha0 (1) = h2

2me2ra1

e2

40ra1(170)

Second atom: Hb0(2) = h2

2me2rb2

e2

40rb2(171)

Molecule: H = Ha0 (1) +Hb0(2)

e2

40

(1

ra2+

1

rb1 1r12

1R

)(172)

The Hamiltonian of the molecule gives an solvable Schrodinger equation. Thus we shall make anapproximation using physically reasonable wavefunctions.

Exchange Symmetry of 2 quantum particles: (Quantum Mechanics 3) The 2-boson wavefunction obeys the following under exchange:

boson(1, 2)12 boson(2, 1) = boson(1, 2) (173)

The 2-fermion wavefunction obeys the following under exchange:

fermion(1, 2)12 fermion(2, 1) = fermion(1, 2) (174)

16

Now we make the guess for the spatial part of the 2-electron wavefunction.

+ =12

(a(1)b(2) + a(2)b(1)

)(175)

=12

(a(1)b(2) a(2)b(1)

)(176)

It appears that + does not obey the 2-fermion exchange symmetry and indeed it does not, but wekeep + anyway.

For the spin part, we recall the results of the addition of the two spin-12 angular momenta.

s1 =1

2, s2 =

1

2= s = 0, 1 (177)ms1 = 12 ,ms2 = 12 = | 12ms1 = 12 ,ms2 = 12 = | 12ms1 = 12 ,ms2 = 12 = | 12ms1 = 12 ,ms2 = 12 = | 12

|s = 0,ms = 0 = |0, 0 = 12(| 12 | 12) = S(1, 2)|s = 1,ms = 1 = |1, 1 = | 12 = T (1, 2)|s = 1,ms = 0 = 12(| 12+ | 12) = T (1, 2)|s = 1,ms = 1 = | 12 = T (1, 2)

where S(1, 2) is the singlet state and S(1, 2)12 S(2, 1) = S(1, 2). And T (1, 2) are the triplet

spin states with T (1, 2)12 T (2, 1) = T (1, 2).

Now we can construct the full spatial + spin wavefunctions obeying 2-fermion exchange symmetry.Recall that

symmetric antisymmetric = antisymmetric (178)symmetric symmetric = symmetric (179)

antisymmetric antisymmetric = symmetric (180)S = +S (181)

=12

(a(1)b(2) + a(2)b(1)

)S(1, 2) (182)

T = T (183)

=12

(a(1)b(2) a(2)b(1)

)T (1, 2) (184)

and S and T obey the exchange symmetry, i.e. S(1, 2)12 S(2, 1) = S(1, 2) and T (1, 2) 12

T (2, 1) = T (1, 2).The energy expectation values are, (note that these are not eigenvalues!)

ES =

d1d2SHS (185)

ET =

d1d2THT (186)

Note the difference between the 2 energies are,

ES ET =d1d2 (SHS THT ) (187)

| recall that S and T are orthogonal to each other (188)| and expand out the spatial parts (189)=

d1d2a(1)b(2)Ha(2)b(1) +

d1d2a(2)b(1)Ha(1)b(2) (190)

J (191)Now we want to make an effective Hamiltonian that treats the wavefunctions S and T as eigenstates

and ES and ET as the respective eigenvalues.

17

We first need to note this manipulation:

Consider,(~S1 + ~S2

)2= ~S21 +

~S22 + 2~S1 ~S2 (192)

~S1 ~S2 = 12

[(~S1 + ~S2

)2 ~S21 ~S22

](193)

| note that ~S1 + ~S2 = ~S (194)=

1

2

[~S2 ~S21 ~S22

](195)

| recall that the ~S2 = s(s+ 1) (recall that they are dimensionless)(196)=

1

2[s(s+ 1) s1(s1 + 1) s2(s2 + 1)] (197)

| and for s1 = 12, s2 =

1

2we have s = 0 and s = 1 (198)

=1

2

[s(s+ 1) 3

2

](199)

=

{ 34 for s = 014 for s=1

(200)

Thus if we define the effective Hamiltonian,

Heff =1

4(ES + 3ET ) (ES ET ) ~S1 ~S2 (201)

it has 2 eigenvalues,

for s = 0, Heff|s = 0 =(1

4(ES + 3ET ) (ES ET )

(34

))|s = 0 = ES |s = 0 (202)

for s = 1, Heff|s = 1 =(1

4(ES + 3ET ) (ES ET )

(1

4

))|s = 1 = ET |s = 1 (203)

Using the notation of exchange integral J ,

Heff =1

4(ES + 3ET ) J ~S1 ~S2 (204)

The first term is a constant and can always be absorbed by a shift. This gives the HeisenbergHamiltonian for a 2-spin system,

HHeisenberg J ~S1 ~S2 (205)A few comments are in order:

What we have done is to show that Coulomb interaction (not explicitly shown here actually) andPaulis Exclusion principle (via exchange symmetry) can give rise to singlet spin states (preferringopposite spins) and triplet spin states (preferring parallel spins) having different energy values. Thesinglet state turns out to be lower in energy and thus in this case, the opposite spin arrangementis preferred.

This is not really a derivation of Heisenberg Hamiltonian. This is just a justification that evenwithout the presence of a magnetic field, other interactions can still bring about a preferred spinarrangement.

For nearest-neighbour interaction in a multispin system, we can write the Heisenberg Hamiltonianas

HHeisenberg = i,j

Jij ~Si ~Sj (206)

18

where i, j means over nearest neighbours and the exchange integral Jij is usually taken to bea parameter.

The comparison between the different models are:

Heisenberg model: H = i,j

Jij(Sxi S

xj + S

yi S

yj + S

zi S

zj

)(207)

Ising model: H = i,j

JijSzi Szj (208)

XY model: H = i,j

Jij(Sxi S

xj + S

yi S

yj

)(209)

The other types of exchange interactions are called indirect exchange interactions. The differentclasses of indirect exchange interactions are:

Superexchange interaction

This exchange interaction is mediated by a non-magnetic ion in-between the magnetic ions.

RKKY exchange interaction

This is devised by Ruderman, Kittel, Kasuya and Yosida. Exchange is mediated by valenceelectrons and thus the exchange integral is distance dependent.

Double exchange

Magnetic ions can have different oxidation states and electron hopping between ions resultsin an extended ferromagnetic (i.e. parallel spins) exchange interaction.

6.2 Ferromagnetic Order

Ferromagnetism refers to spontaneous magnetization, without an applied field, such that the magneticmoments all point in the same direction. Beyond the so-called Curie temperature TC , the interactionbreaks down and the material becomes paramagnetic.

There are 2 aspects of ferromagnetism:

1. within a domain, the magnetic moments are all aligned

2. domains interact to produce other magnetic effects such as hysteresis

We will discuss the first aspect: interactions within a domain.

6.2.1 Weiss Mean Field theory of Ferromagnetism (Simplest theory of ferromagnetism)

A theory of ferromagnetism (and collective magnetism) must be able to at least show a phase transitionbetween paramagnetism and ferromagnetism. The transition temperature should be calculable from thetheory.

In Weiss Mean field theory, he took the paramagnetic Hamiltonian Hpara = geffB ~B ~J and wrote~B ~B+ ~BMF where ~B is the external field and ~BMF is some kind of internal mean field (or exchangeinteraction or molecular field).

He further assumed that the mean field is proportional to the magnetization.

~BMF = ~M (210)

where is a parameter and ~M is the magnetization. The Hamiltonian for the theory is then

HWeiss =Zi=1

geffB ~Ji ( ~B + ~BMF ) =Zi=1

geffB ~Ji ( ~B + ~M) (211)

19

Since this Hamiltonian is so similar to the one used for Curie Paramagnetism, we can immediatelycarry the results over. We carry over the magnetization expression

M = geffBJBJ(xJ) (212)

and BJ(xJ) is the Brillouin function with the modification:

x =geffB(B + M)

kBT(213)

For external field B = 0,

M = geffBJBJ

(geffBM

kBTJ

)(214)

Thus solving for M is not trivial but can be done graphically. Let the equations be written as 2equations

y = M (215)

y = geffBJBJ

(geffBM

kBTJ

)(216)

The graphical solution of M will mean finding the intersection points of these 2 graphs.M = 0 is always a solution but it is trivial. We shall now discusss the non-trivial solutions. We see

that for different values of temperatures y = geffBJBJ(. . .) can change and we have 3 possible cases:

Figure 8: The 3 cases of the solutions of M.

1. For case (a):

The only solution for M = 0 which is the paramagnetism situation (recall that there is no externalfield). So it represents T > TC .

20

2. For case (c):

The non-trivial solution is marked by the cross. This non-zero M represents a spontaneous mag-netization (ferromagnetism) and this occurs for T < TC .

3. For case (b):

The second intersection point is about to appear. Thus it makes sense to define this temperatureas the transition temperature TC . We can actually use this definition to solve for TC . The conditionis that: for small M , the gradient of y = geffBJBJ(. . .) should be 1, therefore (write T = TC),

d

dMgeffBJBJ(. . .)

M=0

= 1 (217)

d

dMgeffBJBJ

(geffBM

kBTCJ

)M=0

= 1 (218)

since M is small, we recall the expansion BJ(xJ) J + 13

x | (219)

geffBJd

dM

(J + 1

3

geffBM

kBTC

)= 1 (220)

g2eff2BJ(J + 1)

3kBTC= 1 (221)

TC =g2eff

2BJ(J + 1)

3kB (222)

Next, we work out the behaviour of M near TC . We recall the expression of magnetization M .

M = geffBJBJ

(geffBM

kBTJ

)(223)

| substitute geffBJkB

=3TC

geffB(J + 1)(224)

= geffBJBJ

(M

T

3TCgeffB(J + 1)

)(225)

| since M is small near TC , we can expand BJ as a Taylor series (226)| this time we need to retain one more order: coth y = 1

y+y

3 y

3

45+ (227)

| we already know the first 2 terms give BJ(xJ) J + 13

x (228)

= geffBJ

[J + 1

3J

M

T

3TCgeffB(J + 1)

2J + 12J

(2J + 1

2J

)3 145

(M

T

3TCgeffB(J + 1)

)3(229)

+1

2J

(1

2J

)3 145

(M

T

3TCgeffB(J + 1)

)3](230)

| remove one common factor of M (231)

1 =TCT

[1 +

1 (2J + 1)445(2J)4

(TCT

)2 27M2Jg2eff

2B(J + 1)

3

](232)

T

TC 1 = 1 (2J + 1)

4

45(2J)427J

g2eff2B(J + 1)

3

(TCT

)2M2 (233)

T

TC 1 = 3(2J

2 + 2J + 1)

10g2eff2BJ

2(J + 1)2

(TCT

)2M2 (234)

M2 =10g2eff

2BJ

2(J + 1)2

3(2J2 + 2J + 1)

(1 T

TC

)(T

TC

)2(235)

21

M2 (T

TC

)2(1 T

TC

)(236)

| for T TC , we can set(T

TC

)2 1 (237)

M 1 T

TC (238)

Figure 9: The relationship between magnetization M near TC as given by Weiss mean field theory.

Finally we can discuss the susceptibility of the ferromagnet in Weiss Mean field theory. We need tohave a (small) magnetic field in the system. Recall the magnetization expression with an externalfield.

M = geffBJBJ(xJ) (239)

| with modification x = geffB(B + M)kBT

and writegeffBkB

=3TC

geffBJ(J + 1)(240)

= geffBJBJ

(3TC

geffBJ(J + 1)

B + M

TJ

)(241)

| at T > TC , the external field causes a small magnetization (242)| thus recall the expansion BJ(xJ) J + 1

3x =

J + 1

3JxJ (243)

geffBJ(J + 1

3J

3TCgeffBJ(J + 1)

B + M

TJ

)(244)

=TC

B + M

T(245)

M =TCT B

(1 TCT

) (246)

0M

B= = 0

TCT

(1 TCT

) (247) =

0TC

1

T TC (248)

The expression 1TTC is the famous Curie-Weiss Law. Do not confuse this with the Curies Lawof paramagnetism.

A few comments are now in order:

22

Recall that Weiss mean field theory is based on the assumption that there is some internal fieldcreating interactions between the magnetic moments. Weiss had no idea of the exchange interaction!Thus it is already remarkable that it contains the phase transition between paramagnetism andferromagnetism. Measuring TC will give a fit for .

Further analysis shows the failures of Weiss theory: For T 0, the expression of magnetization from Weiss theory and that from experiments donot agree.

The T TC expression obtained earlier: M 1 TTC also does not agree with the experi-

mental expression: M (1 TTC

)1/3.

The specific heat for T > TC predicted by Weiss theory is zero and it is completely incorrect.Experimental results seem to indicate a logarithmic singularity as T TC for the specificheat. Weiss theory could not even be used to provide any explanation for that.

Weiss Mean field theory can be derived from Heisenberg Hamiltonian by the following argument:Consider the Heisenberg Hamiltonian with nearest neighbour (nn) and next nearest neighbour(nnn) interactions, HHeisenberg = J

i~Si ~Si+1J

i~Si ~Si+2. Then we take average of the nn

and nnn spins to get HHeisenberg J

i~Si ~Si+1 J

i~Si ~Si+2 and the average spin ~S

is proportional to the magnetization. Thus, HHeisenberg and HWeiss have a similar form.

6.2.2 Magnons (Quantized Spin Waves)

At low temperatures, a low energy excitation called a spin wave propagates among the interacting spins.Quantized spin waves are called magnons. This is extremely similar to phonons which are quantizedlattice waves.

We avoid any semiclassical derivations as this is really a quantum mechanical effect. This is describedby the Heisenberg model and we shall give a simplified derivation of spin waves.

We consider a 1D chain of N interacting spins 2. Thus each spin has 2 nearest neighbours except forthe first and last spin. The Heisenberg Hamiltonian can be written as

HHeisenberg = N1i=1

Ji,i+1~Si ~Si+1 (249)

| take Ji,i+1 = Ji+1,i = J (250)

= JN1i=1

~Si ~Si+1 (251)

= JN1i=1

Szi Szi+1 + S

xi S

xi+1 + S

yi S

yi+1 (252)

| write Sx and Sy in terms of raising and lowering operators (253)| thus recall Sxi =

1

2(S+i + S

i ) and S

yi =

1

2i(S+i Si ) (254)

= JN1i=1

Szi Szi+1 +

1

2(S+i + S

i )

1

2(S+i+1 + S

i+1) +

1

2i(S+i Si )

1

2i(S+i+1 Si+1)(255)

= JN1i=1

Szi Szi+1 +

1

2(S+i S

i+1 + S

i S

+i+1) (256)

2Strictly speaking, we require translational invariance in this derivation. We can achieve this by either taking N or by joining the head and tail of the spin chain together (Periodic Boundary condition). We neglect this detail in thisderivation and a better derivation will be done in the tutorial.

23

Now we specialise to spin-12 for a simplified discussion thus there are only 2 states: ms =12 up and

ms = 12 down. The T = 0 ground state of the ferromagnet can be written as

|0 = | 1 N (257)

The lowest excited state is one spin flip at site j.

|1j = | 1 j1jj+1 N (258)

The total spin of the system changed by 1 (hence the notation). Thus this is a bosonic excitation.We want to find the energy eigenvalue of such 1-spin-flip states.

First consider,

HHeisenberg|0 =(J

N1i=1

Szi Szi+1 +

1

2(S+i S

i+1) + S

i S

+i+1

)| 1 N (259)

| note that S+i | i = 0 and Szi | i =1

2| i (260)

= JN1i=1

Szi Szi+1| 1 N (261)

= J N 14

|0 (262)

And also,

HHeisenberg|1j =(J

N1i=1

Szi Szi+1 +

1

2(S+i S

i+1 + S

i S

+i+1)

)| 1 j N (263)

| note the only relation we need is S+j | j = | j (264)

= JN1i=1

(Szi S

zi+1

1

2(| 1 jj+1 N + | 1 j1j N )

)(265)

= J((Sz1S

z2 + + Szj1Szj + SzjSzj+1 + + SzN1SzN )|1j+

1

2(|1j+1+ |1j1)

)| note that Szj | 1 j N =

1

2| 1 j N (266)

= J(N 34

14 14

)|1j 1

2J (|1j+1+ |1j1) (267)

= J N 54

|1j 12J (|1j+1+ |1j1) (268)

Now consider the plane wave state |q = 1N

j e

iqRj |1j (this is inspired from phonons)

HHeisenberg|q = HHeisenberg 1N

j

eiqRj |1j (269)

| where Rj is the distance from the first spin to j-th spin (270)| let a be the spacing between adjacent spins, so Rj = ja (271)=

1N

j

eiqRj(J N 5

4|1j 1

2J (|1j+1+ |1j1)

)(272)

=1N

j

(J N 5

4eiqRj |1j 1

2J (eiqRjeiqaeiqa|1j+1+ eiqRjeiqaeiqa|1j1)

)

24

| note Rj + a = Rj+1, Rj a = Rj1 and 1N

j

eiqRj1 |1j1 = |q (273)

| the last relation requires translational invariance which we ignore here (274)=

(J N 5

4 12J (eiqa + eiqa)

)|q (275)

=

(J N 1

4+ J (1 cos(qa))

)|q (276)

Thus |q is an eigenstate of HHeisenberg. The state |q is a sum of all 1-spin flip states |1j with phasefactors eiqRj , thus |q is a plane wave of 1-spin flips.

If we shift up the Heisenberg Hamiltonian by J N14 , we have,

H Heisenberg = HHeisenberg + JN 14

(277)

then, H Heisenberg|0 = 0|0 (278)H Heisenberg|q = J (1 cos(qa))|q (279)

Thus the dispersion of a plane wave of 1-spin flips or spin wave is E(q) = J (1 cos(qa)). (Seefigure)

We will not carry on to quantize the spin waves into magnons as this will involve second quantizationnotation and Holstein-Primakoff transformation which is beyond the scope of the module.

Figure 10: The dispersion relation of Ferromagnetic spin waves

Figure 11: The semiclassical picture of spin waves. Do not take it too seriously.

6.2.3 Ferromagnetic Domains

[Why do domains form?] There are 4 types of energy competing in the ferromagnet. This resultsin why the ferromagnet is not made of a single domain of aligned spins but domains of aligned spins indifferent directions. Such a structure of domains actually results in a lower energy for the material.

25

1. Exchange Energy:

This short-ranged interaction was modelled by the Heisenberg Hamiltonian H = J ~Si ~Sj andparallel alignment lowers the energy.

2. Magnetostatic Energy:

The simplest way to describe this long-ranged energy is to think that this energy is used to main-tain the external magnetic field of the magnetized material. If we break the material, making 2dipoles, and anti-align them, the external field (and thus the energy) will be greatly reduced. 3

This favours anti-alignment and the formation of domains with anti-aligned moments.

Figure 12: The formation of domains (from left to right) resulting in a reduction of magnetostatic energy.The rightmost diagram shows closure domains.

3. Anisotropy Energy:

If we only consider the first 2 types of energies, then the most favourable condition is that themagnetic moments gradually flip over from one end of the material to the other end, i.e. the widthof the domain wall is the width of the material! Due to the crystalline structure of the material,there are certain directions that cost less energy for the magnetic moments to align with (easyaxes). The hard axes are defined oppositely. The width of the domain would be a balance ofexchange and anisotropy energies as most of the rotated spins would not be along the easy axisand this costs anisotropy energy.

4. Magnetoelastic energy:

Change in magnetization results in (small) elastic distortions of the material. The larger thedomains, the more elastic energy is needed. Thus the formation of smaller domains (such asclosure domains) is favoured by magnetoelastic energy.

[Types of Domain Walls]

Bloch Wall: magnetization rotates in a plane parallel to the plane of the domain wall. See Figure13(a).

Neel Wall: magnetization rotates in a plane perpendicular to the plane of the domain wall. SeeFigure 13(b).

90o or 180o change in magnetic moments between adjacent domains. See Figure 14.3One can also think of the fact that a broken magnet will also tend to anti-align.

26

Figure 13: (a) Bloch Wall and (b) Neel Wall.

Figure 14: (a) 1800 domain wall and (b) 900 domain wall.

[Estimation of a 180o Blochs Walls energy:] Assume the change in moments occurs over Nsites and thus each spin rotates N radians with respect to its neighbour. The exchange energy Eex isestimated using the Heisenberg Hamiltonian.

HHeisenberg = Ji

~Si ~Si+1 (280)

roughly speaking= E = J

i

SiSi+1 cos (281)

= JNS2 cos( N

)(282)

| assume the angle is small and expand cos x 1 x2

2(283)

= JNS2 + JNS2( N

)2 12

(284)

The first term is for parallel alignment. The exchange energy is thus the second term

Eex 12JNS2

( N

)2(285)

For the anisotropy energy density, we assume a simple form.

E = K sin2 K > 0 (286)

where K is an anisotropy constant and this model would mean that each spin prefers = 0 or = .Thus the total anisotropy energy is

Eani =

Ni=1

K sin2 i (287)

| assume a large number of spins and take continuum limit (288)=

N

0K sin2 d (289)

| the factor N

is needed since =

N(290)

=N

K

0

1

2 12cos 2d (291)

27

=N

K

2(292)

=NK

2(293)

Total energy in the domain wall:

Etotal = Eex + Eani =J S222N

+NK

2(294)

Thus this expression is in line with the earlier discussion that exchange energy favours a wide domainwall N . We minimise Etotal with respect to N ,

dEtotaldN

= 0 =J S22

2N2+K

2= 0 (295)

N = S

JK

(296)

Thus the width of the domain wall depends on the ratio between the exchange constant and theanisotropy constant.

[Magnetization and Hysteresis] The most obvious consequence is that the Curie temperature ofiron TC 1000K so why isnt iron magnetized at room temperature? The answer is that each domainis magnetized in one direction and these directions are quite random so there is no net magnetization.

The other consequence is that for soft magnetic materials, it only requires a small magnetic field(as low as 106T) to attain saturation magnetization. The short answer is that since there is alreadyalignment within a domain, only a small magnetic field is needed to align the domains.

We now consider the process of magnetization of a ferromagnet with domains.

Figure 15: The so-called hysteresis loop. The 7 parts in the hysteresis loop are discussed in the text. Ms- saturation magnetization, Mr - remanent magnetization and Hc - coercive field.

1. We assume there is no net magnetization in the material to begin with. Magnetization M increaseswith applied field H. There is a linear region where the process is reversible: removing the field

28

results in no magnetization. Increasing the field beyond the linear region brings the magnetizationto a maximum saturated value Ms. Microscopically, domains aligned parallel to the field will growin size and this is reversible for small domain wall motion. Saturation magnetization is reachedwhen domains that are aligned to H are at their largest size.

2. From Ms, when the external field is decreased to zero, there is a remanent magnetization Mr.Microscopically, domains that are aligned to H decrease in size when H is decreased. However,domain walls may get pinned by strains and impurities resulting in net magnetization even whenH = 0.

3. An opposite field is applied to kill Mr. That amount of opposite field is called Hc, the coercivefield. Microscopically, Hc forces the pinned domain walls to move so that net magnetization canbe zero. The size of Hc thus depends on the material preparation and properties.

4. A large enough opposite field will cause an opposite saturation magnatization.

5. This is the repeat of process 2.

6. This is the repeat of process 3.

7. This is the repeat of process 4.

6.3 Antiferromagnetic and Ferrimagnetic Order

Antiferromagnetism and ferrimagnetism are simply cases where adjacent spins prefer to anti-align witheach other. In antiferromagnetism, the antiparallel moments are of the same magnitude and they cancelto give no net magnetization. In ferrimagnetism, the antiparallel moments are unequal in magnitudeand this results in a net magnetization.

We will discuss Weiss mean field theory for antiferromagnetism. Weiss mean field theory for ferri-magnetism is set as a tutorial problem.

We should also note that the Heisenberg model can be modified to describe antiferromagnetism andferrimagnetism. To favour anti-alignment, the exchange energy simply needs to be negative.

HHeisenberg = Ji=1

~Si ~Si+1 = |J |i=1

~Si ~Si+1 J < 0 (297)

6.3.1 Weiss Mean Field Theory of Antiferromagnetism

We assume only nearest neighbour (nn) interactions so we have 2 kinds of effective fields.

B = ||M +B and B = ||M +B (298)

where

Term ||M is the mean field as seen by the up-spins. Since we assume nn interactions, up-spinssees a down-spin magnetization. The opposite case is for the term ||M. We also assumedthat there is only one mean field constant . is negative because the exchange interaction isnegative.

Term B is the external field.The Hamiltonian is then

HWeiss =

Zi=1

geffB ~Ji ~B +Zi=1

geffB ~Ji ~B (299)

29

Thus we can again carry over the Curie Paramagnetism expression, (set B = 0)

M = geffBJBJ

(geffB ||M

kBTJ

)(300)

M = geffBJBJ

(geffB ||M

kBTJ

)(301)

But in an antiferromagnet below TC (and B = 0), M = M, so we write |M| = |M| = M . Weget 2 equations that are exactly the same

M = geffBJBJ

(geffB||M

kBTJ

)(302)

Since this expression is so similar to the one in ferromagnetism, we simply carry over the results fromferromagnetism and replace with ||. First the transition temperature,4

Neel Temperature: TN =g2eff

2B||J(J + 1)3kB

(303)

Next, the susceptibility for T > TN , (put back the external field B)

M = geffBJBJ

(geffB(B ||M)

kBTJ

)(304)

| write geffBkB

=3TN

geffB||J(J + 1) and expand BJ(xJ) J + 1

3JxJ (305)

geffBJ(J + 1

3J

3TNgeffB||J(J + 1)

B ||MT

J

)(306)

=TN||

B ||MT

(307)

Similarly, the other equation becomes

M TN||B ||M

T(308)

Solving these 2 equations give,

M = M =BTN||

1

T + TN(309)

= = 0MB

=0(M +M)

B=

20TN||

1

T + TN (310)

A more realistic calculation will be to include mean fields from next nearest neighbours (nnn) but itwill not be carried out here.

4Neel temperature fits the experimental values badly actually.

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7 Tutorial Problems

1. Numerical comparison of Diamagnetism and Paramagnetism

Estimate the diamagnetic orbital susceptibility of a gas of hydrogen atoms (with number density1 1020m3) in ground state (n = 0), and compare this with the paramagnetic spin susceptibilityat 100K.

2. Exchange Interaction is not magnetic dipole-dipole interaction

Estimate the ratio of the exchange and dipolar coupling of two adjacent Fe atoms in metallic Fe.(The exchange constant in Fe can be crudely estimated by setting it equal to kBTC where TC isthe Curie temperature. For Fe, TC = 1043K.) The dipolar energy is the usual expression fromelectromagnetism.

E =04r3

[~1 ~2 3

r2(~1 ~r)(~2 ~r)

](311)

3. Corrected Derivation of Ferromagnetic Spin Waves

By imposing periodic boundary condition on the chain 1 N , we need to add one term toHHeisenberg. Show that this modification gives a more satisfying derivation of ferromagnetic spinwave dispersion, i.e. we get

HHeisenberg|0 = JN4

, HHeisenberg|q =[JN

4+ J (1 cos(qa))

]|q (312)

4. Ground state of an antiferromagnet is a difficult problem

Consider the system of 4 spin-12 particles with antiferromagnetic arrangement |0 = | 1234.This is commonly called the Neel state. The theoretical model is the antiferromagnetic HeisenbergHamiltonian HHeisenberg = |J |

4i=1

~Si ~Si+1. Use periodic boundary conditions where ~S5 = ~S1.(a) Show that |0 is not an eignestate of HHeisenberg. Then show that the expectation value

0|HHeisenberg|0 = |J |.(b) The (unnormalised) ground state state is

|0 = (| 12 | 12) (| 34 | 34) (313)= 2| 1234 | 1234 | 1234+ 2| 1234 | 1234 | 1234

(314)

Check that indeed this is an eigenstate with eigenvalue 2|J |, i.e. HHeisenberg|0 = 2|J ||0. Donot tackle the 16-dimensional problem directly.

5. Weiss Mean Field theory for Ferrimagnetism

Consider the following model of ferrimagnetism:

Here, we include next neighbours (nnn) interactions. Denote as the mean field constant associatedwith exchange constant J = J and () as the mean field constant associated with exchangeconstant J(J).

(a) Write down the 2 effective field including external field B.

(b) Write down the Hamiltonian and magnetization expressions M, M. Note that the Landeg-factor and angular momentum J of the 2 sublattices may not be the same.

(c) For T < TC , and no external field,

31

i. Solve for the critical transition temperature TC by requiring non-trivial solution for mag-netizations M, M.

ii. Give a reason why the positive term is chosen to be TC .

(d) For T > TC and a small external field,

i. Solve for the susceptibility = 0MB =0B (M +M).

(e) Treat some special cases: (in each case write down TC and )

i. Ferrimagnetism with nn mean field.

ii. Antiferromagnetism with nnn mean field.

iii. Antiferromagnetism with nn mean field.

iv. Ferromagnetism with nn mean field.

32