Jacques Klaasse- Spin Density Waves: Seen Through 20th Century Glasses

8/3/2019 Jacques Klaasse- Spin Density Waves: Seen Through 20th Century Glasses

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Presentation Groupmeeting June 3rd, sorry 10th, 2009

by Jacques Klaasse


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Spin Density Waves

This talk is based on a book-chapter on antiferromagnetism,

written by Anthony Arrott

in Rado-Suhl, Volume IIB, 1966.

Contents:

-

Exchange interactions

-

Spin Density Waves

-

Neutron diffraction

-

Chromium

- Conclusions


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Starting point of the (historical) discussion:

-

atoms with intrinsic localised

magnetic moment.

-

exchange interaction (Heisenberg) because of direct overlap.

-

Weiss molecular field.

The interaction can be ferromagnetic or antiferromagnetic, dependent on thesign of the exchange parameter.

Kramers

introduced also superexchange mediated by electrons on

intervening non-magnetic atoms (oxygen!).

In 1946 Stoner questioned the picture in case of metals.

He presented the collective electron picture: itinerant

(ferro)magnetism

by mutual exchange of d-band electrons.

Stoner Criterion: ferromagnetism occurs if D(F

) * IS

> 1

where D(F

) is the density of states at the Fermi level,

and IS

is

the Stoner exchange parameter.


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The two subbands are shifted inenergy because of the exchangeinteraction.The shift is based on the Hubbard

Hamiltonian Unn , which can berewritten, with n = n

+ n

, as

(U/4) { n2

(n - n)2 }.

The exchange potential is not a

fixed potential but is governed bythe other electrons.

Moment and potential are both

given by the same medium.

This results in a non-zero ferromagnetic moment even in zero applied field,as long as the gain in exchange energy is larger than the loss

in kinetic energy.

The Coulomb interaction in metals seems to favourferromagnetic coupling!!


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Let D be the average DOS per spin direction

around F

and let the splitting beE.Then (n

- n) = D.E

Let s = ( n - n ) / N , andLet Is

be the exchange parameter.

We calculate now the splitting energy.

We do Ekin

first.

Ekin

= 0

E/2

(2) D d

= (D/4)E2

= (D/4) (Ns/D)

2

= (Ns/2)

2

(1/D).

For the exchange we found Eexch

= -(Is

/4) (Ns)2, so, for the total splitting

energy we find: Es

= (N2s2/4) { 1/D

Is

} = (N2s2/4D) { 1

DIs

} .

From this formula follows the Stoner Criterion. This shows to work for

ferromagnets

like Fe and Ni, but not for much more. Obviously,

a high D (or low density in real space) favours

ferromagnetism.

Above Tc

the material should be a Pauliparamagnet. This is not seen!


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E/2

D()

dN

= D d

n-n

= DE= N s

Dd

E

)2(

2/

0

Help screen for calculating Ekin


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Another class of materials contain localised

moments apart from

conduction electrons.

It is shown by Ruderman, Kittel, Kasuya, and Yosida

that the interaction can be

formulated in a way that a Heisenberg-like picture is simulated without directoverlap (RKKY interaction, published between 1954 and 1957).

This makes the problem similar to the non-metallic problem.

The coupling can be FM as well as AFM, and is strongly oscillating withdistance.

This work followed upon a discussion by Zener

on the role of the conduction

electrons in providing (ferro)magnetic

interactions.

Zener

also revived the old suggestion of Nel

that Cr and Mn

as metals were

antiferromagnets, where so far antiferromagnetism

seemed to be only a

property of non-metals.

What happens here. Is Cr metal an RKKY magnet, or do we have something

special? The moments of Cr++

and Cr+++

are 4.9 and 3.8 Bohrmagnetons

per atom respectively and the Cr++

saturation moment should be 4 B

.


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Stimulated by this discussion Shull and Wilkinson proved that

these metals are weakly antiferromagnetic.

See Rev. Mod. Phys25 (1953), p100.

Moreover, the wavelength was not equal to the length

of the cubic unit cell (Corliss et al., PRL 3 (1959), p211) .

These results came on the moment that neutron

diffraction techniques became a suitable tool for

determining magnetic structures.

We come to that later.Anyhow, there was a problem how to

explain these results.


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SDWs

It was Overhauser

(around 1960) who showed that in the one-

dimensional Hartree-Fock approximation the antiferromagnetic

state could exist and may have lower energy.The AF periodicity is not given by the lattice but by the wavevector equal to the diameter of the volume of occupied states in

k-space.With the help of neutron diffraction techniques, it is shown thatthese Spin Density Waves indeed exist.

Main conclusions from Overhausers

work:

a) Spin density waves are allowed states.

b) SDWs

may be ground state.

c) SDWs with wave vector q = 2kF

are most

likely to minimise

energy.


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SDWs

Source: A.W. Overhauser,PRL 3,9 (1959) 414

Source: [Overhauser (1962)]


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SDWs

Here, we will not give all the mathematical details of the HF

procedure.

We only give some flavour

of what was going on here, in particular

we will show some pictures to elucidate the situation.

For detailed information we recommend the following paper:

A. W. Overhauser, Phys. Rev. 128 (1962) p 1437 1452.

You need a reasonable starting wave function (for fermions this is a

Slater determinant) and a smart trial potential.

Then you have to solve the Schrdinger equation by a variational

method until you find an internally consistent solution whereyour potential is stable under continued iterations.With a proper starting set, the procedure is generally convergent,

but it is not sure your solution is the real ground state.


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SDWs

Some citations on the HF method:

far from being straightforward

coupled integral equations are thoroughly nonlinear

and require an iteration technique for their solution.

repeated until a self consistent set of solutions is obtained

convergence dependent on initial guess of the

one-particle states.

Overhauser

started his calculations with a helical polarization

( this leads to an off-diagonal contribution to the one-

electron exchange potential.)


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SDWs

Overhauser

showed that for spin up

and spin down a gap opens at the

Fermi wavevector, but for the two at

a different sign of kF

(=q/2).

The two waves at kF

and -kF

give

together two charge density waves

at q=2kF

with a certain spin polarization,

resulting in a static spin density wave withconstant charge density.

It has some resemblance with the opening

of the gap at the Brillouin

zone, but

there the potential is fixed, here the

potential is determined by the electron gas,

with largest effect near the Fermi wave

vector.Source: A.W. Overhauser,PRL 4,9 (1960) 462


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SDWs



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SDWs

The spin susceptibility

shows for SDWs todiverge at 2kF

.



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SDWs

In 3 dimensions the problemmay not yet be solved but here

we give an artists impression

of the situation.

The gap causes a lowering of

the D(F

), and thus of the linear

term, , in the specific heat.From this effect the gapped

surface fraction of the Fermi

Surface can be derived.

In order to conserve entropy, the entropy loss by the lower

is recovered

at the transition point to the paramagnetic state, resulting over there in apeak in the observed heat capacity.

In the resistance, a jump should be expected on opening the gap,

caused

by a lowering of the number of available carriers, which is proportional to

the opened Fermi Surface fraction.


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SDWs

In order to see whether in reality SDWs

occur, we have to minimise

the total

energy, being the sum of the totalkinetic energy including the effects of

the SDWs

and the total potential

energy including the total exchange

energy.

The algebra necessary to do this, and

calculate the correct parameters, is

considerable even in the one

dimensional case.


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SDWs

In three dimensions, the so called

nesting vectors play the role of

the 2kF

from the one dimensional

case.

What is told is that SDW vectors

should be in directions where the densityof states is high.

This sounds reasonable.

Source: Wikipedia


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Neutron diffraction

It is not surprising that the discoveryof Spin Density Waves (Shull and

Wilkinson, 1953) goes parallel

with the development of neutron

diffraction techniques.However, also working with neutrons

has its restrictions.

Longitudinal waves cannot be

detected: neutrons see no spinsbut only magnetic field.

A longitudinally polarized magnetic

field wave is incompatible with the

Maxwell equations.Further: domains!!

Domains add ambiguity to the

interpretation of results.

Source: [Shull & Wilkinson (1953)]

( Normal bcc structure: h + k + l = even. )


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Neutron diffraction

Arrott

p333: It is not possible to decide

from the diffraction experimentsbetween the existence of a helical spindensity wave state and the presenceof two types of domains each withtransverse linear spin density waves

but with mutually perpendicularpolarizations.

A magnetic field may unravel the problem.

Other problems: the intensities of theSDW reflections are very weak, and you

dont know where they are.

(needle in hayloft).

With

K = 2G

q, only the G=0

reflections give sufficient intensity.

So, K is, in reciprocal space, not knownIn magnitude, nor in direction. Source: [Shull & Wilkinson (1953)]


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Neutron diffraction

Spin configurations are generally described in terms ofHelical Spin Density Waves:

p(z) = p(ex

cos

qz ey

sin

qz)

p(z) = p(ey

cos

qz ez

sin

qz)

p(z) = p(ez

cos

qz ex

sin

qz), forq // z.

The first are called normal

helical waves, the last two

lines describe end-over-end

helical waves.

This set of functions form a complete set to describe any spin

wave with q =

qez

.

The use of this set instead of plane waves gives some profit

in the analysis of neutron diffraction patterns. The

normal

and the end-over-end

behave differently

(Arrott, p 300).


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Chromium

A review article by E Fawcett, Rev.Mod. Phys. 60 (1988) p209, gives75 pages of material on

chromium properties.

Too much to handle here.

There is agreement on the AF low

temperature state with TN

=311K

and 0.5B

.

In the picture here you see the

behaviour

of the thermal

expansion, resistivity, specific

heat, and thermo electric power.


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Chromium

From neutron results it

follows that there is a

second spin-flip

transition at about

115K.

Below 115K the SDWs

are

longitudinal, and above

it the SDWs

are

transverse.

Source: [Fawcett (1988)]


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Chromium

My interest is in particular the heatcapacity. This here, on this picture,

looks like a second order transition.

However this is cold rolled material.

The results showed to be strongly

dependent on the strain situation.

In the next slide we show results on

a strain-free single crystal.


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Chromium

This is clearly a

First-order

SDW-PM

Transition.

I. S. Williams,E. S. R. Gopal,and R. Street,

J. Phys F: MetalPhys, 9 (1979)P 431.


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Chromium

There is a similarity in the structure of

the SDW HF equations and those of

the BCS model for superconductivity.

The temperature dependence ofthe SDW gap, which is

proportional to the amplitude of

the SDWs

is a lookalike of the

BCS curve.Maybe the similarity is not only

mathematical.

Maybe SDWs and SC aretwo sides of the same coin.Source: [Overhauser, 1962]


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Chromium

The entropy around the transition (obtained from heat capacitymeasurements) amounts to about somewhat less than 0.02J/K.mol. This should be the recovery of the effect of a lower

.

From

=1.4 mJ/K.mol

we can derive the total entropy of the electron

gas at TN

, being about 0.42 J/K.mol. This should point in thedirection that 4.5% of the FS is gapped.

This is in agreement with the resistivity jump of about 5% at TN

.

If Cr should have a permanent moment on-site, the entropy in thepeak should be of the order of R.ln(2) = 5.76 J/K.mol.

The observed entropy is about two orders of magnitude lower,indicating no permanent moment is present on the Cr sites.


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Chromium

From the BCS theory follows that the gap is

(3.5 * kB

TN

)

0.1 eV.

For this energy holds roughly 2R02/2m*

where R0

is (in reciprocal space!!) the

radius of the truncated part of FS

and m* the effective mass.

If p is the number of truncated faces (here 6), then for the total truncated

fraction, t, holds t = p R02

/ 4 kF

2

.

With m* 1.5m, a value for R0

can be derived: 0.2*108

cm

-1.

From q 2kF

follows kF

1.1*108

cm

-1.

Result: t 0.05, in good agreement with the earlier estimates.


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Conclusions

Spin Density Waves are possible solutions for the freeelectron state.

Overwhelming evidence exists that in Cr this SDW solution isground state.

In an SDW state a part of the FS is gapped.

The wave vector of the SDW is determinednot by the lattice but by a nesting vector,

being about 2kF for a simple Fermi sphere.

It is not clear whether, in a real situation,these nesting vectors can be calculated,

or simply follow from experiment.


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Thank you for your attention.

Help, a gap,

Ive to flip over

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Jacques Klaasse- Spin Density Waves: Seen Through 20th Century Glasses