Introduction to Quantum Magnetism

DPS Department of Physical Sciences

Vivek Pareek 10MS46

So you are saying, magnetism ain’t classical?

December 3, 2014

Outline

1 Bohr-van Leeuwen Theorem

2 What is quantum about magnetism?

3 Exchange InteractionDirect ExchangeKinetic exchange

4 Hubbard Model: Towards t-J model

5 Super exchange

6 Band Magnetism

7 Acknowledgment

8 References

Bohr-van Leeuwen Theorem

Consider a system of N electrons with position ri and momentum pi.

H =1

2m

∑i

p2i + V (r1...rN ) (1)

Partition function is given by

Z =

∫ ∏i

d3rid3pi exp (−βH) (2)

In presence of magnetic field, B = ∇×A, the Hamiltonian becomes,

H =1

2m

∑i

(pi +

e

cAi

)2+ V (r1...rN ) (3)

The magnetization of the system can be found by using the followingequation,

M =N

V〈m〉 (4)

=N

V βZ∂Z∂B (5)

The partition function for this system is given by

Z =

∫ ∏i

d3rid3pi exp (−βH)

=

∫ ∏i

d3rid3pi exp

(−β

[1

2m

∑i

(pi +

e

cAi

)2+ V (r1...rN )

])

=

∫V

∏i

d3ri exp (−βV (r1...rN ))

∫ ∞−∞

∏i

d3pi exp

(−β2m

∑i

(pi +

e

cAi

)2)(6)

The integration limit for momentum is the key here

p′i = pi +e

cAi (7)

Hence, we get

Z∗ =

∫V

∏i

d3ri exp (−βV (r1...rN ))

∫ ∞−∞

∏i

d3p′i exp

(−β2m

∑i

(p′i)2)

(8)

There is no dependence of magnetic field in the partition function Z∗. Fromequation 5 we get

M = 0 (9)

This proves Bohr-van Leeuwen Theorem, which essentially shows that wecan not have magnetism classically.

Dipole-dipole interaction

Let us consider two magnetic dipoles with dipole moments m1 and m2.The field generated by m1 is

B(r) =µ0

4π

3(m1 · r)r− r2m1

r5(10)

The energy of another moment at m2 at r is given by,

Vdipole = −m2 ·B(r)

=µ0

4π

3(m1 · r)(m2 · r)− r2m1 ·m2

r5(11)

Can this dipole-dipole interaction explain the magnetic ordering as weobserve it?

Consider the critical temperature corresponding to the nearest neighborinteraction

kBTC = zµ0

4π

m2s

a3

= zgµ0

4π

µ2B

4a3

=µ0

4π

e2~2

4m2ea3

(12)

where z is the nearest neighbor number.

Consider Fe with z = 8 and a = 2.49, we get the TC to be 0.3K !

Experimentally, Fe becomes ferromagnet below 1043K !

Dipolar interaction is too weak to explain this. We need much strongerinteraction.

This is provided by Coulomb interaction in conjugation with Pauli’sexclusion principle. This forms the basis for Exchange interaction.

Exchange Interaction

Fundamental properties of electrons are responsible for the stronginteraction

Electron’s spin.

Electron’s kinetic energy(delocalisation).

Pauli’s Exclusion principle(fermi statistics).

Coulomb repulsion.

Exchange interaction can be categorised into the following types

1 Direct(Potential) exchange Minimize potential energy byintroducing nodes in the spatial wavefunction ⇒ Antisymmetric spatialwavefunction ⇒ Symmetric spins ⇒ Ferromagnetism

2 Kinetic(Super) exchange Minimize kinetic energy by reducinggradient of spatial wavefunction, i.e, delocalize electrons ⇒ Symmetricwavefunction function ⇒ Antisymmetric spins ⇒ Anti ferromagnetism.

Direct Exchange

Consider two electrons in two nearly degenerate single particle orbitals φ1

and φ2 with energies ε1 and ε2. Field operator in the basis of orthonormalWannier function φRm(r) localised at lattice site R can be written as

Ψ†σ(r) =∑Rm

φ∗Rm(r)c†Rmσχσ (13)

where σ =↑, ↓, with χ↑ =

(10

); χ↓ =

(01

)and m = all quantum number

except spin.

HCoulomb =1

2

∑R1m1

· · ·∑

R4m4

⟨R1m1,R2m2

∣∣∣∣ e2

4πε0|r1 − r2|

∣∣∣∣R3m3,R4m4

⟩×∑σ1σ2

c†R1m1σ1c†R2m2σ2

cR3m3σ2cR4m4σ2 (14)

On site Coulomb Interaction

Consider the case R1 = R2 = R3 = R4. If we consider HCoulomb as aperturbation, then non-zero 1st order contribution is possible only ifm3 = m1,m4 = m2 or m3 = m2,m4 = m1.Therefore, we get,

HCoulomb ∼=1

2

∑R

∑m1m2

∑σ1σ2

{Km1m2c

†Rm1σ1

cRm1σ1c†Rm2σ2

cRm2σ2

− Jm1m2c†Rm1σ1

cRm1σ2c†Rm2σ2

cRm2σ1

}(15)

where, Direct coulomb interaction is

Km1m2 =

⟨m1m2

∣∣∣∣ e2

4πε0|r1 − r2|

∣∣∣∣m2m1

⟩(16)

and Exchange coulomb interaction is

Jm1m2 =

⟨m1m2

∣∣∣∣ e2

4πε0|r1 − r2|

∣∣∣∣m1m2

⟩(17)

Using number operator,

nRm =∑σ

c†RmσcRmσ (18)

and spin operator,

sαRm =∑σσ′

c†Rmσ(σασσ′/2)cRmσ′ (19)

where σα are Pauli matrices. We can show that,

HCoulomb =1

2

∑Rm1m2

{(Km1m2 −

1

2Jm1m2

)nRm1nRm2 − 2Jm1m2sRm1 · sRm2

}(20)

We get an isotropic spin - spin interaction, which is analogous toHeisenberg Spin Model. Nature of Jm1m2 will determine the ordering in thesystem. Consider,

Jm1m2 =

∫d3r1d

3r2φ∗m1

(r1)φ∗m2(r2)

e2

4πε0|r1 − r2|φm1(r2)φm2(r1) (21)

Consider the following Fourier transform,

1

|r1 − r2|=

∫d3k

(2π)34π

k2eik·(r1−r2) (22)

Using 21 and 22 we get,

Jm1m2 =

∫d3k

(2π)3e2

ε0k2

∫d3r1φ

∗m1

(r1)φm2(r1)eik·r1 }I(k)

×∫d3r2φm1(r2)φ∗m2

(r2)e−ik·r2 }I∗(k)

=

∫d3k

(2π)3|I(k)|2 > 0 (23)

We can also show that the coulomb term is reduced by this

Km1m2 − Jm1m2 =1

2(Km1m2 +Km2m1 − Jm1m2 − Jm2m1)

=1

2

∫d3r1d

3r2e2

4πε|r1 − r2|[φ∗m1

(r1)φ∗m2(r2)− φ∗m2

(r1)φ∗m1(r2)]

× [φm2(r2)φm1(r1)− φm1(r2)φm2(r1)]

(24)

=1

2

∫d3r1d

3r2e2

4πε|r1 − r2||f(r1r2)|2 ≥ 0 (25)

Spin interaction of the form −2Jm1m2sRm1 · sRm2 for Jm1m2 > 0,prefers parallel spin alignment as their ground state⇒ Ferromagneticorder.

This is the derivation of Hund’s first rule: The total spin of electrons ina partially filled shell of one ion tends to be maximal.

Inter ion exchange interaction

Assumption: Single, non-degenerate orbitals per site. Orbitals fromdifferent sites have negligible overlap,i.e, they are orthogonal.For first order perturbation, R1 = R4 and R2 = R3 or R1 = R3 andR2 = R4. In complete analogy with previous section,

HCoulomb ∼=1

2

∑R1R2

{(K12 −

1

2J12

)n1n2 − 2J12s1 · s2

}(26)

where,

K12 = KR1R2 =

∫d3r1d

3r2|φR1(r1)|2 e2

4πε0|r1 − r2||φR2(r2)|2 (27)

J12 = JR1R2 =

∫d3r1d

3r2φ∗R1

(r1)φ∗R2(r2)

e2

4πε0|r1 − r2|φR1(r2)φR2(r1)

(28)

In an ionic crystal charge should not fluctuate much. In our simple model,ni = 1 is the only interesting case. The interaction becomes−∑

R1R2J12s1 · s2 for J12 > 0. Hence we get ferromagnetic order.

In order to respect the overall anti-symmetry of the total wavefunction,with spins being aligned, the orbital wavefunction should have a node atr1 − r2, where the Coulomb interaction is maximum.

Figure: Orbital wavefunction for Direct Exchange (Fig. credit Mohit)

Kinetic Exchange

Consider the Hubbard model,

H = −∑

RR′σ

t(R−R′)c†R′σcRσ + U∑R

c†R↑c†R↓cR↓cR↑

= −∑

RR′σ

t(R−R′)c†R′σcRσ + U∑R

c†R↑cR↑c†R↓cR↓ (29)

For simplicity, let us consider a dimer toy model

H = −t∑σ

(c†1σc2σ + c†2σc1σ

)−µ

∑σ

(c†1σc1σ + a†2σc2σ

)+U

∑i=1,2

c†i↑ci↑c†i↓ci↓

(30)Since each site can have 4 states, i.e, empty, spin-up, spin-down, doublyoccupied, the dimension of the fock space is 42 = 16. The hamiltonianconserves the total electron number.

We consider sector of 2 electrons, which corresponds to a hilbert space ofdimension six. In this space, the µ term is of course an irrelevant constant.Suitable basis vectors are {|↑↓, 0〉 , |0, ↑↓〉 , |↑, ↓〉 , |↓, ↑〉 , |↑, ↑〉 |↓, ↓〉}. H is a6× 6 matrix in above basis,

H′ =

U 0 t −t 0 00 U t −t 0 0t t 0 0 0 0−t −t 0 0 0 00 0 0 0 0 00 0 0 0 0 0

(31)

We can simplify H′ further by transforming from {|↑, ↓〉 , |↓, ↑〉} onto|↑, ↓〉 − |↓, ↑〉√

2,|↑, ↓〉+ |↓, ↑〉√

2which gives

H′ =

U 0√

2t 0 0 0

0 U√

2t 0 0 0√2t√

2t 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

(32)

with eigen energies

U ,U ±

√U2 + 16t2

2in the first sector.

0, 0, 0 in the second sector which corresponds to spin triplet.

We are interested in the ionic systems for which t << U . Then the firstsector energies become,

U,U ±

√U2 + 16t2

2∼= U +

4t2

U(33)

U ±√U2 − 16t2

2∼=−4t2

U< 0 (34)

For,U

|t| → ∞, the corresponding eigenstate approaches spin singlet

state, i.e, we find anti-ferromagnetic interaction.This results from lowering the kinetic energy for anti-parallel spin.Parallel spin is restricted to honor Pauli exclusion principle, which iswhy t doesn’t appear in the eigenenergies of the triplet state.

Figure: Orbital wavefunction for Kinetic Exchange (Fig. credit Mohit)

Hubbard Model

The minimal Hubbard model for a band structure in atomic limit is givenby the tight binding hopping parameters tij , the on site interaction U , andnumber of electrons,

H =∑ijσ

tijc†iσcjσ + U

∑i

ni↑ni↓ (35)

Let us begin by choosing the zeroth order Hamiltonian to be,

U = U∑i

ni↑ni↓ (36)

U divides the Fock space into two subspace,

S = [|n1↑, n1↓, n2↑ · · · 〉 : ∀i, ni↑ + ni↓ ≤ 1] (37)

D = [|n1↑, n1↓, n2↑ · · · 〉 : ∃i, ni↑ + ni↓ = 2] (38)

The hopping term is considered as perturbation,

T = −∑ijσ

tijc†iσcjσ (39)

T couples S to D by hopping an electron into or out of a doubly occupiedsite. U is diagonal and T lifts the degeneracy in the two subspace.

H =

[Ps(T + U)Ps PsTPd

PdTPs Pd(T + U)Pd

](40)

where Ps,d are projections onto subspace S and D. Now consider thefollowing operator,

G(E) =1

E −H(41)

where E are not the eigenvalues of H, and the following identity[(A BC D

)−1]ss

=(A−BD−1C

)−1(42)

The projection of G into the subspace S is,

PsG(E)Ps = Ps[E −H]−1Ps = [E −Heff (E)]−1 (43)

using 42 and PsU = UPs = 0, Heff is,

Heff = PsTPs − PsTPd[E − (U + T )]Pd−1TPs (44)

The eigenenergies of H are given by the zeroes of the characteristicpolynomial.

det|En −Heff (E)| = 0 (45)

Expanding Pd[E − (U + T )]Pd to zeroth order inE

Uand to second order in

t

Uto get an eigenvalue equation of an energy independent Hamiltonian, we

have

Heff → Ht−J [1 +O(E/U) +O(t/U)]

Ht−J =

T − 1

U

∑ijkσσ′

tijtjkc†iσcjσnj↑nj↓c

†jσ′ckσ′

(46)

Rearranging the fermionic operators,

Ht−J = Ps(T + HQHM + J ′)Ps (47)

T = −∑ijσ

tijc†iσcjσ (48)

HQHM =1

2

∑ij

Jij(Si · Sj −ninj

4) (49)

J ′ =−1

2U

∑ijk;i6=k

tijtjk

[∑σ

(c†iσckσnj

)− c†iσck · c

†jσcj

](50)

At half filling ni = 1, Ps annihilates T and J ′.

There are no low energy hopping if every site has one electron. This isMott insulator phase.

Thus we obtain anti-ferromagnetic spin order near half filling inHubbard Model.

Super exchange

In real ionic crystal, nonmagnetic anions separate the magnetic cations.

Direct and kinetic exchange between cations is weak.

Electron hopping to the other cation via anion gives larger exchange.This is Super exchange. The strength and sign is determined byGoodenough-Kanamori rules.

Half filled orbitals of two cations overlap with same empty or filedorbitals of the intervening anion⇒ Strong anti-ferromagnetic exchangeinteraction.Half filled orbitals of two cations overlap with orthogonal orbitals of theintervening anion⇒ Weak ferromagnetic exchange interaction.

Figure: Super exchange interaction in Copper oxide

Band Magnetism: Stoner Criterion

Heisenberg Model doesn’t completely explains ferromagnetism in metals.Hence we need a band picture to describe magnetism in metals.Assumption:

E↑(k) = E(k)− In↑N

(51)

E↓(k) = E(k)− In↓N

(52)

where I is the Stoner parameter describing the energy reduction due toelectron correlation.

Figure: Pictorial description for Band magnetism(Fig. courtesy wikipedia)

Let R =n↑ − n↓N

∝M . SubtractI(n↑ + n↓)

2Nfrom the energies to get

E↑(k) = E(k)− IR

2(53)

E↓(k) = E(k) +IR

2(54)

where E(k) = E(k)− I(n↑ + n↓)

2N. Now,

R =1

N

∑k

f↑(k)− f↓(k)

=1

N

∑k

1

exp

E(k)− IR

2− EF

kBT

+ 1

− 1

exp

E(k) +IR

2− EF

kBT

+ 1

(55)

Using

f(x− M x

2

)− f

(x+

M x

2

)= − M xf ′(x) (56)

we get

R =1

N

∑k

∂f(k)

∂E(k)· IR (57)

In order to achieve finite magnetization we must have R > 0. We arrive atthe Stoner criterion for ferromagnetism

I · 1

N

∑k

∂f(k)

∂E(k)> 1 (58)

In terms of the Density of states, the Stoner criterion is,

ID(EF ) > 1 (59)

where D(EF ) =V

ND(EF ).

In an external magnetic field,

R = D(EF ) (IR+ 2µBB) (60)

Magnetization is given by,

M = µBN

VR = D(EF )

(IM + 2µ2

BN

VB

)(61)

= 2µ2BN

V

D(EF )

1− ID(EF )B (62)

we know χ =µ0M

B. Hence we get the Stoner enhancement,

χ =χ0

1− ID(EF )(63)

Acknowledgment

I would like to thank our course instructor Dr. Siddharth Lal for allowingme to work on a wonderful topic and for his guidance. I would like tospecially thank Mohit for listening to this entire topic and giving his inputsfor improvement. I would also like to thank all my friends for their help andinput. Finally, I would like to thank Two steps from hell for their wonderfuland inspiring music, that helped me complete this term paper.

References

1 Interacting electrons and Quantum magnetism - Asa Auerbach

2 Theory of Magnetism - Carsten Timm

3 Quantum magnetism - Jonathan keeling

4 Quantum theory of magnetism - Wolfgang Nolting, AnupuruRamakanth