heisenberg spin wave

Heisenberg Magnets: An Introduction Holstein-Primakoff representation Spin-wave theory of ferromagnets Spin-wave theory of Anti-ferromagnets Spontaneous Symmetry Breaking and Goldstone Modes Goldstone Modes:Gapless Excitations

Heisenberg Spin Wave Theory

Abhilash Sahoo (11MS040)Arunav Bordoloi (11MS071)

Soumik mitra (13IP032)

Indian Institute of Science Education and Research Kolkata

April 22, 2015

Abhilash Sahoo (11MS040) Arunav Bordoloi (11MS071)Soumik mitra (13IP032) IISER Kolkata



Overview

1 Heisenberg Magnets: An Introduction

2 Holstein-Primakoff representation

3 Spin-wave theory of ferromagnets

4 Spin-wave theory of Anti-ferromagnets

5 Spontaneous Symmetry Breaking and Goldstone Modes

6 Goldstone Modes:Gapless Excitations




Heisenberg Magnets: An Introduction

Hesisenberg magnets are basically defined by:

H =1

2

∑i 6≡j

JijSiSj

such that[Sαj ,S

βj ] = i

∑γ

εαβγSγj

The coefficients Jij are termed as exchange interaction. We further adhere to acertain set of assumptions in our discussions.

Jij =

{J if i,j are neighbouring positions0 Otherwise

At high temperature, there are strong thermal fluctuations. Thus, the system isin a disordered phase or 〈Si 〉 = 0

The sign of J determines whether the behavior of the system is ferromagnetic oranti ferromagnetic.

sgn(J) =

{+1 Anti ferromagnetic behaviour−1 Ferromagnetic behaviour




Ferromagnetic order:

Anti-ferromagnetic order:




Holstein-Primakoff representation

Holstein-Primakoff representation expresses the spin operators on a site j in termsof canonical boson creation and annihilation operators i.e. a and a†

S+j =

√2S − njaj

S−j =√

2S − nja†j

Szj = S − nj

Here nj = a†j aj ;

S+ = Sx + iSy and S− = Sx − iSy

The a†j and aj being bosonic creation and annihilation operators, follow bossonic

commutation relations.

[ai , a†j ] = δij

In form of the bosonic operators, the spin commutation relations as mentionedearlier remains intact i.e.

[Sαj ,Sβj ] = i

∑γ

εαβγSγj




Spin-wave theory of ferromagnets

Considering only nearest neighbour interaction,

H = J∑〈ij〉

Sxi S

xj + Sy

i Syj + Sz

i Szj

H = J∑iδ

1

2(S+

i S−i+δ + S−i S+i+δ) + Sz

i Szi+δ

We define a state with all spins along z-direction i.e.Szj = S to be the ground

state of the heisenberg ferromagnet. This is an eigenstate of the hamiltonian andhas the minimum energy.

Eo = −|J|∑i,δ

max(SiSi+δ)

= −−|J|S2Nz

2




We intend to focus on low energy oscillations about the ordering direction (=spin

waves). These oscillations make⟨Szj

⟩< S or

⟨nj⟩6= 0

The boson number,⟨nj⟩

is much smaller than S or〈nj〉S

<< 1. Expansion in

terms of〈nj〉S

should make sense. The approximation would further make goodsense if S >> 1. This assumption basically leads us to a semi classical limit.

If〈nj〉S

is not small then our basic assumption will break down.

We write the spin operators in form of bosonic operators in the manner asdescribed earlier by Hp representation. We further set,√

2S − nj =√

2S

√[1−

nj

2S

]and apply the assumption that

〈nj〉S

<< 1




The Hamiltonian is now of the form

H = −|J|NS2z/2− |J|S∑iδ

a†i+δai + a†i ai+δ − a†i ai − a†i+δai+δ + o(S0)

Now introducing fourier transformation of bosonic operators,

ak =1√N

∑i

e−ikri ai

Hamiltonian in fourier space is given by

H = E0 +∑k

wka†kak

where,E0 = −|J|SN2z/2

and,wk = 2|J|S

∑δ

1− cos(k.δ) ≡ S |J|z(1− γk )




The new hamiltoniandescribes a Hamiltonian which is just a bunch of independentharmonic oscillators, each labeled by a wavevector k. The quanta of the harmonicoscillators are called magnons; they are the quantized spin wave excitations (justlike phonons are the quantized lattice vibrations in a crystal) with energy wk

In the limit k tend to 0,wk ≈ |J|S |k|2

They are basically bosons, therefore the mean number of magnons withmomentum k is given by

〈nk 〉 =1

eβwk − 1

Magnetization,M ≡1

N

∑i

〈Si z 〉 = S −1

N

∑i

〈ni 〉 ≡ S −∆M

We first introduce an artificial wave vector cut off k0 which is the smallest wavevector in the k sum; the real system is described by the limit k0→ 0. We alsointroduce another wave vector k ′ > k0 which is chosen such thatwk′ << kBT << |J|S; this means in particular that for |k| < k ′ the quadraticform is valid.




∆M =1

N

∑k0<|k|<k′

1

e|J|Sk2/kBT − 1+∑|k|>k′

1

ewk/kBT − 1

Converting the sum to an integral and expanding the exponential (using|J|S2k2 << kBT )

δM ∝∫ k′

k0

dkkd−1 kBT

|J|Sk2∝

kBT

|J|S.

{1/k0 + ...(d = 1)−log(k0) + ...(d = 2)

Therefore at non-zero temperatures in 1D and 2D, δM diverges as k0 is sent to 0.Thus our assumption as was initially put was wrong.

Note that the quantity δM/(2S) is the expectation value of the average over allsites of our original expansion parameter nj/(2S), which we assumed to be smallwhen we expanded the square roots in the HP expression.

Thus we conclude that M = 0 (i.e. there is no ferromagnetic order) at finite (i.e.nonzero) temperatures for the Heisenberg model in one and two dimensions. For3D, ferromagnetic order is stable at sufficiently low temperatures.




Spin-wave theory of Anti-ferromagnets

Anti-ferromagnets : J > 0

In case of classical spins, the ground state is given by spins pointing in oppositedirection on neighboring atoms. The state can be written as∏

j∈A|S〉j

∏l∈B|−S〉l ≡ |N〉

where, A and B are two sublattices such that spins on site A and B haveeigenvalues +S and -S with eigenstates |+S〉 and |−S〉 of Sz respectively.

But for the Heisenberg model, this state is neither the ground state nor aneigenstate of the Heisenberg Hamiltonian.

Hence quantum fluctuations terms involving the raising and lowering spinoperators play an important role in the anti-ferromagnetic case. The GS energy ischanged away from the classical result.

Still the real ground state may have anti-ferromagnetic order, so |N〉 captures thetrue ground state in a qualitative sense.




Holstein-Primakoff Representation : Sublattice A has spin projection +S, so therepresentation are :

S+Aj =

√2S − njaj

S−Aj =√

2S − nja†j

SzAj = S − nj

Sublattice B has spin projection -S, so the representation are :

S+Bl =

√2S − nlb

†l

S−Bl =√

2S − nlbl

SzBl = −S + nl

Here, j and l runs over sites in sub-lattices A and B respectively.

Hence the Hamiltonian upto order S is given by :

H = J∑j∈A

∑δ

[S(ajbj+δ + h.c.) + S(a†j aj + b†j+δbj+δ)− S2]+

J∑l∈B

∑δ

[S(blal+δ + h.c.) + S(b†l bl + a†l+δal+δ)− S2]




Now introducing inverse fourier transformation of bosonic operators,

aj =1√NA

∑k

e ik.rj ak

bl =1√NB

∑k

e ik.rl bk

where NA = NB = N2

is the number of lattice sites in each sub-lattice.

Hamiltonian after the fourier transform and using∑

j exp[i(k − k′).rj ] = NAδk,k′

is :H = −NJS2z/2 + JSz

∑k

[γk (akb−k + a†kb†−k ) + a†kak + b†kbk ]

Unlike ferromagnetic case, fourier transform alone does not diagonalize theHamiltonian.




Hence Bogoliubov transformation is used that is defined by :

αk = ukak − vkb†−k

βk = ukbk − vka†−k

uk and vk are real functions of k.

From [αk , α†k′ ] = [βk , β

†k′ ] = δ

k,k′ , we get

u2k − v2

k = 1

The inverse transformation is :

ak = ukαk + vkβ†−k

bk = ukβk + vkα†−k

Using the above the Hamiltonian is :

H = −NJS2z/2 + JSz∑k

[(2γkukvk + u2k + v2

k )(α†kαk + β†kβk ) + 2(γkukvk + v2k )

+[γk (u2k + v2

k ) + 2ukvk ](αkβ−k + α†kβ†−k )]




To diagonalise, coefficients of (αkβ−k + α†kβ†−k ) must be zero, so :

[γk (u2k + v2

k ) + 2ukvk ] = 0

We take uk = cosh θk and vk = sinh θk , where θk is given by inserting into aboveequation as :

tanh 2θk = −γkHence, the final diagonalized Hamiltonian is :

H = −NJS2z/2− NJSz/2 +∑k

ωk (α†kαk + β†kβk + 1)

H = E0 +∑k

ωk (α†kαk + β†kβk )

where,

ωk = JSz√

(1− γ2k )

E0 = −NJS2z/2− NJSz/2 +∑k

ωk




The antiferromagnetic dispersion is linear and given by :

ωk ∝ |k| as k → 0

The Ground state of H, denoted by |G〉, is defined by the relation :

αk |G〉 = 0, βk |G〉 = 0, for all k

Hence H |G〉 = E0 |G〉, i.e. the ground state energy is E0. The first term−NJS2z/2 is the GS energy of a classical nearest-neighbor anti-ferromagnet withspin length S.

The other part is the quantum correction ∆E which is negative as shown below :

∆E = E0 − Eclass =∑k

ωk − NJSz/2 = JSz∑k

[√

1− γ2k − 1]

Hence, the quantum fluctuations lower the energy of the system.

Order parameter for anti-ferromagnets : sublattice magnetization whosemagnitude is given by :

MA =1

NA

∑j∈A

⟨Szj

⟩= S −

1

NA

∑j∈A

⟨a†j aj

⟩= S −

1

NA

∑k

⟨a†kak

⟩




Writing MA = S −∆MA, the correction ∆MA to the classical result S is given by

∆MA =1

NA

∑k

⟨a†kak

⟩Writing in terms of α and β operators we get

∆MA =2

N

∑k

[u2k

⟨α†kαk

⟩+ v2

k

⟨β−kβ

†−k

⟩+ ukvk 〈αkβ−k + h.c.〉]

The expectation value of the last term is zero. Hence :

∆MA =2

N

∑k

[u2k

⟨α†kαk

⟩+ v2

k

⟨β†kβk

⟩+ v2

k ]

∆MA =2

N

∑k

[nk cosh 2θk +1

2(cosh 2θk − 1)]

∆MA = −1

2+

2

N

∑k

(nk +1

2)

1√1− γ2

k

where⟨α†kαk

⟩=⟨β†kβk

⟩= 1/(exp (βωk )− 1) ≡ nk and cosh 2θk = 1/

√1− γ2

k




At T = 0, nk = 0; hence in one dimension the k-sum becomes (note γk = cos k) :

1

N

∑k

1√1− γ2

k

∝ limk0→0

∫ π/2

k0

dk√

1− cos2 k

The most inmportant contribution comes from small k-region, cos k = 1− k2/2,hence the leading term gives :∫

k0

dk/k = − log k0 →∞ as k0 → 0

Hence, ∆MA diverges and in one dimension the system is not magneticallyordered at zero temperature. In 2 and 3 dimensions, ∆MA is nonzero and thesystem is ordered at zero temperature.

At finite non-zero temperature, no anti-ferromagnetic order exists in 1-dimensionsince none existed at zero temperature. In two-dimension, the magnetic orderdoes not survive at finite temperature, while in 3-dimension the system is orderedat sufficiently low temperatures.




Spontaneous Symmetry Breaking and Goldstone Modes

It is a phenomenon where the Hamiltonian has a symmetry and hence has aconserved quantity due to Noether’s theorem, but the ground state of the systemdoes not respect the symmetry.

The Hamiltonian is rotationally symmetric. If all the spins are rotated by thesame angle,Hamiltonian does not change at all. According to Noether’s theorem,a symmetry of the system must lead to a conserved quantity.

~S =∑i

~si

The total spin ~S commutes with H; [~S ,H] = 0 hence conserved.

The rotation generator is given by the global spin rotation,

U(~θ) = e i~S.~θ/~




Its useful to write the Hamiltonian as

H = −J∑<ij>

(szi szj +

1

2(s+

i s−j + s−i s+j ))

Using the above we can see that one eigenstate oh the Hamiltonian is |.... ↑↑ ....|i.e. all spins up.

This is called Spontaneous Symmetry Breaking ; though the Hamiltonian is

symmetric under spin rotation,the ground state spontaneously choose a particularorientation and is not invariant under that symmetry transformation. So theground state spontaneously breaks the global spin rotation symmetry of theHamiltonian.




For classical spins there must be infinite possible orientations of spin.Startingfrom the state where all spins are up,we perform finite rotations of all spins

e i~s~θ/~ = e i~σ.

~θ/2 = cos|~θ|2

+ i~σ.~θ

|~θ|sin|~θ|2

operating it on the up state | ↑ | we get

e i~σ.~θ/2| ↑ | = cos

|~θ|2| ↑ |+ isin

|~θ|2

( θz|~θ|| ↑> +

θx + iθy

|~θ|| ↓>

)fro the above we can calculate to find,

< .. ↑↑ ..|U(θ)|.. ↑↑ .. >= 0

So the rotation of a ground state yields another ground state which is orthogonalto the original ground state.




Goldstone Modes:Gapless Excitations

Symmetries of a Hamiltonian can be classified as discrete or continuous.Asymmetry is discrete if the associated symmetry transformations form a discreteset. In contrast,a symmetry is continuous if the associated symmetrytransformations form a continuous set.The global spin-rotation symmetry of theHeisenberg model, on the other hand, is a continuous symmetry, since therotations form a continuous set.

In the case of spontaneously broken continuous symmetry, there is always alow-lying excitation whose energy E(~p) does not have a gap E(~p) = 0 as itsmomentum goes to zero.




Consider an excited state |~k >=∑

n |.. ↑ .. ↑ ... ↓ ... ↑ ... > e ikna

We perform a translation on this state by a lattice constant, e ipa/~

e ipa/~|~k > =∑n

|... ↑ ... ↓︸︷︷︸n−1

... ↑ ... > e ikna

=∑n

|... ↑ ... ↓︸︷︷︸n

... ↑ ... > e ik(n+1)a

= e ika|~k >

The energy of this state we get as

H|~k >= −J∑m

szmszm+1 +

1

2(s+

m s−m+1 + s−m s+m+1)

∑n

| ↑ .. ↓︸︷︷︸n−1

.. ↑> e ikna




The action of this Hamiltonian yields

H|~k > = −J∑m

szmszm+1 +

1

2(s+

m s−m+1 + s−m s+m+1)

∑n

| ↑ .. ↓︸︷︷︸n−1

.. ↑> e ikna

= −J(N~2

4+

~2

2(e−ika + e ika)

)|~k >

= −J(N~2

4+ ~2(coska− 1)

)|~k >

Thus the excitation energy becomes

E(k) = −J(coska− 1)

So as the momentum goes to zero the excitation energy goes to zero and is

thus gapless .




At T = 0, nk = 0; hence in one dimension the k-sum becomes (note γk = cos k) :

1

N

∑k

1√1− γ2

k

∝ limk0→0

∫ π/2

k0

dk√

1− cos2 k

The most inmportant contribution comes from small k-region, cos k = 1− k2/2,hence the leading term gives :∫

k0

dk/k = − log k0 →∞ as k0 → 0

Hence, ∆MA diverges and in one dimension the system is not magneticallyordered at zero temperature. In 2 and 3 dimensions, ∆MA is nonzero and thesystem is ordered at zero temperature.

At finite non-zero temperature, no anti-ferromagnetic order exists in 1-dimensionsince none existed at zero temperature. In two-dimension, the magnetic orderdoes not survive at finite temperature, while in 3-dimension the system is orderedat sufficiently low temperatures.




References

221B Lecture Notes: Spontaneous Symmetry Breaking.

Michael Plischke and Michael Bergersen.Equilibrium statistical physics.World Scientific, 2006.



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