How the Ground State in a Material Will Be Affected by the Spin-Phonon Interactions Between Nuclei in Diatomic Molecular Structures

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How the ground state in a material will be affected by the spin-phonon interactions between nuclei in diatomic molecular structures

Isabel Roca Vich

Institutionen för fysik och astronomi Uppsala Universitet

Examensarbete C Kandidatprogrammet i Fysik Supervisor: Jonas Fransson Subject reader: Lars Nordström

June 09, 2016 Uppsala

Abstract

Wave-like phonons are often used to describe the heat capacity in materials. In this report the spin-phonon interaction between nuclei in a diatomic molecular structure is introduced by looking at the Hamiltonian in its ground state. The corresponding Green's functions are computed in order to investigate how this interaction affects the phonons. When calculating the spin, pseudo fermions and tensor products are introduced to make the calculation easier because the spin statistics could be a bit tricky to deal with. Three different cases of how the total interaction Hamiltonian behaves are investigated i.e. when the phonon is coupled to the spin. It turns out that in two of these cases an effect on the phonons can be seen but not in the other case.

Sammanfattning

I denna rapport betraktas två system där det ena är ett spinnsystem och det andra ett fononsystemet och målet att se hur fononerna påverkas av kopplingen till spinnet. Varje fonon kan ses som ett " kollektivt läge" som orsakas av varje atoms rörelse i gittret. En fonon kan precis som en foton betraktas som en våg eller en partikel. Vågliknande fononer används ofta för att beskriva värmekapaciteten hos ett material. Bakgrundsdelen i denna rapport beskriver den bakomliggande teorin för vibrationer som matematiskt beskrivs av harmoniska oscillationer.

I avsnittet ”Method” används Greensfunktioner vilket oftast används i materialteori.

I resultatavsnittet kommer den växelverkande delen av fononens och spinnets Hamiltonianer slutligen att förenas i en ny Greensfunktion och betraktas för tre olika fall. Tanken var att från början även titta på processen från andra hållet d.v.s. hur spinnet skulle ha påverkats av kopplingen till fononen men då beräkningarna visade sig bli allt för långdragna så kortades projektet ner till vad det är nu. Resultaten visar att det faktiskt uppstår en viss effekt på fononerna på grund av kopplingen till spinnet mellan de båda atomkärnorna. Det kan ses som förändringar i den totala energin för systemet, om det tillförts mer energi eller mindre.

Det kommer att visa sig att i det första fallet så finns det alltså en koppling mellan spinnet och fononen samt mellan de två atomkärnorna. I det andra fallet så erhålls enbart bidrag från spinndelen. I det tredje fallet är kriteriet från början satt till att vissa utav kopplingskonstanterna kommer att var satta till noll vilket gör att vissa termer försvinner i matrisen och den resulterande matrisen kommer att se ut som den gjorde i sitt okopplade grundtillstånd d.v.s. att det enbart sker en lokal växelverkan mellan spinnet och fononen hos en atomkärna och ej mellan de två atomkärnorna. I okopplade system så kan fononer styras med elektriska fält medan spinnet styrs med magnetiska fält, vid kopplade system som i det första scenariot här, blir det tvärtom d.v.s. att man kan kontrollera fononerna genom magnetisk styrning och spinnet med elektrisk styrning. Detta kan möjligen användas för att styra magnetiska egenskaper hos vissa material i den mån att göra exempelvis kylskåp mer energisnåla än vad de är idag.

Contents

1 Introduction 1 1.1 Purpose 1

1.2 Problem 2

2 Background 3 2.1 Vibrations 3

2.2 Second quantization 4

2.2.1 Harmonic oscillator in second quantization 4

2.2.2 Second quantization for particles 4

2.2.3 Fermions 7

2.3 Green’s Functions 8

2.4 The spin Hamiltonian 9

2.4.1 Homogeneous spin system 9

3 Method 10

3.1 The spin Hamiltonian 10

3.1.1 Introducing pseudo fermions 11

3.2 The phonon Hamiltonian 14

4 Results 16

4.1 The interaction Hamiltonian 17

4.2 The spin interaction Hamiltonian 19

4.3 The total Interaction Hamiltonian 20

5 Discussion H in + Hin 25

5.1 Outlook 25

6 Conclusion 26

References 27

Appendix 28

1 Introduction In this report two systems are considered, a spin system and a phonon system, and the goal is to see how the phonons are affected by the coupling to the spin.

Each phonon can be viewed as a “collective mode” which is caused by each atom’s movement in the lattice [9].

If an electron is coupled to the spin there is a so-called spin-spin coupling and if the electron is coupling to the mechanical movement in a metal it is called a phonon-phonon coupling[9]. The electron also experience a relativistic magnetic effect relative the motion of the ion, this effect is called spin-orbit interaction[8]. These interactions are indirect couplings and are not included here.

In the method part of the report the second quantization and Green’s functions will be used because phonons can be viewed as collective excitations of the elastic medium which are particle-like objects[5]. Green’s functions are often used in Many-Body theory and are representing the propagators which denote the probability amplitude for a particle traveling with a certain amount of energy, momentum or from one point in space to another within in a certain time interval. When it comes to spins, pseudo fermions and tensor products are introduced because the spin statistic can be complicated to use.

In the result section the interaction parts of the phonon’s and the spin’s Hamiltonians will finally merge into a new Green’s function for three different cases. The results show that there indeed is an effect on the phonons due to the coupling to the spin. It can be seen as alterations in the Green's functions. It will turn out that two of the three considered cases are similar to each other in the sense that for the first two cases, there actually will be contributions from both the spin’s interaction part of the Hamiltonian as well as from the new modified part of the Hamiltonian. Together those contributions will merge into a total interaction of the system and give a contribution of more or less energy of the whole system. In the first case, the phonons will propagate in the x –and y-plane while the spin will be directed in the z direction i.e. the phonons are incoming orthogonal to the spin. Indeed there is a spin-phonon coupling between the two nuclei while in the second case there are only some contributions from the spin’s interaction Hamiltonian with the phonons as well as the spin’s vector directed in the z direction i.e. spin and phonons are parallel to each other which only gives a very small amount of contribution to the energy of the system. In the third and last case some of the interaction constants will be put to zero from the beginning which gives an uncoupled result i.e. there will be no contribution to the energy of the system. This kind of couplings can be used to control the behavior in magnetic materials. If the lattice structure is changed then also the total energy of the system is changing and this can maybe be used in the purpose of making refrigerators more energy-efficient. 1.1 Purpose The aim is to determine how the ground state of the Hamiltonian is deviating compared to the original ground state. The ground state of the Hamiltonian for this system with two nuclei includes couplings between phonon and spin interactions. Phonons with wave-like behavior are described by normal modes and the number of normal modes is the same as the number of particles. The character of the particle and the wave can be combined if the ”particle” is described as a superposition of waves which are localized within a certain volume in space. Phonons with a wave-like behavior are often used to describe the heat capacity in materials.

1.2 Problem Spin-spin coupling may be indirect coupling between the spins via the electrons while phonon-phonon coupling may be indirect coupling between the mechanical movements via the electrons. If the indirect coupling is valid the question is, whether a spin-phonon coupling also would be valid and if so, what would the effect be? This will be investigated by studying the ground state of two coupled nuclei, each with their own spin and mechanical motion. Questions concerning whether the spin and mechanical motion of the nuclei are coupled and what effects on the ground state of the system such couplings have, will be investigated in this report. This will be done by first looking at the ground state of the Hamiltonian in absence of such a coupling and then adding the spin-phonon interaction.

2 Background The solid contains ions and they are magnetic. The spin of the ions and the mechanical movement disturb the electrons in a way such that a coupling will arise between the spin, the electrons and the mechanical shift. In this case, only the coupling between two atoms are considered and not these indirect spin-spin and spin-orbit interactions. Lattice vibrations arises from the atoms oscillations around its equilibrium and these vibrational modes are called phonons. Vibrations from neighboring atoms are not independent from each other and phonons, just like photons, can have both particle and wave-like behavior. Furthermore, phonons do not obey the Pauli exclusion principle because they are bosons[2]. In addition, the vibrational mode of the atom in solids is quantized because of the first quantization and phonons are these quantized vibrational modes. For example, photons transfers the electromagnetic force and photons are also quantized which leads to second quantization. The second quantization describes the induced interaction between electrons i.e. the phonon interacts with an electron, the phonon has to move but can vibrate until the next electron comes by[3]. 2.1Vibrations The potential function between atoms is given by:

(1)

is the position for the atom.∑ Furthermore,�R − R if the atom is vibrating becomes the equilibrium position for the atom and is the displacement from the equilibrium position of R R the atom. Q (2) The potential energy of theR phonon = R is+ found Q by first plugging in equation (2) into eq. (1) and Taylor expand the potential function around the equilibrium. There will be two terms left, the first and the third one. However, the third term is quadratic in the displacements and the one of importance here because it will denote the potential energy of the phonons. The vanishing term is the one which is linear in displacement and it is vanishing due to the fact that the equilibrium position is defined as the position where the sum of all forces on the atom j becomes zero. R The potential energy of phonons is given by:

(3) 2 ∂ p ∑ υ ∂Rμ ∂Rυ . � = (Q − Q (Q − Q �R − R

[]

2.2 Second quantization

2.2.1 Harmonic oscillator in second quantization

In first quantization the one dimensional Harmonic oscillator contains the two conjugate variables momentum p and position x which are included in the Hamiltonian:

(4) In second quantization the HamiltonianH = � can+ bem� rewritten� with ladder operators which are satisfying the boson commutation relations. and denote the creation and the annihilation operators. † � � The bosonic commutation relations are:

[a , a ] = † † [a , a ] = (5) † They are not Hermitian but their product [a , ais. ] x = and p can be rewritten with these ladder operators: † � �

(6) ℏ † � = √ � + � (7) ℏ † The ladder operators are the ones which� = do −� √the work� −and � that is why they are used in second quantization. The Hamiltonian H of the Harmonic oscillator can be rewritten by inserting eq. (6) and (7) into eq. (4):

(8) † [1] � = ℏ�� +

2.2.2 Second quantization for particles

A intrinsic basis of second quantization resulting from the derivation of a Lagrangian is the following commutation relation:

(9) † ′ ′ is the wave function and [�, t is, its� conjugate., t] = � The − commutation � relations (see eq. (5)) for creation and annihilation† ′ operator lead to the field commutator in eq. (9). � † � � �

and are expanded in this way: † ′ � � (10)

� = ∑ atϕ � (11) † ′ † ∗ ′ Letting H have eigenstate and� eigenvalue: = ∑ �� ϕ �

(12)

� � � �� = � � (13) 2 ℏ The commutation relationH for= − and∇ + U� is then: † ′ � � (14) ′ [�, t, � , t] = (15) † † ′ [ �, t, � , t] = (16)

† ′ ′ †′ ∗′ ′ [�, t, � , t ] = ∑�� [at, �� ] ϕ�ϕ� � (17) ∗ ′ ′ Now expand the wave= function∑� ϕ� inϕ terms� =of �states − �where is the set of states:

ϕ � (18)

The annihilation and creation operators� = ∑ havebϕ the� same commutation relation as the one above (see eq. (5)).

The number operator i.e. the number of particles in the system:

(19) † And that the Hamiltonian is: ∑ �b = �

(20) † Where � = ∑ �b �

(21) ∗ A way of solving this �Hamiltonian = ∫ � �� is to �use�� the� equation of motion for the destruction operator:

(22) This method of letting−� the� equation= [�, � ]of= motion − ∑ �for an� operator act on the Hamiltonian will often be used in the Green’s functions later.

Furthermore, it is assumed that the operator is developing over time so that the solution of the problem can be written as:

(23)

Where the eigenvalues E is the =solution∑� to − ��

(24)

Many-body theory is not just�� about|� being− �� able |to= diagonalize matrices, which would be true for Hamiltonians with only bilinear operators but it is also about studying Hamiltonians with extra terms. The equation below is a many-particle Hamiltonian with an extra term which can be interactions with phonons for example, spin effects or particle-particle interactions.

(25) ≠ The second term H is = the∑ Hparticle-particle+ ∑ �� −interaction � term and as below, it is written with two creation and two destruction operators. This takes part in a two particles scattering �� − � process i.e. that one particle which is in state n scatters to state m and the other one in state l will scatters to state k. A particle is scattered from n to m when k = l while the other one does not change its state at all. The ordering of the operators is important, here the destruction operators are to the right and the creation operators to the left in order to avoid processes when a particle interacts with itself.

(26) † † † The ordering of theH operators = ∑ H forb a bparticle+ ∑ which� interactsbb withb b itself would look like this:

(27) † † and if m = l, all , andbb b bis the vacuum state, by plugging in , which just contains one particle in the state , into the second term in the Hamiltonian above† would give nothing butn zero= because b |⟩ two= particles|⟩ cannot interact with each other whenb the|⟩ system α only contains one particle. But on the other hand, this would not be zero if would be inserted with the operators expressed in the Hamiltonian s second term eq. (11† ), this would ’ mean that the particle would be interacting with itself two times. b |⟩

(28) † † † † However, this is avoided ∑ by� writingbb themb b pairwise.b |⟩ = ∑ � b |⟩

2.2.3 Fermions

Fermions obey Pauli’s exclusion principle. The fields anti-commute and that means that there is either one or zero particles in each state of the system. Here c and are the destruction and creation operators and the anti-commutation relations which are denoted† by curly brackets are: �

(29)

{c�σ, c�σ} = c�σc �σ = (30)

′ {c, c } = (31)

† †′ {c , c } = (32)

†′ ′ Because two particles{c cannot, c } = be destroyed from the same state, there is one or no particle in the system and the same is for the creation operators -two particles cannot be created at the same time.

The different commutation relations of the operators give rise to the differences of the way bosons and fermions act. The Hamiltonian looks the same for the fermions as for the bosons:

(33) † † † Once againH the = ordering∑ H ofc thec operators+ ∑ is� important.cc c cIf two creation or two destruction operators switch places there will be a switch of the sign as well. Plane waves are a popular set of bases for electrons. The eigenstates are described by q and the spin by the index . Fermions are often associated with electrons and as a consequence of that describes the direction of the spin, in that case would be spin up or spin down and have the values � . The Hamiltonian is of the form: , � � ± (34) † ′ ′ † †′ ′ ′ ′ �� H =is ∑the eigenenergy c c + of∑ theU particle.�ρ + vThe∑ density operatorc�+��c� for−�� thec electronsc is . the interaction between the electron and the atom is described by the second �� ρ� = term. †The last term is the electron-electron interaction which is just a Coulomb potential. ∑�� c�+�� c�� As a final part of describing the background of phonons and their interaction with particles in general it is time to introduce the phonon Hamiltonian:

(35) † The sum over the numbers of Hnormalp = ∑ modes� � arisea�a �from the fact that the solid is vibrating. i.e. phonons, each with a eigenfrequency .

�

2.3 Green’s functions

There are many different types of Green’s functions and they are used as building blocks to get solutions, solutions for the Schrödinger equation as well as solutions to other differential equations but they are indeed solutions of equations of motion. They include the relevant information about the problem i.e. only a part of the wave function’s full information is included. In this case, it is used to find the solution of the Hamiltonian. The one presented here are the so called time-ordered Green’s function[1], [9].

′ ′ † ′ ′ † ′ † ′ (36) Gt, t = θt − t −it t ∓ θt − t −i t t = −iTt t The minus sign is for fermions and the plus sign for bosons[9]:

Fermions: (37) † ′ † ′ † ′ ′ {t, t } = t t + t t = t − t Bosons: (38) † ′ † ′ † ′ ′ And as an extension[t, of thet ] commutation= t t relations − tfrom tthe = section t − tabout fermions are[9]:

(39) † † † , The general Green[c ,’ s functionc c ] = here({c is,[9]: c }c − c {c c } = c

(40) ′ † ′ Taking the derivativeGt, t of = the −i GreenTt’s functiot n[9]:

′ ′ † ′ ′ † ′ † (41)′ Whichi ∂Gt, can t be= rewrittent − t tas: t ∓ −t − t t t + −iTi ∂t t

(41) ′ ′ † ′ † ′ At this stagei ∂ Gthet, second t = ∂ termt − twill[ oftent, appeart ] as+ a−i new TiGreen’s ∂ t functiont because there will, especially in the case of the phonon Hamiltonian (see section 3), be ladder operators left after the derivation. This is done by using the method of equation of motion (see eq.(22)) for each ladder operator by letting each ladder operator act on the Hamiltonian again. The Heisenberg equation of motion is defined as

(42)

i ∂ = [, H]

2.4 The Spin Hamiltonian

The spin’s raising and lowering operators are going to be frequently used when it comes to the calculations later on in this report. Not more than one atom is at each position and that is the exclusion principle which is represented by the anti-commutation rules for spins.

(43) + S = S + iS (44) − With the commutation relations:S = S − iS

(45) + − [S ,S ] = S (46) + + [S ,S ] = S (47) − − [S ,S ] = −S

2.4.1 Homogeneous spin system

The Heisenberg Hamiltonian for spin is of the form:

(48) ,+δ +δ ,+δ This HamiltonianH = −J is ∑only exactly� ∙ � solvable= −J ∑ in oneS dimension.S+δ + S S +δThe+ spins S S+δ are inclining to line up parallel to each other for values where J > 0. When J < 0 the ordering of the spin is both up and down. From the mixing terms follows this[3]:

(49) + † S = CC (50) − † S = CC

3 Method The quantized vibrational movements for an atom are mathematically described by a harmonic oscillation. The parameters may have their origin from the electron interactions but in this case, there is only the spin-phonon coupling between nuclei in diatomic molecular structures which is investigated by using the methods of quantum field theory and statistical quantum mechanics where the Green’s functions will be frequently used. However, in the end, the solution of the Hamiltonian will be reduced to a diagonalized matrix. 3.1 The spin Hamiltonian

Here the interaction is just between two nuclei i.e. the nearest spin. This Hamiltonian is called the Heisenberg Hamiltonian and its spin interacts with interaction vectors[3].

(51) † † The first twoH terms = ∑ inσ= the equationσdσdσ contain+ σd ladderσdσ operators,− JS ∙ S the annihilation and creation operator. where i = 1, 2 … is the energy of the particle and is the spin index. The dσ last† term is the interaction part of the spin Hamiltonian with J as the Heisenberg exchange σ �� dconstant. Here� the commutation relation for spin is the same as introduced� earlier in section 2.4.

(52) + − − + 3.1.1 IntroducingS ∙ S = SPseudoS + Fermions S S + S S = S S + S S + S S

Spin operators have an arduous spin statistic so it is easier to use a tool for that so called pseudo fermions which obey Fermi statistics.

Abrikosovs pseudo-fermion representation of spin operator can be used in many-body perturbation theory to relate the spin to its Green’s function [6].

The impurity spin operator can be replaced by the pseudo-fermion by a bilinear combination which is[9]:

(53) † Were the pseudo-fermion field i.e.S =is ∑a locald �fermiond field with the same anti-commutation relations as the fermion field and obey the usual commutation relations for spin[7]. d † + And for spin ½, is[9]: dd

� (54) Where are the Pauli matrices. The physical = �and unphysical states are uncoupled because the total occupation number operator is a constant of motion so the pseudo- � fermion Hamiltonian do not induce any transition† between the three first states and therefore � = ∑ dd possible to relate it to the spin Hamiltonian because commutes with the pseudo-fermion

� 10

Hamiltonian and it is possible to define the local fermion field as a super selection rule in Hilbert space[7].

For a spin half system[9]:

(55) s = s , s , s (56) † † s = d d + d d (57) † † s = −id d + id d (58) † † The spin operator can be up or down fors each= d spind + component d d x, y, z.

(59) † † † † † † Theys have = /d the same d + structure d d, −id as dthe + Pauli id d matrices:, d d − d d

(60) −i σ = , σ = , σ = And with spinors[9]: i −

(61) d = ( ) d (62) † = d d (63) † � =

† † d † † −i d † † d s = [(d d ( ),(d d ( ),(d d ( )] d d d i − † † −i d † † † = (d d [ , , ] ( ) = (σ, σ, σ = � = Here we will use i − d

+ = (σ + iσ = (64) ± and = ± i

− = (65) = σ 11

Interaction part of the Hamiltonian is obtained by using the method of tensor products. A tensor product maps from two Hilbert spaces to a new, higher-dimensional Hilbert space. Taking the tensor product of two 2x2 matrices will result in a 4x4 matrix. The new matrix will have the “form” of the second matrix but will contain the first matrix. This means that a new Hilbert space is created[9].

σ σ ⊗ σ = ( ) −σ

σ ⊗ = ( ) = ( ) The tensor product of and : −σ − + − + − ⊗ = ⊗ = + The new 4x4 matrix has the “shape” of the -matrix but contains the -matrix. − +

− − + ⊗ = ⊗ = Now one has to be careful when adding the matrices together. We have to consider that we have:

(69) + − − + −JS ∙ S = −J S S + S S + SS

+ − − + = −J + + − = −J [ + + + ( )] The tensor products can now be added into one matrix and finally − give the matrix for the spin interaction Hamiltonian:

− J − ( + ) Now the resulting matrix for looks like this: −

��

J A = −J − = − B −

� = − − Taking the eigenvalues of the matrix B:

The eigenvalues are given by (see appendix|� −A��): | =

, =

This means that the eigenvalues are one =triplet and and a= singlet. − Inserting all the eigenvalues back into the spin Hamiltonians diagonalized matrix:

−J/ −J/ ) −J/ 3.2 The phonon Hamiltonian J/ The phonon Hamiltonian is:

(70) † † † † is the numberH ofp particles= aa in +that particularaa + va state. + aa + a † Thea aGreen’s function for the phonon Hamiltonian[3]:

(71) ′ ′ Where Dt, t = −iTAtAt

(72) † Taking the derivative of this Green’sA function = a + D[9]: a

(73) ′ ′ ′ If i = j theni the ∂D firstt, tterm = is zerot − tbecause[A,A it] commute. + −iTi Using∂A thetA equationt of motion for the ladder operators: , the derivative can be rewritten it as[9]:

i ∂A = [A,Hp] ′ ′ i ∂ D t, t = −iTi ∂ A tA t (74) ′ † ′ = −iT[A,Hp]tAt = −iT(a − a tAt

and a new Green’s function is defined because has to act on the Hamiltonian again. The new Green’s function E (see Appendix A)[9]: † a − a (75) ′ † ′ � t, t = −iT(a − a t A t (76) † † † p [A ,H ] = [a + a , a a ] = a − a (77) † † † † For i = j = 1[a − a , Hp] = [a − a , a a] = (a + a = A

′ † ′ Taking the derivative of E: �t, t = −iTa − a tAt

′ ′ † † † ′ i ∂ � t, t = t − t [′ a − a , a ′+ a ] + −i′ Ti ∂ a − a A t Here one can easily see= thatt the − tderivative + D oft, D t is +nothingvD elset, t but Green’s function E times

. ′ ′ i ∂Dt, t = �t, t ′ ′ ′ ′ And for i = j = 2: i ∂�t, t = t − t + Dt, t + vDt, t

′ ′ i ∂Dt, t = �t, t ′ ′ ′ ′ This can be rewritteni ∂ as� a matrix(seet, t = Appendix):t − t + Dt, t + vDt, t

� �� = ( ) � � � Taking the Fourier transform we get�� that[9]: = + � �

μDμ = �μ

μ�μ = + Dμ + vDμ

Solving out the E:s: μ�μ = + Dμ + vDμ

�μ = + Dμ + vDμ μ �μ = + Dμ + vDμ And plug in into and intoμ and solving out the D:s: � D � D μ − Dμ − vDμ = Which can be written as a matrixμ − in componentsDμ − of D[9]:vDμ =

μ − − v �μ = ( ) −v μ − μ − v v μ − ⇒ �μ = μ − μ − − v

An uncoupled case when v = 0:

The relation between D and E can be written like this when v = 0:

μ − Dμ ( ) = − μ �μ Dμ μ ( ) = = �μ μ μ μ − μ − Dμ = = ⋯ = − μ − μ − μ + μ �μ = μ − dμ = , dμ = v = μ − μ −

4 Results The aim was to determine the ground state of the Hamiltonian for nuclei in a diatomic molecular structure and later include the phonon-spin interaction to this for comparison of these two i.e. investigate the deviation of the ground state and then compare it to the intrinsic ground state. For these different cases[9]:

1. When 2. When � ± � = , � ≠ 3. ± � � = , � ≠ ∆ = � ∆ z Figure 1: This figure illustrates the first case when . The phonon is propagating in the x – and� y-plane± (the pink arrow) orthogonal to the spin direction� = , �which≠ is aligned in the z direction (the aqua S colored arrow). In the second case, the scenario will y look almost the same except that the spatial operator for the phonon also will be directed in the z direction i.e. parallel to . x � S ± � The Green’s function below is the same as the one from the phonon part E (see section 3.2) but here the interaction part will be added. The derivative of the Green’s function E will now look like this[9]:

′ (78) i � t,′ t = ′ ′ ′ Wheret − t +is �an Dinteractiont, t + parametervDt, and t + is∑ an ∆ operator ∙ −i whichTS describestAt the spatial direction of the displacement of the phonon. Sorting out all the terms in the equation above starting∆ with the interaction Hamiltonian:

(79)

Here the coupling betweenH = the∑ phonon∆ and∙ S theA spin takes place via their operators[9]:

(80) + − − + And plugging this back∙ S into= (the Sinteraction+ S Hamiltonian:+ S

(81) + − − + is the operatorH = from ∑ the∆ phonon( S part+ (seeS +3.2)and S S isA the spin operator from the spin part (see 3.1). � This looks like a new Green’s function. Defining the new Green’s function as F:

(82) ′ ′ � t, t = −i T ∙ S tA16 t

Taking the derivative of F gives:

(83) ′ ′ ′ ′ Fromi th ∂e� phonont, t part= thet − equation t [ ∙ Soft, motion At for] the+ −iladderTi operators� ∙ S wastA used:t and the same method is used here except that in this case will give contribution �� = from both the interaction part of the new interaction Hamiltonian as well from the spin [�, ��ℎ] part . H H (84)

i ∂ S = [S,H + H ] (85)

At this point, I have decided toH divide= −J�this part∙ � of the result section into two parts. In the first part, I will only calculate when is acting on will be considered and in the second part I will be calculating when the spin operator is acting on . In the end, they will all be ∙ S H put together and back into the equation for E in order to solve out D, which in turn will be used to find E and F. H

4.1 The Interaction Hamiltonian

Here and the spin operator are acting on the interaction Hamiltonian:

� (86)

Some commutation rules for spin[ be ∙ used S,H thorough] the calculations in this sections and the next (see appendix).

Calculate these values for each . ′  Case 1: when � andt, t m = 1 or 2 and n = 1 or 2: ± � ≠ �� =

The values forH = ∆ ∙ SA + ∆ ∙ SA + ∆ ∙ SA + ∆ ∙ SA

�:

[ ∙ S,H] = [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ]

= ∆[ ∙ S, ∙ S ]A + ∆[ ∙ S, ∙ S ]A + ∆[ ∙ S, ∙ S ]A Here the only contribution+ ∆[ ∙ S will, come∙ S ] Afrom the third term because the first term will be zero due to the fact that the spin operators have the same index and they will cancel out. The second and the last term will also be zero because the spin operators have different indices so they are different spins and commute.

From the third term, the following contribution is given (see Appendix A):

− + + − ∆[ ∙ S, ∙ S ]A = ∆ − SA

�Only: the last term will give a contribution:

− + + − ∆[ ∙ S, ∙ S ]A = ∆� � − � � ��

�Only: the first term will give a contribution:

+ − − + ∆[ ∙ S, ∙ S ]A = ∆� � − � � ��

�Only: the second term will give a contribution:

+ − − + ∆[ ∙ S,Case ∙ S 2: ] Awhen = ∆� � and − � � � �m� = 1 or 2 and n = 1 or 2: ± Here every value for each �F will≠ be zero� �so= there will not be any contribution from this scenario. The lattice vibrations are determined by the result in section 3.2.

 Case 3: when and

∆≠ , ∆≠ ∆= ∆=

�:

∆[ ∙ S, ∙ S ]A + ∆[ ∙ S, ∙ S ]A =

�The: only contribution given here is from the second term:

∆[ ∙ S, ∙ S ]A − + + − + + − − − + = ∆ � � − � � �� + � − � � + � − ��

�The: only contribution from here is from the first term:

∆[ ∙ S, ∙ S ]A + − − + + + − − − + = ∆ � � − � � �� + � − �� + � − � � �

�Also: here everything will be zero.

4.2 The Spin Interaction Hamiltonian

The same procedure as in section 4.1 but only with the interaction part from the spin Hamiltonian .

H (87)

 Case 1: when [ and∙ S ,H ] m = 1 or 2 and n = 1 or 2: ± � ≠ �� =

�:

[ ∙ S, −JS ∙ S] − − + + + − = J SS − S S + S S − SS

�:

−J[ ∙ S,S ∙ S ] − − + + + − = J S S−SS + SS − S S

�:

−J[ ∙ S,S ∙ S ] − − + + + − = J SS −S S + S S − SS

�:

−J[ ∙ S,S ∙ S ] − − + + + − = JSS Case−S S2: when + SS − and S S m = 1 or 2 and n = 1 or 2: ± � = �� ≠

�:

[ ∙ S, −JS ∙ S] = + − − + = J S S − � S

�:

[ ∙ S, −JS ∙ S] = − + + − = J S S − � S

�:

[ ∙ S, −JS ∙ S] = + − − + = J� S − S S 19

�:

[ ∙ S, −JS ∙ S] = − + + − = JS S + � S Here it looks like there is a coupling between the phonon and spin. However, the phonon is not coupling to the z direction of the spin since the only non-vanishing component of λ is only multiplied with the spin’s lowering and raising operators.

 Case 3: when and

∆≠ , ∆≠ ∆= ∆=

�:

[ ∙ S, −JS ∙ S] = − − + + + − = J SS − S S + S S − SS

�:

−J[ ∙ S,S ∙ S ] − − + + + − = J S S−SS + SS − S S

�:

−J[ ∙ S,S ∙ S ] − − + + + − = J SS −S S + S S − SS

�:

−J[ ∙ S,S ∙ S ] − − + + + − Which= JS areS exactly−SS the same+ S valuesS − given S S for case 1 for all F:s.

4.3 The total Interaction Hamiltonian

The new Green’s function F: �� + ��

′ ′ Now the total interaction Hamiltonian� t, t = will −i beTi put� back∙ S intotA thet derivative of the new Green’s function F.

′ ′ ′ ′ i ∂ � t, t = t − t [ ∙ S t, A t ] + −iTi� ∙ S tA t

�� ∙ St = [ ∙ St,H + H]� 20

Case 1:

′ ′ ′ i ∂ � t, t = t − t [ ∙ S− t,+ A +t −] + −i T∆ − SAt J + − + − + − − + + − − + + − + − [ S ,S S ] + [ S ,S S ] + [ S , SS ] + [ S ,S S ] − + − + − + ′ The last terms belonging+ [ S to,S the S spin ] + interaction[ S , S partS of]A the tHamiltonian are just higher order terms so this equation can be rewritten as:

′ − + + − ′ The first termi ∂ �in thet, derivative t = +−i of FT∆ vanishes because − the S operatorsAt+h.α.t.A in the commutatort commute. and can be rewritten as the expectation value of by using this approximated′ expression′ [3]. AtAt = Dt, t S S + −′ ′ [S� ,S� ] = �� S − + + − − = � + − − + Fourier transforming all of the derivatives� � of− �F: � = −�

�μ = ∆�SDμ + h. α. t. � �μ = ∆�SDμ+h.α.t. � �μ = ∆�SDμ + h. α. t. � �μ = ∆�SDμ+h.α.t. � �μ = −∆�SDμ + h. α. t. � �μ = −∆�SDμ+h.α.t. � �μ = −∆�SDμ + h. α. t. � �μ = −∆�SDμ+h.α.t. �

Plugging the F:s back into each value of E:

For m = n = 1:

′ ′ ′ ′ ′ i�t, t = t − t + �Dt, t + vDt, t + ∑ ∆ ∙ −iTStAt Fourier transforming and solves out E and let (see Appendix):

= and

� � ∆ ∆ = ( + ( ) + ( ) ) � � � ∆ ∆ Plug all the values of F back into each value of E:

� � + �∆ ∆ S + ∆ ∆ S = + � ) ) � � + �−∆∆S − ∆∆S � �

Solving out the D:s

� − �Dμ − v� + �∆∆S + ∆∆SDμ = � � � − �Dμ − v� + �∆∆S + ∆∆SDμ = � � − �Dμ − v� + �−∆∆S − ∆∆SDμ = � � − �Dμ − v� + �−∆∆S − ∆∆SDμ = � Rewrite the expression for E as[9]:�

� � � � = � + ( ) + ( ) is the identity matrix. � � �

�

Add the matrices in front of together:

� − � v� + �∆∆S + ∆∆S � ) v� + �−∆∆S − ∆∆S � − � � (See Appendix A for more details).

The out coming result for are (values for E and F see Appendix):

� �� ( ) + ( ) = ( ) ��

�� − � −�� −�� − � = � − � � − � − ��

Indeed there is an interaction that takes place here between the phonon and the spin. The phonon propagates in the x –and y plane while the spin is directed in the z-direction. Values have contributions from the new Green’s function F. It means, depending on the sign of that value, it will add or remove energy from the system. Case 3: when and :

∆≠ , ∆≠ ∆= ∆=

− + �μ = ∆�S − S + S Dμ + h. α. t. � − + �μ = ∆�S − S + S Dμ + h. α. t. � − + �μ = −∆�S + S + S Dμ + h. α. t. � − + �μ = −∆�S + S + S Dμ+h.α.t. Plug the F:s into E and � : ∆= ∆=

�μ = + �Dμ + vDμ + ∆�μ �

�μ = + �Dμ + vDμ � �μ = �Dμ + vDμ + ∆�μ � �μ = �Dμ + vDμ � �μ = �Dμ + vDμ + ∆�μ � ) �μ = � �Dμ + vDμ

�μ = + �Dμ + vDμ � �μ = + �Dμ + vDμ Plug in the� E:s into the expression of :

� − � v� � = ( ) v� � − � � μ − v v μ − ⇒ = μ − μ − − v

This means that all the values from F in both D and E will be zero and the matrix will look exactly the same as it did in section 3.2 for . Apparently there are no coupling between the two nuclei for this scenario when and but there is a local interaction between the phonon and the spin for each nuclei separately. ∆≠ , ∆≠ ∆= ∆= In the end, this is only for one direction, how the phonon is affected by this spin-phonon coupling. To investigate the coupling in the other direction would be to look at how the spin would be affected by this interaction with the phonon. To do all calculation it would be too long for this report.

5 Discussion In case 1, when is affecting the movement of the phonon, it will propagates orthogonal to the spin which is ±aligned in the z-direction for both of the nuclei. Here D and E have contributions from� the new Green’s function F compared to D and E without the spin-phonon coupling (see section 3). Depending on their sign, they will increase or decrease the values of D and E and therefore affect the ground state of the phonons. That will say that it will contribute with more energy or remove energy from the system. However, with the analytic calculations done here it is only possible to see the changes qualitatively. In order to get a quantitative picture one would have to plug in numbers and use Matlab.

The displacement introduced in section 2.1 can be related to in this way[9]:

� † In case 2 the spin and are aligned�=�−�=��+� but the spin is alternating because the raising and lowering operators are flipping the spin up and down. The only contribution is from the spin interaction Hamiltonian and this will end up in the higher order term (h.o.t.) and can be viewed as an extra but very weak contribution to the phonon. The energy contribution from this scenario is very small. Case 3 gives the same values for and as in case 1 except for an additional contribution from and . In this scenario the Green’s functions D and E look the same as �μ �μ in the case without the− spin-phonon+ interaction, i.e. as in section 3.2. The ground state of the phonon is not changed.� This� means that the spin-phonon interaction is local i.e. that the interaction is only within the nuclei.

5.1 Outlook There are several interesting aspects for future work. First of all, the results for D could be interpreted quantitatively in Matlab by for example letting the phonon's eigenfrequency run from -10 to 10 and then plot figures for the cases when the interaction parameter v is zero and when it is not i.e. when the phonon is coupled to the spin. Then the obtained curves could� be compared. Furthermore, this investigation was only done in one direction, i.e. when the phonon is coupled to the spin, but it could also be investigated how the spin will be affected by the interaction with the phonon. Afterwards, it might also be interesting to consider adding additional interactions or looking at systems with more than two atoms.

6 Conclusion A diatomic system with spin-phonon coupling was considered and it was investigated how this interaction influences the phonons. The corresponding Green's functions of this system were calculated. It turned out that this interaction indeed has an impact on the ground state of the phonons. In particular, the Green's functions D and E were changed and the additional Green's function F appeared. It was possible to see qualitatively how these changes of the Green's functions look like. However, it was not possible to see the influence of the spin- phonon coupling quantitatively because this would have required numerical calculations which could be interesting for future work.

In particular, three different cases were considered which corresponded to different choices of anisotropy in the system. In case one the phonons are directed in the x and y directions while the spin is pointing in the z direction such that there is an interaction between the spins and the phonons of the two nuclei. In the second case the phonons and spins are both directed in the z direction, i.e. they are parallel to each other, and the interaction is very small. In the last case there is only a local interaction between the phonon and the spin and nothing in between nuclei one and two. This investigation was interesting because usually phonons can be controlled by electric fields while spin can be controlled by magnetic fields but after this interaction has taken place the opposite occurs, i.e. the spin can be electrically and the phonons magnetically steered and this behavior can be used for example to control the behavior of magnetic materials. Furthermore, this type of interaction with a criterion as in the first case can add or remove energy from the system so it will change the structure of the lattice. For example, this could be used for manufacturing refrigerators which could be much more energy saving than they are today.

References

[1] Bruus H, Flensberg K. Introduction to Many-body quantum theory in condensed matter Physics. 2. edition. Copenhagen: University of Copenhagen, Technical University of Denmark Press: 2002.

[2] Hofmann P, Solid State Physics: An Introduction. Weinheim: University of Aarhus Press: 2008.

[3] Mahan G. D, Many-Particle Physics. 3. edition. Knoxwille, Tennessee: University of Tennessee, Oak Ridge National Laboratory Press: 2000.

[4] Nayak C, Many-Body Physics. Los Angeles: University of California Press:1999.

[5] Larsen U. A Note on the Pseudo-Fermion Representation of a Spin 1/2 or 1. Z. Physik. June 1972; 256, 65-72. © by Springer-Verlag 1972. Physics Laboratory I, H. C. Ørsted Institute, University of Copenhagen, Copenhagen, Denmark.

[6] Fogedby H.C. On the pseudo-fermion representation of the Kondo Hamiltonian. Physics Letter. September 1972; 41A(2).

[7] Eisberg R. Resnick R. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. 2nd. Edition. John Wiley & Sons Inc. Press: 1985.

[8] Wikipedia. Phonons [webpage]. [Access 2016-05-25; läst 2016-05-25]. Tillgängligt via: https://en.wikipedia.org/wiki/Phonon

[9] Notes from supervisor.

A Appendix

3 Method

3.1 The Spin Hamiltonian

Eigen values from the interaction part of the Spin Hamiltonian:

− − − = − − −

− − − − − = − =0 − − − − − − + − = − = , = 3.2 The phonon Hamiltonian + − = = and = −

Equation of motion [9] :

† p [A , H ] = (a + a † p [A , H ] = (a + a † p [a + a ,H ] = A + vA † [a + a,Hp] = A + vA

′ ′ i ∂Dt, t − �t, t = ′ ′ ′ ′ i ∂�t, t − Dt, t − vDt, t = t − t ′ ′ i ∂Dt, t − �t, t = ′ ′ ′ ′ i ∂ � t, t − D t, t 28− vD t, t = t − t

The derivative of E:

′ ′ † † † ′ i ∂�t, t = t − t [a − a , a + a ] + −iTi ∂a − a At ′ † ′ p = t − t + −iT[a − a ,H ]tA t ′ † † † † † ′ = t − t + −iT[a − a , aa + aa + v(a + a(a + a]tAt

′ † ′ † ′ = t − t + −iTa + a tAt + v−iTa + a tAt ′ ′ ′ = t − t + −iTAtAt + v −iTAtAt ′ ′ ′ [10]= t − t + Dt, t + vDt, t

4. Results

4.1 The Interaction Hamiltonian

Spin commutation relations [9]:

+ − [S ,S ] = S + + [S ,S ] = [S + iS ,S ] = −iS − S = −S − − [S ,S] = [S − iS,S] = −iS + S = S

+ − − + S ∙ S = S S + S S + SS The values for for case 1:

�

∆ + − − + + − − + ∆[ ∙ S, ∙ S ]A = [ S + S , S + S ]A ∆ + − + − + − − + − + + − − + − + = [ S , S ] + [ S , S ] + [ S , S ] + [ S , S ]A ∆ + − − + = −S + S A − + + − = ∆ − SA

�:

[ ∙ S,H] = [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] 29

�:

[ ∙ S,H] = [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ]

�:

[ ∙ S,H] = [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] + [ ∙ S, ∆ ∙ SA ] 4.2 The Spin Interaction+ [ ∙ S Hamiltonian, ∆ ∙ SA ]

For case 1:

�:

−J[ ∙ S,S ∙ S ]

J + − + − + − − + + − − + + − − + − + = − [ S ,S S ] + [ S ,S S ] + [ S , SS ] + [ S ,S S ] + [ S ,S S ] − + + [ S , SS ]

J + − + − + − − + − + − + = − [S ,S ]S + [S ,S ]S + [S ,S ]S + [S ,S ]S J − + − + + − + − = − −SS + S S + SS − SS − − + + + − 4.3= J TheS totalS − Interaction S S + Hamiltonian S S − S S

The new equation will now be: �� + ��

′ ′ For n = 1: i ∂�t, t = ∆�SDt, t + h. α. t.

′ ′ For n = 2: i ∂�t, t = ∆�SDt, t + h. α. t.

′ ′ and the rest will follow: i ∂�t, t = ∆�SDt, t + h. α. t.

′ ′ i ∂�t, t = ∆�SDt, t + h. α. t. ′ ′ i ∂�t, t = ∆�SDt, t + h. α. t. ′ ′ i ∂�t, t = −∆�SDt, t + h. α. t. 30

′ ′ i ∂�t, t = −∆�SDt, t + h. α. t. ′ ′ i ∂�t, t = −∆�SDt, t + h. α. t. ′ ′ From page 22-24 i ∂�t, t = −∆�SDt, t + h. α. t.

Fourier transform E:

�μ = + �Dμ + vDμ + ∆�μ + ∆�μ � �μ = �Dμ + vDμ + ∆�μ + ∆�μ � �μ = �Dμ + vDμ + ∆�μ + ∆�μ � �μ = + �Dμ + vDμ + ∆�μ + ∆�μ Plug each �value of F back into each value of E:

�μ = + �Dμ + vDμ + �∆∆S + ∆∆SDμ � � �μ = �Dμ + vDμ + �∆∆S + ∆∆SDμ � � ) �μ = � �Dμ + vDμ + � �−∆∆S − ∆∆SDμ

�μ = + �Dμ + vDμ + �−∆∆S − ∆∆SDμ Plug each �value of E back into all the values� of the first Green’s function D:

� Dμ = + �Dμ + vDμ + �∆∆S + ∆∆SDμ � � � Dμ = �Dμ + vDμ + �∆∆S + ∆∆SDμ � � � Dμ = �Dμ + vDμ + �−∆∆S − ∆∆SDμ � � � Dμ = + �Dμ + vDμ + �−∆∆S − ∆∆SDμ � �

Matrix

Write E as matrix:

� � = ( ) And D: � �

D D = ( ) Here we can see that the terms from the couplingD beforeD F is applied that will say the interaction part:

� − � �� ∆ ∆ S + ∆ ∆ S + � ) �� − � �−∆∆S − ∆∆S � = ( ) And solving for the matrix D[9]:

− = � ( )

Define matrix M as:

� − � v� + �∆∆S + ∆∆S � = � ) v� + �−∆∆S − ∆∆S � − � Defining: �

� = v� + �∆∆S + ∆∆S �

� = v� + �−∆∆S − ∆∆S �

⇒ = � − � � − � − ��

� − � −� × ( ) −� 32 � − �

× ( )

� − �� −�� = ( ) � − � � − � − �� −�� − ��

Plug in the values for each D in each E by using:

� D = � The values for E (see Appendix): �

�� − � � = � − � � − � − ��

� �� = − � � − � � − � − ��

�� − � � = � − � � − � − ��

And finally the results for each F:

�� μ = − ∆�S + h. α. t. � � − � � − � − ��

�� − � �μ = ∆�S + h. α. t. � � − � � − � − ��

�� μ = − ∆�S + h. α. t. � � − � � − � − ��

�� − � �μ = ∆�S + h. α. t. � � − � � − � − ��

�� − � �μ = −∆�S + h. α. t. � � − � � − � − ��

�� μ = ∆�S + h. α. t. � � − � � − � − �� 33

�� − � �μ = −∆�S + h. α. t. � � − � � − � − ��

�� μ = ∆�S + h. α. t. � � − � � − � − ��