Contents

1 Introduction 2 1.1 Goals in this course ...... 2 1.2 of free Fermi ...... 2 1.2.1 T = 0 Fermi sea ...... 2 1.2.2 T > 0 Free energy...... 4 1.2.3 Avg. number...... 4 1.2.4 at low T...... 5 1.2.5 Classical limit...... 8 1.3 Second quantization ...... 8 1.3.1 Symmetry of many-particle wavefunctions ...... 9 1.3.2 Field operators ...... 11 1.3.3 2nd-quantized Hamiltonian ...... 14 1.3.4 Schr¨odinger,Heisenberg, interaction representations . . . . 16 1.4 ...... 18 1.4.1 Review of simple harmonic oscillator quantization . . . . . 18 1.4.2 1D harmonic chain ...... 19 1.4.3 ...... 21 1.4.4 Anharmonicity & its consequences ...... 23

1 1 Introduction

1.1 Goals in this course These are my hopes for the course. Let me know if you feel the course is not fulfilling them for you. Be free with your criticism & comments! Thanks. • Teach basics of collective phenomena in systems • Make frequent reference to real experiment and data • Use 2nd quantized notation without field-theoretical techniques • Get all students reading basic CM journals • Allow students to practice presenting a talk • Allow students to bootstrap own research if possible

1.2 Statistical mechanics of free Fermi gas

1.2.1 T = 0 Fermi sea Start with simple model of in , neglecting e− − e− interac- tions. Hamiltonian is 2 2 ∑ h¯ ∇ Hˆ = − j , j = 1,...N particles (1) j 2m Eigenstates of each −(¯h2∇2/2m) are just plane eik·r labelled by k, with ki = 2πni/Li in box with periodic B.C. Recall electrons are , which means we can only put one in each single-particle state. Including we can put two particles (↑↓) in each k-state. At zero the of N-electron system is then formed by adding particles 2 2 until we run out of electrons. Energy is εk =h ¯ k /2m, so start with two in lowest state k = 0, then add two to next states, with kx or ky

2 or kz = 2π/L, etc. as shown. Energy of highest particle called “” εF , magnitude of corresponding vector called kF . Typical 4 Fermi energy for metal εF ≃ 1eV ≃ 10 K. At T = 0 only states with k < kF occupied (Fermi “sea” or Fermi sphere), so we can write of electrons as 2*# occupied states/Volume (2 is for spin): 3 3 2 ∑kF ∫ d k 1 ∫ k k n = ≃ 2 = F k2dk = F (2) 3 k

Figure 1: States of Fermi gas with parabolic spectrum, ε = k2/2m.

The total ground state energy of the Fermi gas must be of order εF , since there is no other energy in the problem. If we simply add up the energies of all particles in states up to get   ∫ 2 2 2 5 E 1 kF h¯ k  h¯ kF = dk k2   = (4) L3 π2 0 2m 10π2m and the ground state energy per particle (N = nL3 is the total number) is E 3 = ε . (5) N 5 F

3 1.2.2 T > 0 Free energy. Reminder: partition function for free fermions in grnd conical ensemble is Z = Tr e−β(Hˆ −µNˆ) (6) ∑ −β(Hˆ −µNˆ) = ⟨n1, n2...n∞|e |n1, n2...n∞⟩ (7) n1,n2...nk∞ ∑ ∑ −β( i[εini−µni]) = ⟨n1, n2...n∞|e |n1, n2...n∞⟩ (8) n1,n2...n∞ where i labels single-fermion state, e.g. i = k, σ, and ni runs from 0 to 1 for fermions. Since many-fermion state in occ. no. representation is simple product: |n1, n2...n∞⟩ = |n1⟩|n2⟩...|n∞⟩, can factorize:     ∑ − − ∑ − − Z =  e β[ε1n1 µn1] ···  e β[ε∞n∞ µn∞] , (9) n1 n∞ so

( ) ∞ −β(εi−µ) Z = Πi=0 1 + e (10)

Since the free energy (grand canonical potential) is Ω = −kBT log Z, we get

( ) ∑∞ − − − β(εi µ) Ω = kBT i=1 log 1 + e (11)

1.2.3 Avg. fermion number. We may want to take statistical averages of quantum operators, for which we need the statistical operator ρˆ = Z−1e−β(Hˆ −µNˆ). Then any operator Oˆ has an expectation value ⟨O⟩ˆ = Tr(ˆρOˆ). For example, avg. no. of particles

⟨Nˆ ⟩ = Tr(ρNˆ ) (12) Tr(e−β(Hˆ −µNˆ)Nˆ ) = (13) Tr(e−β(Hˆ −µNˆ))

4 Now note this is related to the derivative of Ω wrt chem. potential µ: ∂Ω ∂ log Z −k T ∂Z = −k T = B (14) ∂µ B ∂µ Z ∂µ = −Tr(ρNˆ ) = −⟨Nˆ ⟩ (15) and using Eq. 11, we see ∞ ∞ ⟨ ˆ ⟩ ∑ 1 ≡ ∑ 0 N = β(ε −µ) ni (16) i=1 1 + e i i=1 i where n0 is the avg. number of fermions in a single-particle state i in equilibrium at temperature T . If we recall i was a shorthand for k, σ, but εk doesn’t depend on σ, we get Fermi-Dirac distribution function

0 1 nkσ = − (17) 1 + eβ(εk µ)

1.2.4 Fermi gas at low T. Since the Fermi energy of is so high (∼ 104K), it’s important to understand the limit kBT ≪ εF , where the Fermi gas is nearly degen- erate, and make sure the classical limit kBT ≫ εF comes out right too. Let’s calculate, for example, the entropy and specific heat, which can be obtained from the thermodynamic potential Ω via the general thermody- namic relations      ∂Ω   ∂S  S = − ; CV = T (18) ∂T V,µ ∂T V,µ From (11) and (16), and including spin, we have ∑ ( ) −β(εk−µ) Ω = −2kBT log 1 + e k ∫ ( ) 3 −β(ε−µ) = 2kBTL dεN(ε) log 1 + e ⇒ ( ) ∫ ∞ − CV 1 ∂f 2 cV ≡ = 2 dεN(ε) (ε − µ) 3 0 L kBT ( )∂ε 1 ∫ ∞ −∂f = 2 dξ N(ξ) ξ2 (19) − kBT µ ∂ξ

5 ∑ −3 − where I introduced the density of k-states for one spin N(ε) = L kδ(ε εk). The Fermi function is f(ε) = 1/(1+exp β(ε−µ)), & I defined shifted energy variable ξ = ε−µ. In general, the degenerate limit is characterized by k sums which decay rapidly for energies far from the , so the game is to assume the varies slowly on a scale of the thermal energy, and replace N(ε) by N(εF ) ≡ N0. This type of Sommer- feld expansion1 assumes the density of states is a smoothly varying fctn., i.e. the V → ∞ has been taken (otherwise N(ε) is 2 2 too spiky!). For a parabolic band, εk =h ¯ k /(2m) in 3D, the delta-fctn. can be evaluated to find2

  3 n ε 1/2 N(ε) =   θ(ε). (22) 2 εF εF This can be expanded around the Fermi level:3 1 N(ξ) = N(0) + N ′(0)ξ + N ′′(0)ξ2 + ... (24) 2 (In a horrible misuse of notation, N(0), N(εF ), and N0 all mean the

1If you are integrating a smooth function of ε multiplied by the Fermi function derivative −∂f/∂ε, the derivative restricts the range of integration to a region of width kBT around the Fermi surface. If you are integrating something times f(ε) itself, it’s convenient to do an integration by parts. The result is (see e.g. Ashcroft & Mermin appendix C)

∫ ∫ ∞ µ ∑∞ 2n−1 2n d | dεH(ε)f(ε) = dεH(ε) + an(kBT ) 2n−1 H(ε) ε=µ (20) −∞ −∞ dε n=1 ( ) 2(n−1) 2 4 where an = 2 − 1/2 ζ(2n)(ζ is Riemann ζ fctn., ζ(2) = π /6, ζ(4) = π /90, etc.). 2Here’s one way to get this: ∫ √ ∫ ( ) √ ( ) ∑ 3 2 1/2 −3 − → d k | dε |−1 − 2mε k dk m − 2mε 3 n ε N(ε) = L δ(ε εk) 3 δ(k ) = 2 2 δ(k ) = (21) (2π) dk ¯h 2π ¯h k ¯h 2 εF εF k

3When does the validity of the expansion break down? When the approximation that the density of states is a smooth function does, i.e. when the thermal energy kBT is comparable to the splitting between states at the Fermi level, 2 | ≃ ¯h kF δk ≃ δk ≃ a δεk εF εF εF , (23) m kF L 4 where a is the lattice spacing and L is the box size. At T = 1K, requiring kBT ∼ δε, and taking εF /kB ≃ 10 K says that systems (boxes) of size less than 1µm will “show mesoscopic” effects, i.e. results from Sommerfeld-type expansions are no longer valid.

6 Figure 2: Density of states for parabolic spectrum, ε = k2/2m density of states at the Fermi level). The leading order term in the low-T specific heat is therefore found directly by scaling out the factors of T in Eq. (19):     1 ∫ ∞ −∂f ∫ ∞ −∂f ≃ 2   2 2   2 cV 2kB N0 −∞ dξ ξ = 2kBTN0 −∞ dx x (25) T ∂ξ | {z ∂x } π2/3 So 2π2 c ≃ N k2 T + O(T 3). (26) V 3 0 B This is the famous linear in temperature specific heat of a free Fermi gas. 4

4Note in (26), I extended the lower limit −µ of the integral in Eq. (19) to −∞ since it can be shown that the is very close to εF at low T . Since we are interested in kBT ≪ εF , and the range in the integral is only several kBT at most, this introduces neglible error. Why: 0 → − At T =0, the Fermi function nk step function θ(µ εk), so we know µ(T = 0) must just be the Fermi energy 2 2 2/3 εF = ¯h (3π n) /2m.

∞ N ∑ n = = 2L−3 n0 L3 k ∫ k = 2 dεN(ε)f(ε) ∫ µ 2 π 2 ′ ≃ dεN(ε) + (kBT ) N (ε)|ε=µ (continued on next page) −∞ 6

7 1.2.5 Classical limit. I won’t calculate the classical limit. All the standard results for a Boltzman statistics gas, e.g. cV (T ≫ εF ) = (3/2)NkB follow immediately from noticing that the Fermi function reduces to the Boltzman distribution, f(ε) → e−β(ε−µ),T → ∞. (27)

(You will need to convince yourself that the classical result µ/(kBT ) → −∞ is recovered to make this argument.)

1.3 Second quantization The idea behind the term ”second quantization” arises from the fact that in the early days of , forces between particles were treated classically. , position and other observables were represented by operators which do not in general commute with each other. Particle number is assumed to be quantized as one of the tenets of the theory, e.g. Einstein’s early work on blackbody radiation. At some point it was also realized that forces between particles are also quantized because they are mediated by the exchange of other particles. In Schr¨odinger’streatment of the H-atom the force is just the classical static Coulomb force, but a more complete treatment includes the interaction of the H-atom and its constituents with the radiation field, which must itself be quantized (“”). This quantization of the fields mediat- ing the interactions between matter particles was referred to as “second” quantization. In the meantime, a second-quantized description has been developed in which both “matter fields” and “force fields” are described ∫ εF π2 ≃ dεN(ε) +(µ − ε )N(ε ) + (k T )2N ′(ε)| F F 6 B ε=µ | −∞ {z } n 2 ′ π 2 N (εF ) ⇒ µ ≃ εF − (kBT ) 6 N(εF )

′ 2 Since N /N is typically of order 1/εF , corrections are small.

8 by second-quantized field operators. In fact, modern condensed matter usually does go backwards and describe particles interacting via classical Coulomb forces again,5 but these particles are described by field operators. Once the calculational rules are absorbed, calculating with the 2nd-quantized formalism is easier for most people than doing 1st- quantized calculations. They must, of course, yield the same answer, as they are rigorously equivalent. I will not prove this equivalence in class, as it is exceedingly tedious, but merely motivate it for you below. I urge you to read the proof once for your own edification, however.6

1.3.1 Symmetry of many-particle wavefunctions Quantum mechanics allows for the possibility of indistinguishable parti- cles, and nature seems to have taken advantage of this as a way to con- struct things. No electron can be distinguished from another electron, except by saying where it is, what it is in, etc. Inter- nal quantum-mechanical consistency requires that when we write down a many-identical-particle state, we make that state noncommittal as to which particle is in which single-particle state. For example, we say that we have electron 1 and electron 2, and we put them in states a and b re- spectively, but exchange symmetry requires (since electrons are fermions) that a satisfactory wave-function has the form

Ψ(r1, r2) = A[(ψa(r2)ψb(r1) − ψa(r1)ψb(r2)]. (28)

If we have N particles, the wavefunctions must be either symmetric or antisymmetric under exchange:7

B B Ψ (r1 ... ri ... rj ... rN ) = Ψ (r1 ... rj ... ri ... rN ) (29)

5Q: when does this approximation break down? 6See, e.g. Fetter & Wallecka, Quantum Theory of Many-Particle Systems 7This is related to the spin-statistics theorem first formulated by Fierz and Pauli. The proof requires the PCT theorem, profed by Schwinger, L¨udersand Pauli, which says that PCT (parity-charge conjugation-time reversal) is a good symmetry for a system described by a Lorentz-invariant field theory). Recently, a generalization of Bose & Fermi statistics to particles called has been intensely discussed. Under exchange an wavefunction behaves as A iθ A Ψ (r1 ... ri ... rj ... rN ) = e Ψ (r1 ... rj ... ri ... rN ) for some 0 ≤ θ ≤ 2π.

9 F F Ψ (r1 ... ri ... rj ... rN ) = −Ψ (r1 ... rj ... ri ... rN ) Fermions (30)

Given a set of single-particle wave functions ϕEi(r), where Ei is a quan- tum number, e.g. energy (N.B. it can be any set of quantum numbers!) we can easily construct wave fctns. which satisfy statistics, e.g.

( ) ··· 1/2 ∑ B n1!n2! n∞! sgnP N Ψ F (r1,... rN ) = (±1) P Π ϕE (ri), (31) n1...n∞ N! i=1 i P∈{E1,E2...EN } where the factor in parentheses is the number of ways to arrange N objects in boxes, with n1 in the first box, ... Remarks on Eq. (31): • 8 sum all permutations of the Ei’s in product ϕE1(r1)ϕE2(r2) . . . ϕEN (rN ).

• # distinct Ei’s occuring may be less than N, some missing because of multiple occupation in case. Example:

Fig. 2. Possible state of 3 noninteracting Bose particles

B Ψ20100...0(r1, r2, r3) = 1 = √ {ϕE (r1)ϕE (r2)ϕE (r3) + 3 0 0 2

+ϕE2 (r1)ϕE0 (r2)ϕE0 (r3) + } +ϕE0 (r1)ϕE2 (r2)ϕE0 (r3)

• Completely antisymmetric Fermionic wavefunction called Slater de-

8You might think the physical thing to do would be to sum all permutations of the particle labels. This is correct, but actually makes things harder since one can double count if particles are degenerate (see example of 3 bosons below.) The normalization factor is chosen with the sum over all permutation of the Ei’s in mind.

10 terminant:

  ϕ (r ) . . . ϕ (r ) 1/2 Emin 1 Emax 1 1   . . Ψn ...n∞(r1,... rN ) = . . (32) 1 N!

ϕEmin(rN ) . . . ϕEmax(rN )

where there are N eigenvalues which occur between Emin and Emax, inclusive, corresponding to N occupied states.

1.3.2 Field operators 2nd quantization is alternative way of describing many body states. We describe a particle by a field operator ˆ ∑ ψ(r) = aiϕEi(r) (33) i where i runs over the quantum numbers associated with the set of eigen- states ϕ, ai is a “coefficient” which is an operator (I’m going to neglect the hats (ˆ ) which normally denote an operator for the a’s and a†’s), and

ϕEi is a “1st-quantized” wavefunction, i.e. a normal Schr¨odingerwave- function of the type we have used up to now, such that (for example)

HϕEi = EiϕEi. Now we impose commutation relations [ ] ˆ ˆ† ′ − ′ ψ(r), ψ (r ) ± = δ(r r ) (34) [ ] [ ] ˆ ˆ ′ ˆ† ˆ† ′ ψ(r), ψ(r ) ± = ψ (r), ψ (r ) ± = 0, (35) which implies † † † [ai, aj]± = δij ;[ai, aj]± = [ai , aj]± = 0. (36) The upper sign is for fermions and the lower for bosons in Eqs. (35) and (36). Now construct many-body states from vacuum (no particles) |0⟩. a called annihilation operator, a† creation operator (see below).

11 Examples & comments (all properties follow from commutation rela- tions): Bosons: † • one particle in state i: ai |0⟩ ≡ |1⟩i

• annihilate vacuum ai|0⟩ = 0 9 † † • (Bosons) ai ai |0⟩ ≡ |2⟩i † † † • (Bosons) two in i and one in j: ai ai aj|0⟩ ≡ |2⟩i|1⟩j † • ai ai ≡ nˆi is number operator for state i. Proof: (bosons) † | ⟩ † † n| ⟩ (ai ai) n i = (ai ai)(ai ) 0 † † † n−1| ⟩ | ⟩ † 2 † n−1| ⟩ = ai (1 + ai ai)(ai ) 0 = n i + (ai ) ai(ai ) 0 | ⟩ † 3 † n−2| ⟩ | ⟩ = 2 n i + (ai ) ai(ai ) 0 ... = n n i

Similarly show (bosons):10

† 1/2 • ai |n⟩i = (n + 1) |n + 1⟩i

1/2 • ai|n⟩i = n |n − 1⟩i • many-particle state  1/2 1 † †   n1 n2 (a1) (a2) · · · |0⟩ ≡ |n1, n2 . . . n∞⟩ (37) n1!n2! . . . n∞! ⋆ occupation numbers specify state completely, exchange symmetry included due to commutation relations! (normalization factor left for problem set)

9Analogous state for fermions is zero, by commutation relations-check! 10By now it should be clear that the algebra for the bosonic case is identical to the algebra of simple harmonic oscillator ladder operators.

12 Fermions

• Anticommutation relations complicate matters, in particular note 1 a2 = (a†)2 = 0 ⇒ n = (Pauli principle) (38) i 0 • so a†|0⟩ = |1⟩ a|1⟩ = |0⟩ (39) a†|1⟩ = 0 a|0⟩ = 0 • many-particle state † † n1 n2 (a1) (a2) · · · |0⟩ ≡ |n1, n2 . . . n∞⟩ (40) ⋆ note normalization factor is 1 here.

• action of creation & annilation operators (suppose ns = 1): † † † | ⟩ − n1+n2+...ns−1 n1 ··· ··· n∞| ⟩ as . . . ns ... = ( 1) (a1) asas (a∞) 0 † † † = (−1)n1+n2+...ns−1(a )n1 ··· (1 − a a ) ··· (a )n∞|0⟩ 1 | s{z s} ∞ =0 n1+n2+...ns−1 = (−1) | . . . ns − 1 ...⟩

also as| ... 0 ...⟩ = 0 (41) and similarly   − n1+n2+...ns−1| ⟩ †| ⟩ ( 1) . . . ns + 1 ... ns = 0 as . . . ns ... =  (42) 0 ns = 1

13 1.3.3 2nd-quantized Hamiltonian ⋆ Point: Now “it can be shown”11 that state vector ∑ |Ψ(t)⟩ = f(n1, n2 . . . n∞, t)|n1, n2 . . . n∞⟩ (43) n1,n2...n∞ satisfies the Schr¨odingerequation (our old friend) ∂ ih¯ |Ψ(t)⟩ = Hˆ |Ψ(t)⟩ (44) ∂t if we take the ”2nd-quantized” form of Hˆ ˆ ∑ † 1 ∑ † † H = ai ⟨i|T |j⟩aj + ai aj⟨ij|V |kℓ⟩aℓak (45) ij 2 ijkℓ ∫ 3 ˆ† ˆ = d rψ (r)T (r, ∇r)ψ(r) (46) 1 ∫ ∫ + d3r d3r′ψˆ†(r)ψˆ†(r′)V (r, r′)ψˆ(r′)ψˆ(r) (47) 2 where the 1st quantized Hamiltonian was H = T + V . Here i indexes a complete set of single-particle states. Translationally invariant system It may be more satisfying if we can at least verify that this formalism “works” for a special case, e.g. translationally invariant systems. For such a system the momentum k is a good , so choose single-particle plane wave states −3/2 ik·r ϕi(r) = ϕkσ(r) = L e uσ, (48) ( ) 0 −∇2 12 where uσ is a spinor like u↓ = 1 , etc. 1st quantized T is /(2m), so 2nd-quantized kinetic energy is   2 ∑ † ∑  k  † ˆ ≡ ⟨ | | ⟩   T ai i T j aj = akσakσ . (49) ij kσ 2m

11Normally I hate skipping proofs. However, as mentioned above, this one is so tedious I just can’t resist. The bravehearted can look, e.g. in chapter 1 of Fetter and Wallecka. 12I’ll set ¯h = 1 from here on out, unless required for an honest physical calculation.

14 Figure 3: Feynman diagram for 2-body interaction showing momentum conservation.

† Since we showed akσakσ is just number operator which counts # of par- ticles in state kσ, this clearly just adds up the kinetic energy of all the occupied states, as it should. For general two-particle interaction V , let’s assume V (r, r′) = V (r − r′) only, as must be true if we have transl. invariance. In terms of the Fourier transform 1 ∫ V (q) = d3r eiq·r V (r) (50) L3 we have (steps left as exercise)

1 ∑ † † ˆ ′ V = akσak′+qσ′V (q)ak σak+qσ (51) 2 k,k′,q σ,σ′ Note that if you draw a picture showing k′ and k + q being destroyed (disappearing), and k and k′ + q being created, you realize that the no- tation describes a scattering process with two incoming particles and two outgoing ones, where momentum is conserved; in fact, it is easy to see that momentum q is transferred from one particle to another. This is called a Feynman diagram.

15 1.3.4 Schr¨odinger,Heisenberg, interaction representations Here we just give some definitions & reminders on different (equivalent) representations of quantum mechanics. In different reps., time depen- dence is associated with states (Schr¨odinger),operators (Heisenberg), or combination (interaction).

• Schr¨odingerpicture

ˆ state |ψS(t)⟩, operators OˆS ≠ OˆS(t) ∂ i |ψˆ (t)⟩ = Hˆ |ψˆ (t)⟩ ∂t S S has formal solution − ˆ − ˆ iH(t t0) ˆ |ψS(t)⟩ = e |ψS(t0)⟩ (52)

ˆ − Note Hˆ hermitian ⇒ time evolution operator Uˆ ≡ eiH(t t0) is uni- tary. • Interaction picture (useful for pert. thy.) ′ Hˆ = Hˆ0 + Hˆ (where Hˆ0 usually soluble)

ˆ ˆ iH0t ˆ Def. in terms of Schr. picture: |ψI(t)⟩ = e |ψS(t)⟩ iHˆ0t −iHˆ0t OˆI(t) = e OˆSe

∂ ⇒ i |ψˆ (t)⟩ = Hˆ ′(t)|ψˆ (t)⟩ ∂t I I

′ ˆ ′ − ˆ with Hˆ (t) = eiH0tHˆ e iH0t Remarks:

16 – states and ops. t-dependent in interaction picture, but time de- pendence of operators very simple, e.g. ˆ ∑ † H0 = εkakak k

∂ ˆ − ˆ i a (t) = eiH0t[a , Hˆ ]e iH0t ∂t kI k 0 = εkakI(t)

−iεkt ⇒ akI(t) = ake

– Time evolution operator determines state at time t: ˆ ˆ |ψI(t)⟩ = Uˆ(t, t0)|ψI(t0)⟩ From Schr¨odingerpicture we find

iHˆ0t −iHˆ (t−t0) −iHˆ0t0 Uˆ(t, t0) = e e e (53)

(Note ([H,ˆ Hˆ0] ≠ 0!) • Heisenberg picture ˆ state |ψH⟩ t-independent, iHtˆ −iHtˆ operators OˆH(t) = e OˆSe so operators evolve according to Heisenberg eqn. of motion ∂ i Oˆ (t) = [Oˆ (t),H] (54) ∂t H H ⋆ Note–compare three reps. at t = 0: ˆ ˆ ˆ |ψH⟩ = |ψS(0)⟩ = |ψI(0)⟩ (55)

OˆS = OˆH(0) = OˆI(0) (56)

17 1.4 Phonons

1.4.1 Review of simple harmonic oscillator quantization I will simply write down some results for the standard SHO quantization from elementary QM using “ladder operators”. We consider the Hamilto- nian p2 K H = + q2 (57) 2M 2 and extract the relevant dimensions by putting K ω2 = M   Mω 1/2 ξ = q   h¯ ∂ −i = p(¯hMω)−1/2 (58) ∂ξ so   2 hω¯  ∂  H = − + ξ2 . (59) 2 ∂ξ2 −ξ2/2 We recall soln. goes like e Hn(ξ), where Hn are Hermite polynomials, and that eigenvalues are

En =hω ¯ (n + 1/2) (60) Define ladder operators a, a† as   1 ∂ a = √ ξ +  (61) 2 ∂ξ   1 ∂ a† = √ ξ −  (62) 2 ∂ξ Ladder ops. obey commutation relations (check) [a, a†] = 1 ; [a, a] = 0 ; [a†, a†] = 0 (63)

18 eq eq xl x l+1

K K K

x i x i+1

≡ − eq Figure 4: Linear chain with spring couplings K. Dynamical variables are qi xi xi & then H may be written (check)   1 H =hω ¯ a†a +  . (64) 2 a, a† connect eigenstates of different quantum nos. n, as (a†)n |n⟩ = |0⟩, (65) (n!)1/2 where |0⟩ is state which obeys a|0⟩ = 0. Operating on |n⟩ with a† may be shown with use of commutation relations to give a†|n⟩ = (n + 1)1/2|n + 1⟩ ; a|n⟩ = n1/2|n − 1⟩ (66) so with these defs. the ladder operators for SHO are seen to be identical to boson creation and annihilation operators defined above in Sec. 1.3.2.

1.4.2 1D harmonic chain If we now consider N atoms on a linear chain, each attached to its neighbor with a “spring” of spring constant K as shown in figure. First let’s consider the problem classically. The Hamiltonian is 2 ∑ pℓ K 2 H = + (qℓ − qℓ+1) , (67) ℓ 2m 2

19 where the qℓ’s are the displacements from atomic equilibrium positions. Now Hamilton’s eqns. (or Newton’s 2nd law!) yield 2 −Mq¨j = Mω qj = K(2qj − qj−1 − qj+1). (68)

A standing sinusoidal wave qj = A cos(kaj) satisfies this equation if the eigenfrequencies have the form K ω2 = 2(1 − cos ka), (69) k M where if a is the lattice constant, k = 2π/λ. Note that for small k, ωk is 1/2 13 linear, ωk ≃ (K/M) ka. This is the classical calculation of the normal modes of oscillation on a 1D chain. To quantize the theory, let’s impose canonical commutation relations on the position and momentum of the ℓth and jth atoms:

[qℓ, pj] = ihδ¯ ℓm (70) and construct collective variables which describe the modes themselves (recall k is wave vector, ℓ is position) :

1 ∑ ikaℓ 1 ∑ −ikaℓ qℓ = 1/2 e Qk ; Qk = 1/2 e qℓ N k N ℓ 1 ∑ −ikaℓ 1 ∑ ikaℓ pℓ = 1/2 e pk ; Pk = 1/2 e pℓ, (71) N k N ℓ which to canonical commutation relations in wave vector space:

1 ∑ −ikal ik′am [Qk,Pk′] = e e [qℓ, pm] N ℓ,m ih¯ ∑ −ial(k−k′) = e = ihδ¯ k,k′. (72) N ℓ Let’s now express the Hamiltonian (67) in terms of the new variables. We have, with a little algebra and Eq. (69), ∑ 2 ∑ pℓ = PkP−k (73) ℓ k K ∑ − 2 K ∑ − ika − −ika M ∑ 2 (qℓ qℓ−1) = QkQ−k(2 e e ) = ωkQkQ−k 2 ℓ 2 k 2 k

13 Note since k = 2πn/(Na), the ωk are independent of system size

20 so 1 ∑ M ∑ 2 H = pkp−k + ωkQkQ−k . (74) 2M k 2 k Note that the energy is now expressed as the sum of kinetic + potential energy of each mode k, and there is no more explicit reference to the motion of the atomic constituents. To second quantize the system, we write down creation and annihilation operators for each mode k. Define  1/2   Mωk   i  ak = Qk + P−k (75) 2¯h Mωk  1/2   † Mω i  k   −  ak = Q−k Pk (76) 2¯h Mωk which can be shown, just as in the single SHO case, to obey commutation relations [ ] † ′ ak, ak′ = δkk (77) [a , a ′] = 0 (78) [ k k ] † † ak, ak′ = 0 (79) and the Hamiltonian expressed simply as   ∑  † 1 hω¯ k akak + (80) k 2 which we could have probably guessed if we had realized that since the normal modes don’t interact, we have simply the energy of all allowed harmonic oscillators. Note this is a quantum Hamiltonian, but the energy scale in H is the classical mode frequency ωk.

1.4.3 Debye Model Let us imagine using the Hamiltonian (80) as a starting point to calculate the specific heat of a (3D) due to phonons. We take the Debye model for the dispersion to simplify the calculation,   ck k < kD ωk =  (81) 0 k > kD

21 2 1/3 where the Debye wave vector kD = (6π n) is obtained by replacing the first Brillouin zone of the solid by a sphere of radius kD which contains N wave vectors, with N the number of in the crystal. The average value of the Hamiltonian is     ∑ † 1 ∑ 1 1 ⟨ ⟩ ⟨ ⟩    U = H = 3 hω¯ k akak + = 3 hω¯ k βhω¯ + (82) k 2 k e k − 1 2 since the average number of phonons in state k is simply the expectation value of a boson number operator ⟨ † ⟩ ≡ † akak Tr(ρakak) = b(¯hωk), (83) where b(x) = (exp(βx) − 1)−1 is the free Bose distribution function. The factors of 3 come from the 3 independent polarizations, which we consider to be degenerate here. Taking one derivative wrt temperature, the spec. heat per unit volume is14

∫ 3 ∑ ∂u ∂ hck¯ ∂ hc¯ kD k c = = 3 = 3 dk V βhck¯ 2 0 βhck¯ ∂T n ∂T k e − 1 ∂T 2π e − 1   ∂ 3(k T )4 ∫ ∞ x3 ∂ π2 (k T )4 2π2 k T 3 ≃ B = B = k  B  (84) 2 3 0 x − 3 b ∂T 2π (¯hc) | e{z 1} ∂T 10 (¯hc) 5 hc¯ π4/15

So far we have done nothing which couldn’t have been done easily by or- dinary 1st-quantized methods. I have reviewed some Solid State I material here by way of introduction to problems of interacting particles to which you have not been seriously exposed thus far in the condensed matter grad sequence. The second quantization method becomes manifestly useful for the analysis of interacting systems. I will now sketch the formulation (not the solution) of the problem in the case of the phonon-phonon interaction in the anharmonic crystal.

14 − ∂u ∂S Recall du = T dS pdV and ∂T V = T ∂T V

22 1.4.4 Anharmonicity & its consequences As you will recall from Solid State I, most thermodynamic properties of insulators, as well as scattering experiments on most materials, can be explained entirely in terms of the harmonic approximation for the ions in the crystal, i.e. by assuming a Hamiltonian of the form (80). There are some problems with the harmonic theory. First, at higher tempera- tures atoms tend to explore the anharmonic parts of the crystal potential more and more, so deviations from predictions of the equilibrium theory increase. The thermal expansion of in the harmonic approxima- tion is rigorously zero.15 Secondly, some important qualitative aspects of transport cannot be understood: for example, the harmonic theory pre- dicts infinite thermal transport of insulators! (See A&M Ch. 25 for a qualitative discussion of these failures). The obvious way to go beyond the harmonic approximation is to take into account higher-order corrections to the real-space crystal potential systematically, expanding16

1 ∑ (2) 1 ∑ (3) U = D (ℓ, m)qℓqm + D (ℓ, m, n)qℓqmqn + ..., (86) 2! ℓm 3! ℓmn where

n

(n) ∂ U D (ℓ, m, . . . n) = (87) ∂q ∂q . . . ∂q ℓ m n ui=0

15This follows from the independence of the phonon energies in harmonic approx. of the system volume. (see above) Since depends on temperature only through the volume derivative of the mode freqs. (see A & M p. 490), ( ) ( ) ∂p ∂V ∂T = ( )V = 0 (85) ∂p ∂T p ∂V T 16I have dropped indices everywhere in this discussion, so one must be careful to go back and put them in for a 2- or 3D crystal.

23 are the so-called dynamical matrices.17 Using Eqs. (71,76) we find

 1/2 1 h¯ ∑ ∑ † √   ikaℓ ikaℓ qℓ = Qke = (ak + a−k)e (88) N 2mω k k Note that the product of 3 displacements can be written

1 ∑ i(k1aℓ+k2am+k3an) qℓqmqn = 3/2 e Qk1Qk2Qk3 (89) (N) k1k2k3 so the cubic term may be written

∑ (3) H3 = V (k1k2k3)Qk1Qk2Qk3 (90) k1k2k3 with

(3) ∑ (3) V (k1k2k3) = D (ℓ, m, n) exp[i(k1ℓ + k2m + k3n)] (91) ℓmn Note now that the indices ℓ, m, n run over all unit cells of the crystal lattice. Since crystal potential itself is periodic, the momenta k1, k2, and k3 in the sum are not really independent. In fact, if we shift all the sums in (91) by a lattice vector j, we would have

∑ (3) (3) i(k1aℓ+k2am+k3an) i(k1+k2+k3)aj V (k1k2k3) = D (ℓ + j, m + j, n + j)e e ℓmn ∑ = D(3)(ℓ, m, n)ei(k1aℓ+k2am+k3an)ei(k1+k2+k3)aj (92) ℓmn where in the last step I used the fact that the crystal potential U in every lattice cell is equivalent. Now sum both sides over j and divide by N to find

(3) ∑ (3) V (k1k2k3) = D (ℓ, m, n) exp[i(k1ℓ + k2m + k3n)]∆(k1 + k2 + k3), ℓmn (93)

17Recall that the theory with only harmonic and cubic terms is actually formally unstable, since arbitrarily large displacements can lower the energy by an arbitrary amount. If the cubic terms are treated perturbatively, sensible answers normally result. It is usually better to include quartic terms as shown in figure below, however.

24 where 1 ∑ ikaj ∆(k) = e = δk,G (94) N j and G is any reciprocal lattice vector. Return now to (90). We have ascertained that V (3)(ℓ, m, n) is zero unless k1 + k2 + k3 = G, i.e. is conserved. If we expand (90), we will get products of 3 creation or annihilation operators with coefficients V (3). The values of these coefficients depend on the elastic properties of the solid, and are unimportant for us here. The momenta of the three operators must be such that momentum is conserved up to a † † reciprocal lattice vector, e.g. if we consider the term ak1a−k2a−k3 we have 18 a contribution only from k1 + k2 + k3 = G. Note this term should be thought of as corresponding to a physical process wherein a phonon with momentum k1 is destroyed and two phonons with momenta −k2 and −k3 are created. It can be drawn ”diagrammatically” `ala Feynman (as the 1st of 2 3rd-order processes in the figure below). p k k q

q p k q

k k p q' p p q' q' q q

Figure 5: Diagrams representing phonon-phonon collision processes allowed by energy and momen- tum conservation in 3rd and 4th order.

18As usual, processes with G = 0 are called normal processes, and those with finite G are called Umklapp processes.

25 Questions:

• How does energy conservation enter? What is importance of processes involving destruction or creation of 3 phonons? • If one does perturbation theory around harmonic solution, does cubic term contribute to thermal averages? • Can we calculate thermal expansion with cubic Hamiltonian?

26