Condensed Matter - HW 6 :: Luttinger Liquid
PHSX 545
Problem 1 Compressibility of matter can be described as a response to a scalar potential: by applying pressure at one point (changing chemical potential, or particle density at that point) we should be able to tell how this change affect particle density at some other point. This response is described by the density-density correlation function χ(r = r1 r2)= δnˆ(r )δnˆ(r ) . In Fourier space this function is (we will derive it in a few weeks) − h 1 2 i +∞ 1 n n − dp n n − χ(q)= p+q/2 − p q/2 = p+q/2 − p q/2 −V ε ε − 2π~ ε ε p p+q/2 p−q/2 p+q/2 p−q/2 X − −∞Z − where np is the occupation number of state with momentum p and energy εp. (a) Show that compressibility, as we defined it earlier, can be directly obtained from density-density correlations
1 ∂n 1 ∂n 1 κ = = = χ(q 0) , T n ∂P n2 ∂µ n2 → irrespective of the distribution function np(εp) (Fermi, Dirac, Boltzmann). 2 (b) Consider one-dimensional free fermionic gas with εp = p /2m. At T = 0 directly compute the response function ∗ ∗ χ(q) and show that it diverges at some wavevector q . Find q in terms of Fermi momentum pf and plot χ(q).
Problem 2 Using the effective Hamiltonian of the charge-spin separated Luttinger model
= [v q aˆ† aˆ + v q aˆ† aˆ ] H c c,q c,q f s,q s,q q>0 X
find the heat capacity of 1D liquid. Hint: think about what values can particle number operatorsn ˆc,s take, and follow the standard route to compute CV at low temperature. If one applies Fermi liquid theory to 1D case, what temperature dependence of CV is expected - is it the same or different? 2
Answer of exercise 1 (a) The distribution function depends on momenta through the energy, which is measured relative the chemical potential
n = n(ε µ) p p − Then we can write in small-q limit
∂np ∂np np+q/2 np−q/2 (εp+q/2 εp−q/2)= (εp+q/2 εp−q/2) − ≈ ∂εp − − ∂µ − and we can relate the susceptibility to derivative of density n = (1/V ) p np with respect to the chemical potential
1 n n − 1 ∂n ∂n P 1 ∂n 1 χ(q 0) = p+q/2 − p q/2 = p = κ = = χ(q 0) . → −V ε ε V ∂µ ∂µ ⇒ T n2 ∂µ n2 → p p+q/2 p−q/2 p X − X (b) At T = 0 the distribution function is either 0 or 1 depending on the value of the momentum:
1 , pf q q 1 , pf 2 q q 1 , pf + 2 +∞ +∞ p> dp n n − dp n n − dp 2m 4m p χ(q)= p+q/2 − p q/2 = 2 p+q/2 − p q/2 = 2 = ln > − 2π~ εp+q/2 εp−q/2 − 2π~ εp+q/2 εp−q/2 2π~ pq 2π~q p< −∞Z − Z0 − pZ< So the susceptibility is pf +q/2 4m ln , q< 2pf χ(q)= pf −q/2 2π~q ln pf +q/2 , q> 2p ( −pf +q/2 f 4m q + 2p 2 2p q + 2p 2p q/2p + 1 = ln f = f ln f = N f ln f hq q 2p hv q q 2p f q q/2p 1 f f f f − − − with logarithmic infinity at q = 2pf . When the external perturbation tries to modulate the density with 2pf wavelength, the response in density∗ is divergent. This is the nature of one-dimensional chain of particles that can be interpreted as particles not passing through each other. 3 Answer of exercise 2 In the Luttinger effective hamiltonian = [v q aˆ† aˆ + v q aˆ† aˆ ] H c c,q c,q f s,q s,q q>0 X the operatorsa ˆ are bosonic, and thus the occupation numbers can be any integern ˆ = 0, 1, 2, 3 ... . For the energy we have the ensemble averge with the Bose distribution functions: ∞ dq vcq vf q E = = [vcq nˆc(q) + vf q nˆc(q) ]= + hHi h i h i 2π~ eβvcq 1 eβvf q 1 q>0 0 X Z − − Performing dimensionalization of the integrals we write 1 1 1 ∞ xdx 1 1 1 E = T 2 + E = T 2 + ζ(2) 2π~ v v ex 1 2π~ v v c f Z0 − c f where we used some integrals from Bose part of statistical mechanics, and ζ(2) = π2/6 is Riemann zeta-function. The answer for specific heat is dE π 1 1 C = = T + dT 6~ v v c f - linear in temperature. The Fermi liquid theory approach gives the result for heat capacity using Sommerfeld expansion with density of dp states for 1D case N(ε) = 2 1 = 2m/h2ε The temperature corrections to the chemical potential and energy are 2π~ dε 2 T and it results in linear-T dependencep of specific heat. The detailed calculation is as follows: the chemical potential follows from εf µ π2 π2 π2T 2 dεN(ε)= dεN(ε)+ T 2N ′(ε ) µ = ε T 2(ln N)′ = ε + 6 f ⇒ f − 6 f f 12ε Z0 Z0 f The energy is ∞ ∞ µ(T ) π2T 2 1 E = dεN(ε)εn(ε)= Nf √εf dε√εn(ε)= Nf √εf dε√ε + 6 2√εf Z0 Z0 Z0 ! and the specific heat dE ∂µ π2T 1 π2 π C = = Nf √εf √εf + = Nf T = T dT ∂T 6 ε 3 3~v √ f f - same answer as with Bose treatment, if we take the non-interacting Luttinger-gas where charge and spin excitations propagate with same speed vc = vf .