Quantum Phase Transitions in Fermi Liquids

Quantum Phase Transitions in Fermi Liquids in d=1 and higher dimensions Mihir Khadilkar Kyungmin Lee Shivam Ghosh PHYS 7653 December 2, 2010

Outline 1 Introduction 2 d = 1: Mean field vs RG Model 3 Higher Dimensions d = 2 Circular Fermi Surface Other Fermi Surfaces 4 Conclusions

Revision Last time Mihir talked about what is a Fermi liquid Replace interacting electron system by weakly interacting quasiparticles Renormalize interactions between quasiparticles Kyungmin did the calculations for complex scalar field We will now talk about a real system

What are spinless fermions? Fermions, but no spin-spin interaction in the Hamiltonian Exclusion principle still applies Can think of it as electrons in a very high external field The instabilities of the Fermi liquid are due to the features of the Fermi surface Actual instabilities are different but crucial feature will turn out to be the surface itself

Model d = 1: Mean field vs RG

Model The Model H = H 0 + H I = 1 2 X j X + U 0 (n j with n j = y (j) (j) j y (j + 1) (j) + y (j) (j + 1) 1 2 )(n j+1 and f y (j); (m)g = mj 1 2 ) We will work at half filling For spinless fermions, this means 1 2 particle per site

Model In the limit U0 = 0 H = H 0 Taking Fourier transform and solving, Z dk y H 0 = (K) (K)E(K) 2 E(K) = cos K Fermi surface consists of two points Π Π 2 Π 2 Π K F = =2

Model In the limit U0! 1 X H = H I = U 0 (n j j 1 2 )(n j+1 1 2 ) Only nearest neighbour repulsion Ground state when there are no nearest neighbours Alternate sites occupied Charge density wave ordering (CDW): hn j i = 1 2 + ( 1)j 1 2

Model 1

Model Mean Field Result, n j = hn j i How does depend upon U 0? = U 0 e 2U 0 This is non-zero for all non-zero U 0 This suggests that we will get CDW at arbitrarily small U 0 Similar result for arbitrarily small negative U 0 Cooper pair formation leads to superconductivity

Model THIS IS WRONG

Model The Answer The system forms a Luttinger liquid Some of the differences from Fermi liquid: Excitations are not quasiparticles, but collective excitations of charge and spin (spinons and holons) Non-universal critical exponents depending on Luttinger liquid exponent Depends upon interaction strength This can be thought of as a system having a line of fixed points No flow along this line

Formulation of problem We will consider excitations near the Fermi surface Different from complex scalar field because there are two points Π Π 2 Π 2 Π Inter-branch excitations can occur Umklapp processes also possible

Formulation of problem X Z H 0 = i=l;r dk 2 y i (k) i(k)k with k = jkj K F Note that E(K) has been linearized about K F. We are looking at excitations within a cutoff

Road map for the analysis Write down partition function Z 0 and free field action S 0 Find RG transformation which lowers cutoff keeping S 0 constant Introduce perturbations to this fixed point Try to find flow equations

Partition function and free field action Z 0 = Z Y Y d i(!; k)d i (!; k)e S0 i=l;r jkj< X Z Z dk 1 d! S 0 = i (!; k)(i! k) i (!; k) 2 i=l;r 1 2 Compare with the complex scalar action: Z S 2 = d d x (R 0 j(x)j 2 + R 2 jrj 2 ) which when Fourier transformed, gives: S 2 = Z jkj< d d k (2) d (r 0 + r 2 k 2 ) (k)(k)

Rescaling and renormalization We want to integrate out all the "fast" modes ( =s jkj ) Since S 0 is quadratic, slow and fast modes separate Coarse-graining reduces! =s To get back original action, k 0 = sk! 0 = s! 0 i (k0 ;! 0 ) = s 3=2 i(k;!) We can now consider perturbations to this fixed point action

Quadratic perturbations S 2 = X i=l;r Z Z dk 1 2 1 d! 2 (!; k) i(!; k) i (!; k) Again, slow and fast modes separate out Coupling rescales as 0 (! 0 ; k 0 ) = s(!; k) Taylor expanding, we see that only the constant term 0 is relevant Readjustment of the chemical potential

Quartic perturbations S 4 = 1 4 Z K! where (i) = (K i ;! i ) Z = K! 2 Y 4 4 j=1 Z dk j 2 (4) (3) (2) (1)u(4; 3; 2; 1) Z 1 1 d! j 2 3 5 [2 (K 1 + K 2 K 3 K 4 )2(! 1 +! 2! 3! 4 )]

Order-u level Analysis At the order-u level, only tree-level and tadpole diagrams survive Tree-level Simple change in cutoff, leading to renormalized coupling u 0 (k 0 ;! 0 ) = u(k;!) Thus, we get that only the constant term u 0 is marginal; rest irrelevant Tadpole As before the tadpole graph graph renormalizes the quadratic coupling d = u0 d` 2 TADPOLE TREE LEVEL

Order-u 2 Analysis Feynman Diagrams with one Loop

3 4 3 4 1 2 ZS 3 4 1 2 ZS 1 2 BCS

Order-u 2 Analysis We must consider the three graphs, ZS, ZS and BCS On doing the integrals, we see that du = 0 d` ZS du ZS0 = u2 d` du d` = BCS Thus, we get the flow equation du d` = 0 2 u 2 2

So... du d` = 0 ; d d` = u 0 2 Interaction strength does not flow Quadratic coupling gives change in chemical potential Kept constant to keep number density of particles constant Luttinger liquid

Higher Dimensions

d = 2 Circular Fermi Surface d = 2 Now you have a Fermi surface instead of a discrete number of points We must integrate over all of these directions In 1-d, we had u LRLR Now we have u[ 4 ; 3 ; 2 ; 1 ] in 2-d But only two angles are independent Two distinct cases: u[ 2 ; 1 ; 2 ; 1 ] = F ( 1 ; 2 ) and u[ 3 ; 3 ; 1 ; 1 ] = V ( 1 ; 3 )

d = 2 Circular Fermi Surface Flows Convenient to examine flows of V l = R 2 0 dv l dt = cvl 2, c>0 If even one coupling is negative, we get flow d 2 eil v() Cooper pair formation leading to BCS fixed point

Other Fermi Surfaces Other Fermi Surfaces For 3D spherical Fermi surface, same result For non-circular Fermi surfaces, result depends on whether there is time-reversal invariance Nested Fermi surfaces are special New coupling constant W flows to CDW state

Other Fermi Surfaces Π Π 0 K N Π, Π 0 Π Π

Conclusions For system of 1D spinless Fermions Mean field analysis gives wrong result for small U 0 Luttinger liquid scale invariant system formed For higher dimensions, we see flows to BCS instabilities Nested Fermi surfaces produce charge density wave Now Shivam will talk about spin-spin interactions