Week 6 - Quantum phase transitions (in electronic chains)

I. QUNTUM PHASE TRANSITIONS AND BOSONIZATION

So far, we were discussing phase transition in finite temperature. Quantum systems, however, could be describedusing the same techniques, it seems as statistical mechanical systems, except, they need the extra imaginary time.

It is not surprising then, that quantum systems can undergo zero temperature transitions.Let’s see a specific example: The single particle in a double well. Quantum mechanically, at zero temperature, this

is the same as what? A path integral with the the coordinate of the particle as a function of time summed on. Thisis, then, just like a rope that is in a double well potential, waiting to decide where to go along its length. Indeed, thepartition function looks like the sum of all configurations of domain walls of a 1d Ising model [NEED FIGURE].

So the upshot is that quantum systems in d dimensions and zero temperature, are just like classical systems atd+ 1 dimensions. Therefore the lower critical dimension for symmetry breaking quantum phase transitions are 0 fordiscrete symmetry, and 1 for continuous. Also, for 1d systems, we expect K-T transitions if topological defects areinvolved.

One thing we could do is talk about the Transverse field Ising model, which is the mother of all quantum phasetransitions:

H = −J∑i

σzi σzi+1 − h

∑i

σxi (1)

Actually, the 1d TFIM is also the 2d - Ising model, and the key to its exact solution.Instead of talking about this work horse of quantum magnetism, I wanted to talk directly about electrons. They

are in any case responsible for all quantum phenomena in materials. Using the electronic example, we’ll see what itmeans to have a QPT, and how it relates tot he universality of classical phase transitions. We will then use it todevelop the bosonization framework.

A. Dimerization in electron chains

Let’s think of a simple electron chain, with nearest neighbor hopping. Without interactions this is:

Eψn = −J(ψn+1 + ψn−1) (2)

[ It is rather convenient to rewrite this in terms of creation and annihilation operators... definition√L in the def.

transform.] We know how to solve this quantum problem. Brea it into k-modes. We start by writing:

ψk =1√L

∑n

ψne−ikn, ψn =

1√L

∑k

ψkeikn (3)

with k ∈ {−π,−π(L− 2)/L,−π(L− 4)/L, . . . , π(L− 2)/L, π},i.e., going from π to π with 2π/L increments. Puttingthis in, we get:

Ekψk = −J(eik + e−ik)ψk → Ek = −2J cos k. (4)

This gives the standard cosine band. And if the chemical potential µ is 0, half of them are filled, and half are empty.If the system is coupled to phonons, it may want to break the symmetry, and distort. In this case we will have an

additional term:

Eψn = −J(ψn+1 + ψn−1)− δ · (−1)n(ψn+1 − ψn−1) (5)

What would be the spectrum? Let’s fourier transform this by summing over e−ikn and sum over n:

E∑n

e−iknψn = −∑n

e−iknJ(ψn+1 + ψn−1)−∑n

e−iknδ · e−πn(ψn+1 − ψn−1) (6)

by writing: e−iknψn±1 = e−ik(n±1)ψn±1e∓ik, and realize that eikneiπn = ei(k+π)n, we reach the conclusion that:

Eψk = −Jψk(eik + e−ik)− δ(ψk+π)(ei(π+k) − e−i(k+π)) (7)

2

k

E

FIG. 1: Half filled cosine band

So this just becomes a 2× 2 problem:

E

(ψkψk+π

)=

(−2J cos k −2iδ sin k

−2iδ sin(k + π) −2J cos(k + π)

)(ψkψk+π

)=

(−2J cos k −2iδ sin k2iδ sin(k) 2J cos(k)

)(ψkψk+π

)(8)

This is like H = −2J cos kσz + 2δ sin kσz. We know the eigenvalues:

Ek = ±2√J2 cos2 k + δ2 sin2 k (9)

We see that the spectrum becomes:

Ek = ±2√J2 cos2 k + δ2 sin2 k (10)

There is now a gap, and it is found in k = ±π/2:

∆ = 2δ (11)

k

E

FIG. 2: New band structure.

The opening of a gap is a quantum phase transition. You can oven consider a transition as δ goes from positive tonegative. Let us now consider the universal properties of this transition.

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First, where do we expect a singularity? What plays the role of the free energy when the temperature is zero?

F = −T lnZ (12)

So in the limit of T → 0, it is really the ground state energy:

= −T ln e−E0/T = E0 (13)

And now we can ask, what is the singularity with respect to the tuning parameters?Let’s calculate the energy. The filled states are all the negative energy states. So the energy is:

Etot = −2L

π/2∫−π/2

dk

2π

√J2 cos2 k + δ2 sin2 k (14)

Where we changed the sum to an integral. This is a hard integral to do. Let’s ask not what the ground state energyis, but rather, how does it change when let δ vary:

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L∆Etot = −2

π/2∫−π/2

dk

2π(√J2 cos2 k + δ2 sin2 k − J cos k) (15)

We’re looking for contributions near k = π/2. So let’s expand this in terms of p = k − π/2:

∆EtotL

≈ −2

pmax∫0

dp

2π(√J2p2 + δ2 − Jp) (16)

Instinctively ,we want to expand this in small δ. That won’t work, since we need p values of order of pmax � δJ . We

can do the integral with a substitution:

δ sinh v = Jp (17)

and then:

1

L∆Etot ≈ −2

δ2

J

vmax∫0

dv cosh2 v + Jp2m (18)

with vmax = arcsinhJpmaxδ . But cosh2 v = 12 cosh 2v + 1 allows us a quick evaluation:

1

L∆Etot ≈ −2

δ2

J

[1

4sinh 2v + v

]vmax0

+ Jp2m (19)

using 12 sinh 2v ≈ sinh2 v for large v, we see that the sinh just gives the δ = 0 contribution, and what remains is the

v term. For it we can approximate:

vmax ≈ ln2Jpmax

δ(20)

And the end result is:

1

L∆Etot ≈ −2

δ2

Jln

2Jpmaxδ

(21)

pmax could be set to π/2 or so with impunity. The end result is:

∆Etot = −Cδ2 log δ (22)

Think of δ lika a thermodynamic variable - it is the tuning parameter for a transition. Derivatives with respect to thisparameter are equivalent to the derivatives of the free energy. The singularity of E0 - the total ground state energywith respect to changes in δ determine the order of the transition.

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If ∂E0

∂δ jumps, it is a first order QPT. This is the case when there is a level crossing. Meaning it so happen that asa function of the parameter δ some state crossed the ground state. This is trivial. For instance you could think of amagnetic field that just crosses 0:

H = bσz. (23)

As a function of b, we have:

E0 = −|b| (24)

b

E

FIG. 3: Level crossing as a function of parameter b. The ground state switches between the two spins at b = 0

In our case, the situation is more subtle. It is the second derivative that diverges at the transition:

∂2Etot∂δ2

∼ ln δ (25)

This makes this transition a 2nd order transition. The same kind that has nice scaling properties, and is amenableto an RG analysis.

B. Correlations in the dimerized model

If indeed a second order transition, then we need to also consider its correlations. The correlation function is verysimple for a non-interacting system. It is simply:

C(x1;x2) =∑

λ;Eλ<0

ψλ(x1)ψ∗λ(x2) (26)

For the critical point this is easy:

Cδ=0(x1;x2) =

π/2∫−π/2

dk

2πeik(x2−x1) =

1

π

sin(π/2(x2 − x1))

x2 − x1(27)

These are actually the Friedel oscillations you may have heard about for metals (well - once we square this, and itbecomes the density-density correlation).

But what is it for the dimerized model? This is a good exercise for using resolvents. The green’s function for thesingle particle schrodinger equation with energy E is:

(E − H) |ψ〉 = δ(x) (28)

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For this we think of the Hamiltonian as a matrix:

H =∑n

−J(δn, n+1 + δn, n−1)− (−1)nδ(δn, n+1 − δn, n−1) (29)

and try to invert it:

GE(x, 0) =[(E − H)−1

]x,0

=∑λ

ψλ(x)ψλ(0)

E − Eλ(30)

If we want the correlations between 0 and x, though, we need to sum over the denominator only for energies that arenegative. No problem:

Cδ(x, 0) =

∫C

dE

2πi

ψλ(x)ψλ(0)

E − Eλ(31)

with a contour C that wraps around the entire E < 0 half plane of the complex extension of E. [DRAW] This willgive us the needed sum.

Why is this helpful? Because we can choose the basis. It doesn’t have to be the eigenbasis to play this trick. InFourier space we can write:

H(n, n′) =1

L

π∑k=0

(e−ikn

′e−i(k−π)n

)( −2J cos k −2iδ sin k2iδ sin(k) 2J cos(k)

)(eikn

ei(k−π)n

)(32)

Since the hamiltonian Hk is simply the hamiltonian that connects the k and k − π portions of the dispersion. Andthe sum is on positive k not to double count the −π < k < 0 part. So similarly:

G(n, n′) =

[1

E − H

]n,n′

=1

L

π∑k=0

(e−ikn

′e−i(k−π)n

) 1

(E − Hk)

(eikn

ei(k−π)n

)(33)

Now for some Pauli-matrix tricks. We can invert Hk really easily. Check this out:

1

E − Hk=

E + Hk(E − Hk)(E + Hk)

=E + Hk

E2 + (4J2 cos2 k + 4δ2 sin2 k)(34)

how come? Hk is a vector multiplying the Pauli matrices. Since the Pauli Matrices anti-commute, the square of Hkis proportional to the identity. In fact, we can write:

=E + Hk

(E − Ek)(E + Ek)(35)

with Ek =√

4J2 cos2 k + 4δ2 sin2 k.From here on, our job is easy. The energy integral lives us with only the negative energies:

Cδ(n, 0) =

∫C

dE

2πiGE(x, 0) =

π∫0

dk

2π(1, 1)

−Ek − 2J cos kσz + 2δ sin kσy

−2Ek

(eikn

ei(k−π)n

)(36)

Where I allowed myself to turn the sum over k to an integral, and setting one of the coordinates to 0. Looking at thenumerator, there is the Ek term which becomes:

π∫0

dk

2π(1, 1)

1

2

(eikn

ei(k−π)n

)= δn,0 (37)

Not useful since tells us nothing about correlations with n 6= 0. The rest is:

=

π∫−π

dk

2π

−2J cos k + 2iδeiπn sin k√4J2 cos2 k + 4δ2 sin2 k

eikn (38)

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I’ll let you finish the work at home on this version, for finding the correlation length. For now, let’s take a continuumapproximation, and notice that whatever happens, it will be decided by the singularities in the denominator.

The denominator is closest to zero at |k| = π/2. Linearizing near there, k = p+ π/2 we have:

4J2 cos2 k + 4δ2 sin2 k ≈ 4J2p2 + 4δ2 (39)

Indeed, a gap of 4δ arises. But if we allow complex p, we easily get the denominator to vanish if:

p = iδ/J (40)

This will make the correlation function fall of as:

Cδ(n) ∼ e−δn/J (41)

with a correlation length:

ξ ∼ J

δ(42)

and a

ν = 1. (43)

C. Temporal correlations

But with quantum phase transitions, there is another dimension: time! The correlations along imaginary timecorrespond to another important diagnostic of quantum phase transitions. For our example, we can easily find thesecorrelations:

Cδ(t, 0;x) =∑

λ;Eλ<0

ψλ(x, 0)ψ∗λ(x, t) (44)

where now we need to think of the time-dependent Schrodinger equation, and:

ψλ(x, t) = e−iEλtψλ(x) (45)

It is easy to see that this reduces to:

Cδ(t, 0;x) =∑Eλ<0

eiEλt (46)

and with it→ τ , we see that :

Cδ(τ) ∼ e−2δτ (47)

since the dominant decay will be obtain from the place where the gap is lowest. And we find that:

ξτ =1

2δt (48)

so that the only difference with with the spatial correlation is the velocity: 2J .Indeed, near k = π/2 we can linearize the hamiltonian and obtain a time dependent SE:

i∂ |ψ〉∂t≈ −2J(

1

i

∂

∂x− π/2σz)σz |ψ〉+ 2δσy |ψ〉 (49)

and we see that x and t must scale the same way.This is not always the case. For this reason, for QPT there is another exponent defined:

ξτ ∼ ξz (50)

This is the space-time exponent. In our example it is z = 1, but that is not always the case.

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D. QPT at a finite temperature

But what happens if we exceed with ξτ the amount of imaginary time we have,

ξτ > 1/T?

Then we are in trouble: there is no imaginary time available for the relevant fluctuating trajectories, and the systembecomes essentially classical. The extra time dimension is out of the game, since from an RG perspective, it is afinite dimension. The system then becomes d dimensional. If the d dimensional system does not exhibit a transition(d ≤ dlcd), then the critical point only exists at zero temperature. Finite temperatures, however, will show a crossover between the two phases.

The range of the crossover will be determined exactly by the question of where ξtau > β = 1/T . Beyond that, thequantum fluctuations subside, and the two phases emerge. So:

TQF ∼ |δ|νz. (51)

This is shown in fig. 4.

δ

Td<dlcd

Quantum Critical Region

Phase I Phase II

|δ| ~Tνz

δ

Td≥dlcd

Phase I Phase II

|δ| ~Tcνz

FIG. 4: Two possibilities of finite temperature behavior around a phase transition. (a) There is no finite temperature transition,and there is a quantum fluctuation region at finite T, extending between TQF ∼ |δ|νz. (b) A finite T transition also exists, andit coalesces to the quantum transition at T = 0.