Ph 127c Problem Set 2 Solutions

May 25, 2016

1. Dimerized fermi chain

a. We can Fourier transfer the Schroedinger equation to:

E∑n

e−iknψn = −∑n

e−iknJ(ψn+1 + ψn−1) +∑n

e−iknhe−iπnψn (1)

Eψk = −J(eik + e−ik)ψk + hψk+π (2)

Which is equal to:

E

(ψkψk+π

)=

(−2J cos k h

h 2J cos k

)(ψkψk+π

)(3)

Then the eigenvalues are:

Ek = ±√

4J2 cos2 k + h2 (4)

Similar to the example in lecture notes, there’s energy gap 2h at k = ±π2 .

b. From the energy spectrum we know the upper and lower band will be connected if h = 0. Toanalysis the phase transition, we’d like to see how the total energy change when varying theh. At low temperature, the lower band is fully filled.

∆Etot = −

π2∫

−π2

dkL

2π(√

4J2 cos2 k + h2 − 2J cos k) (5)

Then from the calculation in lecture notes, the final result will be:

∆Etot ∼ −h2 log h (6)

This manifests a second order transition, as the second order derivative diverges at h = 0.

c.C(t, 0;x) =

∑λ,Eλ<0

ψλ(x, 0)ψ∗λ(x, t) (7)

with ψλ(x, t) = e−iEλtψλ(x). Then

C(t, 0;x) =∑Eλ<0

eiEλt (8)

Consider the dominant term, and substitute it by τ ,

C(τ) ∼ e−hτ (9)

So the correlation time ξτ is 1h .

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d. For correlation length calculation, the process is presented in lecture notes, here

C(n, 0) =

∫ π2

0

dk

2π

−2J cos k(1− e−iπn) + h(e−iπn + 1)

−2Ekeikn (10)

The exponential decay terms can be found from the singularities of the denominator√

4J2 cos2 k + h2.This can be done by using the linear form near the singular point. In the end, we will have

C(n) ∼ e−hn2J (11)

So the correlation length is ξ = 2Jh . By comparing the correlation time and the correlation

length:ξτ ∼ ξz (12)

We find z = 1.

2. The transverse field Ising model in imaginary time

a. By expanding both sides to the second order, equivalently we need to prove:

(1− εH +1

2ε2H2) = (1 +

J

2ε∑

σzi σzi+1 +

1

2(J

2ε∑

σzi σzi+1)2) (13)

(1 + hε∑

σxi +1

2(hε∑

σxi )2) (14)

(1 +J

2ε∑

σzi σzi+1 +

1

2(J

2ε∑

σzi σzi+1)2) (15)

Comparing the LHS and RHS, the zeroth order is the same. And easily we can see the firstorder of ε is also the same. So let’s check the ε2 terms,

LHS(ε2) =1

2(∑

Jσzi σzi+1 +

∑hσxi )2 (16)

And

RHS(ε2) = 2× 1

2(J

2

∑σzi σ

zi+1)2 + (

J

2

∑σzi σ

zi+1)2 (17)

+(h∑

σxi )× (J

2

∑σzi σ

zi+1) + (

J

2

∑σzi σ

zi+1)× (h

∑σxi ) (18)

+1

2(h∑

σxi )2 (19)

SoLHS(ε2) = RHS(ε2) (20)

The equation is proved.

b.

Z =∑

{σzi,τ=±1}Li=1

1/T∏τ=0

eJ2εL∑iσzi,τσ

zi+1,τ

L∏i=1

〈σzi,τ | ehεσxi e

J2εJσzi σ

zi+1 |σzi,τ−ε〉 (21)

=∑

{σzi,τ=±1}Li=1

e

1/T∑τ=0

(J2εL∑iσzi,τσ

zi+1,τ+J

2εL∑iσzi,τ+εσ

zi+1,τ+ε)

1/T∏τ=0

L∏i=1

〈σzi,τ | ehεσxi |σzi,τ−ε〉 (22)

=∑

{σzi,τ=±1}Li=1

1/T∏τ=0

eJε

L∑iσzi,τσ

zi+1,τ

L∏i=1

〈σzi,τ | ehεσxi |σzi,τ−ε〉 (23)

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c. First expand the operator:

ehεσxi = 1 + hεσxi +

1

2h2ε2 + ... (24)

Which means:〈σz1 | ehεσ

x |σz2〉 = 〈σz1 | (cosh(εh) + sinh(εh)σx) |σz2〉 (25)

The σx operator will flip the eigenstates of σz, so from the two different cases σ1 = σ2 andσ1 6= σ2 we have two equations:

cosh(εh) = CeJ (26)

sinh(−εh) = Ce−J (27)

So we can get:

C =√

cosh(εh) sinh(−εh) (28)

J = lncosh(εh)

sinh(−εh)(29)

d. We can take the solution of part c) back into expression in part b), the partition function willbecome:

Z =∑

{σzi,τ=±1}Li=1

CNe

∑i,τ

(Jεσzi,τσzi+1,τ+Jσzi,τσ

zi,τ+ε)

(30)

This is the partition function of a 2d Ising model, with the imaginary time being the seconddimension.

3. K-T transition in superconducting films

a. First, we can expand the partition function in powers of ζ:

Z =

∫D[θ(~x)]e−

∫d2xλ

2(∇θ)2[

1+2ζ

∫d2x cos θ(~x)+

(2ζ)2

2

∫d2x1

∫d2x2 cos θ(~x1) cos θ(~x2)+...

](31)

And Z can be rewrote by Fourier transformation, with each term replaced by:∫D[θ(~x)] →

∫D[θk] (32)

∫d2x

λ

2(∇θ)2 →

Λ∫0

d2k

(2π)2

λk2

2|θk|2 (33)

cos θ(~x) → cos

Λ∫0

d2k

(2π)2θke

ik·x (34)

b. Now we’d like to integrate the momentum components Λ′ < |~k| < Λ. Let’s denote θ>k as thecomponents to be integrated out, and θ<k as the components that are left. The integral limitsshould be understood accordingly. The partition function is now:

Z =

∫D[θ<k ]e

−∫

d2k(2π)2

λk2

2|θ<k |

2∫D[θ>k ]e

−∫

d2k(2π)2

λk2

2|θ>k |

2[1 + (35)

2ζ

∫d2x

1

2(ei∫

d2k(2π)2

θ<k eik·xei∫

d2k(2π)2

θ>k eik·x

+ e−i

∫d2k

(2π)2θ<k e

ik·xe−i

∫d2k

(2π)2θ>k e

ik·x) + ...

](36)

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Let’s divide the partition function by the constant:

C =

∫D[θ>k ]e

−Λ∫

Λ′

d2k(2π)2

λk2

2|θ>k |

2

(37)

This is just a product of Gaussian integrals which does not affect the partition function. Nowlet’s figure out the integration of θ>k . The zeroth order term is unchanged. And first orderterm of θ>k :

ζ∫

[θ>k ]e−

∫d2k

(2π)2λk2

2|θ>k |

2

e±i

∫d2k

(2π)2eik·x∫

[θ>k ]e−

∫d2k

(2π)2λk2

2|θ>k |2

= ζ⟨e±i

∫d2k

(2π)2eik·x⟩

(38)

Since the system is translationally invariant, we can set x = 0, and we’ll use the property:

< e±iθ>k >= e−

12<θ2

k> = e−1

2λk2 (39)

And a more general form of this equation is indeed proved in last quarter’s problems set.

Hence, the coefficient of∫d2xe

i∫

d2k(2π)2

θ<k eik·x

is:

ζe− 1

2

∫d2k

(2π)2<θ>k

2>

= ζe− 1

2

Λ∫Λ′

d2k(2π)2

2λk2

= ζe−1

4πλln( Λ

Λ′ ) (40)

Since Λ′ = e−dlΛ, we have ln ΛΛ′ = dl. So far, we only dealt with the first order term. The

second order term will be discussed in parts e) and f).

c. Rescaling: now we must rescale dx′ = Λ′

Λ dx, the first order term is then:

2ζe−1

4πλdl(

Λ

Λ′)2

∫d2x′

eiθ<(~x) + e−iθ

<(~x)

2= 2ζ ′

∫d2x′ cos θ<(x′) (41)

With ζ ′ = ζe2dle−dl

4πλ ,dζ

dl= (2− 1

4πλ)ζ (42)

d. Now we turn to the second order term. As in the previous parts, let’s separate θ to θ> andθ<. In the expanded form, we will deal with

∫d2x1

∫d2x2, when |x1 − x2| � 1

Λ , we can dothe integration separately. And we simply get the square of the first order term of parts b)and c0, so we were reexponentiating correctly.

But for |~x1 − ~x2| = nΛ , where n is some unit vector,

ζ2

∫d2x1

∫d2x2 cos θ(x1) cos θ(x2) =

ζ2

2

∫d2x1

∫d2x2(cos(θ(x1)+θ(x2))+cos(θ(x1)−θ(x2)))

(43)For x1 ∼ x2, cos(θ(x1) + θ(x2)) ∼ cos(2θ(x1)). If we had a term like this in the action fromthe start, we would find that the RG equation is:

dα

dl= (2− K

222)α (44)

Where α is the coefficient of cos(2θ). Hence it is less relevant that the cos θ term and we canforget it. The cos(θ(x1)− θ(x2)) term:

ζ2

∫d2x1

∫d2x2 cos(θ(x1)− θ(x2)) = ζ2

∫d2xcm

∫dφ(1− 1

2(∇θ · n

Λ)2)

dl

Λ2(45)

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The integral of dφ is over all the possible angles of n, ∇θ · n = cos(φ0 + φ).

2π∫0

dφ(∇θ · n) = π(∇θ)2 · dlΛ

(46)

And for∫d2xcm, from class we know this term gives a simple multiplicative factor proportional

to the area, which we can ignore. So we finally get:

− ζ2

∫d2xcm(∇θ)2 πdl

2Λ4(47)

This term is exactly the same as the λ term, we can reexponentiate and add this to theoriginal λ:

λ′ = λ+π2

Λ4ζ2dl (48)

dλ

dl=

π

Λ4ζ2dl (49)

e.

K =1

2πλ(50)

dK

dl= − 1

2π

1

λ2

dλ

dl(51)

= −2π2ζ2K2

Λ4(52)

4. Josephson Junction

The interaction between phase slips is very similar to the interaction between vortices, which wehave studied extensively by now. The only difference is that phase slips live in one dimension(imaginary time), whereas vortices were in two dimensions. To find the transition resistance, werepeat the arguments given in class. When a pair of vortices unbind, they will be an entropycontribution to the free energy in addition to the interaction energy. The difference between thephase slips case and the vortices case is that the entropy when phase slips unbind is proportionalto the length of the system, L, while in the vortices case the entropy is proportional to the area,L2.

The free energy of a pair of phase slips is:

Funpaired = 2RQ ln(L)− 2R ln(L) (53)

where the resistance R plays the role of temperature in this problem. Obviously, the free energy ofunpaired phase slips changes sign at R = RQ, and that is when the unbinding transition happens.

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