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Page 1: 1 Fermions in one dimensioneduardo.physics.illinois.edu/phys582/582-2019-5-solutions.pdf · 2g( x) (x) (x) 1. To simplify matters even further we consider the case of a heavy Carbon

1 Fermions in one dimension

We will be using the effective continuum Hamiltonian:

H =∑σ=±

∫dx ψ†σ(x)

(− ivf

∂x

)σ3ψσ(x) +

∫dx

[Π2(x)

8Ma2o

+1

2∆2(x)

]+∑σ=±

∫dx√

2g∆(x)ψσ(x)ψσ(x)

1. To simplify matters even further we consider the case of a heavy Carbon atom: M →∞. When we do this, wecan drop the kinetic term, which amounts to disregarding the quantum fluctuations. Therefore, we can workwith a constant classical field, so we replace ∆(x) with ∆o. Our Hamiltonian reduces to:

H =∑σ=±

∫dx ψ†σ(x)

(− ivf

∂x

)σ3ψσ(x) +

∫dx

1

2∆2o +

∑σ=±

∫dx√

2g∆oψσ(x)ψσ(x) (1)

=∑σ=±

∫dx ψ†σ(x)

(− ivf

∂x

)σ3ψσ(x) +

∫dx

1

2∆2o +

∑σ=±

∫dx ψ†σ(x)

(√2g∆o

)σ2ψσ(x) (2)

Let’s group some terms together and write this out in two separate ways:

H =

∫dx

(∑σ=±

iψ†σ(x)

[−vf ∂

∂x −√

2g∆o√2g∆o vf

∂∂x

]ψσ(x) +

1

2∆2o

)(3)

=∑σ=±

∫dxψσ

(− iγ1∂x +

√2g∆o

)ψσ +

∫dx

1

2∆2o (4)

Notice that the second term looks like a Dirac Hamiltonian with a mass term: m =√

2g∆o. We could solvethis as was done in lecture, but we will opt for a straight forward Fourier expansion instead,

ψσ(x) =

∫dp

m

ω(p)ψ±σ (p)e±ipx (5)

Here I used superscripts for the modes, so that we do not confuse them with the ”spin” subscripts. ω(p) willturn out to be the absolute value of our eigen energies. These factors are chosen to make the integrationmeasure Lorentz invariant.

In order for the anticommutation relations to be preserved, we need:

{ψσα(p), ψ†σ′α′(p

′)}

= 2πω(p)

mδσσ′δαα′δ(p− p′)

With all other anti-commutation relations zero.

Throwing in our Fourier transforms, and performing the necessary manipulations, we see the following matrixequation is what concerns us:

(ψ±σ (p)

)† [ ±pvf −i√

2g∆o

i√

2g∆o ∓pvf

]ψ±σ (p) (6)

In either case the egienvlaue equation is the same:

Det

[±pvf − E −i

√2g∆o

i√

2g∆o ∓pvf − E

]= −

[(pvf + E)(pvf − E) + (

√2g∆o)

2]

= −[(pvf )2 − E2 + (

√2g∆o)

2]

= 0 (7)

Which means:

E(p) = ±√

(pvf )2 + 2g∆2o := ±ω(p)

1

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Pushing the analogy between the above relation and the Pythagorean identity further we can define:

1 =

√(pvf/ω(p))2 + (

√2g∆o/ω(p))2 (8)

:=

√sin2(θ) + cos2(θ) (9)

Let’s expand the field about the positive and negative eigenvalues (energies):

ψσ =

∫dp

m

ω(p)

(aσ(p)uσ(p)e−ipx + bσ(p)vσ(p)eipx

)I choose to set:

{aσ(p), a†σ′(p′)} = 2π

ω(p)

mδσσ′δ(p− p′) (10)

{bσ(p), b†σ′(p′)} = 2π

ω(p)

mδσσ′δ(p− p′) (11)

etc. The factor of ω/m in the mode expansion as well as the above anti-commutation relations will set ournormalization of our eigenvectors. Indeed:{ψσ(x), ψ†σ′(x

′)}

=

∫dpdq

(2π)2

m2

ω(p)ω(q)

({aσ(p), a†σ′(q)

}u(p)u†(q)e−i(px−qx

′) +{bσ(p), b†σ′(q)

}v(p)v†(q)ei(px−qx

′))

(12)

= δσσ′

∫dp

m

ω(p)

(u(p)u†(p)e−ip(x−x

′) + v(p)v†(p)eip(x−x′))

(13)

= δσσ′δ(x− x′) (14)

⇒ u(p)u†(p) = v(p)v†(p) =ω(p)

2m(15)

We will remember this for later.

We solve the following matrix equation:

ω(p)u(p) =

[pvf −i

√2g∆o

i√

2g∆o −pvf

]u(p) (16)

−ω(p)v(p) =

[pvf −i

√2g∆o

i√

2g∆o −pvf

]v(p) (17)

Re-expressing in terms of sin and cos:

u(p) =

[sin(θ) −icos(θ)icos(θ) −sin(θ)

]u(p) (18)

−v(p) =

[sin(θ) −icos(θ)icos(θ) −sin(θ)

]v(p) (19)

This suggests (do not confuse a±, b± with our creation and annihilation operators!):

±[a±b±

]=

[sin(θ) −icos(θ)icos(θ) −sin(θ)

] [a±b±

](20)

⇒ ±a± = a±sin(θ)− ib±cos(θ) (21)

±b± = ia±cos(θ)− b±sin(θ) (22)

⇒ b± = a±icos(θ)

sin(θ)± 1(23)

˙. .

[a±b±

]= c±

[sin(θ)± 1icos(θ)

](24)

Here c± is a normalization constant, which can be determined by recalling the normalization we want:

2

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ω(p)

2m=

1

2cos(θ)= |c±|2

[(sin(θ)± 1)2 + cos2(θ)

](25)

= 2|c±|2(1± sin(θ)) (26)

⇒ |c±| =1

2

1√cos(θ)

(1± sin(θ)

) (27)

=1

2

ω(p)√m(ω(p)± pvf )

(28)

Which means:

u(p) =1

2√m(ω(p) + pvf )

[pvf + ω(p)

im

]and

v(p) =1

2√m(ω(p)− pvf )

[pvf − ω(p)

im

]Re-expressing our Hamiltonian in terms of the mode expansion we arrive at:

H =∑σ=±

∫dp

m

ω(p)ω(p)

[a†σaσ − b†σ bσ

]+L

2∆2o (29)

Notice I replaced the integral over x with the length or our polymer chain L = Nao. We can now find the groundstate with the usual prescription for a fermionic system: we define hole creation and annihilation operators:

dσ = b†σ, d†σ = bσ

So we have:b†σ bσ = dσd

†σ

using

{dσ(p), d†σ′(p′)} = 2π

ω(p)

mδσσ′δ(p− p′)

We normal order our Hamiltonian:

H =∑σ=±

∫dp

m

ω(p)ω(p)

[a†σ(p)aσ(p) + d†σ(p)dσ(p)− 2π

ω(p)

mδσσδ(p− p)

]+

1

2

∫dx∆2

o(x) (30)

We now define our ground state in the usual manner:

aσ(p)|gnd〉 = dσ(p)|gnd〉 = 0

Thus, for the ground state we have:

Egnd = −∑σ=±

∫ Λ

−Λ

dpδσσδ(p− p)√

2g∆2o + (pvf )2 +

L

2∆2o (31)

= −Lπ

∫ vfΛ

−vfΛ

dp

vf

√2g∆2

o + p 2 +L

2∆2o (32)

Note: the Dirac delta function in momentum space gives us the factor of L/2π, and we summed over sigma. Λis our momentum cutoff.

3

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We can finally find the distortion of the lattice in the ground state by minimizing the above equation in respectto the classical coordinate ∆o:

1

L

d(Egnd)

)d∆o

= − 1

π

∫ vfΛ

−vfΛ

dp

vf

2g∆o√2g∆2

o + p 2+ ∆o = 0 (33)

⇒ πvf2g

=

∫ vfΛ

−vfΛ

dp1√

2g∆2o + p 2

(34)

Of course ∆o = 0 satisfies this equation, but let’s find a non-trivial solution. The first integral is easy if werealize: ∫ vfΛ

−vfΛ

dp1√

2g∆2o + p 2

=

∫ vfΛ

−vfΛ

dp√2g∆o

1√1−

(i p√

2g∆o

)2 (35)

=1

isin−1

(i

p√2g∆o

)∣∣∣∣∣vfΛ

−vfΛ

(36)

= sinh−1( p√

2g∆o

)∣∣∣∣∣vfΛ

−vfΛ

(37)

= 2sinh−1( vfΛ√

2g∆o

)(38)

Plugging in we can solve for the ∆o which minimizes 1 our ground state energy:

πvf2g

= 2sinh−1( vfΛ√

2g∆o

)(39)

⇒ ∆o =

[√2g

Λvfsinh

(πvf4g

)]−1

(40)

∼√

2

gΛvfe

−πvf4g (41)

This is the spontaneous mass generation which was referred to as dimerization in the original SSH model!

We could plug ∆o back into the ground state energy to find a more explicitly expression, but the integral isnasty, so we only note this in passing.

2. We build our single particle states out of the ground state defined above in the usual manner:

a†σ(p)|gnd〉, b†σ(p)|gnd〉

Recall:

H =∑σ′=±

∫dp′

2πE(p′)

[a†σ′(p

′)aσ′(p′) + d†σ′(p

′)dσ′(p′)]

+ Egnd (42)

= : H : +Egnd (43)

Since:

a†σ(p′)aσ′(p′)

[a†σ′(p)|gnd〉

]= a†σ′(p

′)[a†σ(p)aσ′(p

′) + δσσ′δ(p− p′)]|gnd〉

1We do need to take a second derivative of our expression to show that this ∆o is indeed lower in energy than the trivial solution, butthis homework assignment is long enough

4

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We get:: H : a†σ(p)|gnd〉 = E(p)a†σ(p)|gnd〉: H : b†σ(p)|gnd〉 = E(p)b†σ(p)|gnd〉

This tells us that there is a 4-fold degeneracy of the single particle states corresponding to the particles andantiparticles as well as the spin polarization.

3. Let’s return to our original continuum Hamiltonian without any approximations. We investigate this modelunder the combination of the discrete transformations:

ψ(x)→ γ5ψ(x) = σ3ψ(x)

[1 00 −1

] [RL

]=

[R−L

](44)

∆(x)→ −∆(x) (45)

This obviously leaves:Π2(x),∆2(x)

invariant. Recall:

H =

∫dp

∑σ=±

[R† L†

] [ pvf −i√

2g∆(x)i√

2g∆(x) −pvf

] [RL

]+

∫dx

[Π2(x)

8Ma2o

+1

2∆2(x)

](46)

Thus, our Hamiltonian transforms as:

H →∫

dp

∑σ=±

[R† −L†

] [ pvf i√

2g∆(x)−i√

2g∆(x) −pvf

] [R−L

]+

∫dx

[Π2(x)

8Ma2o

+1

2∆2(x)

](47)

=

∫dp

∑σ=±

[R† −L†

]σ3

[pvf −i

√2g∆(x)

i√

2g∆(x) −pvf

]σ3

[R−L

]+

∫dx

[Π2(x)

8Ma2o

+1

2∆2(x)

](48)

= H (49)

Our Hamiltonian is invariant under the combination of these transformations! Notice the following term hasan odd parity under this combined transformation:

ψψ(x) = ψ†(x)σ2ψ(x)→ ψ†(x)σ3σ2σ3ψ(x) = −ψ†(x)σ2ψ(x) = −ψ(x)ψ(x)

We could easily deduce this from our Hamiltonian being invariant since√

2∆ψψ(x) is present there. Thus, theoperator ψψ(x) is odd under this discrete symmetry, and so it is a good candidate for an order parameter.

How can we interpret these transformations when viewing the original lattice model? The best way to find outis to just do it. First, recall pF = π

2ao, and we set ao = 1. So, starting with the fermion operator we see:

cσ(n) = eipFnRσ(n) + e−ipFnLσ(n) (50)

→ eipFnRσ(n)− e−ipFnLσ(n) (51)

= eiπ/2e−iπ/2eipFnRσ(n) + e−iπe−ipFnLσ(n) (52)

= −ieipF (n+1)Rσ(n)− ie−ipF (n+1)Lσ(n) (53)

The factor of −i is no problem since these operators respect a global U(1) symmetry. The phase factors of thenth site becomes that of the (n+ 1)st site. We interpret this as a shift in the dimerized structure.

A similar result holds for the phonons:

x(n) = δ(n) + e2ipFn∆+(n) + e−2ipFn∆−(n) (54)

→ δ(n)− e2ipFn∆+(n)− e−2ipFn∆−(n) (55)

= δ(n) + eiπe2ipFn∆+(n) + e−iπe−2ipFn∆−(n) (56)

= δ(n) + e2ipF (n+1)∆+(n) + e−2ipF (n+1)∆−(n) (57)

5

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We effectively shift even sites to odd sites, and odd sites to even sites under the combination of this globaldiscrete symmetry. Our original lattice model had a asymmetrical hopping parameter corresponding to shortand long bond lengths, which results in an accumulation of electrons on even or odd sites.

Since the above transformation is a symmetry of our Hamiltonian, we see there is an ambiguity of where ourelectrons reside due to the fact this above symmetry shifts our discrete lattice sites! The two different stateshave a relative polarization however (classified by their Berry phase), and this ambiguity gives rise to a chargepump. Thus, polyacetylene is a sort of conductive polymer!

4. Let’s now find the expectation value of the order parameter of question 1.3 in the ground state. We simplysubstitute in our mode expansion in for our fermion fields:∑

σ

〈gnd|ψσ(x)ψσ(x)|gnd〉 =∑σ

∫dpdq

(2π)2

m

E(q)

m

E(p)v(q)v(p)〈gnd|dσ(q)d†σ(p)|gnd〉ei(q−p)·x (58)

=∑σ

∫dp

m

E(p)v(p)v(p) (59)

= −∫dp

π

√2g∆o√

(vF p)2 + 2g∆2o

(60)

= −∫dp

π

1√( vF p√

2g∆o)2 + 1

(61)

= −2

√2g∆o

vFsinh−1

(vFΛ√2g∆o

)(62)

We see the expectation value is only nonzero if ∆o is non-zero! This order parameter is thus related to aspontaneous generation of mass m =

√2g∆o, or Dimerization. This is observed as a distortion of our lattice:

the so called Peierls instability.

2 Grassmann Stuff

1. Let a and a∗ be a pair of Grassmann variables and consider: g(a∗) = go + g1a∗, f(a) = fo + f1a. Furthermore,

define:

〈f |g〉 =

∫da∗da e−a

∗af(a∗)∗g(a∗)

Recall: a2 = (a∗)2 = 0; thus, e−a∗a = 1− a∗a. Now let’s expand the product of the two functions:

f(a∗)∗g(a∗) =[fo + f1a

∗]∗[go + g1a∗] (63)

=[fo + f1a

][go + g1a

∗] (64)

= fogo + f1goa+ fog1a∗ + f1g1aa

∗ (65)

From here we can see (I set a2 = (a∗)2 = 0 without writing our explicitly) :

e−a∗af(a∗)∗g(a∗) =

[1− a∗a

][fogo + f1goa+ fog1a

∗ + f1g1aa∗] (66)

=[1− a∗a

]fogo +

[f1goa+ fog1a

∗ + f1g1aa∗] (67)

Now we recall (analogous rules exist for a∗): ∫da κa = κ

and, ∫da κ∂aa =

∫da κ = 0

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It follows, after integration, that only terms which survive are those with products aa∗.

Finally, we note that we need to reorder the term −a∗a = aa∗, which uses the anticommutation relations ofGrassmann variables, since the a integral needs to be done first. The resulting expression is:

〈f |g〉 =

∫da∗da

[fogo + f1g1

]aa∗ =

∫da∗

[fogo + f1g1

]a∗ = fogo + f1g1

2. Now let’s consider the matrix equation:

[A11 A12

A21 A22

] [fof1

]=

[gog1

](68)

This vector space is spanned by:

~b1 =

[10

], ~b2 =

[01

](69)

There is a clear isomorphism staring us in the face: ~b1 → 1, ~b2 → a∗.

Then:

~f =

[fof1

]→ f(a∗) = fo + f1a

∗ (70)

We know that the most general function of two Grassmann variables has the form:

A(a∗, α) = A00 +A01α+A10a∗ +A11a

∗α

Using properties of the Grassmann variables we can actually compute this matrix product in an alternativeway; indeed:

(Af)(a∗) =

∫dα∗dαA(a∗, α)f(α∗)e−α

∗α

The integrand can be expanded to (I drop terms which do not have the product α∗α):

[A00 +A01α+A10a

∗ +A11a∗α][fo + f1α

∗][1− α∗α]→ [A00 +A10a

∗]fo(− α∗α)+[A01 +A11a

∗]f1αα∗

(71)

={[A00fo +A01f1

]+[A10fo +A11f1

]a∗}αα∗ (72)

And, since:∫dα∗dααα∗ = 1, we see:

(Af)(a∗) =

∫dα∗dαA(a∗, α)f(α∗)e−α

∗α =[A00fo +A01f1

]+[A10fo +A11f1

]a∗ = g(a∗)

We can also construct a rule for matrix multiplication:

(A B)(a∗, a) =

∫dα∗dαe−α

∗αA(a∗, α)B(α∗, a)

Where: B(α∗, a) = B00 + B01a + B10α∗ + B11α

∗a. Let’s follow the same protocol and expand the integrand(again dropping terms which clearly integrate to zero):

e−α∗αA(a∗, α)B(α∗, a) =

[1− α∗α

][A00 +A01α+A10a

∗ +A11a∗α][B00 +B01a+B10α

∗ +B11α∗a]

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→[A01α+A11a

∗α][B10α

∗ +B11α∗a]− α∗α

[A00 +A10a

∗][B00 +B01a

]=[(A00B00 +A01B10

)+(A01B11 +A00B01

)a+

(A11B10 +A10B00

)a∗ +

(A11B11 +A10B01

)a∗a]αα∗

= C(a∗, a)αα∗

It follows:

(A B)(a∗, a) =

∫dα∗dαe−α

∗αA(a∗, α)B(α∗, a) = C(a∗, a)

3. Let’s now define two operators a∗ and a which satisfy:

a∗f(a∗) = a∗f(a∗), af(a∗) =d

da∗f(a∗)

As a result of these definitions, we see:

a∗a∗f(a∗) = a∗af(a∗) = (a∗)2f(a∗) = 0

Thus, (a∗)2 = 0. Similarly:

(a)2f(a∗) =d

da∗

[ d

da∗f(a∗)

]= 0

This last equality can be viewed as a consequence of differentials of Grassmann variables acting like Grassmannvariables, or the second derivative of any function of Grassmann variables is zero since the most general functionof a Grassmann variable is linear in Grassmann variables. In any case: (a)2 = 0.

Now let’s go ahead and see what happens if we combine these operators; we will need f(a∗) = fo + a∗f1. Sowe have two combinations to calculate:

(a) aa∗f(a∗) = aa∗f(a∗) = dda∗

[foa∗ + f1(a∗)2

]= d

da∗

[foa∗] = fo

(b) a∗af(a∗) = a∗ dda∗ f(a∗) = a∗ d

da∗

[fo + f1(a∗)

]= a∗

[f1

]= a∗f1

Putting these together we see:

{a, a∗}f(a∗) = fo + a∗f1 = f(a∗)→ {a, a∗} = 1

These operators have the same properties as creation and annihilation operators for Fermions!

4. Now let’s consider the set of 2N Grassmann variables:{{ξi}, {ξi}

}. Let’s compute the following integral:

Z =

∫ N∏j=1

[dξjdξj

]e−

∑k,l ξkMklξl

The integrand can be expanded in a Taylor series, which results in terms of the following form:

e−∑k,l ξkMklξl =

∑n≥0

1

n!

(−∑k,l

ξkMklξl

)nUsing properties of Grassmann variables we can greatly simplify this expression. For one, certain products willcontain terms with squares of identical Grassmann variables, so these will automatically go to zero. Further-more, we recall: ∫

dη 1 = 0

Thus, the only nonzero combinations, after integration, are those terms which have exactly one of each Grass-mann variable.

It is not hard to see we are forced to look at the term:

8

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1

N !

(−∑k,l

ξkMklξl

)N=

(−1)N

N !

(∑l1

ξ1M1l1ξl1 +∑l2

ξ2M2l2ξl2 + · · ·+∑lN

ξNMNlN ξlN

)N(73)

= (−1)N∑l1...lN

M1l1M2l2 . . .MNlN ξ1ξl1 ξ2ξl2 . . . ξNξlN (74)

This last line comes about from noting that we have N products of terms of the form in the parentheses andif we multiply any identical sum, then these integrate to zero. Thus, we drop those terms. Furthermore, thereare N ! ways to achieve these products.

The determinant is now staring us right in the face! We can recast the above sum into the following form:

1

N !

(−∑k,l

ξkMklξl

)N= (−1)N

∑l1...lN

M1l1M2l2 . . .MNlN ξ1ξl1 ξ2ξl2 . . . ξNξlN (75)

= (−1)N

[ ∑σ∈SN

sgn(σ)M1σ(1)M2σ(2) . . .MNσ(N)

]ξ1ξ1ξ2ξ2 . . . ξNξN (76)

Here σ is a element of the symmetry group of N elements. The sgn(σ) term comes from reordering theGrassmann variables in the order indicated outside the brackets, so the accumulation in minus signs will haveto be the same as the parity of the symmetry element.

We are now ready to insert this into the partition function. The integral is trivial after we permute theGrassmann variables to match the order of the differentials for the integration. We just end up with anadditional factor of (−1)N ; thus, an overall factor of 1! Our desired result is almost immediate:

Z =

∫ N∏j=1

[dξjdξj

]e−

∑k,l ξkMklξl (77)

=

∫ N∏j=1

[dξjdξj

](−1)N

[ ∑σ∈SN

sgn(σ)M1σ(1)M2σ(2) . . .MNσ(N)

]ξNξN . . . ξ2ξ2ξ1ξ1 (78)

=∑σ∈SN

sgn(σ)M1σ(1)M2σ(2) . . .MNσ(N) (79)

= det(M) (80)

3 Dirac Fermions

Consider the following Lagrangian density:

L = ψ(i/∂ −m

1. As before, let’s consider the amplitude (h = 1):

〈Ψf , tf |Ψi, ti〉 = 〈Ψf |e−iH(tf−ti)|Ψi〉

:=

∫DψDψeiS(ψ,ψ) × projection operators

Where:

S(ψ, ψ) =

∫ tf

ti

dt[ihψ∂tψ −H(ψ, ψ)

]From our Lagrangian we see that the momentum canonically conjugate to ψ is iψ; thus:

S(ψ, ψ) =

∫d4xL

Now let’s couple our above Lagrangian to some external sources η and η:

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L = ψ(i/∂ −m

)ψ + ψη + ηψ

As was the case with a scalar field theory, we can now find the Generating functional:

Z[η, η] =1

〈0|0〉

∫DψDψeiS

For simplicity, let’s define A = i/∂ −m. There are several ways to go about solving this problem; for instance,we could complete the square. I will opt to shift the fields as we did in the previous homework, but these twomethods are essentially the same. Let’s define:

ψ(x) = ψo(x) + δ(x)⇒ ψ(x) = ψo(x) + δ(x)

This means:L =

[ψoAψo + ψo(x)η(x) + ψo(x)η(x)

]+ δ(x)Aδ(x)

+[δ(x)Aψo(x) + δ(x)η(x)

]+[ψo(x)Aδ(x) + δ(x)η(x)

]Using AG(x− y) = δ(x− y) we note that:

ψo(x) = −∫d4yG(x− y)η(y)⇒ δ(x)Aψo(x) + δ(x)η(x) = 0 (81)

Since the Greens function is symmetric in x and y and we can integrate by parts, it follows for:

ψo(x) = −∫d4yG(x− y)η(y)

that,ψo(x)Aδ(x) + δ(x)η(x) = 0

Substituting these solutions back into our Lagrangian yields the uncoupled (or completed square) version ofour original Lagrangian (these steps are trivial since Aψo = −η(x)):

L = −∫d4yη(x)G(x− y)η(y) + δ(x)Aδ(x)

Finally, we find:

Z[η, η] =

[∫DδDδei

∫d4xδ(x)Aδ(x)

]e−i

sd4xd4yη(x)G(x−y)η(y)

We now note this looks a lot like the functional integral we met in the last problem except the summation isnow an integration; this here is just the continuum case of what we already know. Thus, we say the functionalintegral evaluates to the determinate of our operator. Thus:

Z[η, η] = Det(i/∂ −m

)e−i

sd4xd4yη(x)G(x−y)η(y) (82)

= Det(i/∂ −m

)e−i

sd4xd4yηα(x)Gαβ(x−y)ηβ(y) (83)

2. Let’s go ahead and find the Feynman propagator which is defined by:

iSαβF (x− y) = 〈0|Tψα(x)ψβ(y)|0〉

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Notice we can achieve our goal if we just use the generating functional for our Dirac theory; I’ll copy its originalform below for convenience:

Z[η, η] =

∫DψDψ exp

{i

∫d4z(ψ(z)

(i/∂ −m

)ψ(z) + ψ(z)η(z) + η(z)ψ(z)

)}

First, let’s note that when taking derivatives of products of distinct Grassmann variables, we must keep inmind for F [η, η] =

∫d4zη(z)η(z) we have:

δF [η, η]

δη(x)=

∫d4z

δη(z)

δη(x)η(z) = η(x)

but,

δF [η, η]

δη(x)=

δ

δη(x)

∫d4zη(z)η(z) (84)

= − δ

δη(x)

∫d4zη(z)η(z) (85)

= −∫d4z

δη(z)

δη(x)η(z) (86)

= −η(x) (87)

That is, we must permute the Grassmann variable that we are differentiating to match the order in which weare differentiating said variables. This will result in an accumulation of various minus signs. Recall that eachof these guys have four components, so we say:

η(z)ψ(z) = ηγ(z)ψγ(z)

Subscripts will be used from here on out. We can make our lives easier if we perform the followingtrick:

δ2Z[η, η]

δηα(x)δηβ(y)=

1

i

δ

δηα(x)i

δ

δηβ(y)Z[η, η] (88)

(89)

That is, we accompany derivatives in respect to our Grassmann variables with an imaginary number in aspecified way (indicated above). Why? Let’s see it in action:

iδZ[η, η]

δηβ(y)= i

∫DψDψ

∫d4zi

δ(ψγ(z)ηγ(z)

)δηβ(y)

eiS (90)

= −i2∫DψDψ

∫d4z δβγδ(y − z)ψγ(z)eiS (91)

=

∫DψDψ ψβ(y)eiS (92)

There is no accumulation in minus sign, or an i, as a result of our convention: pairing a i withderivatives in respect to η! The η definition kills imaginary numbers too:

δ2Z[η, η]

δηα(x)δηβ(y)=

1

i

δ

δηα(x)

∫DψDψ ψβ(y)eiS (93)

= −∫DψDψ ψβ(y)ψα(x)eiS (94)

=

∫DψDψ ψα(x)ψβ(y)eiS (95)

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Finally, we set η = η = 0; thus:

1

Z[0, 0]

δ2Z[η, η]

δηα(x)δηβ(y)

∣∣∣∣∣η=η=0

= 〈0|Tψα(x)ψβ(y)|0〉

Now that we have this expression it is time to use our result from the previous part of this question; this willgive an expression for the Feynman propagator in terms of our Green function for the Dirac operator. Let’sget going:

iδZ[η, η]

δηβ(y)=

[det(i/∂ −m)

]i

δ

δηβ(y)e−i

sd4zd4z′η(z)G(z−z′)η(z) (96)

=

[x

d4zd4z′δ

δηβ(y)

[η(z)G(z − z′)η(z′)

]]Z[η, η] (97)

Let’s take a closer look at our functional derivative in the integral. First, let’s note that this function must bea scalar. We perform the following contraction:

η(z)G(z − z′)η(z′) =∑κγ

[[η(z)]κ[G(z − z′)]κγ [η(z′)]γ

]:= ηκ(z)Gκγ(z − z′)ηγ(z′)

That is repeated indices contract.

δ

δηβ(y)

[η(z)G(z − z′)η(z′)

]=

δ

δηβ(y)

[ηκ(z)Gκγ(z − z′)ηγ(z′)

](98)

= −ηκ(z)Gκγ(z − z′)δ(y − z′)δβγ (99)

The overall minus sign comes from my spiel on taking derivatives of a product of Grassmann variables. Thus,we see:

iδZ[η, η]

δηβ(y)= −

[∫d4zηκ(z)Gκβ(z − y)

]Z[η, η]

Differentiating this expression again:

δ2Z[η, η]

δηα(x)δηβ(y)= −1

i

δ

δηα(x)

[∫d4zηκ(z)Gκβ(z − y)

]Z[η, η] (100)

=

[i

∫d4z

δηκ(z)

δηα(x)Gκβ(z − y)

]Z[η, η] +

[i

∫d4zηκ(z)Gκβ(z − y)

]δZ[η, η]

δηα(x)(101)

η,η→0−−−−→ iGαβ(x− y)Z[0, 0] (102)

Thus:

SαβF (x− y) = − i

Z[0, 0]

δ2Z[η, η]

δηα(x)δηβ(y)

∣∣∣∣∣η=η=0

(103)

= Gαβ(x− y) (104)

= 〈x, α| 1

i/∂ −m|y, β〉 (105)

3. Now we want to find:

S4F,αβγδ(x1, x2, x3, x4) = 〈0|Tψα(x1)ψβ(x2)ψγ(x3)ψδ(x4)|0〉 (106)

=1

Z[0, 0]

δ4Z[η, η]

δηα(x1)δηβ(x2)ηγ(x3)δηδ(x4)

∣∣∣∣∣η=η=0

(107)

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Because of the accumulation of minus signs, We will have to be extra careful here! Let’s get going:

iδZ[η, η]

δηδ(x4)= −

[∫d4zηκ(z)Gκδ(z − x4)

]Z[η, η]

i2δ2Z[η, η]

δηγ(x3)δηδ(x4)= i

[∫d4zηκ(z)Gκδ(z − x4)

]δZ[η, η]

δηγ(x3)(108)

= −

[∫d4zηκ(z)Gκδ(z − x4)

][∫d4zηκ(z)Gκγ(z − x3)

]Z[η, η] (109)

For now on any derivatives of Z[η, η] will be neglected since this will give a factor of η, and we only differentiatein respect to η from here on out:

iδ3Z[η, η]

ηβ(x2)δηγ(x3)δηδ(x4)= −1

i

δ

δηβ(x2)

[∫d4zηκ(z)Gκδ(z − x4)

][∫d4zηκ(z)Gκγ(z − x3)

]Z[η, η] (110)

→1

i

∫d4z

[Gβγ(x2 − x3)ηκ(z)Gκδ(z − x4)−Gβδ(x2 − x4)ηκ(z)Gκγ(z − x3)

]Z[0, 0]

(111)

Finally, we get:

S4F,αβγδ(x1, x2, x3, x4) = 〈0|Tψα(x1)ψβ(x2)ψγ(x3)ψδ(x4)|0〉 (112)

=1

Z[0, 0]

δ4Z[η, η]

δηα(x1)δηβ(x2)ηγ(x3)δηδ(x4)

∣∣∣∣∣η=η=0

(113)

= Gβδ(x2 − x4)Gαγ(x1 − x3)−Gβγ(x2 − x3)Gαδ(x1 − x4) (114)

4 Functional Determinants and the Casimir Effect

Consider the free scalar field φ(x, t) := φ(x) in 1+1 space-time dimensions; we know:

L =1

2∂µφ(x)∂µφ(x)− 1

2m2φ(x)2

Furthermore, suppose for all time t:φ(x, t) = φ(x+ L, t)

1. Let’s begin by looking for the classical ground state energy of the system; thus, let’s find the Hamiltoniandensity:

H =δL

δ∂oφ(x)∂oφ(x)− L = Π(x)2 − L =

1

2

[Π(x)2 +

(∇φ(x)

)2+m2φ(x)2

]In general, to minimize a Hamiltonian of the form:

H =1

2

[Π(x)2 +

(∇φ(x)

)2+ V (φ(x))

]we see that the first two terms should be zero since they are positive definite; thus, φ(x) =constant. We thenchoose φ(x) in such a way that V (φ(x)) = Vmin. In our case this is very simple: φ(x) = 0. Thus, our groundstate energy is zero; this is believable since our Hamiltonian is positive definite.

2. Here we are asked to use the path integral methods to derive a energy density for our real scalar field, butwe did this in lecture. Furthermore, this procedure is very similar to the complex scalar field we did in theprevious homework. Because of this, I will go through the derivation in a telegraphic way:

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(a) We start with the usual prescription (equation 5.109 in Fradkin’s notes):

J〈{φ(~x, xo)}|{φ′(~y, yo)}〉J = N∫b.c.

Dφe ihS(φ,∂µφ,J)

Where: S(φ, ∂µφ, J) =∫d4x[

12 (∂µφ)2 − V (φ) + Jφ

], and N is a normalization constant.

(b) We then relate this to the Vacuum Persistent Amplitude by projecting out the excitations with the Gell-Mann-Low Theorem. We thus work with our best friend Z[J ], the vacuum persistent amplitude, beingthe generating functional of our theory:

Z[J ] = J〈0|0〉J = N∫Dφe ihS(φ,∂µφ,J)

(c) We now perform a wick rotation:

L t→−iτ−−−−→ LE =1

2(∇µφ)2 +

1

2m2φ2 − Jφ

So then we work with:

ZE [J ] = J〈0|0〉J = N∫Dφe− 1

hSE(φ,∂µφ,J)

Where: SE(φ, ∂µφ, J) =∫d4xLE .

(d) Next we integrate by parts to put the Euclidean Lagrangian density in a bilinear form:

LE →1

2φ[−∇2 +m2]φ− Jφ

(e) We now shift the field and use the Green’s function, GE(x− x′) of our operator 12 [−∇2 +m2] to express

our shifted field in terms of our source, which decouples the source from the field. The result is:

ZE [J ] = ZE [0]e12

∫dDxdDx′J(x)GE(x−x′)J(x′)

where:

ZE [0] =

∫Dδe− 1

2

∫dDxδ(x)[−∇2+m2]δ(x)

(f) Finally, we expand δ(x) in terms of the eigen functions of our operator −∇2 +m2. The result of this is justa product of a bunch of Gaussian integrals with exponents of the form: 1

2Anc2n where An is the eigenvalue

of the nth eigenfunciton and cn is the nth expansion coefficient of δ(x). The resulting integration is simplygiven by:

ZE [0] =∏n

A−1/2n := det

[−∇2 +m2

]−1/2

(g) In conclusion:

ZE [J ] = det[−∇2 +m2

]−1/2e

12

∫dDxdDx′J(x)GE(x−x′)J(x′)

Now let’s go ahead and use our partition function to calculate the ground state energy. This can be found withthe following equation:

EG = − limβ→∞

1

βln tre−βH = − lim

β→∞

1

βlnZE [J ]

We know from lecture, and the last homework, that we can expand our Lagrangian around the classical pathand calculate the Quantum corrections. For the case of a field φc(x) + φ(x), where φc(x) is the classical pathand φ(x) is the Quantum correction, a Lagrangian of the form L = 1

2 (∂µφ)2 − V (φ) takes the following form(to leading order):

L = L[φc] +1

2φ(x)

[− ∂2 − V ′′[φc]

]φ(x) + . . .

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The term with the quantum corrections is bi-linear in this field. Thus, from our previous experience with realscalar fields we see:

Z =

∫DφeiS[φ] = N

[det(− ∂2 − V ′′[φc]

)]−1/2

eiS[φc][1 + . . . ]

In our case we have V (φ) = 12m

2φ2(x), so then: V ′′[φc] = m2

We however prefer to work in imaginary time, so our corresponding Euclidean action can be achieved with thefollowing substitutions:

(a) iS[φc]→ −S[φc]

(b) 12φ(x)

[− ∂2 − V ′′[φc]

]φ(x)→ − 1

2φ(x)[−∇2 +m2

]φ(x)

Here: ∇2 = ∂2τ + ∂2

x. From these substitutions we see:

Z = N[det(−∇2 + V ′′[φc]

)]−1/2

e−S[φc][1 + . . . ] = N e−βEo[det(−∇2 +m2

)]−1/2

[1 + . . . ]

Where I made the following identification between the classical action and the ground state energy for a freefield :

S[φc] = −β−1Eo

Our equation for our ground state energy above becomes:

EG = Eo +1

2βln det(−∇2 +m2) + . . .

Thus, the term which arises from our quantum corrections amounts to evaluating a determinant for the operator :(−∇2 +m2) := A.

3. We can find an explicit expression for the energy density via a number of methods. Here I will use the methodcovered in class which uses the generalized ζ-function associated with A:

ζA(s) =∑n

1

asn

These an’s are just the spectrum of eigenvalues for A; that is:

Afn(x) = anfn(x)

As a quick reminder, I write out the use of the ζ-function:

lims→0+

dζAds

= − lims→0+

∑n

ln anasn

= −ln∏n

an := −ln detA

We thus need to find our generalized ζ-function to compute the functional determinant. As we did in class, wecan accomplish this by using the generalize heat kernel :

GA(x, y; τ) =∑n

e−anτfn(x)f∗n(y) := 〈x|e−τA|y〉

Here τ > 0. The Heat Kernel has the following initial condition:

limτ→0+

GA(x, y; τ) =∑n

fn(x)f∗n(y) = δ(x− y)

We now relate our Heat Kernel to our generalized ζ-function in the following way:∫dDx lim

y→xGA(x, y; τ) =

∑n

e−anτ∫dDx fn(x)f∗n(x) =

∑n

e−anτ := tr e−τA

We know that:Γ(s)

asn=

∫ ∞0

dτ τ s−1e−anτ

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So then we see by the above procedure:∑n

Γ(s)

asn=

∫ ∞0

dτ τ s−1∑n

e−anτ =

∫ ∞0

dτ τ s−1

∫dDx lim

y→xGA(x, y; τ)

We finally have:

ζA(s) =1

Γ(s)

∫ ∞0

dτ τ s−1

∫dDx lim

y→xGA(x, y; τ)

There are two ways we can proceed:

(a) We find the eigenfunctions for A to build the generalized heat kernel.

(b) We use the eigenvalues of A to build the generalized ζ-function.

I will find the eigenfunctions, but we will construct the generalized ζ-function. Also, recall we are in 1 + 1-D,so we only have two components.

To begin, we note that our eigenfunctions need to satisfy periodic boundary conditions in our spatial coordinate;i.e. (two labels for the two coordinates) fln(x + L, xo) = fln(x, xo). Here xo is the label I choose to use forimaginary time since τ is already taken. This function also needs to be periodic in imaginary time with periodβ, which we will take to infinity in due time.

Our eigenfunctions are very simple here! They are just complex exponentials. We could separate variables andshow the imaginary time and the spatial components are both complex exponentials. If we did that, we’d getfor fln(~x) := fln(x, xo) = hl(xo)gn(x):

Afln(x, xo) = anfln(x, xo)⇒1

hl(xo)∂2ohl(xo) +

1

gn(x)∂2xgn(x) = aln −m2 := k2

We then define k2x + k2

o = k2, and do all the things we learned way back in intro to ODE’s. I’m just going tocut to the chase and say our eigenfunctions are:

fln(~x) =ei~x·

~k

√Lβ

with eigenvalues: aln = k2 +m2. Note: ~k = (kn, ωl) = ( 2πL n,

2πβ l).

Since we know our eigenvalues, we can now write out our expression for the generalized ζ-function:

ζA(s) =∑l,n

1

a2ln

(115)

=1

Γ(s)

∑l,n

∫ ∞0

dττ s−1e−alnτ (116)

Γ(s)

∑n

∫ ∞0

∫ ∞−∞

dτdω

2πτs−1e−(ω2+k2

n+m2)τ (117)

Notice that we replaced the sum with an integral in the last step by using the density of states. We can easilyperform the integral over ω since it is just Gaussian:

ζA(s) =β

Γ(s)

∑n

∫ ∞0

dτdω

2πτs−1

√π

τe−(k2

n+m2)τ (118)

Γ(s)

∑n

∫ ∞0

2√πτs−3/2e−(k2

n+m2)τ (119)

We are now in a position to use Poisson’s Summation formula:∑n

f(n) =∑m

∫ ∞−∞

dye2πimyf(y)

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We identify: f(n) = exp(−k2nτ). Thus, substituting y in for n in kn explicitly we find:

ζA(s) =β

Γ(s)

∑m

∫ ∞−∞

dy

∫ ∞0

2√πe2πimyτs−3/2e−

(( 2πL )2y2+m2

)τ (120)

=βL

4πΓ(s)

∑n

∫ ∞0

dττ s−3/2

∫ ∞−∞

dk

[1√πeinkLe−

(k2+m2

](121)

=βL

4πΓ(s)

∑n

∫ ∞0

dττ s−2e−m2τ− (nL)2

4τ (122)

The penultimate was achieved simply from a change of variables, k = (2π/L)y, and I sent m → n to avoidconfusion with our mass term (but can potentially be confused with the eigenvalue, which it is not!). Thefinal step just relies on completing the square to find the Gaussian integral.

We need to consider the massless limit, but there will be issues at n = 0; essentially, we have no regulator totame the term τs−2. Thus, we imagine we have a very, very small mass which allows us to neglect it for allnonzero n. Applying this line of reasoning, and recalling the Euler γ-function:

Γ(s) =

∫ ∞0

dzzs−1e−z

allows us to finish the calculation. The n = 0 term is (almost) automatic, but the n 6= 0 terms only require a

substitution of variables: z = (nL)2

4τ . Furthermore, this is even in n, so we just sum over positive values. Thus:

ζA(s) =βL

4πΓ(s)

[∫ ∞0

dττ s−2e−m2τ + 2

∑n=1

∫ ∞0

dττ s−2e−(nL)2

](123)

=βL

4πΓ(s)

[Γ(s− 1)

(−m2)s−1+ 2

∑n=1

( 2

nL

)2(1−s) ∫ ∞0

dτz−se−z

](124)

=βL

4πΓ(s)

[Γ(s− 1)

(−m2)s−1+ 2( 2

L

)2(1−s)ζ(2− 2s)Γ(1− s)

](125)

Now we need to consider what happens in the limit as s→ 0 2. The Riemann ζ-function is easy:

ζ(2− 2s)→ ζ(2) =∑n>0

1

n2=π2

6

For the Euler Γ-function we recall:

γ(z) =1

z− γ +

1

2

(π2

6+ γ2

)z +O

(z2)

where γ is the Euler–Mascheroni constant. Thus:

Γ(1− s)Γ(s)

≈ s

We can also use the relation which is used to analytically continue the Γ-function to the negative real numbers:

Γ(s+ 1) = sΓ(s)

which implies:Γ(s− 1)

Γ(s)=

1

s− 1≈ −1

This first term does indeed blow up as expected! We can actually handle this infinity with some renormalizationscheme, some of which we will learn about next semester. For now, we just ignore it and say we defined anew reference of energy, which is a naive renormalization procedure. This term is the divergent extensive part

2This means I am assuming the derivative and the limit commute. This just makes the differentiation easier!

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of the energy. All other limits are trivial. Note: this minus one above makes the two terms in our functionaldeterminant have equal signs as s→ 0, which makes the overall bosonic energy negative.

Now that we have all our cards on the table let’s play them:

ln detA = − lims→0+

dζAds

(126)

⇒ Efluc = EG − Eo =1

2βln det(A) + . . . (127)

= − π

6L+ . . . (128)

We note that this term arises from the finite length of our system, L! It also goes to zero as L→∞.

4. Finally! After all that hard work, we will just take a derivative of this expression to find the force exerted onthe walls:

F = −dEfluc

dL= − π

6L2+ . . .

This is the famous Casmir effect ! Notice we did not use the usual vanishing boundary conditions here; thiswould require a linear combination of our eigenfunctions, which makes the math a bit more hairy.

5 Weakly interacting Bose Gas

Consider the Hamiltonian:

H =

∫d3x(φ†(~x)

[ ~p 2

2m− µ

]φ(~x) +

1

2

∫d3x′n(~x)V (~x− ~x′)n(~x′)

)We’ll take: V (~x− ~x′) = λδ3(~x− ~x′).

1. In this problem we will be using the Bose coherent states:

|{φ(~x)}〉 = e∫d3xφ(~x)φ~x)|0〉

As was the case with creation and annihilation operators, we replace the field operators with complex scalarfields (after doing the necessary work of course). Slicing up the time intervals in the usual way (I outline thisin the appendix after this problem), one can show:

〈f |e− ih H∆t|i〉 =

∫DφDφ exp

{i

h

∫ tf

ti

dt

(∫d3x

h

i

[φ(~x, t)∂tφ(~x, t)− φ(~x, t)∂tφ(~x, t)

]−H[φ, φ]

)}

×ψf (φ(~x, tf ))ψi(φ(~x, ti))e12

∫d3x

(|φ(~x,tf )|2+|φ(~x,ti)|2

)We can write this in a simpler form as was done in lecture:

S =

∫dxφ(x)

(ih∂t +

h2

2m~∇2 + µ

)φ(x)− 1

2

∫dx

∫dy|φ(x)|2|φ(y)|2V (x− y)

Where, V (x− y) = λδ3(~x− ~y)δ(tx − ty), and dx = dtxd3x.

We can use the coherent-state path integrals to represent the partition function at finite temperature T:

Z = tre−βH

We’d also like to work with the Euclidean version of the partition function. What we need to do is:

(a) Set: |i〉 = |f〉(b) Sum over all possible intermediate states.

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(c) Perform a Wick rotation t → −iτ . Since we are eventually going to derive a partition function, we aregoing to need a trace of our path integral. This is the same thing as saying our particle starts and ends inthe same state; i.e. periodic boundary conditions in time. As we know from chapter 5, equation (5.130),periodic boundary conditions in imaginary time will always hold for Bosonic theories. That is:

φ(x, τ) = φ(x, τ + β)

This is a compactified space-time for our Euclidean theory.

The Wick rotated action is achieved in the usual way:

t→ −iτ (129)

⇒∂t → i∂τ (130)

⇒dtd3x→ −idτd3x (131)

This is easy enough, so I will just go ahead and write everything out putting what is new in small square bracesfirst:

iS →i∫ β

0

[−idτ ]

∫d3xφ(x)

(ih[i∂τ ] +

h2

2m~∇2 + µ

)φ(x)− i1

2

∫ β

0

[−idτ ]

∫d3x

∫d3y|φ(x)|2|φ(y)|2V (x− y)

(132)

= −

[∫ β

0

∫d3xφ(x)

(h∂τ −

h2

2m~∇2 − µ

)φ(x) +

1

2

∫ β

0

dτd3x

∫d3y|φ(x)|2|φ(y)|2V (x− y)

](133)

Thus,

SE =

∫ β

0

∫d3xφ(x)

(h∂τ −

h2

2m~∇2 − µ

)φ(x) +

1

2

∫ β

0

dτd3x

∫d3y|φ(x)|2|φ(y)|2V (x− y) (134)

=

∫ β

0

∫d3xφ(x)

(h∂τ −

h2

2m~∇2 − µ

)φ(x) +

λ

2

∫ β

0

∫d3x|φ(x)|4 (135)

On the last line I just used the delta function to collapse the y-integral.

The trailing product in our above expression, which depends on the initial and final states, drops out because itis just a constant due to our boundary conditions. We arrive at our integral formula for our partition function:

Z =

∫DφDφe−SE(φ,φ)

2. Now that we have a form for our generating functional let’s go ahead and find our saddle points for ourEuclidean action above. We can hand pick the Lagrangian density from the above expression for SE :

LE = φ(x)(h∂τ −

h2

2m~∇2)φ(x) +

[λ2|φ(x)|2 − µ

]|φ(x)|2

= φ(x)(h∂τ −

h2

2m~∇2)φ(x) + Veff

(φ(x)

)This guy is clearly invariant under the transformation:

φ′(x) = φ(x)eiθ

So we have a local U(1) symmetry; thus, our ground state is not unique (just change the phase). Furthermore,we can view the Lagrangian as the energy density for the Euclidean theory. The first term is a kinetic term,so it is positive definite. To find the ground state energy we can say φ(x) is a constant.

To figure out what constant we minimize Veff . This is implies:

φ(x) =

õ

2λeiθo

Where θo is an arbitrary constant phase we will take θo = 0 below.

Note: we’d arrive at this result if we took functional derivatives instead.

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3. Let’s analyze the Green’s function in the semi-classical limit; that is, consider:

φ(x) = φc(x) + δ(x)

Here δ(x) is a small, but arbitrary, fluctuation about the classical path. Furthermore, since the fluctuationsare small (semi-classical), we will Taylor expand the action to quadratic order, and then find the corresponding(real time) Green’s function. First, recall:

S =

∫dzφ(z)

(ih∂t +

h2

2m~∇2 + µ

)φ(z)− λ

2

∫dz|φ(z)|4

Expanding about the classical path:

S = S(φc) +

∫dx

[δS

δφ(x)

∣∣∣∣∣φc

δ(x) + δ(x)δS

δφ(x)

∣∣∣∣∣φc

]+

∫ ∫dxdyδ(x)

δ2S

δφ(x)δφ(y)

∣∣∣∣∣φc

δ(y) + . . . (136)

= S(φc) +

∫dxδ(x)

[ih∂t +

h2

2m∇2 + µ− 2λ|φc|2

]δ(x) + . . . (137)

= S(φc) +

∫dxδ(x)

[ih∂t +

h2

2m∇2

]δ(x) + . . . (138)

In the last line I plugged in for the classical field. We now use this result to find our Green’s function:

G(x− y) = −i〈T(φ(x)φ†(y)

)〉 (139)

= −i〈T(φc(x)φc(y)

)〉 − i〈T

(δ(x)δ(y)

)〉 (140)

The first term is easy to find:

〈T(φc(x)φc(y)

)〉 =

µ

Recall the phase was also constant, so this cancels out. We can deduce the Green’s function for the fluctuatingfield by considering its effective action:

Sδ =

∫dxδ(x)

[ih∂t +

h2

2m∇2]δ(x) (141)

⇒ Gδ(x− y) = 〈x| 1

ih∂t + h2

2m∇2 + iε|y〉 (142)

The ε is introduced here to make sure the Green’s function converges. We can find the long distance behaviourof Gδ(x− y) with a Fourier transform:

Gδ(x− y) =

∫d4p

(2π)4

1

po − 12m |~p|2 + iε

e−ip(x−y) (143)

= iθ(xo − yo)∫

d3p

(2π)3e−ε(xo−yo)exp

(i(~p · (~x− ~y)− 1

2m|~p|2(xo − yo)

))(144)

xo → yo = i

∫d3p

(2π)3exp(i(~p · (~x− ~y)

)) ~x−~y→∞−−−−−→ 0 (145)

This tells us our (real time) Green’s function behaves in the following way in this limit:

G(x− y)→ −i µ2λ

This is known as off diagonal long range order and is the signature of a superfluid!

4. We expand the field as:

φ(x) =√ρo + δρ(x)eiθ(x)

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so then,φ(x) =

√ρo + δρ(x)e−iθ(x)

Here ρo = µ2λ , and δ(x) and θ(x) are small fluctuations. Let’s expand our Lagrangian out in terms of these

fields:

LE = φ(x)(h∂τ −

h2

2m~∇2)φ(x) +

[λ2|φ(x)|2 − µ

]|φ(x)|2

:= φ(x)(h∂τ −

h2

2m~∇2)φ(x) + Veff (φ(x))

We note that since our system is 1+1 dimensional: ∇2 = ∂2x. For the first term, the kinetic part, we expand

by direct substitution. First, let’s look at:

∇2φ(x) = ∇2√ρo + δρ(x)eiθ(x)

= eiθ(x)∇2√ρo + δρ(x) +

√ρo + δρ(x)∇2eiθ(x) + 2∇eiθ(x) · ∇

√ρo + δρ(x)

= eiθ(x)

{∇2δρ(x)

2√ρo + δρ(x)

− ∇δρ(x) · ∇δρ(x)

4(√

ρo + δρ(x))3 + i

∇θ(x) · ∇δρ(x)√ρo + δρ(x)

+ [i∇2θ(x)− (∇θ(x))2]√ρo + δρ(x)

}Thus, to second order:

φ(x)∇2φ(x) =∇2δρ(x)

2− ∇δρ(x) · ∇δρ(x)

4(ρo + δρ(x)

) + i∇θ(x) · ∇δρ(x) + [i∇2θ(x)− (∇θ(x))2][ρo + δρ(x)] (146)

≈ ∇2δρ(x)

2− ∇δρ(x) · ∇δρ(x)

4ρo+ i∇θ(x) · ∇δρ(x) + i∇2θ(x)[ρo + δρ(x)]− (∇θ(x))2ρo (147)

Quickly notice that if we do a quick integration by parts, and invoke our boundary conditions, that the followingterms drop out:

i∇θ(x) · ∇δρ(x) + i∇2θ(x)δρ(x)→ 0

In a similar manner:

φ(x)∂τφ(x) =√ρo + δρ(x)

(∂τδρ(x)

2√ρo + δρ(x)

+ i√ρo + δρ(x)∂τθ(x)

)(148)

=∂τδρ(x)

2+ i(ρo + δρ(x)

)∂τθ(x) (149)

We finally have:

φ(x)(h∂τ −

h2

2m~∇2)φ(x) ≈1

2

[∂τ −

h2

2m∇2]δρ(x) + iρo

[∂τ −

h2

2m∇2]θ(x) (150)

+ iδρ(x)∂τθ(x) +h2

2m

[∇δρ(x) · ∇δρ(x)

4ρo+ (∇θ(x))2ρo

](151)

Now let’s look at what the effective potential looks like:

Veff (√ρo + δρ(x)) =

[λ2

(ρo + δρ(x))− µ](ρo + δρ(x)) (152)

2(ρo + δρ(x))(ρo + δρ(x))− µ(ρo + δρ(x)) (153)

2(ρ2o + 2ρoδρ(x) + (δρ(x))2)− µ(ρo + δρ(x)) (154)

2

(( µ2λ

)2

+ 2µ

2λδρ(x) + (δρ(x))2

)− µ

( µ2λ

+ δρ(x))

(155)

→ λ

2(δρ(x))2 (156)

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The right arrow is just dropping constants from our Lagrangian. Now to make some arguments on how thelinear terms in δρ(x) fall out in the kinetic expression. This linear term is:

1

2

[∂τ −

h2

2m∇2]δρ(x)

and is being integrated, so we can integrate by parts to move the derivatives to the constant on the other side,and use the boundary conditions to throw this term out.

In a similar fashion the term linear in θ falls out, which is:

iρo

[∂τ −

h2

2m∇2]θ(x)

Thus, our Lagrangian reduces to:

LE(δρ, θ)→ iδρ(x)∂τθ(x) +h2

2m

[∇δρ(x) · ∇δρ(x)

4ρo+ (∇θ(x))2ρo

]+λ

2δρ(x))2

So our path integral:

Z =

∫DδρDθe−SE(δρ,θ)

has the following effective action:

SE =

∫dτ

∫d3x

[iδρ(x)∂τθ(x) +

h2

2m

[∇δρ(x) · ∇δρ(x)

4ρo+ (∇θ(x))2ρo

]+λ

2(δρ(x))2

](157)

=

∫dτ

∫d3x

[iδρ(x)∂τθ(x) +

h2

2m

[−δρ(x)∇2δρ(x)

4ρo+ (∇θ(x))2ρo

]+λ

2(δρ(x))2

](158)

=

∫dτ

∫d3x

[δρ(x)

[λ2− h2

2m

∇2

4ρo

]δρ(x) + iδρ(x)∂τθ(x)− h2

2mρoθ(x)∇2θ(x)

](159)

We have seen this sort of thing before and we know what to do: we decouple the fields by defining a new

reference with δρ(x) = δo(x) + ξ(x). Let’s call i∂τθ(x) = J(x), A =[λ2 −

h2

2m∇2

4ρo

], and its Green’s function

GA(~x− ~x′), then we simply choose:

δo(x) = −1

2

∫d3x′GA(~x− ~x ′)J(τ, ~x ′)

Then we can see:

SE(δρ, θ) =

∫dτ

∫d3x

[δρ(x)

[λ2− h2

2m

∇2

4ρo

]δρ(x) + iδρ(x)∂τθ(x)− h2

2mρoθ(x)∇2θ(x)

](160)

→ SE(ξ, θ) =

∫dτ

∫d3x

[ξ(x)

[λ2− h2

2m

∇2

4ρo

]ξ(x) +

1

4

∫d3x′∂τθ(τ, ~x)GA(~x− ~x ′)∂τθ(τ, ~x ′)−

h2

2mρoθ(x)∇2θ(x)

](161)

We are almost done! We use the invariance of our integration measure to rewrite our partition function, andthen integrate out ξ, which will just be a normalization constant:

Z =

∫DδρDθe−SE(δρ,θ) (162)

=

∫DξDθe−SE(ξ,θ) (163)

= N∫Dθe−SE(θ) (164)

Where:

SE(θ) =

∫dτ

∫d3x

[− h2

2mρoθ(x)∇2θ(x) +

1

4

∫d3x′∂τθ(τ, ~x)GA(~x− ~x ′)∂τθ(τ, ~x ′)

]We see the kinetic term is non-local as it stands. Man that was a lot of work! The reward was worth it however.

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5. Now we consider slowly varying configurations of our field θ(τ, ~x). We can resolve the issue on non-locality ifwe look at the Fourier transform of our Green’s function:

GA(~x− ~x′) = 8mρo

∫d3~p

(2π)3

ei~p·(~x−~x′)

h2|~p|2 + 4mλρo(165)

≈ 8mρo

∫d3~p

(2π)3

ei~p·(~x−~x′)

4mλρo(166)

=2

λδ(~x− ~x′) (167)

Thus, our action becomes:

SE(θ) =

∫dτ

∫d3x

[ρoh

2

2m

(∇θ(x)

)2

+1

(∂τθ(x)

)2

+ . . .

]

After Wick rotating back, we have:

S(θ) =

∫dt

∫d3x

[ρoh

2

2m

(∇θ(x)

)2

− 1

(∂tθ(x)

)2

+ . . .

](168)

=

∫dt

∫d3x θ(x)

[− ρoh

2

2m∇2 +

1

2λ∂2t + . . .

]θ(x) (169)(

θ →√

m

ρoh2 θ

)→∫dt

∫d3x θ(x)

1

2

[−∇2 +

m

h2ρoλ∂2t + . . .

]θ(x) (170)

=

∫dt

∫d3x θ(x)

1

2

[−∇2 +

1

v2∂2t + . . .

]θ(x) (171)

This is just the action of a massless Klein-Gordon field with the speed of light equal to v2 = ρoh2λ

m = µh2

2m .Indeed, we can see that the propagator for this scalar field is of the form:

G(x− x′) = i⟨T(θ(x)θ(x′)

)⟩(172)

=

∫d2p

(2π)2

i

p2 − ω2

v2 + iεeip(x−x

′) (173)

Which satisfies the following equation of motion:[−∇2 +

1

v2∂2t

]⟨T(θ(x)θ(x′)

)⟩= −iδ(x− x′)

Both are these equations are that of a relativistic massless Klein-Gordon field.

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Appendix: Derivation of Bose Coherent state integral

Let’s find the path integral using the Bose coherent states:

|{φ(x)}〉 = e∫dxφ(x)φ†(x)|0〉

With the coherent states we know that the Resolution of the Identity takes the form:

I =

∫DφDφe−

∫dx|φ(xxx)|2 |{φ}〉〈{φ}|

From here we can repeat the general prescription of slicing up time and inserting the resolution of the identity inbetween each segment. Let’s take: ∆t = εN ; what I am about to do with this should be all too familiar. Breakingthe propagator up into tiny segments and inserting our resolution of our identity between each segment, we find:

〈f |e− ih H∆t|i〉 = 〈f |

[1− i

hH∆t

]N |i〉=

∫ N∏j

DφjDφj e−∑j

∫dx|φj(xxx)|2

[N−1∏k

〈{φk+1}|[1− i

hH∆t

]|{φk}〉

]

×〈f |[1− i ε

hH]|{φN}〉

[1− i ε

hH]〈{φ1}|i〉

Let’s suppose the Hamiltonian is a normal ordered operator, then we know that:

〈{φk+1}|[1− i ε

hH(a†, a)

]|{φk}〉 = 〈{φk+1}|{φk}〉

[1− i ε

hH({φk+1}, {φk}

]Our path integral now becomes:

〈f |e− ih H∆t|i〉 =

∫ [ N∏j

DφjDφj]e−

∑Nj

∫dx|φj(x)|2e−

∑N−1k

∫dxφk+1(x)φk(x)

[N−1∏k

(1− i ε

hH({φk+1}, {φk})

)]

×〈f |[1− i ε

hH]|{φN}〉〈{φ1}|

[1− i ε

hhH]|i〉

Finally, let’s expand the initial and final states using our coherent states; we find:

〈f | = 〈f |I =

∫DφfDφfe−

∫dx|φf (xxx)|2〈f |{φf}〉〈{φf}| =

∫DφfDφfe−

∫dx|φf (xxx)|2 ψf (φ(x, tf ))〈{φf}|

and,

|i〉 = I|i〉 =

∫DφiDφie−

∫dx|φi(xxx)|2 |{φi}〉〈{φi}|i〉 =

∫DφiDφie−

∫dx|φi(xxx)|2 |{φi}〉ψi(φ(x, ti))

We then see that the last term becomes:

〈f |[1− i ε

hH]|{φN}〉〈{φ1}|

[1− i ε

hhH]|i〉 =

∫ ∫DφfDφfDφiDφi ψf

(φ(x, tf )

)ψi(φ(x, ti)

)e−

∫dx

(|φf (xxx)|2+|φi(xxx)|2

)(174)

×[1− i ε

hH({φf}, {φN})

][1− i ε

hH({φ1}, {φi})

]〈{φf}|{φN}〉〈{φ1}|{φi}〉 (175)

The remaining bits are the same as the coherent state path integral derived in the chapter: we expand theoverlaps, group like terms, and integrate by parts. Our final result is:∫

DφDφ exp

{i

h

∫ tf

ti

dt

(∫ddx

h

i

[φ(x, t)∂tφ(x, t)− φ(x, t)∂tφ(x, t)

]−H[φ, φ]

)}

×ψf (φ(x, tf ))ψi(φ(x, ti))e12

∫d3x

(|φ(x,tf )|2+|φ(x,ti)|2

)

24