Problem Set 3 - SolutionsDue: Sept. 21PHY 342 Fall 2006

1 Schwinger-Dyson Equations

As we mentioned in class, writing the path integral with the translational-invariant measure

Dφ ∼∏

i

dφi (1)

and the classical action is not the correct quantum mechanical formulation ofQFT. Starting from that, however, we can calculate the corrections to the action(and therefore to the equations of motion of the fields), and use the quantumaction (or the “effective” potential) in writing down the path integral. Anotherstrategy is to use the classical action, and to stick the quantumness into thepath integral measure. Assume that you can write the measure as the productof a translational-invariant part and a field-dependent functional part µ[φ], i.e.have a partition function of the form

Z[J ] =∫Dφµ[φ]eiS[φ]+i(J,φ). (2)

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(a) Compute the quantum equation of motion for φ for correlation functionsof a functional F [φ] in the presence of a source as we did in class, i.e.demanding invariance of the correlator

〈F [φ]〉J (3)

under a simple shift of variables

φ(x) → φ′(x) = φ(x) + ε(x), (4)

where ε(x) is a function that is infinitesimal everywhere.

Under the infinitesimal shift of variables, the functional expectation value becomes

〈F [φ]〉J ≡∫

Dφµ[φ]eiS[φ]+i(J,φ)F [φ]

=∫

Dφ′ µ[φ′]eiS[φ′]+i(J,φ′)F [φ′]

=∫

Dφµ[φ′]eiS[φ′]+i(J,φ′)F [φ′]

=∫

Dφ

(µ[φ] +

∫dx

δµ[φ]δφ(x)

ε(x))

× eiS[φ]+i(J,φ)

(1 + i

∫dx

{δS[φ]δφ(x)

+ J(x)}

ε(x))

×(

F [φ] +∫

dxδF [φ]δφ(x)

ε(x))

+ [ε2].

(5)

In the second line we relabeled φ(x) → φ′(x) = φ(x) + ε(x), and in the thirdline we used translational invariance of the measure. Since ε(x) is arbitrary andinfinitesimal, we have∫

Dφ

(δµ[φ]δφ(x)

F [φ]µ[φ]

+ i

(δS[φ]δφ(x)

+ J(x))

F [φ] +δF [φ]δφ(x)

)µ[φ]eiS[φ]+i(J,φ) = 0,

and thus we find⟨[δ log(µ[φ])

δφ(x)+ i

(δS[φ]δφ(x)

+ J(x))]

F [φ] +δF [φ]δφ(x)

⟩= 0 (6)

or ⟨[Eφ(x) + i

δ log(µ[φ])δφ(x)

F [φ] + +J(x)]

F [φ]⟩

= i

⟨δF [φ]δφ(x)

⟩, (7)

where we remember that the expectation value is defined with the modified measureµ[φ] included. (This is what creates the logarithm from the variation of µ[φ]). Wesee that in the limit where the source is set to 0 and µ set to 1 we recover thetraditional Schwinger-Dyson equations.

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(b) Do the same for an infinitesimal symmetry transformation, i.e. assumeinvariance of the correlator

〈F [φ]〉J (8)

under the change of variables

φ(x) → φ′(x) = φ(x) + αf(φ), (9)

where f(φ) is a function of the field φ (but not derivatives of the field).Find the appropriate Schwinger-Dyson equations for F [φ] in the presenceof a source.

We are to assume that the transformation of the field given by Eq.(9) is a symmetry,and thus product of the exponential of the action and the quantum measure Dφµ[φ]transform as

Dφµ[φ]eiS[φ] → Dφµ[φ]eiS[φ](−iα)∂µjµ, (10)

which really is the statement ⟨∂µjµ(x)

⟩= 0. (11)

We thus only need to take into account the change of the functional F [φ]:

F [φ] → F [φ] +∫

dx αδF [φ]δφ(x)

f(φ(x)). (12)

We find ∫dx

⟨iα(x) [−∂µjµ(x) + Jf(φ)]F [φ] + α(x)

δF [φ]δφ(x)

f(φ)⟩

= 0.

We simply rewrite this as the modified conservation equation for the current:⟨∂µjµ(x)F [φ]

⟩=

⟨ [J(x)− i

δF [φ]δφ(x)

]f(φ)

⟩. (13)

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2 Symmetries - General Considerations

Show that if we have two conserved charges Q1 and Q2 that generateinfinitesimal transformations

δαiφ = αi[Qi, φ] = αifi(φ), i = 1, 2 (no summation) (14)

their commutator[Q1, Q2] (15)

generates symmetry transformations as well. Assume that the Jacobi iden-tity for triple commutators holds.

It is intuitively clear that the composition of two symmetry variations should giveanother symmetry variation. After all, if we subject the system to two transfor-mation that does not change the physics, the physics, well, should be unchanged.Likewise we should find an invariance if we combine the transformations to a singleone. This is cumbersome language for the simple statement that the symmetrygenerators form a (closed) algebra under the Poisson bracket.The statement that the charges generate symmetries of the action is equivalent to

[Qi,L] = ∂µKµi (16)

for some four-vectors Kµi . Then, using the Jacobi identity, the commutator of those

charges gives

[[Q1, Q2],L] = [Q1, [Q2,L]]− [Q2, [Q1,L]]= [Q1, ∂µKµ

2 ]− [Q2, ∂µKµ1 ]

= ∂µ

([Q1,K

µ2 ]− [Q2,K

µ1 ]︸ ︷︷ ︸

Kµ3

) (17)

Thus [[Q1, Q2],L] = ∂µKµ3 , which implies that it also generates a symmetry of the

action. Since we have found K3, the current jµ3 can be easily computed as well.

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3 Symmetries - Scalings

In massless φ4 in four, consider an infinitesimal spacetime rescaling

xµ → x̃µ = xµ + αxµ

with infinitesimal constant parameter α.

(a) How does the field φ transform under this scaling? How would it trans-form if the dimension of spacetime was two? How does it transform in dspacetime dimensions? What is the finite form of scale transformations,both on xµ and φ in d dimensions?

We have to be careful to include bot

4 Supersymmetry

Do problem 3.5 in Peskin & Schroeder, with (c) for extra credit.h the transformationin field space due to the field’s scaling dimension and the coordinate transformationitself. The term 1

2∂µφ∂µφ determines the scaling dimension, ∆φ, of the field φ sincethe action is dimensionless:

2 + 2∆φ = d ⇒ ∆φ =d

2− 1

Thus the infinitesimal transformation of the field is as follows:

φ′(x) = (1 + α)−∆φφ((1 + α)−1x)= ((1− α(∆φ + xµ∂µ))φ) (x)

For d = 4,∆φ = 1. For d = 2,∆φ = 0, and so in d = 2,the coordinate transformationis the only part of the transformation; the operation on the field itself is the identity.(That is, in d = 2 we can put the transformation in the form F (φ)(x′) = φ(x), withthat F (φ) = φ.) The finite form of this transformation is

x′µ = αx

φ′(x′) = α−∆φφ(αx)

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(b) In four dimensions, construct the Noether current sµ and the conservedcharge. Check that the charge is conserved.

The Lagrangian density of the system varies as follows under an infinitesimal dila-tion with parameter α:

S′ =∫

d4xL(φ′(x))

=∫

d4xL((1− α(1 + xµ∂µ))φ)(x)

S′ − S = α

∫d4x − ∂µ((1 + xν∂ν)φ)∂µφ +

λ

3!φ3(1 + xν∂ν)φ

= α

∫d4x ∂ν

(−xν

(12∂µφ∂µφ− λ

4!φ4

))L′ − L = α∂ν(−xνL)

jµ =∂L

∂(∂µφ)δφ

α+ xµL

= −∂µφ(1 + xν∂ν)φ + xµL= xν (ηνµL − ∂νφ∂µφ)− φ∂µφ

= −xνT νµ − φ∂µφ

(To verify the form of Tµν , see the fourth part.)This charge Q =

∫space

dx s0 is conserved for configurations obeying the equationsof motion (⇒ ∂µTµν = 0, ∂2φ + λ

3!φ3 = 0):

∂µjµ = −(Tµ

µ + ∂µφ∂µφ + φ∂2φ)

= −(−∂µφ∂µφ +

λ

3!φ4 + ∂µφ∂µφ + φ∂2φ

)= −φ

(∂2φ +

λ

3!φ3

)= 0

provided the fields obey appropriate spatial boundary (convergence) conditions.

Aside: In QFT, the current is no longer conserved due to anomalies. In fact,classically any theory that only has dimensionless couplings is scale (or dilation)invariant.

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(c) Add a mass term and repeat the analysis. What goes wrong?

Adding a mass m2φ2 term breaks the scaling symmetry. The scaling dimension ofφ remains the same, but when the mass term is scaled it changes the action by

∆S =∫

ddx 2m2φ(∆ + xµ∂µ)φ

∂µ(xµm2φ2) = dm2φ2 + 2m2φxµ∂µφ

=∫

ddx (2∆− d)m2φ2 + ∂µ(xµm2φ2)

=∫

ddx (2∆− d)m2φ2

for boundary conditions where the fields go to zero at infinity. 2∆ = d− 2 6= d forall values of d, and so adding a mass term must change the action for all values ofd.

(d) Construct the energy-momentum tensor Tµν for massless φ4.

Tµν =∂L

∂(∂µφ)∂νφ− ηµνL

= ∂µφ∂νφ− ηµν

(12∂µφ∂µφ− λ

4!φ4

)

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5 Symmetries - Conformal Transformations

(a) In a large class of theories it is possible to define a an energy momentumtensor θµν such that

sµ = xνθµν . (18)

What is the relation between scale invariance and properties of this energy-momentum tensor?

Conservation of the scale invariance Noether current will imply that this new stressenergy tensor is traceless (when it is also conserved):

∂µsµ = ∂µ(xνθµν)= θµ

µ + xν∂µθµν

= θµµ = 0

(b) We can construct from the energy momentum tensor for a scale invarianttheory not only the conserved scale current, but four other conservedcurrents:

Kλµ = xνxνθλµ − 2xλxνθµν . (19)

Check that∂µKλµ = 0. (20)

∂µKλµ = ∂µ

(xνxνθλµ − 2xλxνθµν

)= 2xµθλµ +������:0

xνxν∂µθλµ − 2xνθλν −����*0

2xλθµµ

= 0

wherein we used conservation and tracelessness of the new stress energy tensor.

Thus it appears that for theories for which Eq.(18) holds, exact scale in-variance implies invariance under four other (at the moment mysterious) in-finitesimal transformations (remember that Noether’s theorem goes both ways:for every conserved current there is a symmetry). What are these mysterioustransformations? Consider the simplest theory, the theory of a massless scalarfield in four dimensions. Since it is a free theory, its complete quantum dynamicsis determined by Green’s function of the kinetic term operator, the two-pointfunction

〈0|φ(x)φ(y)|0〉. (21)

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(c) Given that the equation of motion is

∂µ∂µφ = 0, (22)

what is the two-point function Eq.(21)?

The two-point function is given by the four-dimensional solution of Laplace’s equa-tion

〈φ(x)φ(y)〉 =1

2π2

1(x− y)2

. (23)

The functional dependence can simply be guessed by dimensional analysis. We canalso use the Schwinger-Dyson equations for φ:⟨(

δ

δφ(x)

∫d4x′ 1

2∂µφ∂µφ

)φ(0)

⟩= i 〈δ(x)〉

∂2xG(x) = −iδ(x)

where G(x) ≡ 〈φ(x)φ(0)〉. Then we Fourier transform to obtain:

−p2G̃(p) = −i

G̃(p) =i

p2→ i

p2 + iε

In coordinate space, this is

〈φ(x)φ(y)〉 =∫

d4p

(2π)4e−ip·(x−y) · i

p2 + iε

wherein the translation invariance of 〈φ(x)φ(y)〉 was used.

The wave equation Eq.(22) is similar to the three-dimensional Laplace equa-tion. This equation possesses geometrical symmetries beyond the usual Eu-clidean transformations; for instance, it is invariant under the three-dimensionalinversion of coordinates:

I : ~x → ~x′ =~x

~x2. (24)

This suggests the idea that the theory of a free scalar field may be invariantunder the Minkowski-space inversion

I : xµ → x′µ = −xµ

x2, (25)

with x2 =∑

ν xνxν .

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(d) How does the field φ have to transform for the two-point function to beinvariant if under the inversion

I : (x− y)2 → (x− y)2

x2y2? (26)

Under the inversion the two point function is invariant if we define the field totransform as

I : φ(x) → 1x2

φ

(−x

x2

)(27)

The inversion is a discrete transformation and thus of no use obtaining con-served currents; for these we need continuous transformations. However, we canuse the inversion and apply it to other continuous transformations to obtain newcontinuous ones.

(e) For example, take the four-parameter group of translations

a : xµ → x′µ = xµ + aµ. (28)

Apply the inversion of Eq.(25) to both sides of the equation to obtain thefinite form of conformal transformations,

xµ → x′µ =xµ − 2aµx2

1− xµaµ + a2x2. (29)

What is the infinitesimal form of these transformations?

Applying xµ → −xµ/x2 to both sides of the transformation

x′µ = xµ + aµ

we obtain

x′µ

(x′)2=

xµ

x2− aµ

1/(x′)2 =1− 2a · x + a2x2

x2

x′µ =x2

1− 2a · x + a2x2

xµ − x2aµ

x2

=xµ − x2aµ

1− 2a · x + a2x2

The infinitesimal form of this transformation is calculated by making a infinitesimal:

x′µ = (xµ − x2aµ)(1 + 2a · x) + [a2]

= (1 + 2(a · x))xµ − x2aµ + [a2]

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No new transformations are obtained by applying the inversion to infinites-imal Lorentz transformations or dilations. The set of infinitesimal variations ofthe conformal algebra, Lorentz transformations, translations, conformal trans-formations and dilations gives a fifteen parameter group upon exponentionatingthe generators of the algebra: the conformal group. A Lorentz invariant theorythat is invariant under conformal transformations is automatically scale invari-ant, however the converse need not be true.

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