III Advanced Quantum Field Theory - Symmetries of the path integral

3Symmetries of the path integral

III Advanced Quantum Field Theory

3.1 Ward identities

The derivation of Noether’s theorem used the classical equation of motion. But in

a quantum theory, the equation of motion no longer holds. We must re-examine

what happens in the quantum theory.

Suppose a transformation φ 7→ φ

of the fields has the property that

Dφ e

−S[φ]

= Dφ

−S[φ

]

In theory, the whole expression D

φ e

−S[φ]

is what is important in the quantum

theory. In practice, we often look for symmetries where D

and

−S[φ]

are

separately conserved. In fact, what we will do is that we look for a symmetry

that preserves

[

], and then try to find a regularization of D

that is manifestly

invariant under the symmetry.

Often, it is not the case that the obvious choice of regularization of D

manifestly invariant under the symmetry. For example, we might have a

[

]

that is rotationally invariant. However, if we regularize the path integral measure

by picking a lattice and sampling

on different points, it is rarely the case that

this lattice is rotationally invariant.

In general, there are two possibilities:

(i)

The symmetry could be restored in the limit. This typically means there

exists a regularized path integral measure manifestly invariant under this

symmetry, but we just didn’t use it. For example, rotational invariance is

not manifestly present in lattice regularization, but it is when we do the

cut-off regularization.

(ii)

It could be that the symmetry is not restored. It is said to be anomalous,

i.e. broken in the quantum theory. In this case, there can be no invariant

path integral measure. An example is scale invariance in QED, if the mass

of the electron is 0.

Sometimes, it can be hard to tell. For now, let’s just assume we are in a situation

where Dφ = Dφ

when ε is constant. Then for any ε(x), we clearly have

Z =

Dφ e

−S[φ]

Dφ

−S[φ

]

since this is just renaming of variables. But using the fact that the measure is

invariant when

is constant, we can can expand the right-hand integral in

and again argue that it must be of the form

Dφ

−S[φ

]

Dφ e

−S[φ]



1 −

∂

ε d



in first order in

. Note that in general,

can receive contributions from

[

]

and D

. But if it doesn’t receive any contribution from D

, then it would just

be the classical current.

Using this expansion, we deduce that we must have

Dφ e

−S[φ]

∂

ε d

x = 0.

Integrating by parts, and using the definition of the expectation, we can write

this as

−

ε∂

(x)i d

for any

with compact support. Note that we dropped a normalization factor

in the definition of

(

)

, because

times zero is still zero. So we know

that hj

(x)i is a conserved current, just as we had classically.

Symmetries of correlation functions

Having a current is nice, but we want to say something about actual observable

quantities, i.e. we want to look at how how symmetries manifest themselves with

correlation functions. Let’s look at what we might expect. For example, if our

theory is translation invariant, we might expect, say

hφ(x)φ(y)i = hφ(x −a)φ(y − a)i

for any a. This is indeed the case.

Suppose we have an operator

(

). Then under a transformation

φ 7→ φ

our operator transforms as

O(φ) 7→ O(φ

By definition, the correlation function is defined by

hO(φ)i =

Dφ e

−S[φ]

O(φ).

Note that despite the appearance of

on the left, it is not a free variable. For

example, the correlation hφ(x

)φ(x

)i is not a function of φ.

We suppose the transformation

φ 7→ φ

is a symmetry. By a trivial renaming

of variables, we have

hO(φ)i =

Dφ

−S[φ

]

O(φ

)

By assumption, the function D

φ e

−S[φ]

is invariant under the transformation.

So this is equal to

Dφ e

−S[φ]

O(φ

)

= hO(φ

)i.

This is, of course, not surprising. To make this slightly more concrete, we look

at an example.

Example.

Consider (

M, g

) = (

, δ

), and consider spacial translation

x 7→ x

x − a for a constant vector a. A scalar field φ transforms as

φ(x) 7→ φ

(x) = φ(x − a).

In most cases, this is a symmetry.

We suppose O(φ) can be written as

O(φ) = O

(φ(x

)) ···O

(φ(x

)),

where

depends only on the value of

. A canonical example is when

O(φ) = φ(x

) ···φ(x

) is an n-point correlation function.

Then the above result tells us that

(φ(x

)) ···O

(φ(x

))i = hO

(φ(x

− a)) ···O

(φ(x

− a))i

So the correlation depends only on the separations

−x

. We can obtain similar

results if the action and measure are rotationally or Lorentz invariant.

Example.

Suppose we have a complex field

, and we have a transformation

φ 7→ φ

iα

for some constant

α ∈ R/

πZ

. Then the conjugate field

transforms as

φ 7→

= e

−iα

φ.

Suppose this transformation preserves the action and measure. For example,

the measure will be preserved if we integrate over the same number of

and

modes. Consider the operators

(φ,

φ) = φ(x

)

φ(x

)

Then the operators transform as

(φ,

φ) 7→ O

(φ

) = e

−α(r

−s

)

(φ,

φ).

So symmetry entails

i=1

)

= exp

iα

i=1

− s

)

i=1

)

Since this is true for all α, the correlator must vanish unless

i=1

So we need the same number of φ and

φ insertions in total.

We can interpret this in terms of Feynman diagrams — each propagator joins

up a

and

. So if we don’t have equal number of

and

, then we can’t draw

any Feynman diagrams at all! So the correlator must vanish.

Ward identity for correlators

What we’ve done with correlators was rather expected, and in some sense trivial.

Let’s try to do something more interesting.

Again, consider an operator that depends only on the value of

at finitely

many points, say

O(φ) = O

(φ(x

)) ···O

(φ(x

)).

As before, we will write the operators as O

) when there is no confusion.

As in the derivation of Noether’s theorem, suppose we have an infinitesimal

transformation, with

φ 7→ φ

εδφ

that is a symmetry when

is constant. Then

for general ε(x), we have

Dφ e

−S[φ]

(φ(x

)) ···O

(φ(x

))

Dφ

−S[φ

]

(φ

)) ···O

(φ

))

Dφ e

−S[φ]



1 −

(x)∂

ε(x) dx







(φ(x

)) ···O

(φ(x

))

i=1

ε(x

)δO

)

j6=i

)





where

δO(x

) =

∂O

∂φ

δφ.

Again, the zeroth order piece of the correlation function cancels, and to lowest

non-trivial order, we find that we must have

∂

ε(x)

(x)

i=1

)

x =

i=1

ε(x

)

δO

)

j6=i

)

On the left, we can again integrate by parts to shift the derivative to the current

term. On the right, we want to write it in the form

(

)

···

, so that we

can get rid of the

term. To do so, we introduce some

-functions. Then we

obtain

ε(x)∂

(x)

i=1

)

= −

i=1

ε(x)δ

(x − x

)

δO

)

j6=i

)

Since this holds for arbitrary ε(x) (with compact support), we must have

∂

(x)

)

= −

i=1

(x − x

)

δO

)

j6=i

)

This is the Ward identity for correlation functions. It says that the vector field

(x, x

) =

(x)

)

is divergence free except at the insertions x

This allows us to recover the previous invariance of correlations. Suppose

is compact without boundary. We then integrate the Ward identity over all

By Stokes’ theorem, we know the integral of any divergence term vanishes. So

we obtain

0 =

∂

(x, x

) d

x = −

i=1

δO

)

j6=i

)

= δ

i=1

)

Of course, this is what we would obtain if we set ε(x) ≡ 1 above.

That was nothing new, but suppose

⊆ M

is a region with boundary

− N

Let’s see what integrating the Ward identity over

gives us. The left hand

side gives us

∂

(x)

i=1

)

x =

−N

(x)

i=1

)

d−1

Q[N

]

i=1

)

−

Q[N

]

i=1

)

The right hand side of Ward’s identity just gives us the sum over all points inside

Q[N

]

i=1

)

−

Q[N

]

i=1

)

= −

∈M

δO

)

j6=i

)

In particular, if

contains only one point, say

, and we choose the region to

be infinitesimally thin, then in the canonical picture, we have

hΩ|T [

)]

j=2

) |Ωi = −hΩ|T







j=2

)







|Ωi,

where |Ωi is some (vacuum) state. So in the canonical picture, we find that

O = −[

O].

So we see that the change of

under some transformation is given by the

commutator with the charge operator.