III Advanced Quantum Field Theory - QFT in one dimension (i.e. QM)

2QFT in one dimension (i.e. QM)

III Advanced Quantum Field Theory

2.1 Quantum mechanics

= 1, there are two possible connected, compact manifolds

— we can

have M = S

or M = I = [0, 1].

M = I

We will mostly be considering the case where

. In this case, we need to

specify boundary conditions on the field in the path integral, and we will see that

this corresponds to providing start and end points to compute matrix elements.

We let

t ∈

1] be the worldline coordinate parametrizing the field, and we

write our field as

x : I → N

for some Riemannian manifold (

N, g

), which we call the target manifold. If

U ⊆ N

has coordinates

with

= 1

, ··· , n

dim

(

), then we usually write

(t) for the coordinates of x(t). The standard choice of action is

S[x] =



g( ˙x, ˙x) + V (x)



dt,

where ˙x is as usual the time derivative, and V (x) is some potential term.

We call this theory a non-linear

-model. This is called

-model because

when people first wrote this theory down, they used

for the name of the field,

and the name stuck. This is non-linear because the term

(

˙x, ˙x

) can be very

complicated when the metric

if it is not flat. Note that +

(

) is the correct

sign for a Euclidean worldline.

Classically, we look for the extrema of

[

] (for fixed end points), and the

solutions are given by the solutions to

+ Γ

˙x

= g

∂V

∂x

The left-hand side is just the familiar geodesic equation we know from general

relativity, and the right hand term corresponds to some non-gravitational force.

In the case of zero-dimensional quantum field theory, we just wrote down

the partition function

Z =

−S

and assumed it was an interesting thing to calculate. There wasn’t really a

better option because it is difficult to give a physical interpretation to a zero-

dimensional quantum field theory. In this case, we will see that path integrals

naturally arise when we try to do quantum mechanics.

To do quantum mechanics, we first pick a Hilbert space

. We usually take

it as

H = L

(N, dµ),

the space of square-integrable functions on N.

To describe dynamics, we pick a Hamiltonian operator

H → H

, with the

usual choice being

H =

∆ + V,

where the Laplacian ∆ is given by

∆ =

√

∂

(

√

∂

As usually, the

√

g refers to the square root of the determinant of g.

We will work in the Heisenberg picture. Under this set up, the amplitude for

a particle to travel from x ∈ N to y ∈ N in time T is given by the heat kernel

(y, x) = hy|e

−HT

|xi.

Note that we have

−HT

instead of

−iHT

because we are working in a Euclidean

world. Strictly speaking, this doesn’t really make sense, because

|xi

and

|yi

are not genuine elements of

(

). They are

-functions, which aren’t

functions. There are some ways to fix this problem. One way is to see that the

above suggests K

satisfies

∂

∂t

(y, x) + HK

(y, x) = 0,

where we view

as a fixed parameter, and

is a function of

, so that

can

act on K

. The boundary condition is

lim

t→0

(y, x) = δ(y − x),

and this uniquely specifies

. So we can define

to be the unique solution to

this problem.

We can reconnect this back to the usual quantum mechanics by replacing

the Euclidean time

with

, and the above equation gives us the Schr¨odinger

equation

∂K

∂t

(y, x) = HK

(y, x).

We first focus on the case where

= 0. In the case where (

N, g

) = (

, δ

), we

know from, say, IB Methods, that the solution is given by

(y, x) =

(2πt)

n/2

exp



−

|x − y|



For an arbitrary Riemannian manifold (

N, g

), it is in general very hard to write

down a closed-form expression for

. However, we can find the asymptotic form

lim

t→0

(y, x) ∼

a(x)

(2πt)

n/2

exp



−

d(y, x)



where

(

x, y

) is the geodesic distance between

and

, and

(

) is some invariant

of our manifold built from (integrals of) polynomials of the Riemann curvature

that isn’t too important.

Here comes the magic. We notice that since

I =

z |zihz|

is the identity operator, we can write

(y, x) = hy|e

−T H

|xi

z hy|e

−t

|zihz|e

−t

|xi

z K

(y, z)K

(z, x).

For flat space, this is just reduces to the formula for convolution of Gaussians.

This is the concatenation property of the heat kernel.

Using this many times, we can break up our previous heat kernel by setting

∆t =

for some large N ∈ N. Then we have

(y, x) =

N−1

i=1

∆t

, x

i−1

where we conveniently set x

= x and x

= y.

The purpose of introducing these ∆

is that we can now use the asymptotic

form of K

(y, t). We can now say

−HT

= lim

N→∞



2π∆t



nN/2

N−1

i=1

a(x

) exp

−

∆t



d(x

i+1

, x

)

∆t



This looks more-or-less like a path integral! We now dubiously introduce the

path integral measure

≡ lim

N→∞



2π∆t



nN/2

N−1

i=1

a(x

and also assume our map

(

) is at least once-continuously differentiable, so that

lim

N→∞

N−1

i=1

exp

−

∆t



d(x

i+1

, x

)

∆t



≡ exp



−

dt g( ˙x, ˙x)



= exp(−S[x]).

Assuming these things actually make sense (we’ll later figure out they don’t),

we can write

−HT

i =

]

Dx e

−S[x]

where

[

, y

] is the space of “all” maps

I → N

such that

(0) =

and

x(1) = y

Now consider an arbitrary V 6= 0. Then we can write

H = H

+ V (x),

where H

is the free Hamiltonian. Then we note that for small t, we have

−Ht

= e

−H

−V (x)t

+ o(t).

Thus, for small t, we have

(y, x) = hy|e

−Ht

|xi

≈ hy|e

−H

−V (x)t

|xi

∼

a(x)

(2πt)

n/2

exp

−



d(y, x)



+ V (x)

Then repeating the above derivations will again give us

−HT

i =

]

Dx e

−S[x]

Before we move on to express other things in terms of path integrals, and

then realize our assumptions are all wrong, we make a small comment on the

phenomena we see here.

Notice that the states

i ∈ H

and

| ∈ H

∗

we used to evaluate our

propagator here arise as boundary conditions on the map

. This is a general

phenomenon. The co-dimension-1 subspaces (i.e. subspaces of

of one dimen-

sion lower than

) are associated to states in our Hilbert space

. Indeed, when

we did quantum field theory via canonical quantization, the states corresponded

to the state of the universe at some fixed time, which is a co-dimension 1 slice of

spacetime.

The partition function

We can naturally interpret a lot of the things we meet in quantum mechanics via

path integrals. In statistical physics, we often called the quantity

(

−HT

)

the partition function. Here we can compute it as

−HT

) =

y hy|e

−HT

|yi.

Using our path integral formulation, we can write this as

−HT

) =

[y,y]

Dx e

−S

Dx e

−S

where we integrate over all circles. This is the partition function

(

N, g, V

))

of our theory. If one is worried about convergence issues, then we would have some

problems in this definition. If we work in flat space, then

(

y, y

) =

hy|e

−HT

|yi

is independent of

. So when we integrate over all

, the answer diverges (as

long as

(

y, y

) is non-zero). However, we would get a finite result if we had a

compact manifold instead.

Correlation functions

More interestingly, we can do correlation functions. We will begin by considering

the simplest choice — local operators.

Definition

(Local operator)

A local operator

(

) is one which depends on

the values of the fields and finitely many derivatives just at one point t ∈ M.

We further restrict to local operators that do not depend on derivatives.

These are given by functions

N → R

, and then by pullback we obtain an

operator O(x(t)).

Suppose the corresponding quantum operator is

(

ˆx

), characterized by

the property

O|xi = O(x) |xi.

If we want to evaluate this at time t, then we would compute

O(t) |y

i = hy

−H(T −t)

−Ht

But, inserting a complete set of states, this is equal to

x hy

−H(T −t)

|xihx|

−Ht

|yi

x O(x) hy

−H(T −t)

|xihx|e

−Ht

|yi.

Simplifying notation a bit, we can write

O(t) |y

i =

x O(x(t))

[T,t]

]

Dx e

−S[x]

[t,0]

]

Dx e

−S[x]

But this is just the same as

[T,0]

]

Dx O(x(t))e

−S[x]

More generally, suppose we have a sequence of operators

, ··· , O

we want

to evaluate at times

T > t

> t

n−1

> ··· > t

0, then by the same argument,

we find

−HT

) ···

) |y

i =

[0,T ]

]

Dx O

(x(t

)) ···O(x(t

))e

−S[x]

Note that it is crucial that the operators are ordered in this way, as one would

see if they actually try to prove this. Indeed, the

are operators, but the

objects in the path integral, i.e. the

(

)) are just functions. Multiplication

of operators does not commute, but multiplication of functions does. In general,

if {t

} ∈ (0, T ) are a collection of times, then we have

i=1

(x(t

))e

−S[x]

= hy

−HT

i=1

) |y

where T denotes the time ordering operator. For example, for n = 2, we have

T [

)

)] = Θ(t

− t

)

) + Θ(t

− t

)

where Θ is the step function.

It is interesting to note that in this path integral formulation, we see that

we get non-trivial correlation between operators at different times only because

of the kinetic (derivative) term in the action. Indeed, for a free theory, the

discretized version of the path integral looked like

kin

[x] =



i+1

− x

∆t



∆t,

Now if we don’t have this term, and the action is purely potential:

pot

[x] =

V (x

then the discretized path integral would have factorized into a product of integrals

at these sampling points

. It would follow that for any operators

}

that

depend only on position, we have

(x(t

))

(x(t

))i,

and this is incredibly boring. When we work with higher dimensional universes,

the corresponding result shows that if there are no derivative terms in the

potential, then events at one position in spacetime have nothing to do with

events at any other position.

We have already seen this phenomena when we did quantum field theory

with perturbation theory — the interactions between different times and places

are given by propagators, and these propagators arise from the kinetic terms in

the Lagrangian.

Derivative terms

We now move on to consider more general functions of the field and its derivative.

Consider operators

(

x, ˙x, ···

). We might expect that the value of this operator

is related to path integrals of the form

Dx O

(x, ˙x, ···)|

−S[x]

But this can’t be right. We were told that one of the most important properties

of quantum mechanics is that operators do not commute. In particular, for

= ˙x

, we had the renowned commutator relation

[ˆx

, ˆp

] = δ

But in this path integral formulation, we feed in functions to the path integral,

and it knows nothing about how we order

and

˙x

in the operators

. So what

can we do?

The answer to this is really really important. The answer is that path

integrals don’t work.

The path integral measure

Recall that to express our correlation functions as path integrals, we had to take

the limits

= lim

N→∞

(2π∆t)

nN/2

N−1

i=1

a(x

and also

S[x]

= lim

N→∞

N−1

n=1



n+1

− x

∆t



∆t.

Do these actually make sense?

What we are trying to do with these expressions is that we are trying to

regularize our path integral, i.e. find a finite-dimensional approximation of the

path integral. For quantum field theory in higher dimensions, this is essentially

a lattice regularization.

Before we move on and try to see if this makes sense, we look at another way

of regularizing our path integral. To do so, we decompose our field into Fourier

modes:

(t) =

k∈Z

−2πikt/T

and then we can obtain a regularized form of the action as

[x] =

k=−N

Under this decomposition, we can take the regularized path integral measure to

x =

|k|≤N

This is analogous to high-energy cutoff regularization. Now the natural question

to ask is — do the limits

lim

N→∞

x, lim

N→∞

[x]

exist?

The answer is, again: NO! This is in fact not a problem of us not being able

to produce limits well. It is a general fact of life that we cannot have a Lebesgue

measure on an infinite dimensional inner product space (i.e. vector space with

an inner product).

Recall the following definition:

Definition

(Lebesgue measure)

A Lebesgue measure d

on an inner product

space V obeys the following properties

– For all non-empty open subsets U ⊆ R

, we have

vol(U) =

dµ > 0.

– If U

is obtained by translating U , then

vol(U

) = vol(U).

–

Every

x ∈ V

is contained in at least one open neighbourhood

with

finite volume.

note that we don’t really need the inner product structure in this definition.

We just need it to tell us what the word “open” means.

We now prove that there cannot be any Lebesgue measure on an infinite

dimensional inner product space.

We first consider the case of a finite-dimensional inner product space. Any

such inner product space is isomorphic to

for some

. Write

(

) for an

open hypercube of side length

. By translation in variance, the volume of any

two such hypercubes would be the same.

Now we note that given any such hypercube, we can cut it up into 2

hypercubes of side length L/2:

Then since

(

) contains 2

disjoint copies of

(

2) (note that it is not exactly

the union of them, since we are missing some boundary points), we know that

vol(C(L)) ≥

i=1

vol(C(L/2)) = 2

vol(C(L/2)),

Now in the case of an infinite dimensional vector space,

(

) will contain

infinitely many copies of

(

2). So since

vol

(

2)) must be non-zero, as it

is open, we know

vol

(

)) must be infinite, and this is true for any

. Since

any open set must contain some open hypercube, it follows that all open sets

have infinite measure, and we are dead.

Theorem.

There are no Lebesgue measures on an infinite dimensional inner

product space.

This means whenever we do path integrals, we need to understand that we

are not actually doing an integral in the usual sense, but we are just using a

shorthand for the limit of the discretized integral

lim

N→∞



2π∆t



nN/2

N−1

i=1

exp

−



i+1

− x

∆t



∆t

as a whole. In particular, we cannot expect the familiar properties of integrals

to always hold for our path integrals.

If we just forget about this problem and start to do path integrals, then we

would essentially be writing down nonsense. We can follow perfectly logical steps

and prove things, but the output will still be nonsense. Then we would have

to try to invent some new nonsense to make sense of the nonsense. This was,

in fact, how renormalization was invented! But as we will see, that is not what

renormalization really is about.

Note that we showed that the measure D

doesn’t exist, but what we really

need wasn’t Dx. What we really needed was

Dx e

−S[x]

This is no longer translation invariant, so it is conceivable that it exists. Indeed,

in the case of a 1D quantum field theory, it does, and is known as the Wiener

measure.

In higher dimensions, we are less certain. We know it doesn’t exist for QED,

and we believe it does not exist for the standard model. However, we believe

that it does exist for Yang–Mills theory in four dimensions.

Non-commutativity in QM

Now we know that the path integral measure doesn’t exist, and this will solve

our problem with non-commutativity. Indeed, as we analyze the discretization

of the path integral, the fact that

[ˆx, ˆp] 6= 0

will fall out naturally.

Again, consider a free theory, and pick times

T > t

> t > t

−

> 0.

We will consider

Dx x(t) ˙x(t

−

−S

= hy

−H(T −t)

ˆxe

−H(t−t

−

)

ˆpe

−Ht

−

Dx x(t) ˙x(t

−S

= hy

−H(t−t

)

ˆpe

−H(t

−t)

ˆxe

−Ht

As we take the limit t

→ t, the difference of the right hand sides becomes

−H(t−t)

[ˆx, ˆp]e

−Ht

i = hy

−HT

i 6= 0.

On the other hand, in the continuum path integral, the limit seems to give the

same expression in both cases, and the difference vanishes, naively. The problem

is that we need to regularize. We cannot just bring two operators together in

time and expect it to behave well. We saw that the path integral was sensitive to

the time-ordering of the operators, so we expect something “discrete” to happen

when the times cross. This is just like in perturbation theory, when we bring two

events together in time, we have to worry that the propagators become singular.

Normally, we would have something like

x(t) ˙x(t

−

) − x(t) ˙x(t

) = x



−

− x

−

−δt

δt



− x



+δt

− x

δt



In the regularized integral, we can keep increasing

−

and decreasing

, until

we get to the point



− x

t−∆t

∆t



− x



t+∆t

− x

∆t



Now that

have hit

, we need to look carefully what happens to the individual

heat kernels. In general, we stop taking the limit as soon as any part of the

discretized derivative touches

. The part of the integral that depends on

looks like

∆t

t+∆t

, x



− x

t−∆t

∆t

−

t+∆t

− x

∆t



∆t

, x

t−∆t

Using the fact that

∆t

, x

t−∆t

) ∼ exp



−

− x

t−∆t

)

2∆t



we can write the integral as

∂

∂x



∆t

t+∆t

, x

∆t

, x

t−∆t

)



Now integrating by parts, we get that this is equal to

∆t

t+∆t

, x

∆t

, x

t−∆t

) = K

2∆t

t+∆t

, x

t−∆t

So we get the same as in the operator approach.