1QFT in zero dimensions

III Advanced Quantum Field Theory

1.1 Free theories

We consider the simplest possible QFT. These QFT’s are free, and so

S

(

φ

) is at

most quadratic. Classically, this implies the equations of motions are linear, and

so there is superposition, and thus the particles do not interact.

Let φ : {pt} → R

n

be a field with coordinates φ

a

, and define

S(φ) =

1

2

M(φ, φ) =

1

2

M

ab

φ

a

φ

b

,

where

M

:

R

n

× R

n

→ R

is a positive-definite symmetric matrix. Then the

partition function Z(M) is just a Gaussian integral:

Z(M ) =

Z

R

n

d

n

φ e

−

1

2~

M(φ,φ)

=

(2π~)

n/2

√

det M

.

Indeed, to compute this integral, since

M

is symmetric, there exists an orthogonal

transformation

O

:

R

n

→ R

n

that diagonalizes it. The measure d

n

φ

is invariant

under orthogonal transformations. So in terms of the eigenvectors of

M

, this

just reduces to a product of

n

1D Gaussian integrals of this type, and this is a

standard integral:

Z

dχ e

−mχ

2

/2~

=

r

2π~

m

.

In our case,

m >

0 runs over all eigenvalues of

M

, and the product of eigenvalues

is exactly det M.

A small generalization is useful. We let

S(φ) =

1

2

M(φ, φ) + J(φ),

where

J

:

R

n

→ R

is some linear map (we can think of

J

as a (co)vector, and

also write

J

(

φ

) =

J · φ

).

J

is a source in the classical case. Then in this theory,

we have

Z(M, J) =

Z

R

n

d

n

φ exp

−

1

~

1

2

M(φ, φ) + J(φ)

.

To do this integral, we complete the square by letting

˜

φ

=

φ

+

M

−1

J

. In other

words,

˜

φ

a

= φ

a

+ (M

−1

)

ab

J

b

.

The inverse exists because

M

is assumed to be positive definite. We can now

complete the square to find

Z(M, J) =

Z

R

n

d

n

˜

φ exp

−1

2~

M(

˜

φ,

˜

φ) +

1

2~

M

−1

(J, J)

= exp

1

2~

M

−1

(J, J)

Z

R

n

d

n

˜

φ exp

−1

2~

M(

˜

φ,

˜

φ)

= exp

1

2~

M

−1

(J, J)

(2π~)

n/2

√

det M

.

In the long run, we really don’t care about the case with a source. However, we

will use this general case to compute some correlation functions.

We return to the case without a source, and let

P

:

R

n

→ R

be a polynomial.

We want to compute

hP (φ)i =

1

Z(M )

Z

R

n

d

n

φ P (φ) exp

−

1

2~

M(φ, φ)

.

By linearity, it suffices to consider the case where P is just a monomial, so

P (φ) =

m

Y

i=1

(`

i

(φ)),

for

`

i

:

R

n

→ R

linear maps. Now if

m

is odd, then clearly

hP

(

φ

)

i

= 0, since

this is an integral of an odd function. When m = 2k, then we have

hP (φ)i =

1

Z(M )

Z

d

n

φ (`

i

· φ) ···(`

2k

· φ) exp

−

1

2~

M(φ, φ) −

J · φ

~

.

Here we are eventually going to set

J

= 0, but for the time being, we will be

silly and put the source there. The relevance is that we can then think of our

factors `

i

· φ as derivatives with respect to J:

hP (φ)i =

(−~)

2k

Z(M )

Z

d

n

φ

2k

Y

i=1

`

i

·

∂

∂J

exp

−

1

2~

M(φ, φ) −

J · φ

~

Since the integral is absolutely convergent, we can move the derivative out of

the integral, and get

=

(−~)

2k

Z(M )

2k

Y

i=1

`

i

·

∂

∂J

Z

d

n

φ exp

−

1

2~

M(φ, φ) −

J · φ

~

= ~

2k

2k

Y

i=1

`

i

·

∂

∂J

exp

1

2~

M

−1

(J, J)

.

When each derivative `

i

·

∂

∂J

acts on the exponential, we obtain a factor of

1

~

M

−1

(J, `

i

).

in front. At the end, we are going to set

J

= 0. So we only get contributions if

and only if exactly half (i.e.

k

) of the derivatives act on the exponential, and the

other k act on the factor in front to get rid of the J.

We let

σ

denote a (complete) pairing of the set

{

1

, ··· ,

2

k}

, and Π

2k

be the

set of all such pairings. For example, if we have

k

= 2, then the possible pairings

are {(1, 2), (3, 4)}, {(1, 3), (2, 4)} and {(1, 4), (2, 3)}:

In general, we have

|Π

2k

| =

(2k)!

2

k

k!

,

and we have

Theorem (Wick’s theorem). For a monomial

P (φ) =

2k

Y

i=1

`

i

(φ),

we have

hP (φ)i = ~

k

X

σ∈Π

2k

Y

i∈{1,···,2k}/σ

M

−1

(`

i

, `

σ(i)

).

where the

{

1

, ··· ,

2

k}/σ

says we sum over each pair

{i, σ

(

i

)

}

only once, rather

than once for (i, σ(i)) and another for (σ(i), i).

This is in fact the version of Wick’s theorem for this 0d QFT, and

M

−1

plays

the role of the propagator.

For example, we have

h`

1

(φ)`

2

(φ)i = ~M

−1

(`

1

, `

2

).

We can represent this by the diagram

1 2M

−1

Similarly, we have

h`

1

(φ) ···`

4

(φ)i = ~

2

M

−1

(`

1

, `

2

)M

−1

(`

3

, `

4

)

+ M

−1

(`

1

, `

3

)M

−1

(`

2

, `

4

) + M

−1

(`

1

, `

4

)M

−1

(`

2

, `

3

)

.

Note that we have now reduced the problem of computing an integral for the

correlation function into a purely combinatorial problem of counting the number

of ways to pair things up.