Question about Gaussian Integrals and Path Integrals

Jay R. Yablon

I have a question about Gaussian integrals and path integrals. In the
below, I use $ to designate "integral from negative infinity to positive
infinity." I am concerned here only with J=0.

The basic Gaussian integral used in QFT, for J=0, is:

$ dx e^(-.5iax^2) = (2pi i/a)^.5 (2)

When a is promoted to a matrix A, this factor becomes:

$ dx e^(-.5iAx^2) = (2pi i/det A)^.5 (2)

This, as I understand it, works with discretized spacetime. In the
continuum limit, one passes over to the path integral, which, in this
J=0 special case, is:

Z(J=0) = C = Ce^iW(J=0) (3)

This factor C, which corresponds with the determinant in (2), is often
omitted. I would like to know:

Is this factor, once we pass to the continuum, also proportional to
1/det A? Or, is it proportional to the actual inverse of A, that is,
A^-1 (what becomes the propagator)?

Is there any reference that explicitly shows this factor, for, say, a
free scalar field psi with, say:

Z = $ Dpsi e^[i $ d^4x (-.5)psi(d^2+m^2)psi] (4)

Main question: is this a

(d^2+m^2)^-1 = D, i.e., a propagator?

or, is this a

1/det(d^2+m^2)?

In other words, the full propagator of course shows up inside of W(J)
when J <> 0. Does it also end up in C, or rather, does C go with the
reciprocal of the momentum space determinant?

If you have Zee, I am looking at pages 21-22.

Thanks.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Co-moderator: sci.physics.foundations

Igor Khavkine

On 2007-06-17, Jay R. Yablon <jyablon@nycap.rr.com> wrote:
> I have a question about Gaussian integrals and path integrals. In the
> below, I use $ to designate "integral from negative infinity to positive
> infinity." I am concerned here only with J=0.
>
> The basic Gaussian integral used in QFT, for J=0, is:
>
> $ dx e^(-.5iax^2) = (2pi i/a)^.5 (2)
>
> When a is promoted to a matrix A, this factor becomes:
>
> $ dx e^(-.5iAx^2) = (2pi i/det A)^.5 (2)
>
> This, as I understand it, works with discretized spacetime. In the
> continuum limit, one passes over to the path integral, which, in this
> J=0 special case, is:
>
> Z(J=0) = C = Ce^iW(J=0) (3)
>
> This factor C, which corresponds with the determinant in (2), is often
> omitted. I would like to know:
>
> Is this factor, once we pass to the continuum, also proportional to
> 1/det A? Or, is it proportional to the actual inverse of A, that is,
> A^-1 (what becomes the propagator)?

It remains the determinant of A. There is no reason why in the infinite
dimensional limit, det A should suddenly become A^-1. More over, one is
a scalar, while the other is a matrix. They are not interchangeable.

> Is there any reference that explicitly shows this factor, for, say, a
> free scalar field psi with, say:
>
> Z = $ Dpsi e^[i $ d^4x (-.5)psi(d^2+m^2)psi] (4)
>
> Main question: is this a
>
> (d^2+m^2)^-1 = D, i.e., a propagator?
>
> or, is this a
>
> 1/det(d^2+m^2)?

It's definitely the latter. The reason this factor is dropped most of
the time is that it merely corresponds to an overall normalization.
Since we are only interested in the quantity Z(J)/Z(0), it gets
cancelled. Another way to ignore this factor is to invoke the relation
Z(0) = <vac|vac> = 1, which holds if we work with a normalized ground
state.

I can't give a reference off hand to where this determinant is handled
explicitly, but some authors do take this approach. Google for
"functional determinant".

Igor

Roland Franzius

Jay R. Yablon schrieb:
> I have a question about Gaussian integrals and path integrals. In the
> below, I use $ to designate "integral from negative infinity to positive
> infinity." I am concerned here only with J=0.
>
> The basic Gaussian integral used in QFT, for J=0, is:
>
> $ dx e^(-.5iax^2) = (2pi i/a)^.5 (2)
>
> When a is promoted to a matrix A, this factor becomes:
>
> $ dx e^(-.5iAx^2) = (2pi i/det A)^.5 (2)
>
> This, as I understand it, works with discretized spacetime.

Thats a pure classical formula of the norm eg. for the distribution of n
classical gaussian random variables with mean zero and covariance matrix
A. Since A is symmetric diagonalize with std deviations a_i

x = S y, x^+= y^+ S^+
S^+ A S = diag (1/a^2_1,..1/a^2_n)

and you get

int d^nx exp(-x^+ A x/2 ) = int d^ny det S exp(-sum y_i^2/ (a^2_i 2)
= (2pi)^(n/2) det S/det A


--

Roland Franzius