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Functional integral theorem

by Chopin
Tags: functional, integral, theorem
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Apr16-12, 11:09 PM
P: 354
I've been watching Sidney Coleman's QFT lectures ( I've gotten up to his discussion of functional integration, and I have some questions.

He starts out by discussing a finite-dimensional integral of a Gaussian function: [itex]\int{\frac{d^n x}{(2\pi)^{n/2}}e^{-\frac{1}{2}xAx}} = (det A)^{-1/2}[/itex], where [itex]x[/itex] is an n-dimensional vector, and [itex]A[/itex] an n-dimensional symmetric matrix. So far, that makes sense--if you diagonalize [itex]A[/itex], it just turns into the product of [itex]n[/itex] Gaussian integrals. He then goes on to discuss the integral of a polynomial times a Gaussian, [itex]\int{\frac{d^n x}{(2\pi)^{n/2}}P(x)e^{-\frac{1}{2}xAx}}[/itex], where [itex]P(x)[/itex] is a polynomial. Seemingly out of nowhere, he gives the result of this integral as [itex]P(-\frac{\partial}{\partial b})(det A)^{-1/2}[/itex]. I have absolutely no idea where this comes from.

Google is turning up bits and pieces of information on this, but nothing I can make a complete picture out of. The best I've been able to work out is that it's in some way related to differentiating under the integration sign, but I can't quite put the pieces together. This is clearly going to become important in the subsequent sections, where we're going to go on to develop the path integral formulation of QFT, so I'd really like to figure this out. Can anybody shed any light on how this works?
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Apr16-12, 11:45 PM
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Check out

especially the section titled "Integrals of similar form".

Zee's QFT textbook also covers it a bit, iirc.
Physics Monkey
Apr17-12, 07:08 AM
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Near as I can tell, there are some missing pieces to the formula you wrote. Perhaps these were not copied correctly in the notes?

The general expressions is [tex]
f(A,b) = \int \frac{d^n x}{(2\pi)^{n/2}} \exp{\left(-\frac{x A x}{2} + b x\right)} = \left(\det{A}\right)^{-1/2} \exp{\left(\frac{b A^{-1} b}{2}\right)}.
The final equality follows by completing the square under the integral and then shifting the integration variable. Now assuming that you can exchange integration and differentiation you can write

\int \frac{d^n x}{(2\pi)^{n/2}} x^k \exp{\left(-\frac{x A x}{2}\right)} = \int \frac{d^n x}{(2\pi)^{n/2}} \left[\frac{\partial^k}{\partial b^k} \exp{\left(-\frac{x A x}{2}+bx\right)}\right]_{b=0} = \left[\frac{\partial^k}{\partial b^k} f(A,b) \right]_{b=0}.
The expression for general P then follows from the Taylor expansion of P. There is some indexology that I have left schematic, but let me know if you have trouble working it out. Also, note that if P has an infinite Taylor expansion then the exchange of summation and integration may not be valid i.e. one could use P(d/db) inside the integral but not outside it.

Apr17-12, 09:36 AM
P: 354
Functional integral theorem

Ahh, that's where the [itex]b[/itex] comes from. Yeah, he added a linear term earlier on in the lecture when discussing the integral of a generalized Gaussian, but by the time we got to this point, he'd absorbed everything into a general quadratic form [itex]e^{-Q(x)}[/itex], so I didn't realize that was the [itex]b[/itex] we were talking about. I didn't include it in my initial description because I didn't realize it was relevant, but now I see what's going on.

Quote Quote by Physics Monkey View Post
Also, note that if P has an infinite Taylor expansion then the exchange of summation and integration may not be valid i.e. one could use P(d/db) inside the integral but not outside it.
[itex]P(x)[/itex] is given as being a finite-order polynomial, so I can say with reasonable certainty that its Taylor expansion terminates.

Thank you very much--this makes perfect sense now.

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