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  Calculating correlation functions of exponentials of fields

+ 11 like - 0 dislike

In their book Condensed Matter Field Theory, Altland and Simons often use the following formula for calculating thermal expectation values of exponentials of a real field $\theta$:

$$ \langle e^{i(\theta(x,\tau)-\theta(0,0))} \rangle = e^{-\frac12 \langle (\theta(x,\tau)-\theta(0,0))^2 \rangle} $$

An example can be found in chapter 4.5, problem "Boson-fermion duality", part c). (This refers to the second edition of the book, page 185.)

In other words, expectation values of exponentials can be cast as exponentials of expectation values under certain conditions. Unfortunately, I seem to be unable to find an explanation of why this can be done and what the conditions on the Lagrangian of $\theta$ are.

Hence, my question is:

How to derive the above formula? What do we need to know about $\theta$ for it to be valid in the first place?

Ideally, I am looking for a derivation using the path integral formalism. (I managed to rederive a very special case in terms of operators and the Baker-Campbell-Hausdorff formula, but I'd like to gain a more thorough understanding.)

This post has been migrated from (A51.SE)
asked Oct 6, 2011 in Theoretical Physics by Greg Graviton (775 points) [ no revision ]

2 Answers

+ 15 like - 0 dislike

This is just a property of Gaussian averaging analogous to the finite dimensional case:

$\langle e^{ix} \rangle=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty} e^{ix}e^{-\frac{x^2}{2\sigma^2}}=e^{-\frac{\sigma^2}{2}}= e^{-\frac{\langle x^2 \rangle}{2}}$

The field can be decomposed into its independent Gaussian modes and integrated for each mode separately.

This post has been migrated from (A51.SE)
answered Oct 6, 2011 by David Bar Moshe (4,355 points) [ no revision ]
The simplest answers are simply the best.

This post has been migrated from (A51.SE)
+ 4 like - 0 dislike

David Bar Moshe's derivation is of course right. Let me offer you a Taylor-expansion-based alternative proof: $$ \left\langle e^{ix} \right \rangle = \left\langle \sum_{n=1}^\infty \frac{(ix)^n}{n!} \right \rangle = \left\langle \sum_{k=1}^\infty \frac{(ix)^{2k}}{(2k)!} \right \rangle $$ Here, I just used that by some odd-ness, the odd powers have a vanishing expectation value. In the formula above, $x$ is whatever linear function of the elementary fields you want, including your coefficient. But the expectation value of $x^{2k}$ is $$ \left\langle x^{2k} \right \rangle = \frac{ \int_{-\infty}^\infty {\rm d}x \,x^{2k}\exp(-x^2/2x_0^2)}{ \int_{-\infty}^\infty {\rm d}x \,\exp(-x^2/2x_0^2)} = 1\times 3\times \cdots \times (2k-1) \times x_0^{2k} $$ which may be computed by integrating by parts or by converting it to the Euler integral (which is also evaluated by parts) so when you combine it with the $1/(2k)!$ factor, the odd integers cancel, only the product of the even integers which is equal to $2^k k!$ is left, and the original sum from the first line is $$\sum_{k=1}^\infty \frac{(-x_0^2)^k}{2^k k!} = \exp(-x_0^2/2) $$ where $x_0^2$ is the expectation value of $\langle x^2\rangle$ because I used it in the probabilistic distribution. Again, you may set $x=\theta(x_1,t_1)-\theta(x_2,t_2)$ or whatever linear function of variables and my derivation still holds.

This post has been migrated from (A51.SE)
answered Oct 6, 2011 by Luboš Motl (10,278 points) [ no revision ]

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