Calculation of a particular functional (path) integral

[Hypersphere]

Hello!

On p.424 in the second edition of Altland & Simons, they compute an average as part of a renormalization group analysis of a model for dissipative quantum tunneling. I'd like to use their result in another situation, but I would be much happier if I understood how they derived it. I have discussed it with some of my professors, but we don't seem able to get it. Perhaps it is some "trick" that we just don't think of. Anyway, the step I find difficult is

$\langle S_U \left[ \theta_s, \theta_f \right] \rangle = \frac{c}{2} \int d\tau e^{i \theta_s \left( \tau \right) } \int D\theta_f e^{-\frac{1}{4\pi g} \int_f \frac{d\omega}{2\pi} \theta \left( \omega \right) | \omega | \theta \left( -\omega \right) } e^{i \int_f \frac{d\omega}{2\pi} e^{i\omega \tau} \theta \left( \omega \right) } + c.c. \\ = \frac{c}{2} \int d\tau e^{i\theta_s \left( \tau \right) } e^{-\pi g \int_f \frac{d\omega}{2\pi} |\omega|^{-1}} + c.c.$

where $\theta_s$ is the "slow" field and $\theta_f$ is the fast one, $c.c.$ denotes complex conjugate and
$\int_f d\omega \equiv \int_{\Lambda/b < |\omega| < \Lambda} d\omega$
i.e., it denotes integration over the fast modes.

To me, this does not seem as one of the usual formulas for Gaussian integrals as
1. the exponent contains integrals itself
2. I've seen a formula for the case with a completely imaginary term, but here the term seems generally complex
Also, in the first line, the complex conjugate is of the entire term, i.e. it will begin with $e^{-i\theta_s \left( \tau \right)}$ In effect, I don't know how to complete the square properly.

Does anyone have an idea on how to proceed, or perhaps know of a good reference for this kind of problems? Thanks in advance.

EDIT:
By the way, it is a textbook-related question yes, so feel free to move it to the Homework and coursework question if you need to. It is, however, not part of any homework or coursework - rather my own research - so I was not sure on where to put the topic.

 

[eljose79]

Hello there,

Thank you for your question. I can understand your confusion with the derivation of the average in the renormalization group analysis. Let me try to explain it in a simpler way.

Firstly, the exponent in the integrals can be rewritten as:

-\frac{1}{4\pi g} \int_f \frac{d\omega}{2\pi} \theta \left( \omega \right) | \omega | \theta \left( -\omega \right) + i \int_f \frac{d\omega}{2\pi} e^{i\omega \tau} \theta \left( \omega \right) = -\frac{1}{4\pi g} \int_f \frac{d\omega}{2\pi} \theta \left( \omega \right) | \omega | \theta \left( -\omega \right) + \frac{i}{2} \int_f \frac{d\omega}{2\pi} \left( e^{i\omega \tau} + e^{-i\omega \tau} \right) \theta \left( \omega \right)

This allows us to complete the square and rewrite the exponent as:

-\frac{1}{4\pi g} \int_f \frac{d\omega}{2\pi} \left( \theta \left( \omega \right) - \frac{i}{2\pi g} e^{i\omega \tau} \right) | \omega | \left( \theta \left( -\omega \right) - \frac{i}{2\pi g} e^{-i\omega \tau} \right)

Now, using the definition of the average, we can write:

\langle S_U \left[ \theta_s, \theta_f \right] \rangle = \frac{c}{2} \int d\tau e^{i \theta_s \left( \tau \right) } \int D\theta_f e^{-\frac{1}{4\pi g} \int_f \frac{d\omega}{2\pi} \left( \theta \left( \omega \right) - \frac{i}{2\pi g} e^{i\omega \tau} \right) | \omega | \left( \theta \left( -\omega \right) - \frac{i