# Partition function of an interacting gas

By reading an article, I found a partition function that, according to the author, describes an interacting with random variables as coupling constant.

$$Z =\int \mathrm{d} \lambda_i e^{i(K^{ij}\lambda_i\lambda_j + V^{ijk}\lambda_i\lambda_j\lambda_k)}\mathrm{exp}(e^{iS_{eff}(\lambda)})$$

This expression is totally unfamiliar to me. Could someone show me how to derive that, providing a reference (online course, textbook, etc.) if necessary?

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could you provide a link to the article? –  DJBunk Jul 17 '12 at 11:27
I edited that in to the question for you, toot. (For future reference, editing is the recommended way to make corrections or clarifications, not commenting.) –  David Z Jul 17 '12 at 17:06
Thanks David, I'll keep that in mind ;) –  toot Jul 17 '12 at 17:08

## 1 Answer

This is a quantum partition function, not a statistical mechanical partition function. He is just talking about an idealized self-interacting field. If you have a scalar with cubic self interactions, you write the Lagrangian as

$$\partial_\mu \phi \partial^\mu \phi - \lambda \phi^3$$

If you fourier transform the field variables, this is

$$\int_k k^2 |\phi_k|^2 + \int_{k_1,dk_2,dk_3} \delta(k_1+k_2+k_3) \phi_{k_1}\phi_{k_2}\phi_{k_3}$$

Which, if you think of k as a lattice, can be abstrated to the form Banks writes down. The remaining S_eff term is from renormalization, which changes the low energy theory according to the contributions to the low-energy effective action from high-energy degrees of freedom you are neglecting. This is heuristic, because a real renormalizable model requires a $\phi^4$ term too.

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Thanks for the point Ron Maimon, when I read gas and partition function, my brain directly tilted on statistical mechanics. Just one small point, the random coupling are then the $\lambda_i$ taking any value, that we integrate over? –  toot Jul 18 '12 at 10:38
@toot: Oops, I used incompatible notation--- the K's and the V's are the couplings and $\lambda$ is his field. The coupling distribution is not provided, but in principle you would integrate over K and V (after taking the log of Z) to make a statistical ensemble of couplings you average over. –  Ron Maimon Jul 18 '12 at 16:12