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?

  • 1
    $\begingroup$ could you provide a link to the article? $\endgroup$ – DJBunk Jul 17 '12 at 11:27
  • $\begingroup$ 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.) $\endgroup$ – David Z Jul 17 '12 at 17:06

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.

  • $\begingroup$ 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? $\endgroup$ – toot Jul 18 '12 at 10:38
  • $\begingroup$ @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. $\endgroup$ – Ron Maimon Jul 18 '12 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.