# Help with Correlation/Green's Function of Rotated Variables

## Main Question or Discussion Point

Hello (I'm reposting this from stack exchange, and thought this site may be more appropriate, so if you see it that's why),

I'm working through this paper, and have encountered "a little algebra shows that...", yet I'm not familiar enough with the topic at hand to figure this out. Here is the paper:

https://arxiv.org/abs/cond-mat/0109316

My issues start in sections (5) and (5) (strange numbering system), "Correlation functions" and "Keldysh Rotation" (pages 3 and 4), primarily focusing on equation 11. I'm pretty new to the topic of Green's functions, so perhaps I have missed something. I've come to this paper to try to understand something similar, and encountered this roadblock, so any help understanding how to obtain equation (11) would be greatly appreciated.

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It looks like you should take a choice for $\alpha$,$\beta$, then expand the quantum and classical field variables, $\phi_{cl}$ and $\phi_q$, in the old field variables, $\phi_f$ and $\phi_b$. This will give you a sum of correlation functions which are cited in equation 8. Now combine as necessary using equation 12.

• Christian_K_K
It looks like you should take a choice for $\alpha$,$\beta$, then expand the quantum and classical field variables, $\phi_{cl}$ and $\phi_q$, in the old field variables, $\phi_f$ and $\phi_b$. This will give you a sum of correlation functions which are cited in equation 8. Now combine as necessary using equation 12.
How should the calculations be made in the correlation brackets? Or might you have a reference I could look at as a guide? Also, is the correlation bracket the same as a Green's function?

Also, maybe this isn't part of confusion, do the different matrix elements correspond to different combos of cl and q?

Sorry, for such a late reply. In this case the correlation function is the same as the Green's function. Mathematically, a Green's function is associated with some linear differential operator. In a field theory setting this linear differential operator will be the one appearing in the Euler-Lagrange equations derived from the action functional. In the case where your action functional appearing in the partition function is quadratic in the fields you have the 2-point correlation function and the Green's function (which is associated with the linear differential operator) being identical. The paper you are interested in fits this case. I would caution that the use of "Green's function" is sometimes used more liberally as a synonym for the correlation functions and vice versa. You will have to discover from the context what is intended.

As for calclating inside correlation brackets, you just need to realize that
$\langle(\sum\limits_i \phi_i)(\sum\limits_j \phi_j)\rangle=\sum\limits_i\sum\limits_j\langle\phi_i\phi_j\rangle$

Say you wanted to calculate $-i\langle\phi_{cl}\bar{\phi}_{cl}\rangle$ (which you have said you do), then you would expand these fields in the old fields and get

\begin{align}-i\langle\phi_{cl}\bar{\phi}_{cl}\rangle&=-{i\over 4}\langle\phi_{f}\bar{\phi}_{f}+\phi_{f}\bar{\phi}_{b}+\phi_{b}\bar{\phi}_{f}+\phi_{b}\bar{\phi}_{b}\rangle\\ &=-{i\over 4}\left(\langle\phi_{f}\bar{\phi}_{f}\rangle+\langle\phi_{f}\bar{\phi}_{b}\rangle+\langle\phi_{b}\bar{\phi}_{f}\rangle+\langle\phi_{b}\bar{\phi}_{b}\rangle\right)\end{align}
Now you are free to make substitution for these known correlation functions in the old field variables. I actually did this piece of the calculation and you do get the expected answer:
$-i\langle\phi_{cl}\bar{\phi}_{cl}\rangle={1\over 2}\mathcal{D}^K(t,t')$

As for your last question, yes.

EDIT: I forgot to say that I am not an expert in out-of-equilibrium field theories, so I do not know a reference which would be instructive within that topic. As for equilibrium statistical field theories, there is the second book in a two volume set by Kardar. Perhaps others will chime in with better suggestions.

EDIT#2: In the course of the doing that calculation I found a typo in equation 8 in the paper. The last correlation is for fields both having index $b$

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• Christian_K_K
In a field theory setting this linear differential operator will be the one appearing in the Euler-Lagrange equations derived from the action functional. In the case where your action functional appearing in the partition function is quadratic in the fields you have the 2-point correlation function and the Green's function (which is associated with the linear differential operator) being identical. The paper you are interested in fits this case.
The action functional contains the integrand $\bar{\phi}(t)\mathcal{D}^{-1}\phi(t)$. By quadratic in the fields, do you mean $\bar{\phi},\phi$ or do you mean $\bar{\phi},\mathcal{D}^{-1}\phi$?

For the second half of your response, thank you very much! I didn't know if $\langle ab + cd \rangle = \langle ab \rangle + \langle cd\rangle$, so that is very helpful to know for this paper! I'll check out the Kardar texts. I've been stuck on this issue for awhile, and thankful for figuring it out. It is relieving to see it wasn't so terrible.

I mean the former. I have some general advice for working with field theory and that is to always return to the fundamentals/basics when you get confused. Sometimes the calculations get long and tedious and you might forget what you were trying to do in the first the place. This is probably obvious, and a good practice outside of field theory, but it is worth actively recalling as you do work in field theory.