# When can the source term of a partition function be put in?

More specifically, in quantum field theory books, we usually have this:

$$Z = \int D(\bar{\psi}, \psi) e^{-S + \int_0^\beta d\tau \sum_l [\bar{\eta}_l (\tau) \psi_l (\tau) + \bar{\psi}_l (\tau) \eta_l (\tau) ]} \hspace{10mm} (1)$$

where $S = \int_0^\beta d\tau \sum_l (\bar{\psi}_l (\tau) \partial_\tau \psi_l (\tau ) +H(\bar{\psi},\psi))$ is the action, and $H$ is the Hamiltonian of the system.

Can I instead start off with the Hamiltonian in diagonal form and include the source term in, say, a momentum basis? So that

$$Z = \int D(\bar{\psi}, \psi) e^{-S + \int_0^\beta d\tau \sum_k [\bar{\eta}_k (\tau) \psi_k (\tau) + \bar{\psi}_k (\tau) \eta_k (\tau) ]}$$

where $S$ is now $\int_0^\beta d\tau \sum_k (\bar{\psi}_k (\tau) [\partial_\tau +\epsilon_k]\psi_k (\tau ))$.

Or do I have to start off with equation (1) and apply various transformations to the Hamiltonian and the rest of the exponent until $H$ it is in a diagonal basis? - If this is the case, the resulting source term in the momentum basis could be extremely complicated after the transformations are performed.

Thanks!

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I don't see why not. $\sum_k \bar{\eta}_k \psi_k = \sum_k \sum_m \sum_n e^{-ikm} e^{ikn} \bar{\eta}_m \psi_n = \sum_m \sum_n \delta_{mn} \bar{\eta}_m \psi_n = \sum_m \bar{\eta}_m \psi_m$... well except for all the normalizations and such. –  wsc May 19 '11 at 0:12
@wsc Thanks, but what if the transformations needed to make the Hamiltonian diagonal are more complicated than a simple Fourier transform? –  Jane May 19 '11 at 8:28
Yes, the diagonalization procedure mentioned in v1 of the question can at most work if the Hamiltonian $H(\bar{\psi},\psi)=\sum_{ll'}\bar{\psi}_l H_{ll'}\psi_{l'}$ is bilinear in the fields. –  Qmechanic May 19 '11 at 18:10
@Qmechanic I'm not sure what you mean ... Do you mean that I have to start at equation (1) then diagonalise the Hamiltonian plus source term? And that I can't start with the Hamiltonian in diagonal form? –  Jane May 19 '11 at 19:16