$$ Z = \int \mathcal D[\phi^*] \mathcal D[\phi] \exp(\phi^*A\phi + \phi B\phi) $$

where $A$ and $B$ are operators. I know how to solve a Gaussian path integral involving only $\phi^* A \phi$ but I don't know how to handle the other quadratic term.


You just divide $\phi$ to the real and imaginary part, to make things clear: $$\phi = f + ig, \quad \phi^* = f-ig, \quad f,g\in{\mathbb R}$$ Up to some totally universal normalization factor, the integration measure is simply $$\int {\mathcal D} f \,\,{\mathcal D} g $$ and the exponent in the exponential may be written as $$[(f-ig) A + (f+ig) B] (f+ig) $$ Writing the column $(f,g)^T$ as $h$, the bilinear expression above is nothing else than $$ h M h $$ where the matrix $M$ is, in a block-diagonal form, $$ M = \left(\begin{array}{cc}A+B&-iA+iB\\iA+iB&A-B\end{array}\right) $$ Now, I assume you may calculate the integral $$\int {\mathcal D} h\,\exp(hMh) $$ which is completely analogous to the $\exp(\phi^* A \phi)$ integral. However, with the matrix $M$ enough, the integral is infinity because $M$ is singular (infinity, due to flat directions) because the second row (of blocks) is $i$ times the first. However, you will get a nonsingular result if the exponent will also contain the Hermitian conjugate $\phi^* B^\dagger \phi^*$ or something like that.

If there are algebraic mistakes above, it should be possible to fix them.

  • $\begingroup$ Ah, thanks, that makes sense. I knew I was missing something straightforward. And yes, you're right, the integral I was trying to solve did have a $\phi^* B^\dagger \phi^*$ term I forgot to include in the question. $\endgroup$
    – Koaaala
    Jan 23 '15 at 12:21

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