Let us consider common gaussian path integral over some complex random field $\displaystyle \Psi (\mathbf{r})$:
\begin{equation*} N=\int D\Psi ^{*} D\Psi \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right) \end{equation*} Here $\displaystyle \hat{K}$ is some operator, which posses $\displaystyle \hat{K} \langle \Psi ^{*}(\mathbf{r}_{1}) \Psi (\mathbf{r}_{2}) \rangle =\delta (\mathbf{r}_{1} -\mathbf{r}_{2})$, so $\displaystyle \hat{K}^{-1}(\mathbf{r}_{1} ,\mathbf{r}_{2}) =\langle \Psi ^{*}(\mathbf{r}_{1}) \Psi (\mathbf{r}_{2}) \rangle $ and we work with usuall gaussian statistics. Now I want to calculate average ≪intensity≫ of my field in $\displaystyle \mathbf{r} =\mathbf{y}$. So I write:
\begin{equation*} \langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N}\int D\Psi ^{*} D\Psi \ \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right) \end{equation*} This integral is common and we can easily calculate it introducing source terms:
\begin{gather*} \langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N} \partial _{\lambda } \partial _{\eta }\int D\Psi ^{*} D\Psi \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi +\eta \Psi ^{*}(\mathbf{y}) +\lambda \Psi (\mathbf{y})\right)\biggl|_{\lambda ,\eta =0} =\\ =\partial _{\lambda } \partial _{\eta }\exp\left(\int d^{n} r\int d^{n} r'\ \eta \delta (\mathbf{r} -\mathbf{y})\hat{K}^{-1}(\mathbf{r} ,\mathbf{r} ') \lambda \delta (\mathbf{r} '-\mathbf{y})\right)\biggl|_{\lambda ,\eta =0} =\langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle \end{gather*} So we see, that the calculation is self-consistent. Let us now perform it in an other way, using HS transformation:
\begin{gather*} \langle \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \rangle =\dfrac{1}{N}\int D\Psi ^{*} D\Psi \ \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y}) \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right) =\\ \dfrac{1}{N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty }\dfrac{d\xi }{\sqrt{\pi }}\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2} \xi ^{2}\right) =\\ =\dfrac{1}{N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty }\dfrac{d\xi }{\sqrt{\pi }}\dfrac{1}{2\xi }\dfrac{\partial }{\partial \xi }\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\left( \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right)^{2} \xi ^{2}\right) =\\ =\dfrac{1}{2\sqrt{\pi } N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} +\left( \alpha ^{*} -\alpha \right) \Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})\right) =\\ \\ =\dfrac{1}{2\sqrt{\pi } N}\int D\Psi ^{*} D\Psi \int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha } +\beta \Psi ^{*}(\mathbf{y}) +\beta ^{*} \Psi (\mathbf{y})\right) = \end{gather*} Now we deal with common gaussian path integral with outer sources, and we can integrate $\displaystyle \Psi $ out, so:
\begin{equation*} =\dfrac{1}{2\sqrt{\pi }}\int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha }\right) \cdot \exp\left( \beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) = \end{equation*} And now very weird thing. The integral over $\displaystyle \xi $ is simple, we can evaluate it:
\begin{gather*} =\dfrac{1}{2\sqrt{\pi }}\int\limits ^{\infty }_{-\infty } d\xi \dfrac{1}{\xi }\dfrac{\partial }{\partial \xi }\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\dfrac{1}{\xi ^{2}}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i\left( \alpha ^{*} -\alpha \right)}\exp\left( -\dfrac{\alpha ^{*} \alpha }{\xi ^{2}} -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha }\right) \cdot \exp\left( \beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) =\\ =\dfrac{1}{4}\int \dfrac{d\alpha ^{*} \land d\alpha }{2\pi i}\int \dfrac{d\beta ^{*} \land d\beta }{2\pi i}\dfrac{\left( \alpha ^{*} \alpha \right)^{-3/2}}{\alpha ^{*} -\alpha }\exp\left( -\dfrac{\beta ^{*} \beta }{\alpha ^{*} -\alpha } +\beta ^{*} \beta \hat{K}^{-1}(\mathbf{y} ,\mathbf{y})\right) \end{gather*} And here we easily see, that integral over $\displaystyle \alpha $ is suddenly divergent at $\displaystyle \alpha \rightarrow 0$! Why?! How?! I understand, that in this case Hubbard-Stratonovich transformation is not the most rational way to obtain the answer, but I ask this question due to a little bit more complicated problem, where you can't just introduce sources and get the answer. Thanks everybody in advance!
