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 differential operator. 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} =\hat{K}^{-1}(\mathbf{y},\mathbf{y}) \end{gather*}

Such a calculation involving source terms can be performed for any polynomial, not only $\Psi ^{*}(\mathbf{y}) \Psi (\mathbf{y})$. However, if we want to calculate the average of absolute value for some expression, e.g.

\begin{equation*} \int D\Psi ^{*} D\Psi \ |\Psi ^{*}(\mathbf{y}) \Psi (\mathbf{x})-\Psi ^{*}(\mathbf{x}) \Psi (\mathbf{y})| \ \exp\left( -\int d^{n} r\ \Psi ^{*}\hat{K} \Psi \right) \end{equation*}

the trick doesn't work, because you can't get absolute value as derivatives of linear source terms.

So, the question:

is there some regular way to calculate such mean values?

Thanks everyone in advance!

  • 1
    $\begingroup$ Not really. This is why people mostly care about root-mean-squares instead of mean-root-squares. $\endgroup$ Commented May 22, 2021 at 16:10

1 Answer 1


One of many possible ways to deal with it is to represent the absolute value as

\begin{equation} |x|=-\int \dfrac{\mathrm{d} \omega }{2\pi }\dfrac{e^{i\omega x} +e^{-i\omega x}}{(\omega -i0)^{2}} ,\ \ x\in \mathbb{R} \end{equation}

Then you have all the fields in the exponent and deal with a common gaussian path integral, which can be calculated in a usual way.


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