1
$\begingroup$

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!

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.