1
$\begingroup$

I have come across a very difficult integral that I do not know how to go forward with:

$$\int_{\mathbb{R}^3}\frac{1}{2\pi \hbar}\exp\left(\frac{-(x-x')^2-(p-p')^2+2i(p-p')(x+x')+4iyp-2(x-y)^2}{4\hbar}\right)dxdpdy$$

I only want to integrate over $x$ and $p$, and Mathematica tells me it should end up as:

$$\int_{\mathbb{R}}\exp\left(\frac{4iyp'-(x'-y)^2}{4\hbar}\right)dy$$

but I am unsure of how to realize this. I tried to substitute in new variables for $(x-x')$ and $(p-p')$ but the imaginary part of $2i(p-p')(x+x')$ gives me trouble. I tried switching to some polar coordinates in $x$ and $p$ but once again that term gives me trouble. I tried completing the square, and made no headway. I went to wikipedia, and tried to find some gaussian form that this looked like, and again made no headway as that aforementioned term causes trouble. If anyone could provide some insight, I would be most appreciative.

$\endgroup$
5
  • $\begingroup$ Try to put the integrand under the form $\exp{\frac{1}{2}(-u^TAu+v \cdot u+w}$, where $u=(x,p)$, $v$ is a $2$-dimensional vector, $w$ is a scalar, and $A$ a symmetric positive definite matrix, all three not depending on $x,p$). There should be general formulas for the integral over $u$ in that form (otherwise, apply a ln orthogonal transformation so that $A$ is diagonal and then separate coordinates – unless I’m mistaken, there is a clean enough final formula as a function of $A,v,w$. $\endgroup$
    – Aphelli
    Sep 22 at 22:11
  • $\begingroup$ @aphelli I tried, but the aforementioned term makes it difficult to do so. $\endgroup$
    – Chris
    Sep 22 at 22:35
  • $\begingroup$ Sorry for the thoughtless comment. Here’s hopefully a fix: write $I(A,v,w)$ the integral above, for which you hopefully have a formula $F(A,v,w)$. Then let $D=\begin{pmatrix}0&1\\1&0\end{pmatrix}$; check that $I(A+zD,v,w)$ and $F(A+zD,v,w)$ are both well-defined and analytic in $z$ for $z$ in a neighborhood of the imaginary line, and they are also equal uncountably many times, whenever $z$ is real and close enough to zero. So they’re always equal, even for imaginary values of $z$ – which should unstuck you. $\endgroup$
    – Aphelli
    Sep 22 at 23:06
  • $\begingroup$ @Aphelli, I apologize, but I’m not quite following your argument $\endgroup$
    – Chris
    Sep 23 at 14:23
  • $\begingroup$ The point is just this: if you have an analytic (that is, complex-analytic) formula that works when $A$ is symmetric positive definite, then it will work when $A$ is complex and the integral converges as well. In the one-dimensional case, you can see that $\int_{-\infty}^{\infty}{e^{-ut^2}dt}=\sqrt{\pi/u}$ when $u > 0$, bht in fact this holds as soon as the real part of $u$ is positive. $\endgroup$
    – Aphelli
    Sep 23 at 22:39

0

Your Answer

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