How does one diagonalize a density operator that has exponential elements?

Question

What is the diagonal form of the density operator $\hat\rho$, of which I know that $$\langle x\left|\hat\rho\right|x'\rangle\propto \exp\left[{-\frac{\gamma}{2}(x^2+x'^2)+\beta xx'}\right]$$where $\left| x\rangle\right.$ is the position basis, and $\gamma,\beta$ are some real coefficients?

What might be helpful is to note that it can also be written as $$\langle x\left|\hat\rho\right|x'\rangle\propto\exp\left\{-\frac{1}{4}[(\gamma+\beta)(x^-)^2+(\gamma-\beta)(x^+)^2]\right\}$$ with $x^-=x'-x$ and $x^+=x+x'$.

Answer 1

You might have more luck in math.se, but your second expression is an outer product of two vectors in the x+, x- basis. So I would start by writing it as $uv^T$ and work abstractly.

(I think) you can multiply your matrix on the left and on the right by two diagonal matrices in order to rescale x+ and x-, so that in essence you need to diagonalize $uu^T$ only, which is much easier...

Edit:

A ket $|f\rangle$ is just a vector, and in the position representation/basis this vector has components $\langle x | f \rangle = f(x)$. So you can think of functions as vectors. If I take the two vectors $|u\rangle$ and $|v\rangle$ and form their outer product, $O =|u\rangle \langle v |$, then this is a matrix, which, in the position basis has components $O(x,x') = \langle x | O | x' \rangle$. In your case, roughly, $u(x) = \exp\left( -\frac{1}{4} (\gamma + \beta) x^2\right)$, etc. What remains is to find a unitary transformation that takes you from the $x, x'$ basis into the $x^\pm$ basis.

Answer 2

I found the answer in a paper by Srednicki. (At which I was looking before, so I should have known the answer...)

As the author says the answer 'is found most easily by guessing'...

Here it is: the eigenvalues $p_n$ and the eigenfunctions $f_n(x)$ are \begin{align} p_n&=(1-\xi)\xi^n\\ f_n(x)&=H_n(\alpha^{1/2}x)\exp(-\alpha x^2/2), \end{align} with $H_n$ a Hermite polynomial, $\alpha=(\gamma^2-\beta^2)^{1/2}$ and $\xi=\beta/(\gamma+\alpha)$. The label $n$ runs from zero to infinity.