# How does one diagonalise an operator that has exponential elements?

I asked this question before on the Physics StackExchange, but as one commenter noted I might have more luck here. So the question is:

What is the diagonal form of the (density) operator $\hat\rho$, of which I know that $$\langle x,\hat\rho\,x'\rangle\propto \exp\left[{-\frac{\gamma}{2}(x^2+x'^2)+\beta xx'}\right],$$ where $\gamma,\beta$ are some real coefficients? The $x$ basis is continuous, and obeys $\hat x|x\rangle=x|x\rangle$ where $\hat x$ is the operator and $x$ the eigenvalue. The vectors $|x\rangle$ are in an infinite-dimensional Hilbert space (so $\hat\rho$ is in the space of linear operators on this space. It is also Hermitian).

My attempt at a solution:

The expression in the exponent can be written as $y\cdot M \cdot y$, with $M$ diagonal, $y=(x^-,x^+)$, and $x^-=x'-x,\ x^+=x+x'$. Written out, this reads $$\langle x\left|\hat\rho\right|x'\rangle\propto\exp\left\{-\frac{1}{4}[(\gamma+\beta)(x^-)^2+(\gamma-\beta)(x^+)^2]\right\}.$$ Thanks for any help!