3
$\begingroup$

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 $x^+=x+x'$.

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Do you mean that I have an inner product inside the exponential? $\endgroup$
    – Georg
    Nov 10, 2014 at 10:44
  • $\begingroup$ I've edited my answer to clarify $\endgroup$ Nov 12, 2014 at 12:31
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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