I have a quantum state which takes the following form:
$$\rho = \sum_{b, \,c \, = \,0}^\infty \frac{(-igt)^b(igt)^c}{\sqrt{b!c!}}\vert b\rangle\langle c\vert.$$
This is an infinite Hermitian matrix since $\rho = \overline\rho ^{\mathsf T}$, which can be written in the form:
$$\rho = \begin{pmatrix} 1 & -igt & \frac{-g^2t^2}{\sqrt{2!}} & \frac{ig^3t^3}{\sqrt{3!}} & \cdots \\ igt & g^2t^2 & \cdots & \cdots & \cdots \\ \frac{-g^2t^2}{\sqrt{2!}} & \vdots & \frac{g^4t^4}{\sqrt{4}} & \ldots & \ldots \\ \frac{-ig^3t^3}{\sqrt{3!}} & \vdots & \vdots & \ddots & \ldots\\ \vdots & \vdots & \vdots & \vdots & \ddots\\ & & & & & \frac{(gt)^{2n}}{n!} \end{pmatrix}$$
where the element in the corner corresponds to coefficient in the $b=n$, $c=n$ term. I need to diagonalise this matrix so that there are elements only on the diagonal. There are some factors that can influence the choice to achieve this:
- The matrix $\rho$ is Hermitian
- The diagonal terms are suppressed (and are thus the smallest) due to the $n!$ term on the denominator.
How can I achieve a diagonalised matrix and what would be the result in this case?
Set $$|v\rangle := \sum_{k=0}^\infty\frac{(-igt)^k}{\sqrt{k!}}|k\rangle,$$ Then your density matrix is simply given by $$\rho = |v\rangle\langle v|.$$ Hence it is a rank-1 projection, i.e. $\rho$ describes a pure state. The norm of $|v\rangle$ is $e^{\frac12g^2t^2}$, so that if you take $e^{-\frac12g^2t^2}|v\rangle$ and complete it to an orthonormal basis of your Hilbert space, you have the matrix representation $$\rho = \begin{bmatrix}1&0&\cdots\\0&0&\cdots\\\vdots&\vdots&\ddots\end{bmatrix}.$$ Remark Observe that the trace of $\rho$ is precisely $e^{g^2t^2}$ which must be 1 if $\rho$ has to represent a density matrix.