How to diagonalize an infinte dimensional operator

Question

I want to take logarithm of an infinite dimensional operator given by $\rho = \int\int dx_1 dx_1'C(x_1,x_2)C^*(x_1',x_2)|x_1\rangle\langle x_1' |$, where $C(x_1,x_2)$ is a gaussian function in $x_1$ and $x_2$. For that I was thinking of diagonalizing the matrix by unitary transformations. I know how to do it for finite dimensional matrix but don't know how to do it for the infinite dimension case.

Answer 1

You have to proceed in the same way as in the finite-dimensional case, that is by finding its eigenvectors $|\lambda\rangle$ associated to the eigenvalues $\lambda$, so that its diagonal form is given by $$ \hat{\rho} = \int \mathrm{d}\mu_\lambda\, \lambda|\lambda\rangle\langle\lambda| $$ where $\mu_\lambda$ is a discrete or continuous measure, depending on whether the spectrum of $\hat{\rho}$ itself is discrete of continuous.

Now, we would like to find those eigenfunctions in the position basis. In order to do so, we shall start from the eigenvalue problem $\hat{\rho}|\lambda\rangle = \lambda|\lambda\rangle$. Let's denote the wavefunction of $|\lambda\rangle$ in the position basis by $\psi_\lambda(x) = \langle x|\lambda\rangle$, so that $\lambda\psi_\lambda (x) = \langle x|\hat{\rho}|\lambda\rangle$. Next, we get by introducing a closure relation in the right-hand side : $$ \begin{align} \langle x|\hat{\rho}|\lambda\rangle &= \int\mathrm{d}y\; \langle x|\hat{\rho}|y\rangle \langle y|\lambda\rangle \\ &= \int\mathrm{d}y \left(\int\mathrm{d}z_1\int\mathrm{d}z_2\; C(z_1,z)C^*(z_2,z) \langle x|z_1\rangle\langle z_2|y\rangle\right) \psi_\lambda(y) \\ &= \int\mathrm{d}y \left(\int\mathrm{d}z_1\int\mathrm{d}z_2\; C(z_1,z)C^*(z_2,z) \delta(z_1-x)\delta(z_2-y)\right) \psi_\lambda(y) \\ &= \int\mathrm{d}y\; C(x,z)C^*(y,z)\psi_\lambda(y) \end{align} $$ The eigenvalue equation becomes thus : $$ \lambda\psi_\lambda(x) = \int\mathrm{d}y\; C(x,z)C^*(y,z)\psi_\lambda(y) $$ which is a kind of Fredholm integral equation.