Suppose we have a state with continuous spectrum:

Recall that the Von Neumann entropy is defined as:

$S=\textrm{Tr}(\rho \ln \rho)$

where $\rho =\int_I \lvert a\rangle\langle a\rvert da$ and $I \subset \mathbb{R}$

That is,

\begin{align}
S & =\int_I\langle q\rvert\left(\int_I \lvert a\rangle\langle a\rvert da\right)\ln\left(\int_I \lvert b\rangle\langle b\rvert db\right)\lvert q\rangle dq\\
& =\int_I\left(\int_I \langle q\vert a\rangle\langle a\rvert da\right)\ln\left(\int_I \lvert b\rangle\langle b\vert q\rangle db\right) dq\\

@acuriousmind For this expression:

$S=\int_I\langle q\rvert\ln (\lvert q\rangle) dq$

while it does have the required form of summing the diagonal of the density matrix, I am not sure how to deal with ln. Clearly, ln is not an observable thus I cannot really interpret the diagonal matrix elements as some kind of expectation value of ln. $ln(\lvert q\rangle)$ may not be even a well defined notation since the matrix function ln() is defined only for square matrices. so either I have did something wrong when I evaluate those delta distributions or I am missing a concept

I guess it will be clearer when I wrote out that $I=(x,y)$ is some real interval, and the continuous spectrum of eigenvalues will be all real numbers within this interval.

For the trivial case of a free particle $I=(-\infty,\infty)$ which is all reals. For the bounded interval case above, it can arise in spectroscopy for the continuum portion of a spectrum (usually because the states are so closely spaced that their transitions overlap

So technically, I am not really calculating anything, I am just interested in what the expression for von neumann entropy in continous system look like, wh…