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How to calculate von Neumann entropy in continious case?

Consider density matrix elements:

$$ \rho(x,x') $$

Then: $$S = -Tr(\rho \log \rho) = -\int \rho(x,x')\log \rho(x',x)dx'dx $$ Is it true?

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    $\begingroup$ No, this is deeply flawed. The logarithm of a matrix is not the logarithm of the matrix element you wrote. $\endgroup$ Aug 1 '19 at 19:47
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My last answer was wrong on the final equation. I've corrected it below.

Not really. Remember the definition of the trace, it's the sum of the diagonal matrix elements, not all of them. Using a discrete basis $\{|e_i \rangle\}_{i=1}^\infty$, the trace of an operator $\hat A \in \mathcal{L(H)}$ can be written as: $$\mathrm{Tr}(\hat A) = \sum_{i=1}^\infty \langle e_i | \hat A | e_i \rangle$$

For a continuous basis $\{|x \rangle \}_{x\in \mathcal D}$, the analogous equation is: $$\mathrm{Tr}(\hat A) = \int_{\mathcal D} dx \ \langle x | \hat A | x \rangle$$

where $\mathcal{D}$ is the relevant integration domain, e.g. $\mathcal D = \mathbb R$.

So the correct equation would be: $$S = -\mathrm{Tr}(\hat\rho \log \hat\rho) = - \int_{\mathcal{D}} dx \ \langle x | \hat\rho \log \hat\rho |x \rangle \qquad (*) $$ Now if $\hat \rho$ is diagonal in the $\{|x \rangle\}_{x \in \mathcal D}$ basis, so would $\log \hat \rho$, meaning that:

$$S = - \int_{\mathcal{D}} dx \ \langle x | \hat\rho \log \hat\rho |x \rangle = - \int_{\mathcal{D}} dx \ \langle x | \hat\rho |x \rangle \langle x | \log \hat\rho |x \rangle \qquad (**)$$

$$S= -\int_{\mathcal D} dx \ \rho(x) \log \rho(x)$$

with $\rho(x)$ being the diagonal elements $\rho(x) \equiv \langle x | \hat \rho | x \rangle$.

If $\{|x \rangle\}_{x \in \mathcal D}$ is not an eigenbasis of $\hat \rho$, writing a general equation in terms of the matrix elements $\langle x | \hat \rho |x' \rangle = \rho(x,x')$ is going to be difficult. This is because writing the matrix elements of $\log \hat \rho$ in terms of those of $\hat \rho$ is hard in general. You would need to actually calculate the matrix logarithm first, which generally is: $$\log \hat \rho = \sum_{n=1}^\infty \frac{1}{n} (\mathbb{\hat{I}} - \hat \rho)^n$$

So for a general case, using $(*)$, you can formally write $S$ as:

$$S =- \int_{\mathcal{D}} dx \ \langle x | \hat\rho \log \hat\rho |x \rangle = -\sum_{n=1}^{\infty} \frac{1}{n} \int_{\mathcal D} dx \langle x | \ \hat \rho (\mathbb{\hat{I}} - \hat \rho)^n \ | x\rangle$$

Note however that since the trace is basis independent, it usually makes much more sense to first find the spectrum of $\hat \rho$, and calculate the entropy in that basis instead using $(*)$.

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No, you will need an infinity of integrals.

That is, whereas $$S = \operatorname{Tr}\rho \rho = \int \rho(x,x') \rho(x',x)dx'dx ,$$
you are looking at $$S = -\operatorname{Tr}(\rho \log \rho) = \operatorname{Tr}\left ( \rho \sum_{n=1}^\infty (1\!\! 1 -\rho)^n ~/n \right ), $$ which necessitates an integral per matrix multiplication. Note the sign of the Mercator series.

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