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Given a system whose density operator is $\rho$, the (Von Neumann) entanglement entropy is defined as $$S[\rho]=-\text{Tr}(\rho\log\rho)$$ My question: Is $\log\rho$ simply the matrix whose entries are the logarithm of the entries of $\rho$? Or does it denote something more exotic (e.g. related with the matrix exponential)?

My thoughts on it: I know that $S$ must coincide with Shannon's Entropy $-\underset{j}{\sum}p_{j}\log p_{j}$ (if one does the calculations on an eigenbasis $\{|n\rangle\}$ of $\rho$ with eigenvalues $p_{j}$, which are the probabilities of the system being in the state $|\psi_{j}\rangle$). In fact, if one assumes that $\log\rho$ is the matrix whose entries are the logarithm of the entries of $\rho$, this happens (the calculations are not difficult). This indicates (but of course does not prove) that the $\text{log}\rho$ has this simple interpretation. Does it?

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  • $\begingroup$ $\log\rho\equiv A\Leftrightarrow \rho=\mathrm e^A$. In other words: yes, the definition is something more exotic, and it is indeed related to the matrix exponential. You can only evaluate the logarithm entry-wise if the operator is diagonal. In any case, this seems to be a math.SE question ("how is the logarithm of an operator defined?") $\endgroup$ – AccidentalFourierTransform Jan 21 '17 at 20:24
  • $\begingroup$ I really don't understand the downvotes and I would like to ask whomever downvoted to reconsider their decision: the answer might be off-topic, but it is not a bad question at all. $\endgroup$ – AccidentalFourierTransform Jan 21 '17 at 21:44
  • $\begingroup$ @AccidentalFourierTransform "this seems to be a math.SE question"- not really, since the question was merely about notation: I was asking if it was the logarithm related with the matrix exponential. Anyway, thanks for the answer. $\endgroup$ – Soap Jan 21 '17 at 21:45
  • $\begingroup$ @AccidentalFourierTransform That is, it was a question about notation that physicists use when talking about entanglement entropy. $\endgroup$ – Soap Jan 21 '17 at 21:51
  • $\begingroup$ The answer can be instantly found on the Wikipedia entry for "von Neumann entropy": ln denotes the (natural) matrix logarithm. $\endgroup$ – Norbert Schuch Jan 22 '17 at 2:14
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Firstly, we can define the exponential of an $n\times n$ matrix $A$ through the Taylor series, namely,

$$e^A = \sum_{k = 1 }^\infty \frac{A^k}{k!}.$$

If the matrix $A$ is diagonalizable then it can be expressed as $A = M \Lambda M^{-1}$ where $M$ is the matrix of column eigenvectors and $\Lambda = \mathrm{diag}(\lambda_1, \lambda_2, \dots, \lambda_n)$ where $\lambda_n$ are the eigenvalues. It then follows,

$$e^A = \sum_{k= 1}^\infty \frac{1}{k!}M\Lambda^k M^{-1} = M \left( \sum_{k=1}^\infty \frac{\Lambda^k}{k!}\right) M^{-1} = Me^{\Lambda} M^{-1}.$$

In the special case that $A$ is a diagonal matrix, you can show that the definition of the matrix $\exp$ reduces to the exponential of each diagonal element. Therefore, if $B=\ln A$, then we define the logarithm $B$ as being the matrix such that $e^B = A$, in the matrix exponential sense above.

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  • $\begingroup$ The question was merely: what does $\text{log}$ denote in this case. However, I appreciate the fact that you elaborated on it. $\endgroup$ – Soap Jan 21 '17 at 21:50
  • $\begingroup$ @XicoSim If you didn't want an answer which elaborates, why didn't you just check the Wikipedia entry? $\endgroup$ – Norbert Schuch Jan 22 '17 at 2:17
  • $\begingroup$ @NorbertSchuch Of course I did! I did not find it though: en.wikipedia.org/wiki/Entropy_of_entanglement $\endgroup$ – Soap Jan 22 '17 at 11:54
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    $\begingroup$ @XicoSim en.wikipedia.org/wiki/Von_Neumann_entropy. I'm not sure why you keep calling it entanglement entropy. $\endgroup$ – Norbert Schuch Jan 22 '17 at 17:38

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