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I am learning about von Neumann entropy and I understand that it should vanish for a pure state. I tried to calculate it for the pure state whose wavefunction is $$|\psi\rangle = \frac{1}{\sqrt 2} (\lvert\uparrow\rangle + \lvert\downarrow \rangle)$$ and whose density matrix is

$$\rho = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \quad .$$

But what I am getting is

$$S(\rho) := - \mathrm{Tr}[\rho \ln(\rho)] \\ = - \mathrm{Tr}\left[ \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \ln\left(\frac{1}{2}\right) & \ln\left(\frac{1}{2}\right) \\ \ln\left(\frac{1}{2}\right) & \ln\left(\frac{1}{2}\right) \end{pmatrix} \right] \\ = - \mathrm{Tr}\left[ \frac{1}{2} \ln\left(\frac{1}{2}\right) \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \right] \\ = - \frac{1}{2} \ln(\frac{1}{2}) \mathrm{Tr}\left[ \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \right] \\ = 2 \ln(2) \neq 0.$$

Could somebody explain to me what I am doing wrong? Isn't this state

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    $\begingroup$ I suspect that you took the logarithm of each element of the matrix. But you are supposed to compute the matrix logarithm, which is defined via a power series expansion. $\endgroup$
    – march
    Apr 17, 2023 at 18:36
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    $\begingroup$ There is a more suitable method for this case here: The functions of self-adjoint operators are defined via functional calculus... With the (usual) agreement that $0\log 0=:0$ you will immediately find that the von Neumann entropy of a pure state vanishes. $\endgroup$ Apr 17, 2023 at 18:45
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    $\begingroup$ Orthogonally diagonalize the matrix. The trace of the diagonal equivalent is simpler to evaluate, as functions of a diagonal matrix. What do you find? you don't need a reference. $\endgroup$ Apr 17, 2023 at 18:50
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    $\begingroup$ I don't have a particular reference, sorry. But this was asked several times here on PSE already...and basically any book on quantum mechanics should cover this. $\endgroup$ Apr 17, 2023 at 18:50
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    $\begingroup$ lol... somewhere functions of hermitian operators are for sure defined...For example, in Nielsen and Chuang, section 2.1.8 these objects are defined. $\endgroup$ Apr 17, 2023 at 19:18

1 Answer 1

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The eigenvalues of $\rho=\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}$ are $1$ and $0$. Evaluating the Von Neumann entropy: $$S[\rho]=-\mathrm{Tr}[\rho\ln{\rho}]=-\sum_{i}\lambda_i\ln\lambda_i=-0\ln0-1\ln1$$ With the convention $0\ln0=\lim_{x\to0^+}x\ln x=0$, we see that $S[\rho]=0$, as expected.

To give a bit more detail, the result follows from the following useful lemma, for diagonalizable matrices $\rho=Q\Lambda Q^{-1}$: $$\mathrm{Tr}[\rho^n]=\mathrm{Tr}[Q\Lambda^nQ^{-1}]=\mathrm{Tr}[\Lambda^n]=\sum_i\lambda_i^n$$ Then, since the trace is linear, any analytic matrix function $f(\rho)$ satisfies the following relation, as can be seen from its Taylor expansion: $$\mathrm{Tr}[f(\rho)]=\mathrm{Tr}\left[\sum_ia_i\rho^i\right]=\sum_ia_i\mathrm{Tr}[\rho^i]=\sum_ia_i\lambda_i^n=\sum_if(\lambda_i)$$

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