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Consider the example case:$$|\phi\rangle=\frac{1}{\sqrt{2}}(|01\rangle-i|10\rangle)$$ From this we can easily calculate the density matrix:$$\rho=\frac{1}{2}\begin{bmatrix}0&0&0&0\\0&1&i&0\\0&-i&1&0\\0&0&0&0\end{bmatrix}$$Now trying to calculate the von Neumann entropy I could probably say that "it is a pure state, ergo $S(\rho)=0$". However when calculating explicitly the von Neumann entropy $S=-tr(\rho ln \rho)$ having to take the natural logarithm of $\rho$ runs to a problem.

After calculating the eigenvalues and eigenvectors I find: $\lambda_1=1,\lambda_2=0,\lambda_3=0,\lambda_4=0$. These values are the diagonal elements in the well-known D in $\rho=M D M^{-1}$. Here $M$ is the modal matrix and $D$ the diagonal matrix looking like:$$D=diag(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$$

Now taking the natural logarithm of the matrix would require me to take the natural logarithm of the elements of $D$. In my case thought we would have to say that three of the diagonals go towards $-\infty$.

How is this entropy going to end up $0$ then since taking the trace would require to add infinities?

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  • $\begingroup$ This appears equally for classical entropies. $\endgroup$ May 23 '20 at 12:06
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The convention usually followed is to define $0 \ln 0 = 0$.

That is not so bad as it looks, you may calculate $\lim\limits_{x \rightarrow 0} x\ln x$. Using L'Hopital's rules, \begin{array}{lll}\lim\limits_{x \rightarrow 0} x \ln(x) & = & \lim\limits_{x\rightarrow 0} \frac{\ln(x)}{x^{-1}} \\ { } & = & \lim\limits_{x \rightarrow 0} \frac{x^{-1}}{-x^{-2}} \\ {} & = & \lim\limits_{x\rightarrow 0}-x \\ { } & = & 0.\end{array} Noting columns of $M$ as the eignevectors of $\rho$, with $D$ being the diagonal matrix of corresponding eigenvalues, here you will have \begin{align} -\text{tr}\left(\rho \ln(\rho)\right) &= -\text{tr}\left(M DM^{-1} \ln \left(MDM^{-1}\right)\right)\\ & = -\text{tr}\left(MDM^{-1} M \ln (D) M^{-1}\right) \\ & = -\text{tr}\left(M D\ln\left(D\right) M^{-1}\right). \end{align}

In $D \ln \left(D\right)$ one shall consider $0 \ln 0 = 0$. Hence the entropy ends up as $0$.

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