# How could I simplify this von Neumann entropy functional?

I would like to simplify the expression $C=e^{\frac{2}{3}S[\rho]}$, being $S[\rho]=-\mathrm{tr}(\rho\ln{\rho})$ the von Neumann entropy and $\rho=\sum_{k=1}^{d}\lambda_{k}\left|k\right>\left<k\right|$ the density matrix of the quantum system, being $\left|k\right>$ an orthonormal basis.

I have tried the following:

\begin{align} C&=e^{-\frac{2}{3}\mathrm{tr}(\rho\ln{\rho})}\\ &=[\det(e^{\rho\ln{\rho}})]^{-\frac{2}{3}}\\ &=[\det(\rho^{\rho})]^{-\frac{2}{3}}\\ &=\left(\prod_{i=1}^{d}\lambda_{i}\right)^{-\frac{2}{3}\rho} \end{align}

Could you help me to understand if this is correct? It is possible to simplify this expression more?

• Is it $C=\exp\left(+\frac{2}{3}S\right)$ or $C=\exp\left(-\frac{2}{3}S\right)$? Also, is the $\left\{|k\rangle\right\}$ basis orthonormal? Jan 24, 2018 at 20:42
• I made a misprint, thanks for the note, it is $C=e^{+\frac{2}{3}S}$ and $|k>$ is an orthonormal basis Jan 24, 2018 at 20:49
• Ok. I am not sure if the trace-log identity helps you much in this case because that statement is basically the equality $C =\exp \left[\mathrm{Tr} \left(-\frac{2}{3} \rho \ln \rho\right)\right] = \exp \left[ -\frac{2}{3} \sum_k \lambda_k \ln \lambda_k \right] = \Pi_k \exp \left[ -\frac{2}{3} \lambda_k \ln \lambda_k\right] = \det \left[ \exp \left(-\frac{2}{3} \rho \ln \rho\right) \right]$. Is any of those intermediate expressions good enough? I have my doubts on the last expressions you write there for $C$. Jan 24, 2018 at 20:59

The first step is correct: \begin{align} C&=e^{-\frac{2}{3}\mathrm{tr}(\rho\ln{\rho})}\\ &=[\det(e^{\rho\ln{\rho}})]^{-\frac{2}{3}}. \end{align} However, neither the second nor the third equalities work: \begin{align} C&\neq[\det(\rho^{\rho})]^{-\frac{2}{3}}\\ C&\neq\left(\prod_{i=1}^{d}\lambda_{i}\right)^{-\frac{2}{3}\rho}, \end{align} because (i) the expression $\rho^\rho$ (a matrix to the exponent of a matrix) is not well defined, and (ii) even if it were, splitting the product like you did just doesn't work.
The simplification you're looking for just doesn't exist. Keep it as $e^{\frac23 S(\rho)}$ and work with that - that's just the cleanest form for that object.