We define Von Neumann Entropy for a density matrix $\rho$ (hermitian, positively defined, with trace 1) as :

$S(\rho)=-tr(\rho \ln(\rho))$

Considering $\rho = \rho_1 \bigotimes \rho_2$, I want to show that $S(\rho)=S(\rho_1)+ S(\rho_2)$.

I do not see how the following equality can be (where $\mathbb{Id}$ stands for the identity matrix with appropriate dimension):

$\ln(\rho_1\bigotimes \mathbb{Id})= \ln(\rho_1)\bigotimes\mathbb{Id}$

Actually I don't understand how the definition of ln for matrices applies here.

Could someone help me on this step ?

  • $\begingroup$ The tensor product of two matrices is a new matrix. Apply the logarithm to that. $\endgroup$ – Elchanan Solomon Dec 31 '13 at 17:42
  • $\begingroup$ @IsaacSolomon : Ok, so is that correct to look at this like as a (block) diagonal matrix and because of the $ln(\lambda_i \delta_{ij})=ln(\lambda_i)\delta_{ij}$ we apply similarly to the upper form and get our result ? $\endgroup$ – faero Dec 31 '13 at 17:56
  • $\begingroup$ Actually, $\rho\otimes I$ is not block diagonal but $I\otimes\rho$ is. $\endgroup$ – Algebraic Pavel Jan 1 '14 at 15:52
  • $\begingroup$ BTW, the fact that $\ln(\rho\otimes I)=\ln(\rho)\otimes I$ follows from the useful fact in my answer :) $\endgroup$ – Algebraic Pavel Jan 1 '14 at 15:58

First, it is important to know that $f(\rho)$ for a Hermitian positive definite $\rho$ is given by $f(\rho)=U\mathrm{diag}(f(\lambda_1),\ldots,f(\lambda_n))U^*$, where $\rho=U\Lambda U^*$ is the eigenvalue/eigenvector decomposition of $\rho$ with $\Lambda=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$.

A useful fact:

If $\rho_1=U_1\Lambda_1U_1^*$ and $\rho_2=U_2\Lambda_2U_2^*$ are the eigenvalue/eigenvector decompositions of $\rho_1$ and $\rho_2$, respectively, then $$ \rho_1\otimes\rho_2=(U_1\Lambda_1U_1^*)\otimes(U_2\Lambda_2U_2^*) =(U_1\otimes U_2)(\Lambda_1U_1^*\otimes\Lambda_2U_2^*) =(U_1\otimes U_2)(\Lambda_1\otimes\Lambda_2)(U_1^*\otimes U_2^*) =(U_1\otimes U_2)(\Lambda_1\otimes\Lambda_2)(U_1\otimes U_2)^* $$ is the eigen-decomposition of $\rho_1\otimes\rho_2$ (we used here the mixed-product property of the Kronecker product).

If $\rho=U\Lambda U^*$ is the eigen-decomposition of (the Hermitian positive definite) $\rho$ (with $\Lambda=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$), then $$ \rho\ln(\rho)=(U\Lambda U^*)(U\ln(\Lambda)U^*)=U\Lambda\ln(\Lambda)U^* $$ and hence $$ S(\rho)=-\sum_{i=1}^n\lambda_i\ln(\lambda_i). $$

Now consider $\Lambda_1=\mathrm{diag}(\lambda_1^{(1)},\ldots,\lambda_n^{(1)})$ and $\Lambda_2=\mathrm{diag}(\lambda_1^{(2)},\ldots,\lambda_n^{(2)})$. We have $$ \begin{split} \ln(\Lambda_1\otimes\Lambda_2)&=\mathrm{diag}(\ln(\lambda_1^{(1)}\Lambda_2),\ldots,\ln(\lambda_1^{(1)}\Lambda_2))\\ &=\mathrm{diag}(\ln(\lambda_1^{(1)})I_n+\ln(\Lambda_2),\ldots,\ln(\lambda_n^{(1)})I_n+\ln(\Lambda_2))\\ &=\ln(\Lambda_1)\otimes I_n+I_n\otimes\ln(\Lambda_2) \end{split} $$ (here, we used just the "log-of-product-is-a-sum-of-logs" property of $\ln$). Hence (using again the mixed product property) $$ \begin{split} (\Lambda_1\otimes\Lambda_2)(\ln(\Lambda_1\otimes\Lambda_2)) &= (\Lambda_1\otimes\Lambda_2)(\ln(\Lambda_1)\otimes I_n+I_n\otimes\ln(\Lambda_2))\\ &= (\Lambda_1\otimes\Lambda_2)(\ln(\Lambda_1)\otimes I_n)+(\Lambda_1\otimes\Lambda_2)(I_n\otimes\ln(\Lambda_2))\\ &= (\Lambda_1\ln(\Lambda_1))\otimes\Lambda_2+\Lambda_1\otimes(\Lambda_2\ln(\Lambda_2)) \end{split} $$ Therefore, with $\rho=\rho_1\otimes\rho_2$, using the useful fact, and the trace-of-a-Kronecker-product property, $$ \begin{split} S(\rho)&=-\mathrm{tr}((\Lambda_1\ln(\Lambda_1))\otimes\Lambda_2+\Lambda_1\otimes(\Lambda_2\ln(\Lambda_2)))\\ &=-\mathrm{tr}(\Lambda_1\ln(\Lambda_1))\mathrm{tr}(\Lambda_2) -\mathrm{tr}(\Lambda_1)\mathrm{tr}(\Lambda_2\ln(\Lambda_2))\\ &=\mathrm{tr}(\Lambda_1)S(\rho_2)+\mathrm{tr}(\Lambda_2)S(\rho_1)\\ &=\mathrm{tr}(\rho_1)S(\rho_2)+\mathrm{tr}(\rho_2)S(\rho_1). \end{split} $$ If $\mathrm{tr}(\rho_i)=\mathrm{tr}(\Lambda_i)=1$ ($i=1,2$), then $$ S(\rho)=S(\rho_1)+S(\rho_2). $$

  • 1
    $\begingroup$ Thank you a lot, (especially for your useful fact !), your answer clarified many of my uncertainties ! $\endgroup$ – faero Jan 1 '14 at 23:35

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