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Consider a quantum system consisting of two subsystems, $A$ and $B$. Let $\rho$ be the density matrix of the whole system $A\cup B$. Let $|\alpha\rangle$, $\alpha = 1,2\cdots d_B$, be the states of the subsystem $B$. Then $\rho$ can be written as the following: $$ \rho = \sum^{d_B}_{\alpha=1}\sum^{d_B}_{\beta=1}\sigma_{\alpha\beta}\otimes|\alpha\rangle\langle\beta|, $$ where $\sigma_{\alpha\beta}$ are sub-density-matrices for subsystem $A$ of size $d_A\times d_A$. Here $d_A$ is the dimension of the Hilbert space of the subsystem $A$. The reduced density matrix of subsystem $A$ is given by $$ \rho_A = \sum^{d_B}_{\alpha=1}\sigma_{\alpha\alpha}, $$ and the reduced density matrix of subsystem $B$ is given by $$ \rho_B = \sum^{d_B}_{\alpha=1}\sum^{d_B}_{\beta=1}\mathrm{tr}(\sigma_{\alpha\beta})\otimes|\alpha\rangle\langle\beta|. $$

Let us consider a process after which the quantum coherence of subsystem $B$ is lost. The density matrix then becomes: $$ \rho' = \sum^{d_B}_{\alpha=1}\sigma_{\alpha\alpha}\otimes|\alpha\rangle\langle\alpha|. $$ I am interested to know whether it is possible to relate the Renyi entropy of the new density matrix $\rho'$, defined as $$ S_\alpha(\rho')=\frac{\ln\mathrm{tr}(\rho'^\alpha)}{1-\alpha}, $$ to the Renyi entropy of density matrices $\rho$, $\rho_A$, $\rho_B$, or similar quantities. If the quick answer is no, I hope someone could point me to useful references.

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up vote 1 down vote accepted

For Von Neumann entropies you would have:

$H(AB)_\rho\leq H(A)_\rho+H(B)_\rho$

$H(AB)_{\rho'}\leq H(A)_{\rho'} + H(B)_{\rho'}$

whatever happens in $B$ should not affect $A$ so:

$H(A)_\rho=H(A)_{\rho'}$

for $B$ itself:

$H(B)_\rho\leq H(B)_{\rho'}$

with equality if the decoherence process is in the basis that diagonalizes $B$. And:

$H(AB)_\rho\leq H(AB)_{\rho'}$

This is all textbook material, you can find it in Nielsen and Chuang or in Wilde's Quantum Shannon Theory. Now if you replace Von Neumann entropies by Renyi entropies, the third to fifth relations are still valid, but subadditivity doesn't hold and you would have something a little bit weaker for any bipartite system:

$S_\alpha(AB)\leq S_\alpha(A) + S_0(B)$

This relation was proved in (http://arxiv.org/abs/quant-ph/0204093)

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Thank you for your answer. I am actually more interested in the possibility of directly expressing $S_\alpha(\rho')$ in terms of $S_\alpha(\rho)$, $S_\alpha(\rho_A)$ or similar quantities. In other words, I want to know if it is possible to have equality relations. I apologize if I didn't make my question clear. –  Isidore Seville Sep 20 '13 at 21:57
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Maybe someone can help you, but I don't think you can make any more precise statements than these ones except if you restrict your self to special situations: cq-state and decoherence is on the c system on the "c" basis, or the system is product, etc. –  Ando Masahashi Sep 23 '13 at 19:22
    
After thinking for a while, I realized that the answer to my original question was simply no because mathematically what I asked for was to determine the diagonal elements of a matrix from its eigenvalues. Therefore, I choose Ando's answer as it is probably the best of what we can say about my question. –  Isidore Seville Oct 3 '13 at 3:38
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