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I'm trying to understand the steps in a certain proof, where I think the following claim is used: $S( |i \rangle \langle i| \otimes \rho) = S(\rho)$, where $S()$ is the entropy, $|i \rangle$ is a state from some orthonormal basis, and $\rho$ is a density operator. Here's my attempt to see why this holds:

$$ \begin{align} S( |i \rangle \langle i| \otimes \rho) &= -tr\left[\left(|i \rangle \langle i| \otimes \rho\right) \times \log \left(|i \rangle \langle i| \otimes \rho\right)\right]\\ &= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|) \otimes \rho \log \rho]\\ &= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|)] \times tr[\rho \log \rho]\\ &= 0 \times S(\rho) \end{align}$$

so I am clearly making a mistake somewhere, that $0$ should not be there. The first equality comes from the definition of $S()$, the second because $(A \otimes B)(C \otimes D) = AC \otimes BD$, the third because $tr(A \otimes B) = tr(A) \times tr(B)$. Are any of these incorrect?

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    $\begingroup$ You also put $log (A \otimes B) = log A \otimes log B$ which got fed in as $C$ and $D$ in the second. $\endgroup$
    – AHusain
    Commented Jan 17, 2017 at 0:45

1 Answer 1

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$$\log(A\otimes B)=\log A \otimes I + I\otimes \log B$$ You can show this by diagonalizing $A$ and $B$ and using the idea that the matrix log of a diagonal matrix is the log of the elements. In components,

$$\log(A\otimes B)_{ii',jj'}=\log(A_i B_{i'})I_{ij}I_{i'j'}=(\log A_i I_{ij})I_{i'j'}+I_{ij}(\log B_{i'}I_{i'j'})$$

Then if you use your correct rules for tensor product manipulation in your question you will prove the statement.

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