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$\newcommand\ket[1]{|#1\rangle}$ $\newcommand\bra[1]{\langle #1|}$ $\newcommand\mean[1]{\langle #1\rangle}$

Let us consider an entangled state $\mathcal{H} = \mathcal{H_1}\otimes\mathcal{H_2}$. In this state, one defines the density operator $\rho$ as $\rho=\sum_i\rho_i\ket{i}\bra{i}$. It is a well-known result that the mean value of $A_1\otimes\mathcal{I}_2$ (where $\mathcal{I}_2$ is the identity in $\mathcal{H}_2$) would be given by \begin{equation} \mean{\mathcal{A}_1\otimes\mathcal{I}_2} = \mathrm{Tr}\left[\rho\left(\mathcal{A}_1\otimes\mathcal{I}_2\right)\right] = \mean{\mathcal{A}_1}_{\rho_1} = \mathrm{Tr}_{\mathcal{H}_1}\left[\rho_1\mathcal{A}_1\right] \end{equation} where $\rho_1 = \mathrm{Tr}_{\mathcal{H}_2}\rho$ is the reduced density matrix on the subsystem $\mathcal{H}_1$ and $\mathrm{Tr}_{\mathcal{H}_i}$ is the partial trace over the Hilbert space $\mathcal{H}_i$. Let us consider $\rho_1$ and $\rho_2$.

One defines the Von Neumann entropy as \begin{equation} S(\rho) = -\mathrm{Tr}\left[\rho \log\rho\right] \end{equation} One of the many-properties of this entropy is that - for any entangled state, $S(\rho_1\otimes\rho_2) = S(\rho_1)+S(\rho_2)$.

I have proved it the following way. I define ${\ket{1:\alpha,2:a}}$ as the basis of $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. In this basis, one writes \begin{align} S(\rho_1\otimes\rho_2) = - \mathrm{Tr}\left[\rho_1\otimes\rho_2\; ln\left(\rho_1\otimes\rho_2\right)\right] &= -\sum_{\alpha,a}\bra{1:\alpha,2:a}\rho_1\otimes\rho_2\ket{1:\alpha,2:a}\log(\rho_1\otimes\rho_2)\\ \end{align} I use now the fact that the logarithm of a product is equal to the sum of the logarithms: $\log(a b) = \log(a)+\log(b)$. Am I allowed to generalize this to the tensor product of two states? Using this property, I find that \begin{equation} S(\rho_1\otimes\rho_2) = -\sum_\alpha \bra{1:\alpha}\rho_1\ket{1:\alpha} + \left(-\sum_\beta \bra{2:a}\rho_2\ket{2:a}\log(\rho_2)\right) = S(\rho_1)+S(\rho_2) \end{equation} It feels like cheating, yet it seems to work. Any insights?

Additionally, I have trouble interpreting this result. Thanks for your input!

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  • $\begingroup$ Have you checked the definition of the entropy in terms of the eigenvalues of $\rho$? In other words, do you know how the eigenvalues and eigenvectors of $\rho_1$ and $\rho_2$ are related to the eigenvalues and eigenvectors of $\rho$? $\endgroup$ Jan 2, 2022 at 12:55
  • $\begingroup$ @Jakob I am currently editing the post with an idea I had. Looking into your suggestion when it's published. I'll ping you then. $\endgroup$
    – Juian
    Jan 2, 2022 at 12:58
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    $\begingroup$ I dont think it is true. Did you mean for any 'non entangled state'? (rho1 tensor rho2 is not entangled). I mean take a pure bell state, so the entropy (of a pure state) is 0. The reduced density matrices are fully mixed and have entropy. So you have <= $\endgroup$
    – lalala
    Jan 2, 2022 at 13:03
  • $\begingroup$ The Schmidt decomposition may help you with both calculation and interpretation. It lets you express the state of a system $AB$ as a sum over orthonormal states on $A$ and $B$ separately: $|\psi \rangle = \sum_i\lambda_i|i_A\rangle |i_B\rangle$ (it works for density matrices too). Then the Von Neuman entanglement between $A$ and $B$ is $-\sum_i|\lambda_i|^2\log\left(|\lambda_i|^2\right)$, which is zero in a product state (as is the case for you) because there is only one nonzero $\lambda_i$ ( $\lambda_0=1$) . Your question just requires an extension of this from pure states to density matrices $\endgroup$ Jan 7, 2022 at 6:47

1 Answer 1

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Remarks on your calculations $\newcommand{Tr}{\operatorname{Tr}}$

\begin{align} S(\rho_1\otimes\rho_2) = - \mathrm{Tr}\left[\rho_1\otimes\rho_2\; ln\left(\rho_1\otimes\rho_2\right)\right] &= -\sum_{\alpha,a}\langle 1:\alpha,2:a|\rho_1\otimes\rho_2|1:\alpha,2:a\rangle\log(\rho_1\otimes\rho_2)\\ \end{align} I use now the fact that the logarithm of a product is equation to the sum of the logarithms: $\log(a b) = \log(a)+\log(b)$. Am I allowed to generalize this to the tensor product of two states?

The last equality does not make sense, as $\log(\rho_1\otimes \rho_2)$ is an operator : it should come before the ket (see my calculations below).

Given a hermitian operator $\hat \Lambda$ and a function $f$, we can define $f(\Lambda)$ the following way : if $|i\rangle$ is an orthonormal basis such that $\hat\Lambda|i\rangle = \lambda_i|i\rangle$, then we define $f(\lambda)$ as the linear operator such that : $$f(\hat\Lambda) |i\rangle = f(\lambda_i) |i\rangle$$

Since $f(\lambda) = \lambda \log(\lambda)$ is well-defined on $[0,+\infty)$ (by setting $f(0) = 0$, we can define $\hat \Lambda \log \hat \Lambda$ if $\hat \Lambda$ is positive semi-definite (which is true for density matrices).

Using bases of eigenvectors

Let $|i\rangle_1$ and $|x\rangle_2$ be orthonormal bases of eigen-vectors of $\rho_1$ and $\rho_2$ respectively, with : $$\rho_1 |i\rangle_1 = p_i|i\rangle_1 \qquad \text{and}\qquad \rho_2|x\rangle_2 = q_x |x \rangle_2$$

Then, $|i\rangle_1 \otimes |x\rangle_2$ is a basis of $\mathcal H_1\otimes \mathcal H_2$, so we can use it to compute the trace : \begin{align} S(\rho_1\otimes \rho_2) &= -\Tr(\rho \log\rho) \\ &= -\sum_{i,x}\langle i|_1 \otimes \langle x|_2(\rho_1\otimes \rho_2\log(\rho_1\otimes \rho_2)) |i\rangle_1\otimes |x\rangle_2 \\ &= -\sum_{i,x} p_i q_x \log (p_i q_x) \\ \end{align} From this point on, we are only dealing with real numbers, so we get : \begin{align} S(\rho_1\otimes \rho_2) &= -\sum_{i} p_i \log (p_i)-\sum_{x} q_x \log ( q_x) \\ &= S(\rho_1) + S(\rho_2) \end{align}

What is $\log( \hat A\otimes \hat B)$ ?

In the intermediate steps of the calculation above, we showed that if $\hat A$ and $\hat B$ are positive definite hermitian operators on $\mathcal H_1$ and $\mathcal H_2$, then : $$\log (\hat A \otimes \hat B) = \log(\hat A)\otimes \mathbb I_2 + \mathbb I_1 \otimes\log(\hat B) $$

where $\mathbb I_1$ and $\mathbb I_2$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$.

Using this formula, we can redo the calculations directly, without using a basis : \begin{align} S(\rho_1\otimes \rho_2) &= -\Tr(\rho_1\otimes \rho_2 \log(\rho_1 \otimes \rho_2) )\\ &= -\Tr(\rho_1\otimes \rho_2 \cdot( \log(\rho_1) \otimes \mathbb I_2 + \mathbb I_1 \otimes \log \rho_2) )\\ &= -\Tr ( (\rho_1 \log \rho_1) \otimes \rho_2 + \rho_1 \otimes (\rho_2 \log \rho_2)) \\ &= -\Tr ( \rho_1 \log \rho_1))\Tr( \rho_2) - \Tr(\rho_1)\Tr(\rho_2 \log \rho_2)) \\ &= S(\rho_1) + S(\rho_2) \end{align}

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    $\begingroup$ I've edited some signs which I think were typos. Feel fry to undo the edit if necessary. $\endgroup$ Jan 2, 2022 at 14:48
  • $\begingroup$ Thank you very much! It helped me a lot. $\endgroup$
    – Juian
    Jan 3, 2022 at 15:56

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