# Additive property of the Von Neumann entropy

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

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 $$$$\mean{\mathcal{A}_1\otimes\mathcal{I}_2} = \mathrm{Tr}\left[\rho\left(\mathcal{A}_1\otimes\mathcal{I}_2\right)\right] = \mean{\mathcal{A}_1}_{\rho_1} = \mathrm{Tr}_{\mathcal{H}_1}\left[\rho_1\mathcal{A}_1\right]$$$$ where $$\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 $$$$S(\rho) = -\mathrm{Tr}\left[\rho \log\rho\right]$$$$ One of the many-properties of this entropy is that - for any entangled state, $$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 $$$$S(\rho_1\otimes\rho_2) = -\sum_\alpha \bra{1:\alpha}\rho_1\ket{1:\alpha} + \left(-\sum_\beta \bra{2:a}\rho_2\ket{2:a}\log(\rho_2)\right) = S(\rho_1)+S(\rho_2)$$$$ It feels like cheating, yet it seems to work. Any insights?

• 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 $\rho$? Commented Jan 2, 2022 at 12:55
• @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. Commented Jan 2, 2022 at 12:58
• 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 <= Commented Jan 2, 2022 at 13:03
• 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 Commented Jan 7, 2022 at 6:47

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 : $$f(\hat\Lambda) |i\rangle = f(\lambda_i) |i\rangle$$ (It turns out that this operator does not depend on the eigenbasis we chose.)

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}

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