Proving that reduced density matrices $\rho_{\mathrm{out}}$ and $\rho_{\mathrm{in}}$ have the same eigenvalues

Question

This question references an argument appearing as Reference 1 in the paper Entropy and Area, written by Srednicki. The aim of the argument is to demonstrate that, for a bipartite quantum system, one can compute the entropy associated to either side of the partition, and they will be equal.

The argument achieves this by showing that the reduced density matrices, associated to the subsystems on either side of the partition, have the same eigenvalues:

Let $|0\rangle = \sum_{ia} \psi_{ia} > |i\rangle_{\mathrm{in}}|a\rangle_{\mathrm{out}}$, so that $\rho_{\mathrm{in}})_{ij} = (\psi \psi^\dagger)_{ij}$ and $(\rho_{\mathrm{out}})_{ab} = (\psi^T\psi^*)_{ab}$. Now it is clear that $\mathrm{Tr}\rho_{\mathrm{in}}^k = \mathrm{Tr}\rho_\mathrm{out}^k$ for any positive integer $k$. This can only be true if $\rho_{\mathrm{in}}$ and $\rho_{\mathrm{out}}$ have the same eigenvalues, up to extra zeros.

I understand how to get to the relation of $\mathrm{Tr}\rho_{\mathrm{in}}^k = \mathrm{Tr}\rho_\mathrm{out}^k$, but the final statement is giving me trouble. My work so far is to use the fact that the reduced density matrices $\rho_{\mathrm{in}}$ and $\rho_{\mathrm{out}}$ are Hermitian, and so we can consider a basis in which they are diagonal. In such a basis, taking $\alpha_i$ to be the eigenvalues of $\rho_{\mathrm{in}}$ and $\beta_i$ to be the eigenvalues of $\rho_{\mathrm{out}}$, the trace relation becomes, \begin{align} \sum_{i=1}^{N} \alpha_i^k = \sum_{j=1}^M \beta_j^k, \quad k \in \mathbb{Z}^+ \end{align} and generally we suppose $M>N$, i.e. that our `out' subsystem is larger. This feels like it should be sufficient to demonstrate that all the $\alpha_i$'s are equal to the $\beta_i$'s, plus some extra zeros, as this is quite a restrictive constraint for two sets of numbers to possess this equality for any $k$. However, I'm having trouble finding the correct argument for it.

Answer 1

Here's one way to do it. Without loss of generality, I will assume $N\leq M$, and that both sets of eigenvalues are ordered from largest to smallest; i.e. $\alpha_1$ and $\beta_1$ are the largest eigenvalues of $\rho_\text{in}$ and $\rho_\text{out}$, respectively, and so on. If I raise your last equation $$\sum_{i = 1}^N \alpha_i^k = \sum_{i=1}^M \beta_i^k \tag{1}$$ to the power $1/k$, I get $$\Vert \boldsymbol{\alpha}\Vert_k = \Vert \boldsymbol{\beta}\Vert_k, \tag{2}$$ where $\Vert\cdot\Vert_k \equiv \Big(\sum_n (\cdot)^k\Big)^{1/k}$ denotes the $k$-norm, and $\boldsymbol{\alpha} \equiv (\alpha_1,...,\alpha_N)$ and $\boldsymbol{\beta} \equiv (\beta_1,...,\beta_M)$. It is well known that when $k\rightarrow\infty$, the $k$-norm is simply the largest element of the vector. Since $(2)$ is true for any non-negative $k$, we deduce that the largest $\alpha_i$ has to equal the largest $\beta_i$, i.e. $$\Vert \boldsymbol{\alpha}\Vert_\infty = \alpha_1 = \beta_1 = \Vert \boldsymbol{\beta}\Vert_\infty.$$ Now since these two terms are the same, I can cancel them out in $(1)$, leaving me with $$\sum_{i = 2}^N \alpha_i^k = \sum_{i=2}^M \beta_i^k \tag{3}.$$ So we've ended up in the exact same situation for the vectors $\boldsymbol{\alpha} '\equiv (\alpha_2,...,\alpha_N)$ and $\boldsymbol{\beta}' \equiv (\beta_2,...,\beta_M)$, i.e. $\Vert \boldsymbol{\alpha}'\Vert_k = \Vert \boldsymbol{\beta}'\Vert_k$ for any non-negative $k$. You can now repeat the exact same step to conclude that $\alpha_2 = \beta_2$, and so on. If $N = M$, this will go on until $\alpha_i = \beta_i$ for all $i\in\{1,...,N\}$. If $N<M$, we'll end up with $\alpha_i = \beta_i$ for all $i \in\{1,...,N\}$, and $$0 = \sum_{N+1}^M\beta_i^k.$$ Since $\beta_i$ are eigenvalues of a density matrix, they're all non-negative. Non-negative numbers can only add up to zero if all of them are individually zero, i.e. $$\beta_i = 0 \qquad i\in\{N+1,...,M\}.$$ Therefore, all non-zero $\alpha_i$ and $\beta_i$ have to be the same.