Bob applies a projector - what happens to eigenvalues of Alice's reduced state?

Question

Suppose Alice and Bob share a state $\rho_{AB}$. Let us denote the reduced states as $\rho_A = \text{Tr}_B(\rho_{AB})$ and $\rho_B = \text{Tr}_A(\rho_{AB})$. Bob applies a projector so the new global state is

$$\rho'_{AB} = (I_A\otimes \Pi_B)\rho_{AB}(I_A\otimes \Pi_B)$$

Let us denote the new (subnormalized) reduced state on Alice's system as $\rho'_{A}$. I am given two facts about Bob's projector

1. $\Pi_B$ is diagonal in the eigenbasis of $\rho_B$.

2. It is gentle i.e. $\text{Tr}(\Pi_B\rho_B) \geq \text{Tr}(\rho_B) - \varepsilon$ for some small $\varepsilon$.

I would like to know how the eigenvalues of $\rho'_A$ are related to those of $\rho_A$. So far, the only conclusion I have is that $\rho'_A\leq \rho_A$ where $A \leq B$ means that $B-A$ is positive semidefinite.

In particular, I am interested in any inequality relating the smallest nonzero eigenvalue of $\rho_A$ and the smallest nonzero eigenvalue of $\rho'_A$.

Answer 1

Just notice that $$ \text{Tr}(\rho'_{AB}) = \text{Tr}(\Pi_B\rho_{B}). $$ One way to see this is to consider any decomposition $$ \rho_{AB} = \sum_i A_i \otimes B_i, $$ where $A_i, B_i$ just some matrices, not states.
Then $$ \text{Tr}(\rho'_{AB}) = \text{Tr}\big(\sum_i A_i \otimes \Pi_BB_i\Pi_B\big) = $$ $$ = \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i\Pi_B) = \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i) $$ and $$ \text{Tr}(\Pi_B\rho_{B}) = \text{Tr}(\Pi_B \sum_i \text{Tr}(A_i)B_i) = $$ $$ = \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i). $$

Now we can deduce that $$ \text{Tr}(\rho_{AB} - \rho'_{AB}) = 1 - \text{Tr}(\Pi_B\rho_{B}) \le \varepsilon $$ and hence $$ \text{Tr}(\rho_{A} - \rho'_{A}) \le \varepsilon $$ since $\text{Tr}(\rho'_{AB}) = \text{Tr}(\rho'_{A})$.

This bounds $\rho'_{A}$ from below (in contrast to $0 \le \rho_{A} - \rho'_{A}$). It can be seen now that $\rho'_{A} \rightarrow \rho_{A}$ if $\varepsilon \rightarrow 0$.

Update

From trace inequality and positivity of $\rho_A - \rho'_A$ we can deduce $$ 0 \le \rho_A - \rho'_A \le \varepsilon I, $$ so $$ \rho'_A \le \rho_A \le \rho'_A + \varepsilon I. $$ Hence $$ \lambda'_i \le \lambda_i \le \lambda'_i + \varepsilon, $$ where $\{ \lambda_i \}, \{ \lambda'_i \}$ are sorted eigenvalues of $\rho_A, \rho'_A$.

Now if $\lambda_k$ is the first non-zero eigenvalue and $\varepsilon < \lambda_k$ then $\lambda'_k \ge \lambda_k - \varepsilon > 0$, which means $\lambda'_k$ will be the first non-zero eigenvalue of $\rho'_A$. For it we know that $\lambda_k - \varepsilon \le \lambda'_k \le \lambda_k$.