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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 $\rho'_A$.

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    $\begingroup$ Just to be clear: Bob performs a measurement. He gets a result corresponding to projector $\Pi_B$. You're wanting to describe $\rho_A'$ from the perspective of someone who knows what answer Bob got, rather than, for example, from the perspective of Alice who, if she's distantly separated, wouldn't know the measurement result, and would still describe her system by $\rho_A$? $\endgroup$ – DaftWullie Oct 20 '20 at 6:44
  • $\begingroup$ @DaftWullie that's right. Just to clarify, $\rho'_A$ is subnormalized here. The other subtelty here is looking at the smallest nonzero eigenvalue of $\rho_A$ and $\rho'_A$ $\endgroup$ – user1936752 Oct 20 '20 at 12:25
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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 $\lambda_k - \varepsilon \le \lambda'_k \le \lambda_k$.

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  • $\begingroup$ Thank you for the answer. A short follow up: Let $\lambda$ be the smallest nonzero eigenvalue of $\rho_A$. If $\lambda \leq \varepsilon$, can one say anything about the smallest nonzero eigenvalue of $\rho'_A$? So far, it seems it could be anything in $[0, \lambda]$. $\endgroup$ – user1936752 Oct 23 '20 at 2:21
  • $\begingroup$ From $\rho'_A \le \rho_A$ we can deduce that $n$-th smallest eigenvalue of $\rho'_A$ is not greater than $n$-th smallest eigenvalue of $\rho_A$, i.e. $\lambda'_n \le \lambda_n$. But if $\lambda_k$ is the first non-zero value, $\lambda'_k$ can be zero and the first non-zero value will be at some higher index $\lambda'_{k+l}$. But we can't deduce $\lambda'_{k+l} \le \lambda_{k}$. And that trace inequality doesn't help much if $\varepsilon$ is too big. So it's not even in $[0, \lambda]$. $\endgroup$ – Danylo Y Oct 23 '20 at 8:09
  • $\begingroup$ But if $\varepsilon < \lambda$ then we can deduce it must be in $[\lambda-\varepsilon, \lambda]$, check the update. $\endgroup$ – Danylo Y Oct 23 '20 at 9:31

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