Let $A, B$ be (finite-dimensional) Hilbert spaces, and $\rho$ some mixed state of $A \otimes B$. I am trying to show that a measurement performed on the '$A$-subsystem' does not affect $\rho^B = \text{Tr}_A(\rho)$.

I understand a 'measurement performed on the $A$-subsystem' as given by some observable $X \otimes I$, where $X$ is a self-adjoint operator on $A$ which decomposes as $X = \sum m P_m$ (where $P_m$ is the orthogonal projection on the $m$-eigenspace). If this measurement results in outcome $m$, the resulting state should be

$$\rho' = \frac{(P_m \otimes I) \rho (P_m \otimes I)}{\text{Tr}((P_m \otimes I) \rho)}$$ and I wish to see that $\text{Tr}_A(\rho') = \text{Tr}_A(\rho)$. Now I can use the cyclicity of trace to see that

$$\text{Tr}_A(\rho') = \frac{\text{Tr}_A((P_m \otimes I) \rho)}{\text{Tr}((P_m \otimes I) \rho)}$$

but why should this be equal to $\text{Tr}_A(\rho)$? I did check this is the case if $\rho = \rho^A \otimes \rho^B$ decomposes as a product. And I know a general $\rho$ will be a linear combination of such cases; but since the equality desired is not linear, it does not seem to follow that it'll hold (in fact, it seems to indicate it will not hold). Hopefully someone can point to the mistake in my thinking.

  • $\begingroup$ One way would be to compare the expectation value of an arbitrary local operator $M_A$ on $\rho_A$ with that of the global one $M_A \otimes I_B$ and $\rho_{AB}$, see for example, here. $\endgroup$ Nov 1, 2020 at 5:46

2 Answers 2


What you've denoted as $\rho'$ is just an $m$-th possible outcome. We have to write $$ \rho'_m = \frac{(P_m \otimes I) \rho (P_m \otimes I)}{\text{Tr}((P_m \otimes I) \rho)}. $$ Now, since Bob doesn't know the value of $m$ he has to assume that the new state is a mixture of $\rho'_m$ with corresponding probabilities $\text{Tr}((P_m \otimes I) \rho)$. So, the actual $\rho'$ is $$ \rho' = \sum_m \rho'_m \text{Tr}((P_m \otimes I) \rho) = \sum_m (P_m \otimes I) \rho (P_m \otimes I). $$ Finally, you can see that $$ \text{Tr}_A(\rho') = \sum_m \text{Tr}_A((P_m \otimes I) \rho (P_m \otimes I)) = $$ $$ = \sum_m \text{Tr}_A(\rho (P_m \otimes I)) = \text{Tr}_A(\rho \sum_m (P_m \otimes I)) = \text{Tr}_A(\rho). $$


Indeed, this equation does not hold. Take for example $\rho = |\phi^+\rangle\langle \phi^+|$, where $|\phi^+\rangle = \frac1{\sqrt2}(|00\rangle+|11\rangle)$. If Alice obtains outcome 0, then $\rho' = |00\rangle\langle 00|$, and $\rho^B = |0\rangle\langle0|$, and if Alice obtains outcome 1, then $\rho' = |11\rangle\langle 11|$, and $\rho^B = |1\rangle\langle1|$. This is the typical example of the nonlocality of wavefunction collapse.

The statement of the no-communication theorem is that when you average over Alice's outcomes, then the result does not depend on which measurement she made, or indeed if she made a measurement at all. Let then $$\rho_m = \frac{(P_m \otimes I) \rho (P_m \otimes I)}{\text{Tr}((P_m \otimes I) \rho)},$$ the collapsed state after obtaining measurement result $m$, and $$\rho^B_m = \operatorname{tr}_A (\rho_m).$$ The theorem is then that $$\sum_m p(m) \rho^B_m = \rho^B = \operatorname{tr}_A\rho,$$ for all possible projectors $P_m$, or even POVM elements.


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