The following exercise is found on p. 152 in Benjamin Schumacher and Michael Westmoreland's book Quantum Processes, Systems and Information:

"The composite quantum system AB is initially in the state $|\Psi^{(AB)}\rangle$ and B evolves by the unitary operator $V^{(B)}$. Show that, for any A-state $|a^{(A)}\rangle$, $$\left<a^{(A)}|1^{(A)} \otimes V^{(B)} | \Psi^{(AB)}\right> = V^{(B)}\left<a^{(A)}|\Psi^{(AB)} \right>."$$

(If notation is non-standard, $\left<a^{(A)}|\Psi^{(AB)} \right>$ is the partial inner product, which is equal to the (non-normalized) conditional state of system B given that a basic A-measurement on $|\Psi^{(AB)}\rangle$ results in outcome $|a^{(A)}\rangle$).

This is not difficult so show. What I find strange is the following passage interpreting the result:

"We can interpret this exercise by comparing two processes:

  1. First we evolve the joint state $|\Psi^{(AB)}\rangle$ according to $1^{(A)} \otimes V^{(B)}$, then we use the result to find the conditional state for $|a^{(A)}\rangle$.

  2. First we use $|\Psi^{(AB)}\rangle$ to find the conditional state for $|a^{(A)}\rangle$, then we evolve this state according to $V^{(B)}$

Both processes lead to exactly the same final state of B. The dynamical evolution of B commutes with the measurement process on A."

Here is my question: Is it not a bit misleading to say that both processes lead to the same final state of B? Clearly, by both processes, the final state of B depends on the result of the preceding A-measurement - but this result is non-deterministic, so the result of the A-measurement might be different in process 1 and process 2, resulting in different final states of B? The final state of B will only be the same in both processes, if the A-measurement by chance gives the same result in both cases. Or what? Am I missing something?

In case someone else has the book: The reason I'm asking the above question is that the authors, arguing on p. 153 that you can do "Type 1 quantum communication" by doing "Type 2 quantum communication", seems to need that you always get the exact same final B-state by both processes (or purhaps I just don't get their argument).

  • $\begingroup$ Hi, do you know about density matrices? $\endgroup$ Apr 13, 2020 at 17:40
  • $\begingroup$ Yes, I do. I've learned about them since asking the above question (they are introduced in the chapter following the exercise I was asking about). $\endgroup$
    – Kimarokko
    Apr 14, 2020 at 7:03
  • $\begingroup$ I'm reading through your nice answer below, btw., making sure I understand it line by line. I'll also see if your explanation helps me resolve the trouble I had understanding the argument concerning "type 1 and 2 quantum communication". $\endgroup$
    – Kimarokko
    Apr 14, 2020 at 8:07
  • 1
    $\begingroup$ I hope it's clear enough, otherwise, feel free to ask :) $\endgroup$ Apr 14, 2020 at 8:08
  • $\begingroup$ Did you delete your answer? I hope you put it up again :) $\endgroup$
    – Kimarokko
    Apr 15, 2020 at 8:33

1 Answer 1


A weaker statement than that of "both processes lead to the same state" could be:

The statistics of a measurement of an observable $\Lambda$ on system $A$ will be the same whether it is measured before or after the evolution of $B$.

Meaning, the probability of observing each eigenvalue of a given observable is not influenced by the $B$ evolution, nor are the resulting post-measurement states. To see this,let's work with density matrices instead. Call $\rho_{AB}=|\psi\rangle\langle\psi|_{AB}$, then to obtain the probability of observing the outcome $a$ if $|a\rangle_A$ is the eigenvector of some observable the experimentalist on the $A$ side (let's call her Alice) measures, we have to project the state onto the corresponding eigenspace and take the trace

$$ \mathrm{Tr}(|a\rangle\langle a|_A \otimes \mathbb{1}_B|\psi\rangle\langle\psi|_{AB})$$ You can convince yourself that this is equal to

$$ \mathrm{Tr}_A(|a\rangle\langle a|_A \mathrm{Tr}_B(|\psi\rangle\langle\psi|_{AB})=\mathrm{Tr}_A(|a\rangle\langle a|_A \rho_A)$$

Since, as you've shown, the $B$ evolution commutes with the projection, this probability is the same regardless of whether the observable is measured before or after the evolution: even more, you can see that the probability does not depend at all on the state $\rho_B$ (except for how $\rho_A$ itself depends on it).

To understand the somewhat stronger statement "both processes lead to the same final state on $B$", it's better to understand what we usually mean by state after a measurement. Suppose you have an observable $\Lambda$ with eigendecomposition $\Lambda=\sum_k \lambda_k P_k$ where $P_k$ is the projector onto the eigenspace of eigenvalue $\lambda_k$. For simplicity, let's say there are no degenerate eigenvalues, so that $P_k=|k\rangle\langle k|$ where $|k\rangle$ is an eigenvector of $\Lambda$.

$$ \Lambda |k\rangle=\lambda_k|k\rangle$$ Suppose also you have a quantum state $\rho$. The act of measuring $\Lambda$ corresponds to the quantum channel

$$ \rho\mapsto \mathcal M(\rho)=\sum_k \langle k|\rho |k\rangle |k\rangle\langle k|$$

I.e., the state $\rho$ is replaced by a state which is diagonal in the $|k\rangle$ basis, which can be interpreted as a classical probability distribution: the state $|k\rangle\langle k|$ has probability $\langle k|\rho |k\rangle$. If $\rho=|\psi\rangle\langle \psi|$ is a pure state then this probability turns into the familiar $|\langle \psi|k\rangle|^2$.

This is what you can think of as the state after the measurement, if you don't look at the measurement result: you know the state is one of the eigenvectors of $\Lambda$, and you know the probability of observing each one. If the state is bipartite, $\rho=\rho^{(AB)}$, you can very easily verify that

$$\mathcal M_A(V^B\mathcal \rho^{(AB)}V^{B\dagger})=V^B\mathcal M_A(\rho^{(AB)})V^{B\dagger}$$ or with more explicit notation $$\mathcal M^A\otimes \mathrm{id}^B(\mathbb 1^A\otimes V^B\mathcal \rho^{(AB)}\mathbb 1^A\otimes V^{B\dagger})=(\mathbb 1^A\otimes V^B)\mathcal M^A\otimes \mathrm{id}^B(\rho^{(AB)})(\mathbb 1^A\otimes V^{B\dagger})$$ I.e., the final state on is the same regardless of whether you measure $A$ before or after evolving $B$. Notice that this implies both that the statistics of the $A$ measurement aren't influenced by the order of operations, but also that the final state on $B$ is the same regardless of the order.

The argument given in the book is valid for any single measurement result, I guess it's there to state the result before introducing channels, you can think of it not as saying "if we repeat the experiment twice" but more like "for each possible measurement result, we get the same state whether the measurement was performed before or after the evolution".

To give a more physical motivation, suppose that the two different processes somehow led to different states. Then there would be some experiment Alice could make to know whether or not Bob has already performed the unitary evolution, this would allow Alice and Bob to communicate in principle faster than the speed of light would allow, which wouldn't be good for quantum mechanics at all.

  • $\begingroup$ This is a very clarifying explanation! I understand everything but the last equation - should this be clear from the above or is it the other way around, that the above should follow easily from this equation? I know that $\mathrm{Tr}_B(\rho^{AB})$ is equal to $\rho^{A}$, but I'm not comfortable with the concept of a quantum channel or with the notation $\mathcal E^A(\rho^{AB})$. Also, shouldn't it be $\mathcal E^{B}$ in this last equation? $\endgroup$
    – Kimarokko
    Apr 14, 2020 at 12:16
  • $\begingroup$ @Kimarokko yes it should be $\mathcal E^B$, sorry, that was a typo. And neither equation really follows from the other one, I included the last equation just as a more general principle. In fact I removed the last bit as on second thought it isn't so relevant. I also clarified the notation for the previous line $\endgroup$ Apr 14, 2020 at 18:20
  • $\begingroup$ Nice explanation. It's a a lot clearer to me now. I guess you could summarize some of it by saying that both processes lead to exactly the same density matrices $\rho^{A}$ and $\rho^{B}$ (as long as you don't look at the result of the measurements). $\endgroup$
    – Kimarokko
    Apr 15, 2020 at 16:31

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