Why does quantum distinguishability ensure no faster-than-light communication?

Question

On page 56-57 in Nielsen and Chuang, for a proposed scenario, it's said that:

if Bob had access to a device that could distinguish the four states $|0\rangle$, $|1\rangle$, $|+\rangle$, $|−\rangle$ from one another, then he could tell whether Alice had measured in the computational basis, or in the $|+\rangle$, $|−\rangle$ basis.

I'm confused about why this is the case.

Answer 1

In the situation described in the book, Alice and Bob share the state

$$ |\psi\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}}. $$

Using the definition $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$ and simple algebra we can see that $|\psi\rangle$ can also be written

$$ |\psi\rangle = \frac{|{++}\rangle+|{--}\rangle}{\sqrt{2}}. $$

Now, if Alice measures $|\psi\rangle$ in the $|0\rangle$, $|1\rangle$ basis and obtains the $|0\rangle$ outcome then the joint state collapses to $|00\rangle$. Similarly for the $|1\rangle$ outcome, the state collapses to $|11\rangle$. On the other hand, if Alice measures $|\psi\rangle$ in the $|+\rangle$, $|-\rangle$ basis and obtains the $|+\rangle$ outcome then the joint state collapses to $|{++}\rangle$. Similarly, for the $|-\rangle$ outcome, the state collapses to $|{--}\rangle$.

Thus, if Alice chooses to measure in the $|0\rangle$, $|1\rangle$ basis, then Bob's part of the state becomes $|0\rangle$ or $|1\rangle$ and if Alice chooses to measure in the $|+\rangle$, $|-\rangle$ basis, then Bob's part of the state becomes $|+\rangle$ or $|-\rangle$. Therefore, a device to distinguish the four states $|0\rangle$, $|1\rangle$, $|+\rangle$, $|-\rangle$ would enable Bob to instantly discover the basis in which Alice measured her part of the state.