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

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.

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.

• Very nice explanation. +1 Oct 14, 2021 at 11:52