I'm trying to answer the following question about the BB84 protocol from Nielsen and Chuang's Introduction to Quantum Information. Notfound

As I understand it, the string $b$ is determining whether we are measuring in the computational basis or the Hadamard basis. I decided to test the claim in the question with an example.

Let $\vert\phi\rangle = \alpha \vert 0 \rangle + \beta \vert 1\rangle$ and assume we are measuring in the $\{\vert - \rangle, \vert + \rangle\}$ basis. That means our measurement is

$(\vert - \rangle \langle - \vert + \vert + \rangle \langle + \vert)\vert \phi \rangle = \frac{\alpha}{\sqrt{2}}\vert 0 \rangle + \frac{\beta}{\sqrt{2}}\vert 1 \rangle$

To me this is clearly wrong. Not only does it suggest we can figure something out about the state which $\vert \phi \rangle$ was prepared in, but we don't even have that $\frac{1}{\sqrt{2}}(\alpha + \beta) = 1$. Could I have guidance as to what I am doing incorrectly, or pointers about how to go about this proof?

EDIT: Applying these projections yields $$\frac{\alpha + \beta}{\sqrt{2}}\vert + \rangle + \frac{\alpha - \beta}{\sqrt{2}}\vert - \rangle$$

We can take probabilities like $|\langle + \vert \phi \rangle|^{2}$, and I am still unsure how to prove that they are uncorrelated.

EDIT2: I have a solution and will update it soon.


2 Answers 2


To start with, remember that there are only 4 possible states you could be talking about: $$ |\psi_{00}\rangle=|0\rangle,\qquad |\psi_{10}\rangle=|1\rangle,\qquad |\psi_{01}\rangle=|+\rangle,\qquad |\psi_{11}\rangle=|-\rangle $$

It should be obvious that when $b'=b$, $a'=a$. This is because you're measuring the qubit in the same basis that it's prepared in (that's precisely the statement $b=b'$). For example, if the qubit is prepared in $|1\rangle$ ($a=1$) and you measure it in the $Z$ basis (projectors $|0\rangle\langle 0|,|1\rangle\langle 1|$), you definitely get the answer $|1\rangle$ ($a'=1$).

So, the only thing you need to check is, let's say a qubit is prepared in $|+\rangle$ ($b=1$) but you measure it in the $Z$ basis ($b'=0$). You get the two possible outputs with $50:50$ probability. Equally, had the state been in $|-\rangle$, it's 50:50 with that measurement. So, that measurement tells us nothing about which of the two states it was. The outcomes ($a'$) are random, and uncorrelated with the value of $a$.

  • $\begingroup$ Yep this looks correct. As I left in the OP I solved it before these answers I just haven't had the chance to come back and update it. Thanks for your help. $\endgroup$
    – Boots
    May 13, 2021 at 15:55

If you measure a qubit $\psi$ in $\{\vert - \rangle, \vert + \rangle\}$ basis, you get either $\{|+ \rangle\}$ with probability $|\langle+|\psi\rangle|^2$ or $\{|- \rangle\}$ with probability $|\langle-|\psi\rangle|^2$. The projectors are indeed $|+\rangle\langle+|$ and $|-\rangle\langle-|$ but you should use them right.


About projectors on state:

  1. Projector is a Hermitian operator having 2 eigenvalues, 1 and 0. For example, projector $|+\rangle\langle+|$ have eigenstates $|+\rangle$ with eigenvalue 1 and $|-\rangle$ with eigenvalue 0.
  2. We call $\hat{P}$ projector on state $\psi$ if $\psi$ is eigenstate of $\hat{P}$ with eigenvalue 1. For example, $|+\rangle\langle+|$ is projector on state $|+\rangle$.
  3. As a result of measurement in the projector basis we obtain one of the eigenstates of a projector.
  4. The expected value of a projector $\hat{P}$ on state $\psi$ is equal to probability of obtaining projector's eigenstate with eigenvalue 1 as a result of measurement in the projector basis, for example for projector $\hat{P}=|+\rangle\langle+|$ $$ \langle\psi|\hat{P}|\psi\rangle=\langle\psi|+\rangle\langle+|\psi\rangle=|\langle+|\psi\rangle|^2 $$

PPS: $\vert + \rangle \langle 0 \vert$ is NOT a projector, for example because any projector $P$ has property $P^2=P$ and $$ |+ \rangle \langle 0|^2 = |+ \rangle \langle 0|+ \rangle \langle 0|= \frac{1}{\sqrt{2}} |+ \rangle \langle 0| \neq |+ \rangle \langle 0| $$

  • $\begingroup$ I don't think those are the projectors I need. If you look, that would leave the sum of my amplitudes to be $\frac{\alpha^2 + \beta^2}{2} = \frac{1}{2}$. If instead we use the projectors $\vert - \rangle \langle 0 \vert$ and $\vert + \rangle \langle 1 \vert$ we get amplitudes that sum to 1 and appear uncorrelated to the original state we prepared. The reason we use $\vert - \rangle \langle 0 \vert$ is because we are unitarily transforming the 0,1 basis into the spin basis. $\endgroup$
    – Boots
    May 12, 2021 at 4:39
  • $\begingroup$ What? I don't see what is being "reinvented"? If you apply the projectors you mention here and then use the Born rule, it's clear your result doesn't have probabilities summing to one since $\frac{\alpha^2}{2} + \frac{\beta^2}{2} = \frac{1}{2}$. If you are sticking to this answer, can you at least elaborate why you think it's more correct? $\endgroup$
    – Boots
    May 12, 2021 at 15:23
  • $\begingroup$ I think I see what you mean, as in, I may be taking probabilities incorrectly. Is the correct route to show there's no correlation? We can say that the probability of measuring a $\vert + \rangle$ is unrelated to say measuring a $\vert 0 \rangle$ in the original basis? $\endgroup$
    – Boots
    May 12, 2021 at 15:48
  • $\begingroup$ could you tell me what the difference in these projectors is? I see the projector $\vert + \rangle \langle 0 \vert$ as casting into another basis. That is, it sends $\vert 0 \rangle$ to $\vert + \rangle$, which is what I think we want? $\endgroup$
    – Boots
    May 12, 2021 at 16:24
  • $\begingroup$ The rudeness is not really appreciated. Besides that, sure $\vert + \rangle \langle 0 \vert$ is not a projector but it is a unitary transformation. Would this not be the same as moving to another basis? Of course, it's not useful for this problem since Bob doesn't know what the 'target basis' is. But technically, he could apply this transformation to move to the $\{\vert +\rangle,\vert -\rangle\}$ basis and then apply the obvious measurement. $\endgroup$
    – Boots
    May 13, 2021 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.