Show when $a_k$ and $b_k$ are correlated when measuring in different bases, in the BB84 protocol

Question

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

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.

Answer 1

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 $a$.

Answer 2

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.

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PS:

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 $\hat{P}=|+\rangle\langle+|$ $$ \langle\psi|\hat{P}|\psi\rangle=\langle\psi|+\rangle\langle+|\psi\rangle=|\langle+|\psi\rangle|^2 $$

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