Let $a'_k$ be Bob's measurement result of qubit $ {\Psi_{a_kb_k}}$, assuming a noiseless channel with no eavesdropping. Show that when $b'_k\neq b_k$, $a'_k$ is random and completely uncorrelated with $a_k$. But when $b'_k=b_k$, $a'_k=a_k$.

I have no idea how to approach this problem, but I was thinking that, for instance, one possibility is that Alice sends bob a Qubit prepared in $ab=00$, and Bob measures it in state $a'b'=01$. Then $a=a'$ but $b=b'$.


1 Answer 1


So first, let's define a bit your notations. I guess (correct me if I'm wrong) that you consider Bob honest, and that what you denote by $\Psi_{a_k,b_k}$ is the BB84 qubit in basis $\{0,1\}$ if $b_k = 0$, and in basis $\{+,-\}$ if $b_k = 1$, whose "value" bit is $a_k$, i.e.:

$$\Psi_{a_k,b_k} = H^{b_k}X^{a_k}|0\rangle$$

Then, Bob will measure in basis $b'_k$ (with the same notation as above), and get the result $a'_k$. And you want to prove that:

  1. if $b'_k = b_k$, then $a'_k = a_k$
  2. if $b'_k \neq b_k$, then $a'_k$ is not correlated with $a_k$, i.e. $\forall a_k, Pr[a'_k = 0 | a_k] = Pr[a'_k = 1 | a_k] = \frac{1}{2}$

First direction

Performing a measurement in the $\{+,-\}$ basis consists of a Hadamard gate, and a measurement in $\{0,1\}$ basis (that we will denote by $M_Z$). Basically measuring in the basis $b'_k$ is like performing the circuit $M_Z H^{b'_k}$. So you just need to apply this on your input qubit: $$a'_k = M_Z H^{b'_k} H^{b_k}X^{a_k}|0\rangle$$ but $b_k = b'_k$ so $$a'_k = M_Z H^{b_k} H^{b_k}X^{a_k}|0\rangle = M_Z (HH)^{b_k}X^{a_k}|0\rangle$$ but $HH$ is identity, so $$a'_k = M_Z X^{a_k}|0\rangle$$ And then it's easy to see that $a'_k = a_k$ (if you are not yet convinced, just try to compute this value for the two possible values of $a_k$)

Second direction

Let's start again from equation

$$a'_k = M_Z H^{b'_k} H^{b_k}X^{a_k}|0\rangle$$

derived above. Then, if $b'_k \neq b_k$, you see that $H^{b'_k} H^{b_k} = H$ (if you are not convinced, then just try to write it for the two values of $b_k$). so the equation becomes

$$a'_k = M_Z H X^{a_k}|0\rangle$$

So then, we will have two cases: if $a_k=0$, then $a'_k = M_Z |+\rangle$ and if $a_k=1$, then $a'_k = M_Z |-\rangle$. But measuring a $|+\rangle$ (or a $|-\rangle$) in the computational basis always gives you a uniform random bit. Indeed, to get the probability of obtaining a $0$ as outcome when measuring a $|+\rangle$, you need to compute $$|\langle 0 | | + \rangle|^2 = |\langle 0 |(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle))|^2$$ So $$|\langle 0 | | + \rangle|^2 = |(\frac{1}{\sqrt{2}}(\langle 0 |0\rangle + \langle 0 |1\rangle))|^2$$ i.e. $$|\langle 0 | | + \rangle|^2 = |(\frac{1}{\sqrt{2}}(1 + 0))|^2$$ i.e.

$$|\langle 0 | | + \rangle|^2 = \frac{1}{2}$$

So $Pr[a'_k = 0 | a_k=0] = \frac{1}{2}$. From that, you have directly $Pr[a'_k = 1 | a_k=0] = 1-\frac{1}{2} = \frac{1}{2}$. And when $a_k=1$, it's the exact same computation, but with a minus sign in front of 0... so it does not really matter.

So when $b_k \neq b'_k$, $a'_k$ is not correlated with $a_k$.


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.