# Show that if $b'$ and $b$ are uncorrelated, then $a'$ and $a$ are uncorrelated

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 $$b=b'$$.

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 $$a_k$$.