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 $b=b'$.

Narek Mamikonian

Posted 2020-02-10T01:46:07.230

Reputation: 11



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 $a_k$.


Posted 2020-02-10T01:46:07.230

Reputation: 242