QKD measuring qubit with wrong bases

Question

I'm trying to end the research work for my master thesis about BB84 QKD (and QBC) and a basic problem of quantum mechanics is blocking me.

I'm trying to do a probability calculus of the action of measuring a qubit in a wrong bases. In bibliography, I've always found the statement:

When Bob chooses the wrong bases for measuring a qubit then the result will be completely random.

But what exactly does that mean? The results will be Non-deterministic and then the probability cannot be calculated or that mean that the probability of the result is exactly $\frac{1}{2}$0 and $\frac{1}{2}$1?

Answer 1

Assuming that you are talking abut the usual formulation of BB84 and Bob (the receiver) is supposed to choose either the basis $\{|0\rangle,|1\rangle\}$ or the basis $\{|+\rangle,|-\rangle\}$, then the probability is exactly 1/2 of each measurement when the wrong basis is chosen.

To see this, recall that $$|+\rangle=\frac1{\sqrt2}|0\rangle+\frac1{\sqrt2}|1\rangle$$ $$|-\rangle=\frac1{\sqrt2}|0\rangle-\frac1{\sqrt2}|1\rangle$$ so that if for example we measure $|-\rangle$ in the $\{|0\rangle,|1\rangle\}$ basis, we obtain $|0\rangle$ with probability $(1/\sqrt2)^2=1/2$ and $|1\rangle$ with probability $(-1/\sqrt 2)^2=1/2$. Likewise $$|0\rangle=\frac1{\sqrt2}|+\rangle+\frac1{\sqrt2}|-\rangle$$ $$|1\rangle=\frac1{\sqrt2}|+\rangle-\frac1{\sqrt2}|-\rangle$$ so that if for example we measure $|0\rangle$ in the $\{|+\rangle,|-\rangle\}$ basis, we obtain $|+\rangle$ with probability $(1/\sqrt2)^2=1/2$ and $|-\rangle$ with probability $(1/\sqrt 2)^2=1/2$.

In all cases with transmission and measurement bases mismatching, 1/2 the time we measure a state corresponding to 0 and 1/2 the time a state corresponding to 1. This probability is independent of the state transmitted.