# QKD measuring qubit with wrong bases

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?

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 $$(1/\sqrt 2)^2=1/2$$.