Finding the measurement basis for single qubit with given probability of outcome $0$

Question

I have the general state of a single qubit $|\psi \rangle = \alpha|0\rangle + \beta|1\rangle $. Assume I am given a probability $p$ such that $0 < p <1$. Now I need to find the basis in which the measurement of the qubit $|\psi \rangle $ gives outcome $0$ with probability $p$.

I understand that the required basis $B$ will have $2$ states, so assume the basis is $B =\{|a\rangle, |b\rangle \}$. Also that, probability of outcome $0$, when measured in basis $B$, is $|\langle a |\psi\rangle|^2 $ which should equal $p$. At the same time, probability of outcome $1$ in basis $B$ is $|\langle b |\psi\rangle|^2 $ and should equal $1-p$.

I do not understand how to solve the above two inner-product equations to obtain states $|a\rangle$ and $|b\rangle$?

Answer 1

You have state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle,$ where $|\alpha|^2+|\beta|^2=1$. You want to convert to the basis $B=[|a\rangle,|b\rangle]$.

One way to change the basis is by following:

Let $|\psi\rangle=\sqrt{p}|a\rangle+\sqrt{1-p}|b\rangle$. Where, $|a\rangle=m|0\rangle+n|1\rangle$ and $|b\rangle=-m^*|1\rangle+n^*|0\rangle$, because new basis should be orthonormal.

After putting $|a\rangle$ and $|b\rangle$ into $|\psi\rangle$, you get the following: $|\psi\rangle=(m\sqrt{p}+n^*\sqrt{1-p})|0\rangle+(n\sqrt{p}-m^*\sqrt{1-p})|1\rangle$=$\alpha|0\rangle+\beta|1\rangle$.

Comparing the coefficients and after solving for $m$ and $n$,

$m=\alpha \sqrt{p}^*-\beta^*\sqrt{1-p}$

$n=\beta \sqrt{p}^*-\alpha^*\sqrt{1-p}$.