Maximum probability of success for distinguishing between two pure states with one measurement Suppose you have the states such that $\langle \psi_1| \psi_2 \rangle = \cos(\alpha)$ and you have one measurement to distinguish between the two. It is claimed that the probability of success at guessing correctly is $$P = \frac{1+\sin(\alpha)}{2}.$$ How does one arrive at this probability? 
 A: This problem is known as maximum likelihood estimation, and it is best done as follows.
Since there are only two states involved, one can work in a two-dimensional subspace. Choose a basis $\{|0\rangle,|1\rangle\}$ such that
$$
\begin{pmatrix}
\langle 0|\psi_1\rangle \\ \langle 1|\psi_1\rangle
\end{pmatrix}
=
\begin{pmatrix}\cos(\alpha/2)\\\sin(\alpha/2)\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}
\langle 0|\psi_2\rangle \\ \langle 1|\psi_2\rangle
\end{pmatrix}
=
\begin{pmatrix}\cos(\alpha/2)\\ -\sin(\alpha/2)\end{pmatrix},
$$
which is always possible. (On the Bloch sphere, $|0\rangle$ lies directly in between $|\psi_1\rangle$ and $|\psi_2\rangle$, and $|1\rangle$ is diametrically oposite.)
The measurement process is described by two projectors $\Pi_1$ and $\Pi_2$, which must satisfy $\Pi_1\Pi_2=0$ and $\Pi_1+\Pi_2=\mathbb I$; whenever $\Pi_1$ is observed one pronounces for $|\psi_1\rangle$, and vice versa.
Therefore, the probability of success equals
\begin{align}
P&=
\tfrac12\langle\psi_1|\Pi_1|\psi_1\rangle+\tfrac12\langle\psi_1|\Pi_1|\psi_1\rangle
\\&=\tfrac12\operatorname{Tr}\left[\Pi_1|\psi_1\rangle\langle\psi_1|\right]
+\tfrac12\operatorname{Tr}\left[\Pi_2|\psi_2\rangle\langle\psi_2|\right]
\\&=\tfrac12\operatorname{Tr}\left[\Pi_1|\psi_1\rangle\langle\psi_1|\right]
+\tfrac12
-\tfrac12\operatorname{Tr}\left[\Pi_1|\psi_2\rangle\langle\psi_2|\right]
\\&=\tfrac12+\tfrac12\operatorname{Tr}\left[\Pi_1\left(|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|\right)\right].
\end{align}
Here the combination of state projectors can be worked out to give
\begin{align}
|\psi_1\rangle\langle\psi_1|-|\psi_2\rangle\langle\psi_2|
&=
\begin{pmatrix}\cos(\alpha/2) & \sin(\alpha/2)\end{pmatrix}
\begin{pmatrix}\cos(\alpha/2) \\ \sin(\alpha/2)\end{pmatrix}
\\&\quad-
\begin{pmatrix}\cos(\alpha/2) & -\sin(\alpha/2)\end{pmatrix}
\begin{pmatrix}\cos(\alpha/2) \\ -\sin(\alpha/2)\end{pmatrix}
\\&=
\begin{pmatrix}\cos^2(\alpha/2) & \sin(\alpha/2)\cos(\alpha/2)\\
\sin(\alpha/2)\cos(\alpha/2)& \sin^2(\alpha/2)\end{pmatrix}
\\&\quad-
\begin{pmatrix}\cos^2(\alpha/2) & -\sin(\alpha/2)\cos(\alpha/2)\\
-\sin(\alpha/2)\cos(\alpha/2)& \sin^2(\alpha/2)\end{pmatrix}
\\&=
\sin(\alpha)\begin{pmatrix}0&1\\1&0\end{pmatrix}
\\&=\sin(\alpha)\sigma_x.
\end{align}
With this, the probability equals
$$
P=\tfrac12+\tfrac12\sin(\alpha)\operatorname{Tr}\left[\Pi_1\sigma_x\right].
$$
Here the trace $\operatorname{Tr}\left[\Pi_1\sigma_x\right]$ needs to be optimized by an appropriate choice of projector. The optimal choice is the $+1$ eigenprojector of $\sigma_x$, so $\Pi_1=|+\rangle\langle+|$ is the best possible measurement. 

For that projector the trace equals 1, which means that the overall probability is
$$
P=\tfrac12+\tfrac12\sin(\alpha)
$$
as given in the question.
