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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?

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up vote 4 down vote accepted

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.

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

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Awesome! thanks! That is exactly what I was looking for. – Catherine Holloway Apr 15 '12 at 16:09
I cannot see the link can you update the link. thanks – user57409 Aug 16 '14 at 19:36
@MariGachechiladze see edited answer. Please mark this comment as obsolete (by clicking the flag icon to the left) when you receive it. Enjoy. – Emilio Pisanty Aug 16 '14 at 23:14

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