How to find the projection optimally distinguishing the quantum states $|0\rangle$ and $|+\rangle$? I have a qubit. The probabality $p$ to find it in either of the two possible states $\{|0\rangle,|+\rangle\}$ is $p = 1/2$. I now want to measure it to maximize the probability that the measurement is correct.
From this presentation (page 5), I know that I can determine the probability as $$\frac{1}{2}+\frac{1}{4}\left\||0\rangle-|+\rangle\right\|_{\rm tr}.$$
But how do I find the associated projection?
 A: I made a silly mistake in the comments. The measurement you are looking for consists of the two projectors unto the positive and negative subspaces of $|0\rangle\langle 0 | - |+\rangle \langle + |$. In particular, the matrix $|0\rangle\langle 0 | - |+\rangle \langle + |$ can be diagonalised as $\sum_i \lambda_i |\psi_i\rangle \langle \psi_i |$, where the $|\psi_i\rangle 
$ form an orthonormal basis and the $\lambda_i$ need not be positive. Let's assume that all the $\lambda_i$ are non-zero, you can check later that this will not really matter.
Now write $$|0\rangle\langle 0 | - |+\rangle \langle + | = \sum_i \lambda_i |\psi_i\rangle \langle \psi_i |\\
= \sum_{\lambda_i>0} \lambda_i |\psi_i\rangle \langle \psi_i | + \sum_{\lambda_i<0} \lambda_i |\psi_i\rangle \langle \psi_i |$$
where I have split up the sum into those where the $\lambda_i$ are positive and negative, respectively.
The projectors unto the so-called positive and negative subspaces are then $\Pi_+ = \sum_{\lambda_i>0} |\psi_i\rangle \langle \psi_i|$ and $\Pi_- = \sum_{\lambda_i<0} |\psi_i\rangle \langle \psi_i|$, and give you the projectors you seek for your measurement.
The above can be generalised to the case where the two states $\rho_1$ and $\rho_2$ are given with unequal probabilities $p_1 \neq p_2$. Then one can repeat the above game, but calculating the positive and negative subspaces of $p_1\rho_1 -p_2\rho_2$ instead.
