Chance of distinguishing between many pure states Helstrom has demonstrated that the maximum probability of any process correctly distinguishing between two pure states $|\psi_0\rangle$ and $|\psi_1\rangle$ is determined by their trace distance:
$$d_{\psi_0, \psi_1} = \frac{1}{2} \text{Tr} \Big | |\psi_1\rangle\langle\psi_1| - |\psi_0\rangle\langle\psi_0| \Big | = \sqrt{1-\lvert\langle\psi_0 | \psi_1\rangle\rvert^2}$$
$$p_{\text{distinguish}} = \frac{1}{2} \cdot \left(1 + \sqrt{1-\lvert\langle\psi_0 | \psi_1\rangle\rvert^2} \right)$$
My question is: what if there are $n$ possible pure states, instead of just two? How do you compute the distinguishability in that case?
 A: If I understand this correctly, the setting is the following:
We have a Hilbert space $\mathcal{H}$ and $n$ states $|\psi_i\rangle\in \mathcal{H}$ that we want to distinguish. We want to distinguish "correctly" i.e. we want a protocol with the following properties:


*

*We know that the state we are given is one of the $n$ states specified before with some probability distribution $p_i$. It cannot be any other state.

*The protocol can output either an index of a measured state $i$ or it can output "I don't know".

*If the protocol says we measured $|\psi_i\rangle$, then this is always correct.

*If the protocol says "I don't know", this means we have failed. The average probability of the outcome "I don't know" will be denoted by $p$.


I assume the question you are asking is: What is the smallest possible $p$?
Note The probability distribution of the states plays a role in practice: If I have for example four states, two of which are orthogonal on all the others, whenever I am given one of those two states, the result will never be "I don't know". For the other states, it might be.
This is well researched in the literature, however it seems unsolved in all generality. A good and short overview is given by Keyes. What's the upshot? 
First of all, it is not hard to see that you can only have such a POVM scheme if the $n$ vectors are linearly independent. For example, if you have three qubits this implies that it's just not possible to construct such a scheme at all. The reason is the following:
To each answer corresponds a POVM element $E_i$. Those are positive semidefinite matrices such that $\sum_{i=0}^n E_i=1$ - the answer "0" will correspond to the "I don't know" answer. In order to have unambiguous state discrimination, this implies that $\langle \psi_i|E_j|\psi_i\rangle=0$ for all $j$ except for $i$ and $0$, because otherwise, it could happen that we measure $j$ although the state was $i$. If the $|\psi_i\rangle$ where linearly independent, you'd necessarily have $\langle \psi_i|E_i|\psi_i\rangle=0$, which would mean that the output is always "I don't know" (this is true because $|\psi_i\rangle$ will be in the kernel of $E_j$, which is a vector space).
Thus we have a no-go theorem. What else can be said? 
This and many related questions such as discrimination for mixed states or the best way to distinguish linearly dependent states are also discussed in the fairly long book chapter by Bergou, Herzog and Hillery and the more recent book chapter by Barnett and Croke (open access).
It seems that there are good upper and lower bounds for the general case (see the Barnett and Croke reference). The most recent result I could find is a result by Bandyopadhyay. It gives a general upper bound as a minimisation problem and the upper bound is shown to be tight in many cases. To state the result for completness, it says that an upper bound on the average success probability (i.e. 1-p) can be given by (in my notation above):
$$1-p\leq \min_{\{\theta_i\}} \left\| \sum_j \sqrt{p_j} e^{i\theta_j} |\psi_j\rangle\right\|^2 $$
Obviously, I haven't checked the math.
