# Convex combinations of states yielding a pure state

Geometrically, when the state is not expressible as a convex combination of other states, it is a pure state.[9] The family of mixed states is a convex set and a state is pure if it is an extremal point of that set.

At the same time, it defines convex combination as

In convex geometry, a convex combination is a linear combination of points (which can be vectors, scalars, or more generally points in an affine space) where all coefficients are non-negative and sum to 1.

I might be very picky, or mistaken, but I'm led to believe that if $|\psi_i\rangle$ are pure states spanning whichever Hilbert space we are dealing with, then (say) $|\psi_1\rangle$ is a pure state and also a convex combination $$|\psi_1\rangle\langle\psi_1|=\sum_ia_i|\psi_i\rangle\langle\psi_i|,$$ with $a_1=1$ and $a_i=0$ if $i\neq1$.

Am I misinterpreting something? Is Wikipedia unclear/wrong?

• Okay, I'll try to make Wikipedia clearer. If you repost your comment as an answer, I'll accept it. – Daniel Jul 23 '15 at 16:09
• It says "not expressible as a convex combination of other states", so you are misinterpreting something. – Norbert Schuch Jul 23 '15 at 16:42
• Yup, you're right. My bad. – Daniel Jul 23 '15 at 18:31

## 1 Answer

I'd rather say you are a bit nitpicking. You are correct, but I think you would agree that your case is not a combination of anything in the proper sense. (Also, you could broaden your statement and consider "combinations" with ANY states) (promoted from comment to answer)