Exa 4
$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ A.4}$
Show that the probability distributions $\mu$ whose corresponding operators $M_\mu$ are projectors are exactly the pure distributions.
page revision: 0, last edited: 18 Nov 2012 20:35
For any probability distribution $\mu \colon A \to \mathbb{R}$ the corresponding operator $M_{\mu}\colon \mathbb{R}^A \to \mathbb{R}^A$ is given by (recall that $\mathbb{R}^A = \{f\colon A \to \mathbb{R}\}$):
(1)If $\mu$ is pure, it is concentrated at one element $a\in A$ and is given by the Kronecker delta function $\mu = \delta_a$. In particular,
(2)so $M_{\mu}^2 = M_{\mu}$ and $M_{\mu}$ is a projector.
(3)If on the other hand, the corresponding operator $M_{\mu}$ is a projector, then $M_{\mu}^2 = M_{\mu}$ implies $\mu^2(a) = \mu(a)$ for all $a \in A$. This means, $\mu(a) = 0$ or $\mu(a) = 1$ for any $a \in A$. But since
there can only be one element $a$ with $\mu(a)=1$. That is, $\mu$ is pure.
This shows that the probability distributions $\mu$ whose corresponding operators $M_{\mu}$ are projectors are exactly the pure distributions.