Measurement
Postulate
Quantum Measurements are descrivbed by a collection of measurement operators $\{M_m\}$﻿ with $p(m)= \langle \psi | M_m^\dagger M_m | \psi \rangle,\ \sum{p(m)} = 1$﻿.
$\vert \psi \rangle \xrightarrow{\text{on measurement}} \frac{M_m \vert \psi \rangle}{\sqrt{p(m)}}$
If $M_i = |i\rangle \langle i|,\ \forall i$﻿ then we are measuring in a computational basis.
Non-distinguishability of arbitrary states
We cannot distinguish any two arbitrary non-orthogonal quantum states.
Proof: Let $|\psi_1\rangle$﻿ and $|\psi_2\rangle$﻿ be two non-orthogonal states.
Then $\exists\ E_1, E_2$﻿ such that $E_1 = \sum_{j:f(j) = 1} M_j^\dagger M_j$﻿ and similarly $E_2$﻿ then $\langle \psi_1 | E_1 | \psi_1 \rangle = 1$﻿ and $\langle \psi_2 | E_2 | \psi_2 \rangle = 1$﻿.
Thus, $\sqrt{E_2}| \psi_1\rangle = 0$﻿ and let $\psi_2 = a\psi_1 + b \phi$﻿ where $\psi$﻿ and $\phi$﻿ are orthogonal.
$\langle \psi_2 | E_2 \psi_2 \rangle = |\beta|^2 \langle \psi_2 | \phi \psi_2 \leq |\beta|^2 < 1$
Projective Measurements
We can use projective measurement formalism for any general measurement too.
In case of projective measurements, $M = \sum mP_m$﻿ and $p(m) = \langle \psi | P_m | \psi \rangle$﻿.
$\vert \psi \rangle \rightarrow \frac{P_m \vert \psi\rangle}{\sqrt {p(m)}}$
$E(M) = \sum mp(m) = \langle \psi | M | \psi \rangle = \langle M\rangle$
$\Delta(M) = \sqrt{\langle M^2\rangle - \langle M \rangle^2}$
Here, $E(M)$﻿ is expectation and $\Delta(M)$﻿ is standard deviation or the square root of variance.
POVM Measurements
POVM Measurements are a formalism where only measurement statistics matters.
$\{E_m\} \rightarrow \sum{E_m} = I,\ p(m) = \langle \psi | E_m |\psi \rangle$
Here, each of $E_m$﻿ are hermitian.
Global Phase doesn't matter
We say $e^{i\theta} |\psi\rangle \equiv |\psi\rangle$﻿ but why? Because, $\langle \psi | M_m^{\dagger}M_m | \psi \rangle = \langle \psi \vert e^{-i\theta}M_m^\dagger M_m e^{i\theta}\vert \psi \rangle$﻿.
However, be aware that the global phase is quite different from the relative phase.