Measurement

Postulate

Quantum Measurements are descrivbed by a collection of  measurement operators {Mm}\{M_m\}{Mm​}﻿ with p(m)=⟨ψ∣Mm†Mm∣ψ⟩, ∑p(m)=1p(m)= \langle \psi | M_m^\dagger M_m | \psi \rangle,\ \sum{p(m)} = 1p(m)=⟨ψ∣Mm†​Mm​∣ψ⟩, ∑p(m)=1﻿.

\vert \psi \rangle \xrightarrow{\text{on measurement}} \frac{M_m \vert \psi \rangle}{\sqrt{p(m)}}

If Mi=∣i⟩⟨i∣, ∀iM_i = |i\rangle \langle i|,\ \forall i Mi​=∣i⟩⟨i∣, ∀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 ∣ψ1⟩|\psi_1\rangle∣ψ1​⟩﻿ and ∣ψ2⟩|\psi_2\rangle∣ψ2​⟩﻿ be two non-orthogonal states. 

Then ∃ E1,E2\exists\ E_1, E_2∃ E1​,E2​﻿ such that E1=∑j:f(j)=1Mj†MjE_1 = \sum_{j:f(j) = 1} M_j^\dagger M_jE1​=∑j:f(j)=1​Mj†​Mj​﻿ and similarly E2E_2E2​﻿ then ⟨ψ1∣E1∣ψ1⟩=1\langle \psi_1 | E_1 | \psi_1 \rangle = 1⟨ψ1​∣E1​∣ψ1​⟩=1﻿ and ⟨ψ2∣E2∣ψ2⟩=1\langle \psi_2 | E_2 | \psi_2 \rangle = 1⟨ψ2​∣E2​∣ψ2​⟩=1﻿.

Thus, E2∣ψ1⟩=0\sqrt{E_2}| \psi_1\rangle = 0E2​​∣ψ1​⟩=0﻿ and let ψ2=aψ1+bϕ\psi_2 = a\psi_1 + b \phiψ2​=aψ1​+bϕ﻿ where ψ\psiψ﻿ and ϕ\phiϕ﻿ are orthogonal.

\langle \psi_2 | E_2 \psi_2 \rangle = |\beta|^2 \langle \psi_2 | \phi \psi_2 \leq |\beta|^2 < 1

Hence, proved by contradiction.

Projective Measurements

We can use projective measurement formalism for any general measurement too.

In case of projective measurements, M=∑mPmM = \sum mP_mM=∑mPm​﻿ and p(m)=⟨ψ∣Pm∣ψ⟩p(m) = \langle \psi | P_m | \psi \ranglep(m)=⟨ψ∣Pm​∣ψ⟩﻿.

\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)E(M)E(M)﻿ is expectation and Δ(M)\Delta(M)Δ(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 EmE_mEm​﻿ are hermitian.

Global Phase doesn't matter

We say eiθ∣ψ⟩≡∣ψ⟩e^{i\theta} |\psi\rangle \equiv |\psi\rangleeiθ∣ψ⟩≡∣ψ⟩﻿ but why? Because, ⟨ψ∣Mm†Mm∣ψ⟩=⟨ψ∣e−iθMm†Mmeiθ∣ψ⟩\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⟨ψ∣Mm†​Mm​∣ψ⟩=⟨ψ∣e−iθMm†​Mm​eiθ∣ψ⟩﻿.

However, be aware that the global phase is quite different from the relative phase.