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Notion
Measurement
Postulate
Quantum Measurements are descrivbed by a collection of measurement operators {Mm}\{M_m\} with p(m)=⟨ψ∣Mm†Mm∣ψ⟩, ∑p(m)=1p(m)= \langle \psi | M_m^\dagger M_m | \psi \rangle,\ \sum{p(m)} = 1.
∣ψ⟩undefinedon measurementMm∣ψ⟩p(m)\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  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 and ∣ψ2⟩|\psi_2\rangle be two non-orthogonal states.
Then ∃ E1,E2\exists\ E_1, E_2 such that E1=∑j:f(j)=1Mj†MjE_1 = \sum_{j:f(j) = 1} M_j^\dagger M_j and similarly E2E_2 then ⟨ψ1∣E1∣ψ1⟩=1\langle \psi_1 | E_1 | \psi_1 \rangle = 1 and ⟨ψ2∣E2∣ψ2⟩=1\langle \psi_2 | E_2 | \psi_2 \rangle = 1.
Thus, E2∣ψ1⟩=0\sqrt{E_2}| \psi_1\rangle = 0 and let ψ2=aψ1+bϕ\psi_2 = a\psi_1 + b \phi where ψ\psi and ϕ\phi are orthogonal.
⟨ψ2∣E2ψ2⟩=∣β∣2⟨ψ2∣ϕψ2≤∣β∣2<1\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_m and p(m)=⟨ψ∣Pm∣ψ⟩p(m) = \langle \psi | P_m | \psi \rangle.
∣ψ⟩→Pm∣ψ⟩p(m)\vert \psi \rangle \rightarrow \frac{P_m \vert \psi\rangle}{\sqrt {p(m)}}
E(M)=∑mp(m)=⟨ψ∣M∣ψ⟩=⟨M⟩E(M) = \sum mp(m) = \langle \psi | M | \psi \rangle = \langle M\rangle
Δ(M)=⟨M2⟩−⟨M⟩2\Delta(M) = \sqrt{\langle M^2\rangle - \langle M \rangle^2}
Here, E(M)E(M) is expectation and Δ(M)\Delta(M) is standard deviation or the square root of variance.
POVM Measurements
POVM Measurements are a formalism where only measurement statistics matters.
{Em}→∑Em=I, p(m)=⟨ψ∣Em∣ψ⟩\{E_m\} \rightarrow \sum{E_m} = I,\ p(m) = \langle \psi | E_m |\psi \rangle
Here, each of EmE_m are hermitian.
Global Phase doesn't matter
We say eiθ∣ψ⟩≡∣ψ⟩e^{i\theta} |\psi\rangle \equiv |\psi\rangle 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.
However, be aware that the global phase is quite different from the relative phase.