Measuring a Quantum State

Before: The system is in a state ∣ψ⟩\left\lvert \psi \right\rangle and you measure the observable A^\hat{A}.

Measure: We get an outcome AnA_n with probability Pn=⟨ψ∣Π^n∣ψ⟩P_n=\left\langle \psi \right\rvert\hat{\Pi}_n\left\lvert \psi \right\rangle

Given Born Rule for a non-degenerate eigenstate ∣ϕn⟩\left\lvert \phi_n \right\rangle,

Pn=∣⟨ϕn∣ψ⟩∣2P_n=|\left\langle \phi_n|\psi \right\rangle|^2

This is because Born Rule. Note that Π^n\hat{\Pi}_n, the projector onto the λn\lambda_n-eigenspace of A^\hat{A}, is

Π^n=∣ϕn⟩⟨ϕn∣\hat{\Pi}_n=\left\lvert \phi_n \right\rangle\left\langle \phi_n \right\rvert ⇒Π^n∣ψ⟩=∣ϕn⟩⟨ϕn∣ψ⟩\Rightarrow\hat{\Pi}_n\left\lvert \psi \right\rangle=\left\lvert \phi_n \right\rangle\left\langle \phi_n|\psi \right\rangle ⇒=⟨ϕn∣ψ⟩∣ϕn⟩\Rightarrow=\left\langle \phi_n|\psi \right\rangle\left\lvert \phi_n \right\rangle

So

⇒∣∣Π^n∣ψ⟩∣∣2\Rightarrow||\hat{\Pi}_n\left\lvert \psi \right\rangle||^2 ⇒=∣∣⟨ϕn∣ψ⟩∣ϕn⟩∣∣2\Rightarrow= ||\left\langle \phi_n|\psi \right\rangle\left\lvert \phi_n \right\rangle||^2 ⇒=∣⟨ϕn∣ψ⟩∣2∣∣∣ϕn⟩∣∣2\Rightarrow= |\left\langle \phi_n|\psi \right\rangle|^2||\left\lvert \phi_n \right\rangle||^2

We know that the Magnitude of the eigenstate ∣ϕn⟩\left\lvert \phi_n \right\rangle is 1.

After: State collapses to

∣ψ′⟩=1PnΠ^n∣ψ⟩\left\lvert \psi' \right\rangle=\frac{1}{\sqrt{P_n}}\hat{\Pi}_n\left\lvert \psi \right\rangle

.

The collapse is true only if this is an idealized measurement that is non-destructive. This means that the measurement must not destroy the system (e.g., photon absorbed by detector). There’s no noise as well.