Measuring a Composite System

For a non-composite system, recall that the Set of measurements you can make is the eigenvalues of the observable.

Recall Measuring a Quantum State.

Entanglement doesn’t always require a interacting Hamiltonian. They can also be created by jointly measuring a product state.

Suppose we start with ∣ψ⟩=∣0⟩⊗∣0⟩\left\lvert \psi \right\rangle = \left\lvert 0 \right\rangle \otimes \left\lvert 0 \right\rangle but measured in the xx axis

(a11a12a21a22)=12(1111)⟶product state\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \longrightarrow \text{product state}

Measure S^x(tot)\hat{S}_x^{(\text{tot})}, get 0 with 12\frac{1}{2} probability. Mathematically, we use a projector to measure the state. Via the projection rule we get

11/2ΠSx=0∣ψ⟩=12(∣+,−⟩+∣−,+⟩)\frac{1}{\sqrt{1/2}} \Pi_{S_x} = 0 \quad \left\lvert \psi \right\rangle = \frac{1}{\sqrt{2}}\left(\left\lvert +,- \right\rangle + \left\lvert -,+ \right\rangle\right) 12(0110)⟶entangled state!\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \longrightarrow \text{entangled state!}

This means that joint measurement can product entanglement. Equivalent to interacting via third party.

Consider a composite system HAlice⊗HBobH_{\text{Alice}} \otimes H_{\text{Bob}} with subsystem observables having eigenvalues qaq_a, rbr_b and eigenvectors ∣a⟩\left\lvert a \right\rangle, ∣b⟩\left\lvert b \right\rangle.

Joint probability for outcome (qa,rb)(q_a, r_b):

P(a,b)=∣⟨a,b∣Ψ(AB)⟩∣2P(a, b) = \bigl|\left\langle a, b | \Psi^{(AB)} \right\rangle\bigr|^2

Marginal probabilities:

P(a)=∑bp(a,b),P(b)=∑ap(a,b)P(a) = \sum_b p(a, b), \qquad P(b) = \sum_a p(a, b)

No-signalling from AA to BB. Expand the joint state in the product basis:

∣Ψ(AB)⟩=∑a,bcab∣a⟩⊗∣b⟩,p(a,b)=∣cab∣2\left\lvert \Psi^{(AB)} \right\rangle = \sum_{a,b} c_{ab} \left\lvert a \right\rangle \otimes \left\lvert b \right\rangle, \qquad p(a, b) = |c_{ab}|^2

Group the sum by bb to isolate Bob’s basis:

∣Ψ(AB)⟩=∑b(∑acab∣a⟩)⏟∣ϕb⟩ (unnormalized!)⊗∣b⟩=∑b∣ϕb⟩⊗∣b⟩\left\lvert \Psi^{(AB)} \right\rangle = \sum_b \underbrace{\Bigl(\sum_a c_{ab} \left\lvert a \right\rangle\Bigr)}_{\left\lvert \phi_b \right\rangle \ \text{(unnormalized!)}} \otimes \left\lvert b \right\rangle = \sum_b \left\lvert \phi_b \right\rangle \otimes \left\lvert b \right\rangle

With this notation, P(a,b)=∣⟨a∣ϕb⟩∣2P(a, b) = |\left\langle a | \phi_b \right\rangle|^2, so Bob’s marginal is

P(b)=∑aP(a,b)=∑a⟨ϕb∣a⟩⟨a∣ϕb⟩=⟨ϕb∣ϕb⟩P(b) = \sum_a P(a, b) = \sum_a \left\langle \phi_b | a \right\rangle \left\langle a | \phi_b \right\rangle = \left\langle \phi_b | \phi_b \right\rangle

The middle step used ∑a∣a⟩⟨a∣=I\sum_a \left\lvert a \right\rangle\left\langle a \right\rvert = I — which holds for any orthonormal basis {∣a⟩}\{\left\lvert a \right\rangle\} on Alice’s side. Therefore p(b)p(b) is independent of the choice of Alice’s measurement basis. Alice cannot influence Bob’s marginal statistics by switching what she measures.

Suppose we apply U(A)U^{(A)} to Alice’s subsystem

∣Ψ(AB)⟩=U(A)⊗I∣Ψ(AB)⟩=∑bU∣ϕb⟩⊗∣b⟩\left\lvert \Psi^{(AB)} \right\rangle = U^{(A)} \otimes I \left\lvert \Psi^{(AB)} \right\rangle=\sum_b U\left\lvert \phi_b \right\rangle \otimes \left\lvert b \right\rangle

This means that the marginal probability distributiuon of Bob’s outcome becomes

P′(b)=⟨ϕb∣U+U∣ϕb⟩=⟨ϕb∣ϕb⟩=P(b)P'(b)=\left\langle \phi_b \right\rvert U^+U \left\lvert \phi_b \right\rangle=\left\langle \phi_b|\phi_b \right\rangle=P(b)

This means Bob’s probabilities are unaffected.