Entanglement and partial trace

Physical question

Where is quantum information stored?

A Bell pair is globally pure, but each qubit has density operator I/2. Its relative phase is invisible to local density operators and detectable by suitable joint measurements. Which predictions can be reconstructed from the Schmidt coefficients alone?

Predict: at θ=45°, φ=0, the joint state is |Φ+〉. What are the one-qubit measurement outcomes along Z, and how do their probabilities compare with simultaneous Z/Z measurements?

Computed state and observable

Live measurement summary

Schmidt decomposition

What remains after tracing out B?

The reduced density operator has eigenvalues λ₀=cos²θ and λ₁=sin²θ. The two bars always sum to one; their balance determines the bipartite entanglement entropy for the overall pure state.

Eigenvalues of ρA=TrB|ψ〉〈ψ|. An equal pair denotes maximal entanglement; one eigenvalue equal to unity indicates a product state.

Joint measurement

Correlated outcomes in chosen bases

Probabilities P(A=s,B=t) with s,t=±1. For both equatorial axes, the local marginals are 1/2 and P(s,t)=[1+stE]/4. In the Z/Z basis the outcomes are cos²θ for ++ and sin²θ for --.

Phase information

Local states versus two-body coherence

E=〈A⊗B〉. The curve is constant E=1 in the Z/Z basis and oscillatory in the equatorial bases. Phase variation leaves ρA unchanged, while the nonlocal correlations retain the phase.

Work through the partial trace

Pure globally, mixed locally

We prepare the normalized state |ψ(θ,φ)〉=cosθ|00〉+e^iφsinθ|11〉, in the explicit basis {|00〉,|01〉,|10〉,|11〉}. The global density operator is ρAB=|ψ〉〈ψ|, so Tr(ρAB²)=1.

ρA = TrB(ρAB) = cos²θ |0〉〈0| + sin²θ |1〉〈1|.

Every off-diagonal term of ρAB contains |0〉〈1| or |1〉〈0| on system B, whose partial trace vanishes. Thus the phase φ disappears from each individual subsystem. The information stored in that phase remains measurable through suitable joint observables.

S(ρA)=−λ₀log₂λ₀−λ₁log₂λ₁; C=|sin 2θ|.

Derive the equatorial correlation

A local equatorial Pauli observable is σ(α)=cosασx+sinασy. Acting on |0〉 gives e^iα|1〉. Therefore 〈σ(α)⊗σ(β)〉=sin(2θ)cos(φ−α−β). Each local equatorial expectation vanishes, so the four joint probabilities are (1+stE)/4 for outcomes s,t∈{±1}. This is testable directly using the four-bar chart.

Why does entropy quantify entanglement here?

For a pure bipartite state, the nonzero spectra of ρA and ρB coincide: they are the squared Schmidt coefficients. If the global state factorizes, only one coefficient remains; if two equally weighted coefficients appear, each qubit has entropy exactly one bit. Entropy of a reduced state alone is not a general entanglement measure when the global state is mixed.

Guided investigations

Recover the physics with three controlled changes

Entanglement versus classical uncertainty. Sweep θ from 0° to 45°. Record the reduced purity and entropy. Explain how the global purity remains exactly one throughout.
Hidden relative phase. Fix θ=45° and measure first Z/Z, then X/X. Change φ by 180°. Which distribution can detect the phase? Why does ρA remain unchanged?
Move the analyzer axes. Set θ=30° and φ=90°. Verify that choosing α+β=90° maximizes E and α+β=−90° minimizes it.

Check: At θ=45°, what is Tr(ρA²)?

1 1/2 0

Meaning, scope and next steps

The model explores a pure two-qubit Schmidt family, complete for demonstrating pure-state two-qubit entanglement up to local unitary changes of basis. The displayed phase-sensitive measurements assume ideal local Pauli projectors; imperfections, general mixed bipartite states and Bell inequality tests are separate laboratories. Pure-state concurrence is |sin2θ| here.

Read the source manuscript: composite Hilbert spaces, partial trace and generalized measurement. Prerequisite: Chapter 1 projective measurement; extension: Chapter 3 Bell correlations.