Part VII · Foundations Revisited — 18

Partial Trace

Alice and Bob share an entangled pair. The joint state is perfectly known — a pure state. But if Alice only has her half, what does her qubit look like alone? The answer is the operation that quietly underlies decoherence, measurement, and noise.

↩ before you start · keep these handy
·From Ch. 6: a Bell pair (|00⟩+|11⟩)/√2 can’t be split into "Alice’s qubit times Bob’s."
·From Mixed States: a mixed state lives inside the Bloch ball (arrow length r < 1); pure is on the surface.
·From Ch. 13: S(ρ) = bits of ignorance, 0 for pure, 1 for the maximally-mixed center.
🔑 symbol decoder · every new mark, in plain words
ρₐthe reduced state — the best description of Alice’s qubit once Bob’s is set aside. Tr₋the partial trace over Bob — "sum over all of Bob’s possibilities." ⟨k|₋one of Bob’s basis states; the sum runs over all of them. coherencesthe off-diagonal cross-terms — the "quantum" part that survives only without entanglement. ½Ithe maximally-mixed state — a perfect 50/50 coin, arrow at dead center.
feel

Ignore Bob — what is left?

Bob takes his qubit to the far side of the galaxy. Alice can never touch it again, so for any experiment she does, Bob’s qubit may as well not exist. Tracing it out is the bookkeeping for that: it sums over all of Bob’s possibilities, leaving the best possible description of Alice’s qubit on its own. The startling part: even though the pair is a sharp, pure state, Alice’s half alone can be a mixed state — genuine ignorance, conjured purely from entanglement.

💃 everyday picture

Watch a perfectly rehearsed dance duet from the balcony — every move is graceful and predictable, the whole thing fully “known.” Now watch only one dancer through a narrow keyhole, with the partner hidden. That dancer’s steps look erratic, unpredictable — because all the meaning lived in the coordination you can no longer see. Tracing out Bob is exactly that keyhole: Alice’s qubit looks random not because the pair is, but because its information hides in the link between them.

recapTracing out a partner gives the best description of your half — and a pure pair can leave each half mixed.
play

Dial up entanglement, lose Alice’s purity

Tune the shared state cosθ|00⟩ + sinθ|11⟩ from a plain product at θ=0 to a maximally entangled Bell pair at θ=45°. Trace out Bob and watch Alice’s Bloch arrow sink from the surface to the dead center.

▸ reduced state of Aliceρₐ = Tr₋(|ψ⟩⟨ψ|)
|0⟩ |1⟩ r = {{ rLen }}
shared pure state
{{ cosV }}|00⟩ + {{ sinV }}|11⟩
θ {{ thDeg }}°
eigenvalues of ρₐ{{ ev0 }}, {{ ev1 }}
purity Tr(ρₐ²){{ purity }}
entanglement S(ρₐ){{ entropy }}
{{ desc }}
recapMore entanglement sinks Alice’s arrow deeper; at a full Bell pair her half is a dead-center coin.
math

Summing over the partner

The partial trace takes the joint density matrix and sums over Bob’s basis, sandwiching out his degrees of freedom and leaving an operator on Alice’s space alone:

ρₐ = Tr₋(ρₐ₋) = Σk ⟨k|B ρₐ₋ |k⟩B
for cosθ|00⟩+sinθ|11⟩:  ρₐ = cos²θ|0⟩⟨0| + sin²θ|1⟩⟨1|
Bell (θ=45°):  ρₐ = ½|0⟩⟨0| + ½|1⟩⟨1| = ½I  // maximally mixed

The cross-terms (the coherences) survive only when the two halves are not entangled. The instant entanglement appears, summing over Bob erases them, and Alice is left holding a classical-looking mixture. Her loss of purity is exactly the entanglement entropy of the pair — the same von Neumann S from chapter 13.

✎ worked example · a partly-entangled state, θ = 30°
1.state is cos30°|00⟩ + sin30°|11⟩. Trace out Bob → ρₐ = cos²30°|0⟩⟨0| + sin²30°|1⟩⟨1|.
2.eigenvalues = cos²30° = 0.75 and sin²30° = 0.25 (they add to 1). ✓
3.so Alice’s arrow has length r = |0.75 − 0.25| = 0.5 — sunk halfway inside the ball.
4.entanglement entropy S = −0.75log₂0.75 − 0.25log₂0.25 = 0.81 bits — partly entangled. ✓
recapPartial trace sums over the partner; entanglement erases the coherences, and S(ρₐ) measures how much.
⚠ common misconception

“If the whole is perfectly known, each part must be too.” The opposite can hold. A Bell pair is a pure state — maximal knowledge — yet each qubit alone is maximally mixed, a 50/50 coin. The information isn’t in either qubit; it lives in the correlations between them. That is the deep strangeness of entanglement, now made quantitative.

And this is precisely decoherence (chapter 09): a qubit entangles with an environment you can’t see, you trace that environment out, and a pure state turns mixed. Same operation, different name. Next: the cleanest way to see how much two qubits are entangled — the Schmidt decomposition.

✓ you can now
take the partial trace of a two-qubit state to get one qubit’s reduced description ρₐ
explain how a pure pair can leave each half maximally mixed — info hiding in the correlations
see decoherence as a partial trace over an unseen environment
← 17 The Bomb Tester next · 19 Schmidt Decomposition