Part VII · Foundations Revisited — 19

Schmidt Decomposition

Entanglement isn’t a yes/no property — states can be a little entangled or a lot. The Schmidt decomposition is the single cleanest lens on how much: it rewrites any two-party pure state so that just one or two numbers tell the whole story.

↩ before you start · keep these handy
·From Ch. 18: tracing out Bob gives Alice’s reduced state ρₐ with a pair of eigenvalues.
·From Ch. 13: feed those eigenvalues to S = −Σ p log₂ p for the entropy.
·From Ch. 6: a product state factors into "Alice’s × Bob’s"; a Bell pair can’t.
🔑 symbol decoder · every new mark, in plain words
λᵢa Schmidt coefficient — the weight on one aligned term; all λᵢ² add to 1. |uᵢ⟩, |vᵢ⟩Alice’s and Bob’s private basis states — the special axes that line the pair up. Schmidt rankhow many λᵢ are non-zero — 1 = product (no entanglement), 2 = entangled. SVDsingular-value decomposition — the standard matrix tool that finds those axes and weights. λᵢ²a Schmidt weight squared — it is a reduced-state eigenvalue (Ch. 18).
feel

Every pair has private axes

A general two-qubit state looks like a messy sum of four terms — |00⟩, |01⟩, |10⟩, |11⟩ all tangled together. Schmidt’s theorem promises that for any such state there exists a special pair of bases — one chosen by Alice, one by Bob — in which the mess collapses to a diagonal sum of at most two aligned terms. In those private axes, Alice’s |u0⟩ always pairs with Bob’s |v0⟩, never crosswise. The weights on those terms are the whole secret.

🪞 everyday picture

A tilted ellipse drawn on graph paper looks complicated — its equation mixes x and y in a messy way. But tilt your head to line up with its long and short axes and it becomes trivially simple: “this far one way, that far the other.” The Schmidt decomposition is that head-tilt for an entangled pair: rotate Alice’s and Bob’s viewpoints to the right axes and the four-term mess collapses to just two clean weights.

recapIn the right private bases, any two-qubit pure state is just two aligned terms with weights λ₀, λ₁.
play

Split the Schmidt weights apart

The two weights λ0, λ1 are pinned to a quarter-circle — λ0²+λ1²=1. Push the state toward product and one weight grabs everything (rank 1); push toward Bell and they even out at 1/√2 each (rank 2, maximal).

▸ Schmidt weights|ψ⟩ = λ₀|u₀v₀⟩ + λ₁|u₁v₁⟩
λ₀ λ₁
λ₀ |u₀v₀⟩{{ lam0 }}
λ₁ |u₁v₁⟩{{ lam1 }}
mix
Schmidt rank{{ rank }}
ρₐ eigenvalues{{ ev0 }}, {{ ev1 }}
entanglement S{{ entropy }}
{{ desc }}
recapOne weight grabbing everything = product (rank 1); equal weights 1/√2 = maximal entanglement (rank 2).
math

One sum, two numbers

Pack the four amplitudes into a 2×2 matrix and take its singular-value decomposition. The singular values λi are the Schmidt coefficients; the left and right singular vectors are Alice’s and Bob’s private bases:

|ψ⟩ = Σi λi |ui⟩A |vi⟩B,   λi ≥ 0,   Σ λi² = 1
Schmidt rank 1  ⇔  product state  (not entangled)
rank ≥ 2  ⇔  entangled;  all λi equal ⇔ maximally entangled
the λi² ARE the eigenvalues of ρₐ — so S = −Σ λi² log₂ λi²

This ties the last three chapters into a knot: the Schmidt coefficients squared are exactly the reduced-state eigenvalues from chapter 18, and feeding them to chapter 13’s formula gives the entanglement entropy. Same two numbers, three points of view. Crucially, both sides always share the same Schmidt rank — entanglement is symmetric.

✎ worked example · weights λ₀ = 0.87, λ₁ = 0.50
1.check normalization: λ₀² + λ₁² = 0.75 + 0.25 = 1. ✓ Two non-zero weights → rank 2, entangled.
2.the squares 0.75, 0.25 are exactly ρₐ’s eigenvalues from Ch. 18.
3.entanglement entropy S = −0.75log₂0.75 − 0.25log₂0.25 = 0.81 bits.
4.between 0 (product) and 1 (Bell) — a partly-entangled pair, graded by one number. ✓
recapThe λᵢ² are the reduced-state eigenvalues; feed them to Shannon and you get the entanglement entropy.
⚠ common misconception

“A state is either entangled or it isn’t.” For pure states the Schmidt weights make entanglement a dial, not a switch: a sliver of λ1 is a little entanglement, equal weights are the maximum. The single number S grades exactly how much — and it can’t be increased by Alice and Bob acting locally, which is why it’s the right measure.

One caveat to carry forward: this clean two-number story holds for pure states. For mixed states, quantifying entanglement gets genuinely hard — and that’s where the general formalism of the final part begins. First, one more tool: measurements more flexible than the sharp projective kind.

✓ you can now
✓read a two-qubit pure state’s Schmidt weights and its rank (1 = product, 2 = entangled)
✓connect λᵢ² to the reduced-state eigenvalues and compute the entanglement entropy S
✓explain why entanglement is a dial for pure states — and why mixed states are harder
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