University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.2

Product and entangled states: the Schmidt decomposition

You have written down a state of two spins. Are they entangled? Counting kets will not tell you, and neither will the correlations in one basis. This lesson turns the question into a calculation you can finish: amplitudes into a matrix, singular values out, and an answer no change of axes can alter.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A pure state of two systems is fixed by a table of amplitudes cᵢⱼ = (⟨i|⟨j|)|ψ⟩, and that table is a matrix C with d_A rows and d_B columns. Row i is the state B is left in, unnormalised, when A is found in |i⟩. A product |a⟩|b⟩ has cᵢⱼ = aᵢbⱼ, so every row is a multiple of one vector and C has rank 1; entanglement is exactly the failure of that.

The singular-value decomposition of C, quoted from linear algebra, finishes the job: it rewrites any |ψ⟩ as a single sum Σₖ √pₖ |uₖ⟩|vₖ⟩ that pairs each member of an orthonormal set on A with exactly one member of an orthonormal set on B, with weights pₖ summing to 1 and at most min(d_A, d_B) terms. Local unitaries turn the two bases but cannot touch the weights, so how many are nonzero, and how large they are, belong to the state and not to the kets you chose; for two spin-halves |det C| alone fixes both. The construction has edges.

It needs a pure state, since a mixture has no single C; exactly two parts, since three spins in general have no single-index form; and distinct weights for its bases to be unique, which is why the singlet, with weights ½ and ½, has a Schmidt form on every common axis.

Simple definition: The Schmidt decomposition writes a pure state of two systems as |ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩, pairing orthonormal kets of A with orthonormal kets of B, with weights pₖ ≥ 0 that sum to 1; the state is a product exactly when a single weight equals 1.
Example: 0.6|↑⟩|↑⟩ + 0.8|↓⟩|↓⟩ is already in Schmidt form, with weights 0.36 and 0.64: entangled, but not maximally, since |det C| = 0.6 × 0.8 = 0.48 falls short of the ½ that equal weights would need.

The coefficient matrix of a two-part state|ψ⟩ = Σᵢⱼ cᵢⱼ |i⟩|j⟩, with Tr(CC†) = Σᵢⱼ |cᵢⱼ|² = 1

Row i of C is the unnormalised state of B when A is found in |i⟩. A product |a⟩|b⟩ has cᵢⱼ = aᵢbⱼ: every row parallel, rank C = 1.

i runs over d_A kets of A, j over d_B kets of B; C is a d_A × d_B matrix of dimensionless amplitudes.

Schmidt decomposition, from the SVD (quoted)|ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩, k = 1 … r, with r = rank C ≤ min(d_A, d_B)

One index where there were two: each |uₖ⟩ has exactly one partner |vₖ⟩. Product means r = 1, entangled means r ≥ 2.

⟨uₖ|uₗ⟩ = ⟨vₖ|vₗ⟩ = δₖₗ; the weights pₖ are dimensionless, positive and sum to 1.

Where the weights come frompₖ = the eigenvalues of CC†, with eigenvectors |uₖ⟩

Diagonalise the smaller of the two for the weights. Given A's eigenvectors |uₖ⟩, the partners follow from √pₖ|vₖ⟩ = (⟨uₖ| ⊗ 1)|ψ⟩.

CC† is d_A × d_A, Hermitian, trace 1, with entries Σⱼ cᵢⱼ c*ᵢ′ⱼ. C†C on B's side has the same nonzero eigenvalues.

Two spin-halves: the determinant decidesp± = ½[1 ± √(1 − 4|det C|²)], with det C = c↑↑c↓↓ − c↑↓c↓↑

Product exactly when det C = 0; weights ½ and ½, the most two spins can be entangled, exactly when |det C| = ½.

Dimensionless amplitudes, with no complex conjugates inside det C; 0 ≤ |det C| ≤ ½, and 2|det C| is called the concurrence.

Local unitaries leave the weights alone(U_A ⊗ U_B)|ψ⟩ has C′ = U_A C U_Bᵀ, so C′C′† = U_A CC† U_A†

Same eigenvalues, same pₖ: rotating either spin, or rewriting it in a new basis, cannot turn a product into an entangled state or back.

U_A acts on A alone and U_B on B alone, both dimensionless unitary matrices; ᵀ is the transpose, not the adjoint.

The singlet referred to x(|↑↓⟩ − |↓↑⟩)/√2 = −(|+x⟩|−x⟩ − |−x⟩|+x⟩)/√2

For the singlet's antisymmetric C, a common basis change W gives W C Wᵀ = (det W) C, only a phase. The triplet (|↑↓⟩ + |↓↑⟩)/√2, whose C is symmetric, changes form: on x it is (|+x⟩|+x⟩ − |−x⟩|−x⟩)/√2.

Kets ordered A then B, |±x⟩ = (|↑⟩ ± |↓⟩)/√2; weights ½ and ½ in both forms.

Put the amplitudes in a matrix

Expand in the product basis of the previous topic: |ψ⟩ = Σᵢⱼ cᵢⱼ |i⟩|j⟩, with cᵢⱼ = (⟨i|⟨j|)|ψ⟩, and arrange the cᵢⱼ as a matrix C, rows labelled by A and columns by B. Each row means something. (⟨i| ⊗ 1)|ψ⟩ = Σⱼ cᵢⱼ|j⟩ is the state B is left in, unnormalised, if A is found in |i⟩. A product |a⟩|b⟩ has cᵢⱼ = aᵢbⱼ, so every row is a multiple of the same vector and C = a bᵀ has rank 1; conversely a rank-1 matrix is a column times a row, and the state factors. That makes the definition computable: |ψ⟩ is entangled when rank C ≥ 2, which says that what B is left with depends on what A was found to be. The number of terms written down is no guide. |+x⟩|+x⟩ = ½(|↑↑⟩ + |↑↓⟩ + |↓↑⟩ + |↓↓⟩) has four, yet both rows of C are (½, ½) and the rank is 1.

The singular-value decomposition gives the Schmidt form

Linear algebra supplies, and this lesson quotes, the singular-value decomposition: any d_A × d_B matrix is C = UΣV†, with U and V unitary and Σ zero except for real σₖ ≥ 0 on its diagonal. Then cᵢⱼ = Σₖ Uᵢₖ σₖ V*ⱼₖ, and the double sum collapses to |ψ⟩ = Σₖ σₖ |uₖ⟩|vₖ⟩ with |uₖ⟩ = Σᵢ Uᵢₖ|i⟩ and |vₖ⟩ = Σⱼ V*ⱼₖ|j⟩, orthonormal because U and V are unitary. Write σₖ = √pₖ: Σₖ pₖ = Tr(CC†) = 1, and there are at most min(d_A, d_B) terms. Since CC† = UΣΣ†U†, its eigenvalues are the pₖ and its eigenvectors the |uₖ⟩; so in practice diagonalise CC†, then project: √pₖ|vₖ⟩ = (⟨uₖ| ⊗ 1)|ψ⟩. For a spin-half and a spin-1, (|↑⟩|+1⟩ + |↑⟩|0⟩ + |↓⟩|0⟩ + |↓⟩|−1⟩)/2 has CC† with ½ on the diagonal and ¼ off it: weights ¾ and ¼ on |±x⟩, partnered with (|+1⟩ + 2|0⟩ + |−1⟩)/√6 and (|+1⟩ − |−1⟩)/√2. Two terms, although B has three levels.

Two spin-halves: one determinant decides

For d_A = d_B = 2 the eigenvalue problem solves itself. CC† is 2 × 2 with trace 1 and determinant |det C|², so its eigenvalues obey p₊ + p₋ = 1 and p₊p₋ = |det C|², which gives p± = ½[1 ± √(1 − 4|det C|²)]. Real weights require |det C| ≤ ½. So one number, det C = c↑↑c↓↓ − c↑↓c↓↑, built from the amplitudes as they stand with no conjugates, settles the question: det C = 0 is a product, |det C| = ½ gives weights ½ and ½, the most two spins can be entangled, and anything between is partial. Limiting checks: |↑⟩|↓⟩ has det C = 0 and weights 1 and 0; (|↑↑⟩ + |↓↓⟩)/√2 has det C = ½. Any sensible measure of how entangled a pure state is must be a function of the pₖ alone. The quantity 2|det C|, running from 0 to 1 and called the concurrence (a name quoted here), is one such measure; the entropy met with the partial trace later in this unit is another.

Local unitaries turn the bases, never the weights

Act with U_A on A and U_B on B. Every amplitude is re-expanded, c′ₖₗ = Σᵢⱼ (U_A)ₖᵢ cᵢⱼ (U_B)ₗⱼ, which is C′ = U_A C U_Bᵀ. Then C′C′† = U_A C (U_Bᵀ U_B*) C† U_A† = U_A CC† U_A†, a unitary change of basis with the same eigenvalues: the pₖ do not move, and |det C′| = |det C| because |det U_A| = |det U_B| = 1. The same algebra covers a passive change of basis on either side. Two consequences follow. Product versus entangled cannot depend on the axis you quantised along. And no sequence of operations on A alone and B alone, such as precessing one spin in a field that acts on it alone, can create entanglement or remove it: that takes a Hamiltonian coupling the two, like hydrogen's hyperfine term, proportional to Sₑ⋅Sₚ. Conversely, two states with the same weights are related by the local unitary that carries one pair of Schmidt bases onto the other.

The singlet keeps its form on every common axis

The singlet (|↑↓⟩ − |↓↑⟩)/√2 has c↑↓ = 1/√2, c↓↑ = −1/√2 and zeros on the diagonal, so |det C| = ½ and its weights are ½ and ½: it is already a Schmidt form. Refer both spins to x, with |↑⟩ = (|+x⟩ + |−x⟩)/√2 and |↓⟩ = (|+x⟩ − |−x⟩)/√2, and expand: the |+x⟩|+x⟩ and |−x⟩|−x⟩ terms cancel, leaving −(|+x⟩|−x⟩ − |−x⟩|+x⟩)/√2, the same form up to the phase −1. The general reason is short. A common change of basis sends C to W C Wᵀ, and for the singlet C is antisymmetric, so W C Wᵀ is antisymmetric too, and its one independent entry is the old one times w₁₁w₂₂ − w₁₂w₂₁ = det W, a phase for unitary W. So any common axis serves as a Schmidt basis; Unit 5 traces this to total spin zero. The case is measured: hydrogen's ground-state F = 0 level is this state of electron and proton spins, h × 1420.405751768 MHz = 5.874 μeV below the F = 1 triplet, a hydrogen-maser frequency quoted here.

Where the decomposition stops

Three limits bound what has just been built. Uniqueness: when the weights are distinct the pairs |uₖ⟩|vₖ⟩ are fixed up to phases, but equal weights leave a whole family of Schmidt bases, as the singlet showed. Purity: the construction starts from one ket and its matrix C. An ensemble, such as half the pairs made in |↑↑⟩ and half in |↓↓⟩, has no single C; the density operator of the next topic describes it, and deciding whether a mixture is entangled is a harder question this course leaves aside. The number of parts: three spins have a coefficient array cᵢⱼₖ, not a matrix, and in general no form Σₖ √pₖ|aₖ⟩|bₖ⟩|cₖ⟩ with three orthonormal sets exists; (|↑↑↓⟩ + |↑↓↑⟩ + |↓↑↑⟩)/√3 has none, a result stated here without proof. Grouping two of the spins as one system restores a two-part problem, and the Schmidt form then describes that split alone.

Change one variable at a time

Make the relationship visible.

Interactive model

Weight angle θ30 °

Angle φ from |b₁⟩ to |b₂⟩60 °

Turn β of spin B's kets20 °

Sweep β: the entries of C change, the parallelogram only turns, and the bars stay put. Set θ = 45°, φ = 90° and β = 90° for the singlet, a square with both weights 0.50. Then slide φ to 0°: the rows fall onto one line, the area and p₂ vanish, and the state is a product.

SCHMIDT WEIGHT p₁0.831

SCHMIDT WEIGHT p₂0.169

|det C| = AREA0.375

OVERLAP ⟨b₁|b₂⟩0.500

Live interpretationSCHMIDT WEIGHT p₁: 0.831. SCHMIDT WEIGHT p₂: 0.169. |det C| = AREA: 0.375. OVERLAP ⟨b₁|b₂⟩: 0.500

Catch the common trap

Explain before calculating.

Misconception check

Students count terms. They call (|↑↑⟩ − |↑↓⟩)/√2 entangled because it needs two kets, and (|↑↑⟩ + |↓↓⟩)/√2 entangled for the same reason, yet the first is the product |↑⟩|−x⟩ and the second is as entangled as two spins can be. Both have two terms in the z basis. The count belongs to the basis you happened to write in, and a local change of basis raises or lowers it freely: (|↑↑⟩ − |↑↓⟩)/√2 collapses to one term once B is referred to x, and the Bell state grows to four terms with A on z and B on x. What no local change alters is the rank of C, which is the smallest number of product terms the state can be written with in any bases at all, orthonormal or not. Entangled means that minimum is at least 2, and the singular values of C compute it.

Two spin-halves are prepared in |ψ⟩ = ½(|↑↑⟩ + i|↑↓⟩ + i|↓↑⟩ − |↓↓⟩), kets ordered A then B. Which statement about the Schmidt decomposition of this state is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo spin-halves leave a source in |ψ⟩ = (3|↑↑⟩ + 6|↑↓⟩ + 2|↓↑⟩ + 4|↓↓⟩)/√65, kets ordered A then B. Decide whether the pair is entangled, and factor the state if it is not. An interaction between the two spins then flips the sign of the |↓↓⟩ amplitude alone, something no operation on one spin by itself can do; find the Schmidt weights of the new state.

Normalisation: 9 + 36 + 4 + 16 = 65, so the squared amplitudes sum to 1. C has rows (3, 6)/√65 for A = ↑ and (2, 4)/√65 for A = ↓.
det C = (3 × 4 − 6 × 2)/65 = 0. The rows are parallel, (3, 6) = 1.5 × (2, 4), so C has rank 1: one Schmidt weight is 1 and the state is a product, four terms notwithstanding.
Factor it. The common row direction (1, 2) is B's ket and the column (3, 2) is A's: |ψ⟩ = [(3|↑⟩ + 2|↓⟩)/√13] [(|↑⟩ + 2|↓⟩)/√5], correctly normalised because 13 × 5 = 65.
After the flip, det C = (3 × (−4) − 6 × 2)/65 = −24/65 = −0.3692, so 4|det C|² = 0.5453 and √(1 − 0.5453) = 0.6743.
p± = ½(1 ± 0.6743) = 0.8371 and 0.1629. Checks: they sum to 1, their product 0.1363 equals |det C|² = (24/65)², and both are dimensionless, as probabilities must be.

AnswerAs prepared, a product: (3|↑⟩ + 2|↓⟩)(|↑⟩ + 2|↓⟩)/√65. With the sign flipped, entangled, with Schmidt weights 0.837 and 0.163.

MediumFind the complete Schmidt decomposition, weights and partner kets, of |ψ⟩ = (5|↑↑⟩ + 7|↑↓⟩ + 5|↓↑⟩ + |↓↓⟩)/10, kets ordered A then B, and say along which axis each spin's Schmidt basis points.

C has rows (5, 7)/10 and (5, 1)/10, and 25 + 49 + 25 + 1 = 100 confirms normalisation. det C = (5 × 1 − 7 × 5)/100 = −0.30, not zero, so the state is entangled.
Weights: 4|det C|² = 0.36, so p± = ½(1 ± √0.64) = ½(1 ± 0.80), giving p₁ = 0.90 and p₂ = 0.10.
A's Schmidt kets are the eigenvectors of CC†, which has 0.74 and 0.26 on its diagonal and 0.32 off it. For p₁ = 0.90, (0.74 − 0.90)x + 0.32y = 0 gives y = x/2, so |u₁⟩ = (2|↑⟩ + |↓⟩)/√5, and orthogonal to it |u₂⟩ = (−|↑⟩ + 2|↓⟩)/√5.
Partners by projection: (⟨u₁| ⊗ 1)|ψ⟩ = [2(5|↑⟩ + 7|↓⟩) + (5|↑⟩ + |↓⟩)]/(10√5) = 15(|↑⟩ + |↓⟩)/(10√5), of norm 0.949 = √0.90, so |v₁⟩ = |+x⟩. Likewise [−(5|↑⟩ + 7|↓⟩) + 2(5|↑⟩ + |↓⟩)]/(10√5) = 5(|↑⟩ − |↓⟩)/(10√5), of norm 0.316 = √0.10, so |v₂⟩ = |−x⟩.
Check by rebuilding one amplitude: c↑↓ = √0.90 (2/√5)(1/√2) + √0.10 (−1/√5)(−1/√2) = 0.60 + 0.10 = 0.70, as given; c↓↓ = 0.30 − 0.20 = 0.10, also as given.
Axes: |u₁⟩ = cos 26.57°|↑⟩ + sin 26.57°|↓⟩ is spin up along the axis in the xz plane at 2 × 26.57° = 53.13° from z, and |u₂⟩ is spin down along it; B's Schmidt kets are spin up and down along x.

Answer|ψ⟩ = √0.90 |u₁⟩|+x⟩ + √0.10 |u₂⟩|−x⟩, weights 0.90 and 0.10, with A's Schmidt axis in the xz plane at 53.1° from z and B's along x.

HardA hydrogen atom in its ground state sits in a field B along z. On the pair |↑↓⟩, |↓↑⟩ (electron first, proton second) the hyperfine and Zeeman terms, quoted from Unit 9, have diagonal elements −A/4 + b/2 and −A/4 − b/2 and off-diagonal element A/2, where A = hν with ν = 1420.405751768 MHz measured, and b = (gₑμ_B + gₚμ_N)B. Find the Schmidt weights of either eigenstate at B = 0.10 T, and the field at which the smaller weight is 0.10.

The block is a real symmetric two-state matrix, so from Quantum Mechanics I its eigenkets are cos χ|↑↓⟩ + sin χ|↓↑⟩ and −sin χ|↑↓⟩ + cos χ|↓↑⟩, with tan 2χ = (A/2)/(b/2) = A/b. A and b are both energies, so the ratio is dimensionless, as an angle's tangent must be.
Each eigenket's C has zeros on the diagonal and ±cos χ, ±sin χ off it, so |det C| = sin χ cos χ = ½ sin 2χ = ½A/√(A² + b²). The two-spin formula then gives p± = ½[1 ± b/√(A² + b²)], the same for both eigenkets.
Numbers: A = 6.62607 × 10⁻³⁴ J s × 1.420406 × 10⁹ s⁻¹ = 9.4117 × 10⁻²⁵ J, or 5.874 μeV. gₑμ_B + gₚμ_N = 1.85695 × 10⁻²³ + 0.00282 × 10⁻²³ = 1.85977 × 10⁻²³ J T⁻¹, so b = 1.8598 × 10⁻²⁴ J at 0.10 T and b/A = 1.976.
Then b/√(A² + b²) = 1.976/√(1 + 1.976²) = 0.8923, so p₁ = 0.946 and p₂ = 0.054: a tenth of a tesla has already undone most of the entanglement.
p₂ = 0.10 needs b/√(A² + b²) = 0.80, so b = (4/3)A and B = (4/3) × A/(gₑμ_B + gₚμ_N) = (4/3) × 0.05061 T = 0.0675 T.
Limits: at B = 0 the weights are ½ and ½, the triplet (|↑↓⟩ + |↓↑⟩)/√2 and the singlet; as B grows p₂ ≈ A²/(4b²) → 0 and the eigenkets become the products |↑↓⟩ and |↓↑⟩. The field scale is small because A, quoted as (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.877 μeV and within 0.1% of the measured value, is the 1.45 meV scale α⁴mₑc² of fine structure cut down by mₑ/mₚ = 1/1836.

AnswerAt 0.10 T the weights are 0.946 and 0.054; the smaller weight falls to 0.10 at B = 0.0675 T. Both eigenstates are maximally entangled only at B = 0.