University Physics V · Introduction to Quantum Information · 20.4

Tensor Products, Entanglement & Schmidt Form

Two qubits is where the linear algebra stops being a rewrite of wave mechanics and starts paying. Learn to build the four-dimensional space with a Kronecker product, to tell a product state from an entangled one by the rank of a 2 × 2 matrix, and to measure how entangled it is by tracing one qubit out.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A composite system lives in the tensor product of its parts' spaces, not their sum, so dimensions multiply: two qubits need C² ⊗ C², four amplitudes, and n qubits need 2ⁿ. A product |a⟩ ⊗ |b⟩ carries four real parameters, a general unit ray in C⁴ carries six, so product states are a thin surface and almost every state carries correlations no local recipe produces. The Schmidt decomposition makes this precise.

Arrange the amplitudes cᵢⱼ as a matrix C and take its singular value decomposition: |ψ⟩ = Σₖ √λₖ |k_A⟩|k_B⟩ in one orthonormal basis per side, with Σ λₖ = 1. The state is entangled exactly when more than one λₖ is nonzero, and the λₖ are the eigenvalues of the reduced state ρ_A = Tr_B |ψ⟩⟨ψ| = C C†, so the entropy S = −Σ λₖ log₂ λₖ measures the entanglement: 1 bit for a Bell pair, 0 for a product. What makes this the right definition is invariance: C → U_A C U_Bᵀ leaves every singular value alone, so no local unitary can create or destroy entanglement, and the term count of a ket, which local rotations change freely, never could have measured it.

The cost is that this is a pure-state tool. A mixed pair can have a maximally mixed ρ_A through plain classical correlation, and deciding whether a mixed state is entangled needs a different criterion — the partial transpose for two qubits, and in general an NP-hard problem.

Simple definition: A pure state of two systems is entangled when it cannot be written as one vector tensor another; equivalently its coefficient matrix has rank above one, its Schmidt decomposition has more than one term, and the reduced state of either half is mixed.
Example: (|00⟩ + |11⟩)/√2 has coefficient matrix I/√2, rank 2: tracing out B leaves ρ_A = I/2 with S = 1 bit, whereas (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2 has the rank-1 matrix ½[[1, 1], [1, 1]] and is just |+⟩|+⟩.

Kronecker product of states and gates(|a⟩ ⊗ |b⟩)ᵢⱼ = aᵢ bⱼ · (A ⊗ B)(|a⟩ ⊗ |b⟩) = A|a⟩ ⊗ B|b⟩

np.kron builds both. Dimension multiplies, so n qubits need 2ⁿ amplitudes and a gate on qubit k alone is I ⊗ … ⊗ U ⊗ … ⊗ I.

|a⟩, |b⟩ ∈ C²; the product sits in C⁴ ordered |00⟩, |01⟩, |10⟩, |11⟩ (row 2i + j); amplitudes dimensionless

Coefficient matrix and the product test|ψ⟩ = Σ cᵢⱼ |i⟩ ⊗ |j⟩ · product ⇔ rank C = 1 ⇔ c₀₀c₁₁ − c₀₁c₁₀ = 0

psi.reshape(2, 2) turns the ket into C. A determinant, not a count of terms, decides whether the state factorises.

C is the d_A × d_B matrix of amplitudes, 2 × 2 for two qubits, with Σ|cᵢⱼ|² = 1; the determinant test is two-qubit only

Schmidt decomposition (the SVD of C)|ψ⟩ = Σₖ √λₖ |k_A⟩ ⊗ |k_B⟩ · C = U diag(√λₖ) V† · Σ λₖ = 1

One orthonormal basis per side puts any pure state in diagonal form, and U_A ⊗ U_B cannot change a single λₖ.

λₖ ≥ 0 are the squared singular values of C; Schmidt rank = number of nonzero λₖ ≤ min(d_A, d_B)

Partial trace and the reduced stateρ_A = Tr_B ρ_AB = Σⱼ (I ⊗ ⟨j|) ρ_AB (I ⊗ |j⟩) = C C†

ρ_A predicts every measurement A can make alone; for a pure pair, ρ_A pure means product and ρ_A mixed means entangled.

ρ_A is d_A × d_A, Hermitian, Tr ρ_A = 1, eigenvalues λₖ; ρ_B = (C†C)ᵀ has the same nonzero spectrum

Entanglement entropyS(ρ_A) = −Σₖ λₖ log₂ λₖ = S(ρ_B)

Counts Bell pairs: n copies convert by local operations and classical communication into about nS Bell pairs, and back.

in bits; 0 for a product state, 1 for any Bell state, at most log₂ min(d_A, d_B); defined for pure ρ_AB

Two-qubit concurrenceC = 2|c₀₀c₁₁ − c₀₁c₁₀| = 2√(λ₁λ₂) · λ± = (1 ± √(1 − C²))/2

One determinant gives both Schmidt weights without an SVD: C = 1 is a Bell pair, C = 0 a product.

0 ≤ C ≤ 1, dimensionless; pure two-qubit states here, Wootters' formula extends it to mixed ρ

Composite systems multiply dimensions: build C² ⊗ C²

A single qubit is a unit ray in C². Two qubits do not live in C² ⊕ C², which also has dimension four but the wrong structure; they live in the tensor product C² ⊗ C², whose basis is the four products |ij⟩ = |i⟩ ⊗ |j⟩ with i, j ∈ {0, 1}, ordered |00⟩, |01⟩, |10⟩, |11⟩ so that |ij⟩ occupies row 2i + j. In NumPy the product of two column vectors is np.kron(a, b), and the same call builds gates: np.kron(X, I) is the 4 × 4 matrix that flips qubit A and leaves B alone, and the mixed-product rule (A ⊗ B)(C ⊗ D) = AC ⊗ BD is what lets circuits compose layer by layer. Dimension multiplies rather than adds. Three qubits need 8 amplitudes, ten need 1024, and a 30-qubit state vector in complex doubles is 2³⁰ × 16 bytes, about 17 GB. Now count parameters. A unit ray in C⁴ has 2 × 4 − 2 = 6 real parameters, one normalisation and one global phase removed; a product |a⟩ ⊗ |b⟩ has 2 + 2 = 4. Product states are a four-dimensional surface inside a six-dimensional space, so almost every state you can write down is not a product, and the working question is how to recognise one.

Reshape the ket into a matrix and take its rank

Write |ψ⟩ = Σ cᵢⱼ |i⟩ ⊗ |j⟩ and lay the amplitudes out as a 2 × 2 matrix C with cᵢⱼ in row i, column j — in code, psi.reshape(2, 2). If |ψ⟩ = |a⟩ ⊗ |b⟩ then cᵢⱼ = aᵢ bⱼ, so C = a bᵀ is an outer product of rank 1; conversely every rank-1 matrix is an outer product, so the state factorises. For two qubits rank 1 means det C = c₀₀c₁₁ − c₀₁c₁₀ = 0. Test it: (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2 gives C = ½[[1, 1], [1, 1]], determinant zero, and indeed it is |+⟩|+⟩ despite its four terms. (|00⟩ + |11⟩)/√2 gives C = I/√2, determinant ½: entangled. The in-between case (2|00⟩ + |01⟩ + |10⟩ + 2|11⟩)/√10 gives det C = (4 − 1)/10 = 0.3, so it is entangled, but how much needs the next step. The test is basis-proof. A local change of basis (U_A ⊗ U_B)|ψ⟩ sends C → U_A C U_Bᵀ, and multiplying by invertible matrices on either side never changes the rank, which is why counting terms fails and the rank test does not.

The Schmidt decomposition is the SVD of that matrix

Every matrix has a singular value decomposition C = U Σ V† with U and V unitary and Σ diagonal with entries σₖ ≥ 0. Substitute it into the ket: |ψ⟩ = Σₖ σₖ |uₖ⟩ ⊗ |vₖ*⟩, where |uₖ⟩ are the columns of U and |vₖ*⟩ the conjugated columns of V — one orthonormal basis on each side in which the state is diagonal. The weights λₖ = σₖ² sum to 1 by normalisation, and the number of nonzero ones, the Schmidt rank, is at most min(d_A, d_B): 1 or 2 for two qubits. Take C = [[2, 1], [1, 2]]/√10. It is real and symmetric, so its SVD is its eigendecomposition: eigenvectors (1, 1)/√2 = |+⟩ and (1, −1)/√2 = |−⟩ with eigenvalues 3/√10 and 1/√10, hence λ₁ = 0.9, λ₂ = 0.1 and |ψ⟩ = √0.9 |+⟩|+⟩ + √0.1 |−⟩|−⟩ — two terms in the right bases, four in the computational one. Since U_A C U_Bᵀ has the same singular values as C, the Schmidt weights are precisely what a local unitary cannot touch, which is what qualifies them to measure a property shared between two parties. In NumPy: u, s, vh = np.linalg.svd(ψ.reshape(2, 2)); lam = s**2.

Trace out B and read the entanglement from ρ_A

An observer who never touches B still needs a state that predicts every local measurement. That is ρ_A = Tr_B |ψ⟩⟨ψ|, defined by Tr(ρ_A M) = ⟨ψ|M ⊗ I|ψ⟩ for every operator M on A; explicitly ρ_A = Σⱼ (I ⊗ ⟨j|)|ψ⟩⟨ψ|(I ⊗ |j⟩), which in terms of the coefficient matrix is ρ_A = C C† and ρ_B = (C†C)ᵀ. In code: ρ = np.outer(ψ, psi.conj()).reshape(2, 2, 2, 2); ρ_A = np.einsum('ijkj→ik', ρ). The eigenvalues of C C† are the σₖ² = λₖ, so ρ_A is pure exactly when the Schmidt rank is 1, and both reduced states share a spectrum even when d_A ≠ d_B. For the Bell pair ρ_A = I/2: every local measurement is a fair coin and S(ρ_A) = −2 × ½ log₂ ½ = 1 bit, although the pair itself is pure. For the [[2, 1], [1, 2]]/√10 state, ρ_A = C C† = [[0.5, 0.4], [0.4, 0.5]] with eigenvalues 0.9 and 0.1, purity 0.82 and S = −0.9 log₂ 0.9 − 0.1 log₂ 0.1 = 0.137 + 0.332 = 0.469 bits. That number is operational, not a score: n copies of the state can be converted by local operations and classical communication into about nS Bell pairs and back again, so S counts the Bell pairs the state is worth.

What Schmidt rank cannot classify: mixed states

Everything above assumed ρ_AB is pure. Drop that and a mixed ρ_A stops meaning entanglement. The mixture (|00⟩⟨00| + |11⟩⟨11|)/2 has ρ_A = I/2 and S = 1 bit, identical to the Bell pair, yet it is a classical coin shared by post — separable, a convex sum of product states — and violates no Bell inequality. Nothing local tells them apart; the difference lives in the off-diagonal terms |00⟩⟨11| that only a joint measurement sees. For mixed states the question becomes separability: can ρ be written as Σ pₖ ρ_A^k ⊗ ρ_B^k? For 2 × 2 and 2 × 3 systems the Peres–Horodecki criterion settles it. Transpose the B indices only, ⟨ij|ρ^T_B|kl⟩ = ⟨il|ρ|kj⟩; the state is entangled exactly when the result has a negative eigenvalue. For |Φ⁺⟩⟨Φ⁺| the partial transpose moves the ½ from |00⟩⟨11| to |01⟩⟨10| and produces the eigenvalue −½; for the classical mixture there is nothing to move and every eigenvalue stays non-negative. In higher dimensions a positive partial transpose is necessary but not sufficient, and general separability is NP-hard — a reminder that Schmidt rank is a pure-state luxury.

Three qubits: choose the cut

The Schmidt decomposition is a two-party statement, and it does not extend to three: no basis makes a generic three-qubit state a single sum Σ √λₖ |k⟩|k⟩|k⟩. What you can always do is cut the register in two and apply everything above to the bipartition. For the GHZ state (|000⟩ + |111⟩)/√2 across A|BC the coefficient matrix is 2 × 4 with rank 2, ρ_A = I/2 and S = 1 bit; trace out A instead and the remaining pair is (|00⟩⟨00| + |11⟩⟨11|)/2, the separable classical mixture of the previous section, so the entanglement is held by all three or by none. The W state (|001⟩ + |010⟩ + |100⟩)/√3 gives ρ_A = diag(2/3, 1/3) and S = 0.918 bits across the same cut, but tracing out one qubit leaves a pair whose Wootters concurrence is 2/3: its entanglement survives the loss of a party. GHZ and W cannot be converted into each other by local operations and classical communication, even probabilistically, so beyond two parties there is no single number and no single kind of entanglement, only what each cut reveals.

Change one variable at a time

Make the relationship visible.

Interactive model

Schmidt angle θ30 °

Local rotation α on qubit A0 °

Local rotation β on qubit B0 °

Put θ at 45° and sweep α from 0 to 180°: the four amplitude bars on the left rearrange completely while λ₁, λ₂ and S stay at ½, ½ and 1 bit, because R_y(α) ⊗ I is local. Then set θ to 0 and no α or β will lift λ₂ off the axis — a product state stays a product state.

λ₁ = cos²θ0.750

λ₂ = sin²θ0.250

S(ρ_A) ENTROPY0.811 bits

CONCURRENCE 2|det C|0.866

Live interpretationλ₁ = cos²θ: 0.750. λ₂ = sin²θ: 0.250. S(ρ_A) ENTROPY: 0.811 bits. CONCURRENCE 2|det C|: 0.866

Catch the common trap

Explain before calculating.

Misconception check

Students judge entanglement by counting terms: a ket with four terms 'must be' entangled and one with two terms 'looks' separable. Both halves are wrong. (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2 has four terms and is the product |+⟩|+⟩; (|00⟩ + |11⟩)/√2 has two and is maximally entangled; and flipping one sign in the first, (|00⟩ + |01⟩ + |10⟩ − |11⟩)/2, produces a Bell pair in disguise, (|0⟩|+⟩ + |1⟩|−⟩)/√2. The number of terms depends on which local bases you expand in, and a local basis change U_A ⊗ U_B is exactly the kind of operation that can neither create nor destroy entanglement. The invariant is the rank of the coefficient matrix — its singular values survive C → U_A C U_Bᵀ — so the test is a determinant or an SVD, never a count. The same reflex misreads a mixed ρ_A as proof of entanglement; for a mixed pair it proves only correlation, which a shared coin supplies just as well.

Which of these two-qubit states is entangled? Decide by the rank of the 2 × 2 matrix of amplitudes [[c₀₀, c₀₁], [c₁₀, c₁₁]], not by the number of terms.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTake the Bell state |Φ⁺⟩ = (|00⟩ + |11⟩)/√2. Write its coefficient matrix, find the Schmidt rank, compute ρ_A by partial trace, and give the entanglement entropy and the purity of ρ_A.

Amplitudes in the order |00⟩, |01⟩, |10⟩, |11⟩ are (1, 0, 0, 1)/√2, so C = [[1, 0], [0, 1]]/√2. Its determinant is ½ ≠ 0: rank 2, so the state is entangled.
C is already diagonal, so the Schmidt form reads off directly: |Φ⁺⟩ = √½ |0⟩|0⟩ + √½ |1⟩|1⟩ with λ₁ = λ₂ = ½; the Schmidt bases are the computational bases themselves.
Partial trace: ρ_A = C C† = ½ I = [[0.5, 0], [0, 0.5]]. Element by element, ⟨0|ρ_A|0⟩ = |c₀₀|² + |c₀₁|² = ½ and ⟨0|ρ_A|1⟩ = c₀₀c₁₀* + c₀₁c₁₁* = 0.
S(ρ_A) = −2 × ½ log₂ ½ = 1 bit, the maximum for a qubit; purity Tr ρ_A² = ¼ + ¼ = ½, the minimum. Alice's qubit alone is a fair coin in every basis, yet the pair as a whole is pure.

AnswerSchmidt rank 2 with λ = (½, ½); ρ_A = I/2; S = 1 bit; Tr ρ_A² = ½.

MediumA two-qubit state is |ψ⟩ = (2|00⟩ + |01⟩ + |10⟩ + 2|11⟩)/√10. Decide whether it is entangled, find its Schmidt decomposition, then compute ρ_A, its purity and the entanglement entropy. Cross-check with the concurrence.

Normalisation: 4 + 1 + 1 + 4 = 10, so the amplitudes (2, 1, 1, 2)/√10 give C = [[2, 1], [1, 2]]/√10. det C = (4 − 1)/10 = 0.3 ≠ 0: rank 2, entangled.
C is real and symmetric, so its SVD is its eigendecomposition: eigenvectors (1, 1)/√2 = |+⟩ and (1, −1)/√2 = |−⟩ with eigenvalues 3/√10 and 1/√10. Schmidt weights λ₁ = 9/10 = 0.9 and λ₂ = 1/10 = 0.1, summing to 1.
Schmidt form: |ψ⟩ = √0.9 |+⟩|+⟩ + √0.1 |−⟩|−⟩. Check one amplitude: ⟨00|ψ⟩ = (3/√10)(½) + (1/√10)(½) = 2/√10, as given.
ρ_A = C C† = (1/10)[[2, 1], [1, 2]]² = (1/10)[[5, 4], [4, 5]] = [[0.5, 0.4], [0.4, 0.5]]; its eigenvalues are 0.9 and 0.1 as required. Purity: 0.5² + 0.4² + 0.4² + 0.5² = 0.82.
S = −0.9 log₂ 0.9 − 0.1 log₂ 0.1 = 0.9 × 0.1520 + 0.1 × 3.3219 = 0.1368 + 0.3322 = 0.469 bits — under half a Bell pair's worth.
Concurrence C = 2|det C| = 0.6, and (1 ± √(1 − 0.36))/2 = (1 ± 0.8)/2 = 0.9 and 0.1 reproduces the Schmidt weights without an SVD.

AnswerEntangled: |ψ⟩ = √0.9 |+⟩|+⟩ + √0.1 |−⟩|−⟩; ρ_A = [[0.5, 0.4], [0.4, 0.5]] with purity 0.82; S ≈ 0.469 bits; concurrence 0.6.

HardThe Werner state ρ_W = p|Φ⁺⟩⟨Φ⁺| + (1 − p) I/4 mixes a Bell pair with white noise. Show that the partial trace cannot detect its entanglement, say why Schmidt rank does not apply, and use the Peres–Horodecki partial transpose to find the range of p for which ρ_W is entangled. Evaluate at p = 0.5.

In the |00⟩, |01⟩, |10⟩, |11⟩ basis, |Φ⁺⟩⟨Φ⁺| has ½ at the four corners (00,00), (00,11), (11,00), (11,11). So ρ_W has diagonal ((1 + p)/4, (1 − p)/4, (1 − p)/4, (1 + p)/4) and off-diagonal entries p/2 at (00,11) and (11,00); the trace is (2 + 2p + 2 − 2p)/4 = 1.
Partial trace: ρ_A = Tr_B ρ_W = p (I/2) + (1 − p)(I/2) = I/2 for every p. The reduced state is maximally mixed whether p = 1 (a Bell pair) or p = 0 (pure noise), so S(ρ_A) = 1 bit decides nothing here.
Schmidt rank needs a pure ρ_AB. The eigenvalues of ρ_W are (1 + 3p)/4 on |Φ⁺⟩ and (1 − p)/4 on the other three Bell states, so Tr ρ_W² = (1 + 3p²)/4 < 1 for p < 1: the state is mixed and the pure-state classification does not apply.
Partial transpose on B swaps the second index of each |ij⟩⟨kl|: the corner elements p/2 at (00,11) and (11,00) move to (01,10) and (10,01) while the diagonal stays. ρ_W^T_B is block diagonal: (1 + p)/4 at 00 and at 11, and the 2 × 2 block [[(1 − p)/4, p/2], [p/2, (1 − p)/4]] on {01, 10}.
That block has eigenvalues (1 − p)/4 ± p/2, i.e. (1 + p)/4 and (1 − 3p)/4. The only eigenvalue that can go negative is (1 − 3p)/4, which does so exactly when p > 1/3. For 2 × 2 systems a negative partial transpose is necessary and sufficient, so ρ_W is entangled for p > 1/3 and separable for p ≤ 1/3.
At p = 0.5 the minimum eigenvalue is (1 − 1.5)/4 = −0.125: entangled, and Wootters' concurrence for this family, max(0, (3p − 1)/2) = 0.25, agrees. At p = 0.25 the eigenvalue is +0.0625 and the state is separable even though ρ_A is still I/2 and the two qubits are still correlated.

Answerρ_A = I/2 for all p; ρ_W is mixed, so Schmidt rank is undefined; ρ_W^T_B has eigenvalue (1 − 3p)/4, negative for p > 1/3. At p = 0.5 it equals −0.125 and the concurrence is 0.25.