University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.1

Tensor-product spaces for two subsystems

Every two-body calculation left in this course, from the triplet and singlet to the 21 cm line and the helium atom, runs on one construction. This lesson builds it by hand: what a joint ket is, why the joint space is so much larger than its parts, and how an operator on one spin becomes a 4 × 4 matrix you can multiply.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I gave one system a ket in a space of dimension d. A pair of systems needs more than two such kets. Measure A in a complete basis |i⟩ and B in |j⟩ and the joint experiment has d_A d_B distinguishable outcomes, one for every pair, so the product kets |i⟩|j⟩ form the joint basis; and because superposition holds for the whole as much as for its parts, every combination Σ cᵢⱼ|i⟩|j⟩ is a state too.

That space is the tensor product H_A ⊗ H_B, and its inner product ⟨a′b′|ab⟩ = ⟨a′|a⟩⟨b′|b⟩ is chosen so that two independently prepared systems give joint probabilities that multiply. None of this is derived. It is the composite-system postulate, motivated by those two demands and confirmed by every count of states, from the six ground sublevels of deuterium to the R ln 2 of magnetic entropy per mole of spin-½ ions.

The consequence is that dimensions multiply where classical coordinates add: N spin-halves need 2ᴺ amplitudes, and two particles need a wavefunction on six dimensions. An operator on one part becomes A ⊗ 1, commutes with everything belonging to the other, and when the parts do not interact their energies simply add. The construction is exact but not the whole story: most kets in the space are not products, which the next lesson sorts out, and for identical particles only a symmetric or antisymmetric slice of it is physical, which waits for Unit 7.

Simple definition: The tensor product H_A ⊗ H_B is the state space of systems A and B taken together: it is spanned by the product kets |i⟩|j⟩ of basis kets of each part, so it has d_A d_B dimensions, and any superposition Σ cᵢⱼ|i⟩|j⟩ of them is a state.
Example: An electron spin (2 states) with a spin-1 deuteron (3 states) spans 2 × 3 = 6 dimensions, not 2 + 3 = 5, and deuterium's ground state does split into six sublevels in a magnetic field. With the electron first, |↑⟩|m_I = 0⟩ is the column (0, 1, 0, 0, 0, 0).

Inner product of product kets⟨a′b′|ab⟩ = ⟨a′|a⟩⟨b′|b⟩

Postulated, then extended to sums by linearity. It makes independent systems give joint probabilities |⟨a′|a⟩|²|⟨b′|b⟩|² that multiply, and makes the kets |i⟩|j⟩ orthonormal.

|a⟩, |a′⟩ in H_A and |b⟩, |b′⟩ in H_B; for normalised discrete kets each factor is a dimensionless amplitude.

Dimensions multiplydim(H_A ⊗ H_B) = d_A d_B

Classical coordinates add, 3 + 3 = 6 for two particles. Quantum dimensions multiply, because the joint basis is every pair |i⟩|j⟩, not one list followed by the other.

d_A, d_B are pure numbers, the counts of independent states. Two spin-halves: 4. Electron with deuteron: 6. N spin-halves: 2ᴺ.

A general joint state|Ψ⟩ = Σᵢⱼ cᵢⱼ|i⟩|j⟩, with cᵢⱼ = ⟨ij|Ψ⟩

A product state has cᵢⱼ = aᵢbⱼ, so its joint probabilities multiply. A general Ψ has d_A d_B free amplitudes and is usually not of that form.

cᵢⱼ is dimensionless and Σ|cᵢⱼ|² = 1; |cᵢⱼ|² is the probability of outcome i on A together with outcome j on B.

Operators on different factors commute(A ⊗ 1)(1 ⊗ B) = A ⊗ B = (1 ⊗ B)(A ⊗ 1)

Derived by acting on |a⟩|b⟩. So [S₁ᵢ, S₂ⱼ] = 0 for every pair of components, and a measurement on one part is always compatible with a measurement on the other.

A acts on H_A, B on H_B, and 1 is the identity on the other factor; S₁z means S_z ⊗ 1 and keeps its unit, J s.

Matrix elements of a Kronecker product⟨ij|A ⊗ B|kl⟩ = ⟨i|A|k⟩⟨j|B|l⟩

Each entry of A multiplies a whole copy of B, so S₁z = S_z ⊗ 1 = (ħ/2) diag(1, 1, −1, −1) while S₂z = 1 ⊗ S_z = (ħ/2) diag(1, −1, 1, −1).

i, k label H_A and j, l label H_B; |ij⟩ sits in row d_B(i − 1) + j, the first index slowest. ħ = 1.055 × 10⁻³⁴ J s.

Uncoupled parts: energies add(H_A ⊗ 1 + 1 ⊗ H_B)|a⟩|b⟩ = (E_A + E_B)|a⟩|b⟩

The two terms commute, so U(t) = U_A(t) ⊗ U_B(t) as well. A coupling such as S₁ ⋅ S₂, a sum of products, is what lets one part's energy depend on the other's state.

H_A|a⟩ = E_A|a⟩ and H_B|b⟩ = E_B|b⟩, with every energy in J.

Two demands that motivate the joint space

Add a second system B to a system A whose kets |a⟩ live in a space of dimension d_A. Two demands point to what the pair's state space should be. First, prepare A in |a⟩ and B in |b⟩ independently, in separate laboratories: the chance of then finding A in |a′⟩ and B in |b′⟩ must be the product |⟨a′|a⟩|²|⟨b′|b⟩|², as for any independent events, and the rule ⟨a′b′|ab⟩ = ⟨a′|a⟩⟨b′|b⟩ delivers exactly that. Second, a complete measurement on A and a complete measurement on B together have d_A d_B distinguishable outcomes |i⟩|j⟩, so these must form an orthonormal basis; and superposition applies to the whole as much as to its parts, so every Σ cᵢⱼ|i⟩|j⟩ is a state. The space so built is H_A ⊗ H_B. This is the composite-system postulate, motivated here and not derived, and no experiment has contradicted it. Limiting case: if B has a single state, d_B = 1, every joint ket is |a⟩|0⟩ and the construction hands back H_A unchanged.

Dimensions multiply where coordinates add

Classically, two particles need 3 + 3 = 6 coordinates and N spins need N directions: descriptions add. Quantum mechanically the amplitudes multiply. Put one particle's wavefunction on a grid of 100 points per axis and it takes 10⁶ complex numbers; two particles need 10⁶ × 10⁶ = 10¹², one for every pair of grid points, not 2 × 10⁶. Fifty spin-halves need 2⁵⁰ = 1.13 × 10¹⁵ amplitudes, about 18 petabytes at 16 bytes each. The count is measurable. N independent spin-halves have 2ᴺ states and a magnetic entropy k_B ln 2ᴺ = N k_B ln 2, extensive precisely because dimensions multiply; per mole that is R ln 2 = 5.76 J mol⁻¹ K⁻¹, the value low-temperature calorimetry finds for salts whose magnetic ions carry an effective spin ½. An additive count of 2N states would give k_B ln 2N, which hardly grows with N. Beware the one case that hides the difference: for two spin-halves, 2 + 2 and 2 × 2 are both 4.

Calculating with product kets

The product |a⟩|b⟩, also written |a⟩ ⊗ |b⟩ or |ab⟩, is linear in each slot separately. A number slides freely between the factors, (λ|a⟩)|b⟩ = |a⟩(λ|b⟩) = λ|a⟩|b⟩, and sums distribute: (|a⟩ + |a′⟩)(|b⟩ + |b′⟩) = |ab⟩ + |ab′⟩ + |a′b⟩ + |a′b′⟩, four terms, not two. With orthonormal bases on each side a general joint ket is |Ψ⟩ = Σ cᵢⱼ|i⟩|j⟩ with cᵢⱼ = ⟨ij|Ψ⟩, and the Born rule makes |cᵢⱼ|² the probability of finding i on A and j on B together. The chance of i on A alone, whatever B shows, is the projector Pᵢ ⊗ 1 averaged, which sums over j to give Σⱼ|cᵢⱼ|². For a product state cᵢⱼ = aᵢbⱼ, so every joint probability is a product and A's probabilities do not depend on B at all. That is special. Two spin-halves have six real parameters in a general normalised state, once the norm and overall phase are spent, but a product has only four, two for each spin's direction; which kets are products is the next lesson's question.

An operator on one part, written on the whole

An operator A on H_A acts on the joint space as A ⊗ 1: (A ⊗ 1)|a⟩|b⟩ = (A|a⟩)|b⟩, extended to every joint ket by linearity. Apply A ⊗ 1 and 1 ⊗ B to a product ket in either order and the result is (A|a⟩)(B|b⟩), so (A ⊗ 1)(1 ⊗ B) = A ⊗ B = (1 ⊗ B)(A ⊗ 1). Operators on different factors always commute, even where they would not on one factor: [S₁ₓ, S₂y] = 0 although [Sₓ, S_y] = iħS_z. This is derived, not assumed, and two consequences follow. In a product state ⟨ab|A ⊗ B|ab⟩ = ⟨a|A|a⟩⟨b|B|b⟩, so correlations factorise. And parts that do not interact have H = H_A ⊗ 1 + 1 ⊗ H_B, whose eigenkets are the products |a⟩|b⟩ with energy E_A + E_B; since the two terms commute, U(t) = U_A(t) ⊗ U_B(t), and each part evolves as if the other were absent. Units check: every term of H is an energy in J, and 1 carries no unit. A coupling such as S₁ ⋅ S₂ = S₁ₓS₂ₓ + S₁yS₂y + S₁zS₂z is a sum of three products, and no split into H_A ⊗ 1 + 1 ⊗ H_B reproduces it.

Two spin-halves as four columns

Choose an order and keep it: spin 1's label first, so the basis is |↑↑⟩, |↑↓⟩, |↓↑⟩, |↓↓⟩. The Kronecker product then stacks copies of the second column, each scaled by one entry of the first: (a, b) ⊗ (c, d) = (ac, ad, bc, bd). Matrices obey the same rule, ⟨ij|A ⊗ B|kl⟩ = ⟨i|A|k⟩⟨j|B|l⟩, which follows from the inner product. So S₁z = S_z ⊗ 1 = (ħ/2) diag(1, 1, −1, −1) while S₂z = 1 ⊗ S_z = (ħ/2) diag(1, −1, 1, −1); S₁ₓ swaps entries 1 ↔ 3 and 2 ↔ 4, and S₂ₓ swaps 1 ↔ 2 and 3 ↔ 4. The total spin S = S₁ + S₂ means S ⊗ 1 + 1 ⊗ S. Its z-component is diagonal here, diag(ħ, 0, 0, −ħ), with the eigenvalue 0 twice. S² = S₁² + S₂² + 2S₁ ⋅ S₂ is not diagonal: in S₁ ⋅ S₂ = S₁zS₂z + ½(S₁₊S₂₋ + S₁₋S₂₊) the flip-flop terms give ⟨↓↑|S₁ ⋅ S₂|↑↓⟩ = ħ²/2, so S² links |↑↓⟩ to |↓↑⟩ with the matrix element ħ². The product basis is not its eigenbasis, and Unit 5 finds the one that is. Checks: each 4 × 4 spin matrix has eigenvalues ±ħ/2, each twice, as any spin-½ component must, and Tr S₁z = Tr S_z × Tr 1 = 0.

Wavefunctions, spinors, and where the rule stops

Continuous variables work alike: two distinguishable particles live in L²(ℝ³) ⊗ L²(ℝ³), the functions Ψ(r₁, r₂) on six-dimensional configuration space, with φ(r₁)χ(r₂) only the product special case. Dimensions check: a one-particle ψ(r) carries m⁻³⁄², so Ψ carries m⁻³, as |Ψ|² d³r₁ d³r₂ being a probability demands. A particle with spin is itself a product of its orbital space with the two-state spin space: the Pauli spinor ψ₊(r)|↑⟩ + ψ₋(r)|↓⟩, as in Unit 3's Stern–Gerlach output. Measured check: hydrogen's n = 2 level holds 4 orbital states times 2 spin states, and spectroscopy resolves them into 2s₁/₂, 2p₁/₂ and 2p₃/₂ levels holding 2 + 2 + 4 = 8. The construction itself carries no coupling constant and no order in α: α arrives with the Hamiltonian put on the space, and only a term linking the factors lets one part feel the other, such as the hyperfine coupling of electron and proton spins, whose splitting (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.88 μeV in hydrogen's ground state is quoted from Unit 9 and measured as 5.874 μeV. Two limits remain: for identical particles only the symmetric or antisymmetric part is physical, as Unit 7 shows, and where particles are created or destroyed no fixed product of factors will do.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin 1 polar angle θ₁60 °

Spin 2 polar angle θ₂150 °

Hold θ₁ and sweep θ₂: bars 1–2 and bars 3–4 stay two copies of spin 2's column, scaled by spin 1's fixed entries, so P(spin 1 ↑) never moves. At θ₁ = 360° every bar is the negative of its θ₁ = 0° value: a global phase, the same state.

P(↑↓)0.700

P(SPIN 1 ↑), ANY SPIN 20.750

⟨S₁z + S₂z⟩-0.18 ħ

⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN0.43

Live interpretationP(↑↓): 0.700. P(SPIN 1 ↑), ANY SPIN 2: 0.750. ⟨S₁z + S₂z⟩: −0.18 ħ. ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN: 0.43

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to treat a two-particle state as an ordered pair of kets: to think that giving each particle its own ket specifies the whole system, so that a two-spin column is spin 1's two entries followed by spin 2's, d_A + d_B numbers in all. For two spin-halves the error hides, because 2 + 2 and 2 × 2 are both 4, and it surfaces only when an operator is written down: in the stacked picture S₁ₓ would act on entries 1 and 2, whereas Sₓ ⊗ 1 swaps entries 1 ↔ 3 and 2 ↔ 4. It is wrong because the joint basis is every pair of outcomes, and superposition acts on that basis: (|↑⟩ + |↓⟩)(|↑⟩ + |↓⟩) expands to four terms, not two, and |↑↑⟩ + |↓↓⟩ equals no product of single-spin kets at all. Add a third spin and the counts part, 6 against 8; pair an electron with a deuteron and they are 5 against 6, and the atom shows six levels.

Two spin-halves are written in the ordered basis |↑↑⟩, |↑↓⟩, |↓↑⟩, |↓↓⟩, spin 1's label first. A state has the column (c₁, c₂, c₃, c₄). What column does S₁ₓ, the x-component of spin 1 alone, return?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA ⁶Li atom in its ground state has an electron spin ½ and a nuclear spin I = 1. Find the dimension of its spin space and write |↑⟩|m_I = −1⟩ as a column, electron first. Then, with the electron along +x and the nucleus in √0.50|1⟩ + √0.30|0⟩ + √0.20|−1⟩, find the probability of electron ↓ together with m_I = 0.

Count states: the electron has 2 (mₛ = ±½), the nucleus 2I + 1 = 3 (m_I = 1, 0, −1). The joint space has 2 × 3 = 6 dimensions, not 2 + 3 = 5, and cold-atom experiments do address six ground sublevels of ⁶Li, labelled |1⟩ to |6⟩.
Order the basis electron first: |↑,1⟩, |↑,0⟩, |↑,−1⟩, |↓,1⟩, |↓,0⟩, |↓,−1⟩. Then |↑⟩|m_I = −1⟩ = (1, 0) ⊗ (0, 0, 1) = (0, 0, 1, 0, 0, 0).
The electron along +x is (1, 1)/√2 = (0.7071, 0.7071) and the nucleus is (0.7071, 0.5477, 0.4472). Every joint amplitude is a product of one from each, so the six-column is (0.5000, 0.3873, 0.3162, 0.5000, 0.3873, 0.3162).
Electron ↓ with m_I = 0 is entry 5: P = 0.3873² = 0.150, which is ½ × 0.30, the product of the two separate probabilities.
Checks: the squares sum to 0.25 + 0.15 + 0.10 + 0.25 + 0.15 + 0.10 = 1.00, as ⟨ab|ab⟩ = ⟨a|a⟩⟨b|b⟩ = 1 requires, and P(m_I = 0) = 0.15 + 0.15 = 0.30, the nucleus's own value whatever the electron does.

Answer6 dimensions; |↑⟩|m_I = −1⟩ = (0, 0, 1, 0, 0, 0); P(electron ↓, m_I = 0) = 0.150.

MediumSpin 1 is prepared pointing 60° from +z towards +x, and spin 2 along −x. Build the pair's four-column, write S₁z, S₂z and S₁ₓS₂ₓ as 4 × 4 matrices by Kronecker products, and compute ⟨S₁z + S₂z⟩ and ⟨S₁ₓS₂ₓ⟩ in SI units. Check both against what each spin gives on its own.

Single-spin columns: spin 1 is (cos 30°, sin 30°) = (0.8660, 0.5000), and spin 2 along −x is (1, −1)/√2 = (0.7071, −0.7071).
Kronecker product, spin 1 first: (0.8660 × 0.7071, 0.8660 × (−0.7071), 0.5000 × 0.7071, 0.5000 × (−0.7071)) = (0.6124, −0.6124, 0.3536, −0.3536). The squares 0.375, 0.375, 0.125 and 0.125 sum to 1.
S₁z = S_z ⊗ 1 = (ħ/2) diag(1, 1, −1, −1) and S₂z = 1 ⊗ S_z = (ħ/2) diag(1, −1, 1, −1). So ⟨S₁z⟩ = (ħ/2)(0.375 + 0.375 − 0.125 − 0.125) = ħ/4 and ⟨S₂z⟩ = (ħ/2)(0.375 − 0.375 + 0.125 − 0.125) = 0, giving ⟨S₁z + S₂z⟩ = ħ/4 = 2.64 × 10⁻³⁵ J s.
S₁ₓS₂ₓ = (ħ²/4) σₓ ⊗ σₓ, the matrix with ones on the anti-diagonal, which swaps entries 1 ↔ 4 and 2 ↔ 3. Hence ⟨σₓ ⊗ σₓ⟩ = 2(c₁c₄ + c₂c₃) = 2(−0.2165 − 0.2165) = −0.8660, and ⟨S₁ₓS₂ₓ⟩ = −0.8660 × ħ²/4 = −2.41 × 10⁻⁶⁹ J² s².
Single-spin check: ⟨S₁z⟩ = (ħ/2) cos 60° = ħ/4, and a spin along −x has ⟨S₂z⟩ = 0. Also ⟨S₁ₓ⟩ = (ħ/2) sin 60° = 0.4330ħ and ⟨S₂ₓ⟩ = −ħ/2, whose product −0.2165ħ² is exactly ⟨S₁ₓS₂ₓ⟩, as ⟨ab|A ⊗ B|ab⟩ = ⟨a|A|a⟩⟨b|B|b⟩ requires. The unit J² s² is that of a product of two spin components.

Answer⟨S₁z + S₂z⟩ = ħ/4 = 2.64 × 10⁻³⁵ J s; ⟨S₁ₓS₂ₓ⟩ = −0.866 ħ²/4 = −2.41 × 10⁻⁶⁹ J² s², equal to ⟨S₁ₓ⟩⟨S₂ₓ⟩ as it must be for a product state.

HardA hydrogen atom in its 1s state sits in B = 0.3400 T. Treat its electron and proton spins as uncoupled, H = 2πfₑ S_z ⊗ 1 − 1 ⊗ 2πfₚ S_z, with fₑ = 28.025 GHz T⁻¹ × B and fₚ = 42.577 MHz T⁻¹ × B. Find the four levels as E/h and the electron's spin-flip frequency for each proton state. Measured spectra show the electron resonance of atomic hydrogen as two lines about 51 mT apart. What must H be missing?

Larmor frequencies: fₑ = 28.025 × 0.3400 = 9.5285 GHz and fₚ = 42.577 × 0.3400 = 14.476 MHz. Each term acts on one factor only, so H is diagonal in the product basis and the two energies add.
E/h is +fₑ/2 or −fₑ/2 for electron ↑ or ↓, plus −fₚ/2 or +fₚ/2 for proton ↑ or ↓: |↑↑⟩ 4.7570 GHz, |↑↓⟩ 4.7715 GHz, |↓↑⟩ −4.7715 GHz, |↓↓⟩ −4.7570 GHz. Dimensions: 2πf S_z has units s⁻¹ × J s = J, as an energy must.
Electron flip with the proton ↑: |↑↑⟩ to |↓↑⟩ costs 4.7570 + 4.7715 = 9.5285 GHz. With the proton ↓: |↑↓⟩ to |↓↓⟩ costs 4.7715 + 4.7570 = 9.5285 GHz. One line: in H_A ⊗ 1 + 1 ⊗ H_B the cost of flipping one factor cannot depend on the other. A proton flip costs 14.476 MHz.
The measured doublet, converted at the electron's 28.025 GHz T⁻¹, is 0.051 T × 28.025 GHz T⁻¹ ≈ 1.4 GHz: the electron's flip frequency differs by about 1.4 GHz between the proton's two states, which the uncoupled H forbids.
So H lacks a term linking the factors: the hyperfine coupling, proportional to Sₑ ⋅ Sₚ = SₑₓSₚₓ + SₑySₚy + SₑzSₚz, a sum of products. Quoted from Unit 9, its size is (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.88 μeV, of order α⁴mₑc² like fine structure, but times the small mass ratio mₑ/mₚ.
Check: 5.877 × 10⁻⁶ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 1.421 × 10⁹ Hz, against the measured 21 cm frequency of 1420.4 MHz and the doublet's roughly 1.4 GHz.

AnswerUncoupled levels at ±4.7570 and ±4.7715 GHz; both electron flips cost 9.5285 GHz, so one line. The measured doublet, about 1.4 GHz, needs a coupling Sₑ ⋅ Sₚ between the factors, of size 5.88 μeV, or 1.42 GHz.