Composite Systems

A composite quantum system is a system modeled in terms of two or more subsystems. The parts may be separate objects, such as two atoms, or distinct degrees of freedom of one object, such as an electron’s position and spin.

For distinguishable subsystems $A$ and $B$, the composition postulate is

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B.$$

The tensor product is the new mathematical rule required by composition. It creates room for independent preparations, joint measurements, correlations, and states that cannot be assigned separate subsystem state vectors. Those nonfactorizing pure states are entangled.

Helpful background. Quantum states supplies the state-vector description, and the Born rule connects states to probabilities. Tensor products develops the basis mechanics used below. Most calculations here are finite-dimensional; the position–spin example illustrates the corresponding Hilbert-space completion.

Systems Made of Subsystems

Calling a model composite means that it identifies parts whose preparations or observables can be discussed separately. The labels $A$ and $B$ can represent:

• two spatially separated spins or particles;
• two qubits in one device;
• position and spin of one particle;
• an atom and a radiation mode;
• a system and a measurement probe;
• a chosen system and its environment.

Subsystems need not be spatial regions, and spatial separation alone is not the definition. What matters is the modeled organization of degrees of freedom and observables.

The subsystem split is additional physical structure. A bare Hilbert space does not always come with one preferred factorization. Statements such as “this state is entangled” or “this operator is local” are meaningful only after specifying the tensor factors relative to which they are being made.

For this chapter, the factors are assumed to be distinguishable and already identified. More subtle decompositions are discussed in Entanglement Depends on Decomposition.

Why the Tensor Product?

Suppose subsystem $A$ can be prepared in $\lvert a\rangle$ and subsystem $B$ in $\lvert b\rangle$. The joint theory must contain a state representing both preparations at once:

$$\lvert a\rangle_A\otimes\lvert b\rangle_B.$$

The construction must respect superposition in each subsystem. For example,

$$\left( \alpha\lvert a_1\rangle + \beta\lvert a_2\rangle \right) \otimes\lvert b\rangle = \alpha\lvert a_1\rangle\otimes\lvert b\rangle + \beta\lvert a_2\rangle\otimes\lvert b\rangle.$$

The same linearity holds in the second factor. The algebraic tensor product is generated by formal symbols $|a\rangle\otimes|b\rangle$ subject to these bilinearity relations. In infinite dimensions, completing this space in the product inner-product norm gives the Hilbert tensor product.

Its inner product satisfies

$$\begin{aligned} \langle a\otimes b \mid a'\otimes b'\rangle &= \langle a\mid a'\rangle \langle b\mid b'\rangle. \end{aligned}$$

Therefore normalized subsystem states produce a normalized joint state, and independent Born probabilities multiply.

Simply storing an ordered pair of subsystem vectors does not describe all joint states. Pairs can themselves form the Hilbert-space direct sum,

$$\mathcal H_A\oplus\mathcal H_B,$$

but that construction describes alternatives or sectors: a vector has one component in the $A$ sector and one in the $B$ sector. It has a different inner product and does not encode simultaneous choices of one basis state from each subsystem.

For finite dimensions, the two constructions scale differently:

$$\begin{aligned} \dim(\mathcal H_A\oplus\mathcal H_B) &= d_A+d_B,\\ \dim(\mathcal H_A\otimes\mathcal H_B) &= d_A d_B. \end{aligned}$$

Thus two qubits have composite dimension $2\cdot2=4$. Their direct-sum dimension also happens to be $2+2=4$, but that numerical equality is accidental; three- and four-level examples immediately distinguish the rules.

Product Bases

Let

$$\left\{ \lvert i\rangle_A \right\}_{i=1}^{d_A}$$

and

$$\left\{ \lvert j\rangle_B \right\}_{j=1}^{d_B}$$

be orthonormal bases. Then

$$\left\{ \lvert i\rangle_A\otimes\lvert j\rangle_B \right\}_{i,j}$$

is an orthonormal basis of the composite space. Orthonormality follows from

$$\begin{aligned} \langle i,j\mid k,\ell\rangle &= \langle i\mid k\rangle_A \langle j\mid\ell\rangle_B\\ &= \delta_{ik}\delta_{j\ell}. \end{aligned}$$

For two qubits, a standard ordering is

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle,$$

where the first slot refers to $A$ and the second to $B$. An ordering convention fixes how pairs $(i,j)$ map to matrix rows and columns. It has no physical content, but inconsistent ordering produces wrong matrices and partial traces. See Tensor-Product Ordering.

A product basis consists of product vectors, but the vectors it spans are not all product states. Superposition creates generic vectors that may not factor.

Pure States of the Whole

A general bipartite pure state can be expanded as

$$\lvert\Psi\rangle_{AB} = \sum_{i,j}c_{ij} \lvert i\rangle_A\otimes\lvert j\rangle_B,$$

with normalization

$$\sum_{i,j}|c_{ij}|^2=1.$$

A product state has the form

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A \otimes \lvert\phi\rangle_B.$$

If

$$\lvert\psi\rangle_A = \sum_i a_i\lvert i\rangle_A, \qquad \lvert\phi\rangle_B = \sum_j b_j\lvert j\rangle_B,$$

then its coefficients factor:

$$c_{ij}=a_i b_j.$$

Equivalently, the coefficient matrix $C=(c_{ij})$ has rank one for a nonzero product state. The dedicated Product States and Bipartite Systems pages develop these tests.

A pure state that cannot be written in product form is entangled. The Bell state

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}$$

is the standard example. Its coefficient matrix has rank two, so no choices of local amplitudes $a_i$ and $b_j$ can reproduce it.

Entanglement is not an extra interaction term or force. It is a property of a joint state relative to a specified subsystem split. Interactions can create entanglement, but a state can remain entangled after the subsystems cease interacting.

Joint Density Operators

A general state of the whole is represented by a density operator

$$\rho_{AB} \ge 0, \qquad \operatorname{Tr}_{AB}\rho_{AB}=1.$$

Three different structures should not be conflated.

A product state has

$$\rho_{AB} = \rho_A\otimes\rho_B.$$

In finite dimensions, a separable state is a convex mixture of product states:

$$\rho_{AB} = \sum_k p_k\, \rho_A^{(k)} \otimes \rho_B^{(k)}.$$

Here $p_k\geq0$ and $\sum_kp_k=1$. Such a state may be a product or may be correlated; an entangled mixed state admits no such decomposition. Infinite-dimensional separability additionally includes trace-norm limits of these mixtures.

For example,

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert$$

is not a product density operator, because its outcomes are correlated. It is nevertheless separable: the displayed expression is already a convex mixture of two product states. Correlation alone is therefore not a sufficient test for entanglement.

The distinction is developed at Classical Correlation Versus Entanglement and Separable Mixed States.

Observables on Subsystems

An observable $M_A$ acting only on subsystem $A$ is embedded into the composite space as

$$M_A\otimes I_B.$$

Similarly, a $B$-local observable is

$$I_A\otimes M_B.$$

The identity factor states that this observable reads only $A$. It does not assert that conditioning on an $A$ measurement leaves the predicted state of a correlated $B$ unchanged; only an unread local trace-preserving operation guarantees that.

Local operators on different factors commute:

$$\begin{aligned} [ M_A\otimes I_B,\, I_A\otimes M_B ] &=0. \end{aligned}$$

Indeed,

$$\begin{aligned} (M_A\otimes I_B) (I_A\otimes M_B) &= M_A\otimes M_B,\\ (I_A\otimes M_B) (M_A\otimes I_B) &= M_A\otimes M_B. \end{aligned}$$

The operator $M_A\otimes M_B$ is a product observable that probes joint correlations. More general joint operators are sums of tensor-product terms. In finite dimensions, operator bases on $A$ and $B$ generate an operator basis on $AB$.

More examples and the no-signaling calculation appear in Subsystems and Local Observables.

Local Measurements and Reduced States

Let $E_a$ be a measurement effect on subsystem $A$. The corresponding effect on the joint system is

$$E_a\otimes I_B.$$

For a joint state $\rho_{AB}$, the probability of outcome $a$ is

$$p(a) = \operatorname{Tr}_{AB}\!\left[ \rho_{AB}(E_a\otimes I_B) \right].$$

All such local probabilities can be reproduced by one operator on $\mathcal H_A$, the reduced state

$$\rho_A = \operatorname{Tr}_B(\rho_{AB}),$$

which is defined by the requirement

$$\begin{aligned} \operatorname{Tr}_{AB}\!\left[ \rho_{AB}(E_a\otimes I_B) \right] &= \operatorname{Tr}_A(\rho_AE_a). \end{aligned}$$

The reduced state contains every prediction available from measurements on $A$ alone. It does not contain all joint correlations with $B$.

For the Bell state $\lvert\Phi^+\rangle$,

$$\rho_A=\rho_B=\frac{I}{2}.$$

The whole is in a pure state, while each subsystem is locally mixed. The mixed local state follows from a known entangled joint preparation, rather than a classical choice of one subsystem ket. Local measurements alone cannot distinguish it from a classical ensemble with the same density operator; joint correlations distinguish the preparations.

See Reduced States and Partial Trace: First Encounter for the calculation.

Composite Hamiltonians

A common Hamiltonian structure is

$$\begin{aligned} H_{AB} &= H_A\otimes I_B + I_A\otimes H_B + V_{AB}. \end{aligned}$$

In finite dimensions with Hermitian terms, the first two terms generate local dynamics and $V_{AB}$ represents interactions. For unbounded Hamiltonians, a self-adjoint total Hamiltonian and appropriate domains must also be specified.

If $V_{AB}=0$, the local Hamiltonian terms commute and the propagator factorizes:

$$U_{AB}(t,t_0) = U_A(t,t_0)\otimes U_B(t,t_0).$$

An initial product state remains a product:

$$U_{AB} \left( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \right) = \left(U_A\lvert\psi\rangle_A\right) \otimes \left(U_B\lvert\phi\rangle_B\right).$$

More generally, local unitaries preserve whether a pure state is entangled and preserve its Schmidt coefficients. They can change local bases, but they cannot create entanglement from a product state.

A nonlocal interaction can create entanglement. For two qubits with local Hamiltonian terms set to zero, take real $J$ and

$$V_{AB} = J\,\sigma_z\otimes\sigma_z.$$

Starting from

$$\lvert{+}\rangle_A\otimes\lvert{+}\rangle_B,$$

the evolved state is

$$\lvert\Psi(t)\rangle = \frac12 e^{-i\theta} \left( \lvert00\rangle+\lvert11\rangle \right) + \frac12 e^{i\theta} \left( \lvert01\rangle+\lvert10\rangle \right),$$

where $\theta=J(t-t_0)/\hbar$. Its coefficient determinant is

$$\det C = -\frac{i}{2}\sin(2\theta).$$

The state is entangled whenever this determinant is nonzero. At special times it refactorizes. This simple example shows how a coupling changes not only local states but the factorization structure of the joint state.

Independent Probabilities and Correlations

For a product state $\rho_A\otimes\rho_B$ and local measurement effects $E_a$ and $F_b$,

$$\begin{aligned} p(a,b) &= \operatorname{Tr}\!\left[ (\rho_A\otimes\rho_B) (E_a\otimes F_b) \right]\\ &= \operatorname{Tr}(\rho_AE_a) \operatorname{Tr}(\rho_BF_b)\\ &= p_A(a)p_B(b). \end{aligned}$$

Product states therefore have factorized statistics for all product measurements.

The converse requires the phrase “for all local measurements.” One factorized distribution in one chosen basis does not prove that the state is a product. Conversely, correlated outcomes do not prove entanglement because separable mixtures can also be correlated.

Entanglement concerns the structure of the state, whereas correlation concerns a selected collection of measurement statistics. The two ideas are related but not interchangeable.

Examples of Subsystem Structure

Two spins

For two distinguishable spin-half particles,

$$\mathcal H_{AB} = \mathbb C^2\otimes\mathbb C^2 \simeq \mathbb C^4.$$

A magnetic field acting only on spin $A$ contributes a term such as

$$H_A\otimes I_B,$$

while a spin-spin coupling has joint terms such as

$$J\,\boldsymbol\sigma_A \mathbin{\boldsymbol\cdot} \boldsymbol\sigma_B.$$

The shorthand on the last line denotes a sum of tensor-product Pauli operators.

Position and spin

For one nonrelativistic spin-half particle,

$$\mathcal H = L^2(\mathbb R^3) \otimes \mathbb C^2.$$

A separated state has the form

$$\lvert\psi\rangle_{\mathrm{space}} \otimes \lvert\chi\rangle_{\mathrm{spin}}.$$

A Stern–Gerlach-type interaction can instead produce

$$\lvert\Psi\rangle = \alpha\lvert\phi_+\rangle \otimes\lvert\uparrow\rangle + \beta\lvert\phi_-\rangle \otimes\lvert\downarrow\rangle.$$

With normalized packets and $|\alpha|^2+|\beta|^2=1$, if both amplitudes are nonzero and the two spatial packets are not proportional, position and spin are entangled even though they belong to one particle.

System and environment

A system $S$ coupled to an environment $E$ is modeled on

$$\mathcal H_S\otimes\mathcal H_E.$$

If a joint pure state becomes entangled, ignoring $E$ leaves a mixed reduced state for $S$. This provides a starting point for decoherence and open-system dynamics. Reduced evolution need not be unitary; describing it by one quantum channel for every system input additionally requires a fixed, initially uncorrelated environment preparation.

Scope and Boundaries

The rule

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B$$

is the correct first framework for distinguishable subsystems and separately modeled degrees of freedom. Several important settings require refinement:

• Identical particles: physical states occupy symmetric or antisymmetric subspaces, and particle labels are not ordinary distinguishable subsystem labels.
• Variable particle number: bosonic or fermionic Fock space organizes sectors with different occupation numbers.
• Gauge theories: constraints can obstruct a naive factorization into spatial regions unless boundary or edge degrees of freedom are handled carefully.
• Quantum field theory: local observable algebras are often more fundamental than a simple tensor factor assigned to each region.
• Infinite families: infinite tensor products require additional analytic choices beyond finite-system notation.

These are not exceptions to linear quantum mechanics; they are warnings that identifying the physically correct subsystems can be subtler than writing labels $A$ and $B$. Continue to Identical Particles and Fock Space for the first extensions.

Common Pitfalls

Confusing a product basis with a product state. The Bell state is a superposition of product-basis vectors but cannot be written as a single tensor product.

Calling every correlation entanglement. A classical mixture of $|00\rangle$ and $|11\rangle$ gives perfectly correlated computational outcomes and is separable. The distinction concerns all possible decompositions of the density operator.

Dropping the subsystem labels. Write identity factors in local operators and state a basis ordering before forming matrices. Entanglement and locality refer to the specified tensor factors.

Requiring a continuing interaction. Local unitaries preserve pure-state Schmidt coefficients. Once created, entanglement can persist after the coupling is switched off.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, chs. 2–3.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, ch. 2.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chs. 1–2.

Exercises

1. Let $\dim\mathcal H_A=2$ and $\dim\mathcal H_B=3$. Compare the dimensions of $\mathcal H_A\otimes\mathcal H_B$ and $\mathcal H_A\oplus\mathcal H_B$, and explain which one describes two simultaneous subsystems.

Solution

The tensor-product dimension is

$$\dim(\mathcal H_A\otimes\mathcal H_B) = 2\cdot3 =6.$$

Its basis contains every simultaneous pair $\lvert i\rangle_A\otimes\lvert j\rangle_B$.

The direct-sum dimension is

$$\dim(\mathcal H_A\oplus\mathcal H_B) = 2+3 =5.$$

A direct sum organizes alternatives or sectors rather than simultaneous independent choices. The tensor product is therefore the composition rule for the two subsystems.

2. Let $\lvert a\rangle$ and $\lvert b\rangle$ be normalized. Prove that $\lvert a\rangle\otimes\lvert b\rangle$ is normalized and that product-measurement probabilities factor.

Solution

The tensor-product inner product gives

$$\begin{aligned} \lVert\lvert a\rangle\otimes\lvert b\rangle\rVert^2 &= \langle a\mid a\rangle \langle b\mid b\rangle\\ &=1. \end{aligned}$$

For local projectors $P$ and $Q$,

$$\begin{aligned} p(P,Q) &= \langle a\otimes b\mid P\otimes Q \mid a\otimes b\rangle\\ &= \langle a\mid P\mid a\rangle \langle b\mid Q\mid b\rangle\\ &= p_A(P)p_B(Q). \end{aligned}$$

The factorization follows from both the state and the measurement being products.

3. Determine whether each two-qubit state is a product state:

$$\begin{aligned} \lvert\Psi_1\rangle &= \frac12( \lvert00\rangle+\lvert01\rangle +\lvert10\rangle+\lvert11\rangle),\\ \lvert\Psi_2\rangle &= \frac{1}{\sqrt2}( \lvert00\rangle+\lvert11\rangle). \end{aligned}$$

Solution

The first state factors:

$$\begin{aligned} \lvert\Psi_1\rangle &= \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \otimes \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}\\ &= \lvert+\rangle\otimes\lvert+\rangle. \end{aligned}$$

It is a product state.

For $\lvert\Psi_2\rangle$, the coefficient matrix is

$$C_2= \frac{1}{\sqrt2} \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$$

It has rank two, whereas a nonzero product state’s coefficient matrix has rank one. Thus $\lvert\Psi_2\rangle$ is entangled.

4. Prove that $M_A\otimes I_B$ commutes with $I_A\otimes M_B$. Then find their product.

Solution

Use

$$(X\otimes Y)(X'\otimes Y') = XX'\otimes YY'.$$

Then

$$\begin{aligned} (M_A\otimes I_B)(I_A\otimes M_B) &= M_A\otimes M_B,\\ (I_A\otimes M_B)(M_A\otimes I_B) &= M_A\otimes M_B. \end{aligned}$$

The two products are equal, so

$$[ M_A\otimes I_B,\, I_A\otimes M_B ] =0.$$

Their product is the joint product operator $M_A\otimes M_B$.

5. For the Bell state $\lvert\Phi^+\rangle$, compute the probabilities of measuring $Z=+1$ and $Z=-1$ on subsystem $A$ only. What local density operator reproduces them in every basis?

Solution

The Bell state is

$$\lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

Measuring $Z$ on $A$ gives $+1$ from the $\lvert00\rangle$ term and $-1$ from the $\lvert11\rangle$ term, each with probability $1/2$.

Tracing out $B$ gives

$$\rho_A = \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert = \frac{I}{2}.$$

The maximally mixed state $I/2$ reproduces all local measurement probabilities, not only the computational-basis probabilities.

6. Consider

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert.$$

Show that it is correlated but separable, and compare it with the Bell-state density operator.

Solution

In the computational basis, the joint outcomes are perfectly correlated:

$$\begin{aligned} p(0,0)=p(1,1)=\frac12, \\ p(0,1)=p(1,0)=0. \end{aligned}$$

Each marginal is uniform, so the joint distribution does not factor into its marginals. The state is correlated.

It is separable because it is explicitly a convex mixture of the product states $\lvert00\rangle$ and $\lvert11\rangle$.

The Bell-state density operator contains coherence terms:

$$\lvert\Phi^+\rangle\langle\Phi^+\rvert = \frac12\bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \bigr).$$

Those off-diagonal joint coherences distinguish the Bell state from $\rho_{\mathrm{cc}}$. The Bell state is entangled; $\rho_{\mathrm{cc}}$ is not.

7. For $V_{AB}=J\sigma_z\otimes\sigma_z$, evolve $\lvert+\rangle\otimes\lvert+\rangle$ for $\theta=J(t-t_0)/\hbar$. At which values of $\theta$ is the resulting pure state a product?

Solution

The states $\lvert00\rangle$ and $\lvert11\rangle$ have eigenvalue $+1$ under $\sigma_z\otimes\sigma_z$, while $\lvert01\rangle$ and $\lvert10\rangle$ have eigenvalue $-1$. Therefore

$$\begin{aligned} \lvert\Psi(t)\rangle &= \frac12 \begin{pmatrix} e^{-i\theta}\\ e^{i\theta}\\ e^{i\theta}\\ e^{-i\theta} \end{pmatrix} \end{aligned}$$

in the standard product ordering. Its coefficient matrix is

$$C= \frac12 \begin{pmatrix} e^{-i\theta}&e^{i\theta}\\ e^{i\theta}&e^{-i\theta} \end{pmatrix}.$$

A two-qubit pure state is a product exactly when $\det C=0$. Here

$$\det C = -\frac{i}{2}\sin(2\theta).$$

Thus the state is a product when

$$\theta=\frac{n\pi}{2}, \qquad n\in\mathbb Z,$$

and is entangled at all other times.

8. Consider the spin-position state

$$\lvert\Psi\rangle = \alpha\lvert\phi_+\rangle\otimes\lvert\uparrow\rangle + \beta\lvert\phi_-\rangle\otimes\lvert\downarrow\rangle.$$

Assume the spatial packets are normalized and $|\alpha|^2+|\beta|^2=1$. Give two distinct conditions under which this state is a product, and explain why the subsystem labels matter.

Solution

The state is a product if one branch is absent, so $\alpha=0$ or $\beta=0$.

It is also a product if the spatial states are proportional:

$$\lvert\phi_-\rangle = e^{i\gamma}\lvert\phi_+\rangle.$$

Then

$$\begin{aligned} \lvert\Psi\rangle &= \lvert\phi_+\rangle \otimes \left( \alpha\lvert\uparrow\rangle + e^{i\gamma}\beta\lvert\downarrow\rangle \right). \end{aligned}$$

Otherwise, with both amplitudes nonzero and linearly independent spatial states, the state is entangled between the position and spin factors.

The conclusion refers specifically to the decomposition

$$L^2(\mathbb R^3)\otimes\mathbb C^2.$$

Entanglement is always stated relative to a chosen subsystem factorization.