Tensor Product Foundations

Tensor products are the composition rule for distinguishable quantum subsystems. This chapter builds the joint Hilbert space, basis, operators, observables, and Hamiltonian before entanglement or reduced-state calculations begin.

Helpful background. State vectors supply the objects being combined, mathematical tensor products define the construction, and subsystem notation fixes the order of factors.

Chapter Map

To learn how to…	Read
specify subsystem labels and tensor notation	Notation and Subsystem Labels
construct the joint state space	Tensor Products of Hilbert Spaces
turn subsystem bases into coordinates	Product Bases
embed and combine operators	Operators on Composite Systems
distinguish measurement locality	Local and Global Observables
fix factor and basis order	Tensor Product Ordering
choose tensor product or direct sum	Direct Sums versus Tensor Products
build the total generator of time evolution	Composite Hamiltonians
analyze how subsystems couple	Interactions and Coupling Terms

Declare the Ordered Factors

For distinguishable subsystems $A$ and $B$, write

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

The labels identify the chosen physical decomposition. They may denote particles, spins, modes, spatial regions, or internal and motional degrees of freedom. The interpretation must be stated.

For finite dimensions $d_A$ and $d_B$,

$$\dim\mathcal H_{AB} = d_A d_B.$$

The factor order $A$ then $B$ controls basis labels and matrix representations. There is a natural swap isomorphism between $\mathcal H_A\otimes\mathcal H_B$ and $\mathcal H_B\otimes\mathcal H_A$, but it is not permission to exchange coordinates silently.

For three factors, declare one order such as

$$\mathcal H_{ABC} = \mathcal H_A\otimes \mathcal H_B\otimes \mathcal H_C.$$

Associativity lets parentheses be suppressed up to canonical isomorphism. Ordering remains operationally important.

Construct the Product Basis

If $\{\lvert a_i\rangle\}$ is an orthonormal basis of $\mathcal H_A$ and $\{\lvert b_j\rangle\}$ is one for $\mathcal H_B$, then

$$\left\{ \lvert a_i\rangle\otimes\lvert b_j\rangle \right\}_{i,j}$$

is an orthonormal product basis of the joint space. Its inner product factorizes:

$$\left( \langle a_i\rvert\otimes\langle b_j\rvert \right) \left( \lvert a_k\rangle\otimes\lvert b_\ell\rangle \right) = \langle a_i\vert a_k\rangle \langle b_j\vert b_\ell\rangle.$$

A general vector expands as

$$\lvert\Psi\rangle = \sum_{i,j}c_{ij} \lvert a_i\rangle\otimes\lvert b_j\rangle.$$

The coefficient array $c_{ij}$ is not automatically a product $u_i v_j$. Factorization of that array is a state property, not a consequence of using a product basis.

Two-qubit ordering

With $A$ written before $B$, the default computational order is

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

Thus

$$\lvert0\rangle_A\otimes\lvert1\rangle_B \longleftrightarrow \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix}.$$

A different code library may use the reverse bit significance. Tensor Product Ordering explains how to document and translate that choice.

Embed Operators

For this section take bounded operators; all finite matrices qualify. An operator $A$ acting only on subsystem $A$ becomes

$$A\otimes I_B$$

on the joint space. Likewise, a $B$-local operator is $I_A\otimes B$.

Product operators obey

$$(A\otimes B)(C\otimes D) = AC\otimes BD,$$

on the completed Hilbert tensor product. Their adjoints satisfy

$$(A\otimes B)^\dagger = A^\dagger\otimes B^\dagger.$$

Operators on disjoint factors commute:

$$[A\otimes I_B, I_A\otimes B] =0.$$

This commutator is an algebraic locality statement. It does not imply that the state has no correlations or that measurements are spacelike separated in a relativistic sense.

Matrix check

In the default two-qubit basis,

$$I\otimes\sigma_z = \begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&1&0\\ 0&0&0&-1 \end{pmatrix}.$$

If a computed matrix instead alternates blocks associated with the first factor, the basis or Kronecker-product order has changed.

Classify the Observable

Locality is relative to the declared factorization.

Type	Form	Example
local to $A$	$A\otimes I_B$	spin component of qubit $A$
local to $B$	$I_A\otimes B$	number operator of mode $B$
product or correlation	$A\otimes B$	$\sigma_z\otimes\sigma_z$
sum of local terms	$A\otimes I_B+I_A\otimes B$	noninteracting energy
interaction	terms not expressible as a sum of one-factor Hamiltonians	exchange or Coulomb term
joint measurement	may require operations beyond separate local measurements	complete Bell-basis measurement

A correlation observable need not be an interaction Hamiltonian. An operator can be measured jointly without generating the state being measured. Keep preparation, dynamics, and measurement roles separate.

Local expectation values depend only on the reduced state:

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(A\otimes I_B) \right] = \operatorname{Tr}_A(\rho_A A),$$

where $\rho_A=\operatorname{Tr}_B\rho_{AB}$. This identity is developed in Local Measurement Statistics.

Assemble the Hamiltonian

A common bipartite Hamiltonian is

$$H = H_A\otimes I_B + I_A\otimes H_B + H_{\mathrm{int}}.$$

The first two terms generate independent dynamics. The interaction term couples factors.

For time-independent self-adjoint subsystem Hamiltonians and $H_{\mathrm{int}}=0$, the self-adjoint sum on the tensor product gives

$$e^{-iHt/\hbar} = e^{-iH_A t/\hbar} \otimes e^{-iH_B t/\hbar}.$$

Local unitary evolution preserves product, separable, and entangled classes, including for mixed states. It can rotate local bases but cannot create entanglement across the same $A|B$ split.

When an interaction is present, factorization usually fails. A useful operator expansion is

$$H_{\mathrm{int}} = \sum_\alpha g_\alpha A_\alpha\otimes B_\alpha.$$

The coefficients $g_\alpha$ set coupling scales; the operators determine selection rules and which states are affected.

Decide Whether the Coupling Entangles

An interaction can be capable of entangling states without entangling every state.

Consider

$$H_{\mathrm{int}} = J\,\sigma_z\otimes\sigma_z.$$

Each computational-basis product state is an eigenstate, so evolution changes only its phase. Starting from $\lvert00\rangle$, this interaction does not generate entanglement.

Starting instead from

$$\lvert+\rangle_A\otimes\lvert+\rangle_B, \qquad \lvert+\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2},$$

the four basis components acquire different phases. At generic times the coefficient matrix no longer has rank one, and the state is entangled.

The initial state and elapsed time both matter. Interactions and Coupling Terms develops the calculation and entangling timescales.

Tensor Product versus Direct Sum

Use a tensor product when systems or degrees of freedom coexist:

$$\mathcal H_A\otimes\mathcal H_B.$$

Use a direct sum for alternatives, sectors, or representation blocks:

$$\mathcal H = \mathcal H_1\oplus\mathcal H_2.$$

Fock space uses both constructions:

$$\mathcal F_\pm(\mathcal h) = \bigoplus_{N=0}^{\infty} \mathcal S_\pm\mathcal h^{\otimes N}.$$

The direct sum separates particle-number sectors. The tensor power constructs $N$ simultaneous one-particle slots before bosonic or fermionic symmetry is imposed.

Infinite-Dimensional Caveat

For Hilbert spaces, the algebraic tensor product is completed in the norm induced by the product inner product. For bounded operators, identities such as $(A\otimes B)(C\otimes D)=AC\otimes BD$ behave as expected.

Unbounded operators require domains. The formal expression

$$H_A\otimes I_B + I_A\otimes H_B$$

does not by itself specify a self-adjoint Hamiltonian. One must state a suitable dense domain, closure, or self-adjoint construction. Domains of Operators and Symmetric versus Self-Adjoint explain the functional-analytic details.

Exercises

Basis and operator order

Using the basis $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$, write the diagonal of $\sigma_z\otimes I$ and $I\otimes\sigma_z$.

Solution

The first operator reads the first bit, so

$$\sigma_z\otimes I = \operatorname{diag}(1,1,-1,-1).$$

The second reads the second bit, so

$$I\otimes\sigma_z = \operatorname{diag}(1,-1,1,-1).$$

Changing the basis order permutes these coordinate matrices even though the abstract operators are unchanged.

Interaction without entanglement

Why does $H_{\mathrm{int}}=J\sigma_z\otimes\sigma_z$ fail to entangle $\lvert01\rangle$?

Solution

The state is an eigenvector:

$$(\sigma_z\otimes\sigma_z) \lvert01\rangle = -\lvert01\rangle.$$

Therefore

$$e^{-iH_{\mathrm{int}}t/\hbar} \lvert01\rangle = e^{iJt/\hbar}\lvert01\rangle.$$

Only an overall phase changes, so the state remains the same product ray. The interaction can still entangle superpositions whose components acquire nonseparable relative phases.

Conserved total excitation

Let

$$H = \hbar\omega\,a^\dagger a + \frac{\hbar\omega}{2}\sigma_z + g(a^\dagger\sigma_-+a\sigma_+).$$

Identify a conserved total-excitation operator suggested by the coupling.

Solution

Take

$$N = a^\dagger a + \sigma_+\sigma_-.$$

The term $a^\dagger\sigma_-$ raises the oscillator occupation while lowering the two-level system, and $a\sigma_+$ does the reverse. Each process preserves the sum, so $[H,N]=0$ on the finite-excitation domain of this rotating-wave model. The commutator also vanishes away from resonance when the two uncoupled frequencies differ: each exchange process still preserves the excitation count. The zero point and energy-offset conventions do not change this commutator.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. F. Werner, “Quantum states with Einstein–Podolsky–Rosen correlations admitting a hidden-variable model,” Physical Review A 40, 4277–4281, 1989.