Operators on Composite Systems

An operator on a composite Hilbert space may act on one subsystem, on both subsystems through a product action, or on the joint system through a sum of such actions. For a bipartite space

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

the basic dictionary is:

$$O_A \longmapsto O_A\otimes I_B, \qquad O_B \longmapsto I_A\otimes O_B.$$

This page explains how to read and build operators on tensor-product Hilbert spaces. In finite-dimensional examples the algebraic identities below are literal matrix identities. For unbounded operators in infinite-dimensional Hilbert spaces, domains must be handled with care.

Helpful background. Product Bases explains the pair of indices used below; Subsystems and Local Observables introduces the measurement interpretation.

Operators Acting on One Subsystem

If $O_A$ acts on $\mathcal H_A$, then its embedded action on the composite space is

$$O_A\otimes I_B.$$

On product vectors,

$$(O_A\otimes I_B) \bigl( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \bigr) = (O_A\lvert\psi\rangle_A) \otimes \lvert\phi\rangle_B.$$

Similarly,

$$(I_A\otimes O_B) \bigl( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \bigr) = \lvert\psi\rangle_A \otimes (O_B\lvert\phi\rangle_B).$$

The identity factor is not decorative. It specifies that the operator does not act on that factor. For a local unitary this also preserves the other subsystem’s reduced state; conditioning on a local measurement result can change the state assigned to a correlated partner, as explained in Conditional States.

Product Operators

A product operator $A\otimes B$ acts on both factors:

$$(A\otimes B) \bigl( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \bigr) = (A\lvert\psi\rangle_A) \otimes (B\lvert\phi\rangle_B).$$

Product operators are building blocks for more general composite operators. In finite dimensions every operator on $\mathcal H_A\otimes\mathcal H_B$ is a finite sum of product operators: expand it in products of local matrix units $|i\rangle\langle k|\otimes|j\rangle\langle l|$. The coefficients are unique after choosing this operator basis. Infinite-dimensional expansions require a specified convergence notion.

Algebraic Identities

The multiplication, adjoint, and commutator identities below hold for bounded operators; the trace identity requires trace-class factors. All these requirements hold automatically for finite matrices.

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

The adjoint obeys

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

For trace-class factors the trace factorizes:

$$\operatorname{Tr}_{AB}(A\otimes B) = \operatorname{Tr}_A(A)\operatorname{Tr}_B(B).$$

The commutator of local operators on different subsystems vanishes:

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

These identities are often the quickest way to check a calculation. They also explain why independent local observables are compatible as tensor-factor observables.

Matrix Elements in a Product Basis

Let

$$A_{ik} = {}_A\langle i\vert A\vert k\rangle_A, \qquad B_{jl} = {}_B\langle j\vert B\vert l\rangle_B.$$

Then

$${}_{AB}\langle ij\vert (A\otimes B) \vert kl\rangle_{AB} = A_{ik}B_{jl}.$$

For a local operator on $A$,

$${}_{AB}\langle ij\vert (A\otimes I_B) \vert kl\rangle_{AB} = A_{ik}\delta_{jl}.$$

The spectator index is preserved by the identity operator.

Sums of Local Operators

Many uncoupled Hamiltonians have the form

$$H_0 = H_A\otimes I_B +I_A\otimes H_B.$$

If

$$H_A\lvert a\rangle_A = E_a\lvert a\rangle_A, \qquad H_B\lvert b\rangle_B = E_b\lvert b\rangle_B,$$

then

$$H_0 \bigl( \lvert a\rangle_A\otimes\lvert b\rangle_B \bigr) = (E_a+E_b) \lvert a\rangle_A\otimes\lvert b\rangle_B.$$

Thus product eigenstates of the local Hamiltonians are product eigenstates of the uncoupled composite Hamiltonian, with additive energies.

Interaction Operators

Relative to a fixed subsystem split, an interaction is a contribution that cannot be absorbed into a sum of one-subsystem Hamiltonians and a constant. Write

$$H = H_A\otimes I_B +I_A\otimes H_B +V_{AB}.$$

A two-qubit coupling might be

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

For spin systems one often writes

$$H_{\text{int}} = \kappa\sum_{\alpha=x,y,z} s_{1,\alpha}\otimes s_{2,\alpha}.$$

Here $s_{1,\alpha},s_{2,\alpha}$ are the single-spin angular-momentum operators. Thus $\kappa$ has units of energy divided by angular momentum squared, whereas $J$ in the Pauli coupling above has units of energy. Such terms can split degeneracies and generate entanglement, but an interaction does not entangle every input at every time.

Global Operators

A global operator is any operator on the full Hilbert space. It need not be local, and it need not be a single product operator. For example, the two-qubit controlled-NOT gate can be written

$$U_{\text{CNOT}} = \lvert0\rangle\langle0\rvert\otimes I +\lvert1\rangle\langle1\rvert\otimes X.$$

This is a sum of product operators. It flips the second qubit in the $|1\rangle$ sector of the first, while preserving coherence between the two control sectors.

For a worked example, apply it to $|+\rangle_A|0\rangle_B$, with $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$:

$$\begin{aligned} U_{\mathrm{CNOT}}|+0\rangle &=\frac{U_{\mathrm{CNOT}}|00\rangle+U_{\mathrm{CNOT}}|10\rangle}{\sqrt2}\\ &=\frac{|00\rangle+|11\rangle}{\sqrt2}. \end{aligned}$$

The output is entangled, whereas the input $|00\rangle$ is unchanged. By contrast, a product operator $A\otimes B$ maps a product vector to another product vector or to zero. Its exponential need not be a product: for Hermitian factors, $e^{-itA\otimes B/\hbar}$ can generate entanglement. If either factor is a scalar multiple of the identity, the evolution is local; even a genuine interaction can leave particular inputs unentangled or factor at special times. See Interactions and Coupling Terms for explicit dynamics.

Operators on Many Subsystems

For

$$\mathcal H = \mathcal H_1\otimes\mathcal H_2\otimes\mathcal H_3,$$

an operator acting only on subsystem $2$ is embedded as

$$I_1\otimes O_2\otimes I_3.$$

A product operator supported on subsystems $1$ and $3$ might be written

$$A_1\otimes I_2\otimes C_3.$$

In spin chains and quantum circuits, it is common to write $O_j$ for the operator that acts as $O$ on site $j$ and as identity elsewhere. This shorthand should be declared before use.

Unbounded composite operators

For unbounded operators the product action initially requires vectors in the relevant domains. In particular, smoothness alone does not settle boundary conditions or self-adjointness; see Hermitian versus Self-Adjoint.

If $H_A,H_B$ are self-adjoint, their spectral projections act on different factors and commute. The joint spectral calculus defines the self-adjoint noninteracting Hamiltonian associated with $H_A\otimes I+I\otimes H_B$. Adding an arbitrary unbounded interaction can require new domain conditions; it does not automatically preserve self-adjointness.

Common pitfalls

Product operators and interaction evolution play different roles. Applying $A\otimes B$ to a product vector does not create entanglement. Evolving under the Hamiltonian $A\otimes B$ can do so because its exponential is generally a sum of product operators.

All terms in a sum need the same state space. Add $H_A\otimes I_B$ to $I_A\otimes H_B$, not the unrelated single-subsystem matrices before embedding them.

Units matter in spin couplings. Pauli matrices are dimensionless; spin angular momenta carry units of $\hbar$. Their coupling constants therefore need different dimensions.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013. Publisher.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapters 2 and 4.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 1. Author manuscript.

Exercises

1. Show that bounded local operators on different factors commute.

Solution

Use the product rule:

$$(A\otimes I)(I\otimes B) = A\otimes B,$$

while

$$(I\otimes B)(A\otimes I) = A\otimes B.$$

The two products are equal, so the commutator is zero.

2. If $H_A$ has eigenvalue $E_a$ on $\lvert a\rangle_A$ and $H_B$ has eigenvalue $E_b$ on $\lvert b\rangle_B$, find the energy of the product state under $H_A\otimes I_B+I_A\otimes H_B$.

Solution

The embedded Hamiltonian gives

$$(H_A\otimes I_B+I_A\otimes H_B) \lvert a\rangle_A\lvert b\rangle_B = (E_a+E_b) \lvert a\rangle_A\lvert b\rangle_B.$$

The energy is $E_a+E_b$.

3. Write the operator that applies $X$ to qubit $3$ in a four-qubit register ordered as $1,2,3,4$.

Solution

The operator is

$$I\otimes I\otimes X\otimes I.$$

The identities are part of the notation: they specify that qubits $1$, $2$, and $4$ are left unchanged.