Operators on Composite Systems

An operator on a composite Hilbert space may act on one subsystem, on several subsystems independently, or on the joint system in a genuinely nonlocal way. 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.

For a shorter Core-level discussion focused on observables and local measurements, see Subsystems and Local Observables.

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 states which part of the composite system is left unchanged.

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. A generic operator on $\mathcal H_A\otimes\mathcal H_B$ can often be expanded as a sum of product operators, though the expansion need not be unique unless a basis of operator space has been chosen.

Algebraic Identities

For compatible finite-dimensional operators,

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

The adjoint obeys

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

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

An interaction operator cannot be assigned to only one subsystem. It couples the factors:

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

A two-qubit coupling might be

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

For spin systems one often writes

$$H_{\text{int}} = J\,\mathbf S_1\cdot\mathbf S_2 = J\sum_{\alpha=x,y,z} S_{1,\alpha}\otimes S_{2,\alpha}.$$

Such terms can split degeneracies, correlate measurement outcomes, and generate entanglement under time evolution.

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 acts conditionally: the second qubit is flipped only in the sector where the first qubit is in the one state.

Global operators are common in time evolution, measurement, error correction, scattering, and effective Hamiltonians. The question to ask is not merely whether an operator is written on $\mathcal H_{AB}$, but which tensor factors it couples.

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

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

An operator coupling subsystems $1$ and $3$ while leaving subsystem $2$ unchanged 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.

Domain Caveat for Infinite Dimensions

For bounded operators, tensor products behave cleanly. For unbounded operators such as position, momentum, and many Hamiltonians, the formal expressions remain useful but domains matter. For example, $H_A\otimes I_B+I_A\otimes H_B$ is not just a symbolic sum: it must be defined on a suitable dense domain and then, when possible, extended to a self-adjoint operator.

Most physics calculations use the formal rules safely in standard bases or on dense domains of smooth wavefunctions. Rigorous pages later in the site will treat the functional-analytic details.

Common Mistakes

• Writing $A+B$ for operators on different subsystems instead of embedding them as $A\otimes I+I\otimes B$.
• Dropping identity factors before the tensor-factor support is clear.
• Assuming every global operator is a product operator.
• Forgetting that interaction terms can generate entanglement even when the initial state is a product.
• Applying finite-dimensional trace identities to unbounded operators without checking domains and trace-class conditions.
• Reversing subsystem order in a Kronecker-product matrix implementation.

Cross-Links

• Tensor Products of Hilbert Spaces
• Product Bases
• Multi-Qubit Gates
• Local and Global Observables
• Subsystems and Local Observables
• Tensor Product Ordering
• Composite Hamiltonians
• Interactions and Coupling Terms
• Tensor Product Exercises
• Product States
• Local Unitary Equivalence
• Notation and Subsystem Labels
• Observables
• Commutators
• Two Spin-1/2 Particles

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1980.

Exercises

1. Show that 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.