Composite Hamiltonians

A composite Hamiltonian is an operator on a tensor-product Hilbert space. For a bipartite system,

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

the standard decomposition is

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

The first two terms generate the independent dynamics of the subsystems. The last term couples the factors. It is the source of energy shifts, transitions, bound states, correlations, decoherence, thermalization, and entangling time evolution in many composite models.

This page explains the basic structure for distinguishable tensor factors. The more specialized many-body version, using creation and annihilation operators, belongs to Many-Particle Hamiltonians.

Noninteracting Hamiltonians

The noninteracting Hamiltonian is

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

The identity factors are part of the definition. They say which subsystem is left untouched by each local term.

The two local terms commute:

$$[H_A\otimes I_B,\ I_A\otimes H_B] = 0.$$

For a time-independent finite-dimensional system, this implies

$$\begin{aligned} U_0(t) &= \exp\left(-\frac{i}{\hbar}H_0t\right)\\ &= \exp\left(-\frac{i}{\hbar}H_A t\right) \otimes \exp\left(-\frac{i}{\hbar}H_B t\right). \end{aligned}$$

Thus noninteracting time evolution is a product of local unitaries. A product state remains product:

$$U_0(t) \bigl( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \bigr) = \bigl(U_A(t)\lvert\psi\rangle_A\bigr) \otimes \bigl(U_B(t)\lvert\phi\rangle_B\bigr).$$

For unbounded Hamiltonians in infinite-dimensional spaces, the same formulas are the correct guide but domains and self-adjointness must be handled carefully.

Energy Eigenstates Without Interactions

Suppose

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

Then

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

The noninteracting spectrum is built by adding subsystem energies. The product eigenbasis is

$$\lvert a,b\rangle = \lvert a\rangle_A\otimes\lvert b\rangle_B.$$

Degeneracies require care. If

$$E_a+\epsilon_b = E_{a'}+\epsilon_{b'},$$

then any superposition in the degenerate eigenspace is also an eigenvector. Some of those superpositions may be entangled. The absence of an interaction means there is a product eigenbasis; it does not mean every possible eigenbasis chosen inside a degenerate subspace consists of product states.

Interaction Terms

An interaction term is the part of the Hamiltonian that cannot be written purely as a sum of operators acting on one factor at a time:

$$H_{\mathrm{int}} \ne K_A\otimes I_B + I_A\otimes K_B$$

for any choice of local operators $K_A,K_B$, up to constants and terms absorbed into $H_A,H_B$.

In finite dimensions, many interaction terms can be expanded as sums of product operators:

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

This does not make the interaction local. A single product $A_\mu\otimes B_\mu$ can still couple measurement outcomes or generate phases depending jointly on both subsystems. Locality is about whether the Hamiltonian is a sum of one-factor terms, not whether it can be expanded in a product-operator basis.

The decomposition into “local” and “interaction” terms is sometimes conventional. Constants can be moved between terms, and mean-field approximations can absorb part of an interaction into effective local Hamiltonians. What matters physically is the full operator $H$ and the tensor-factor structure relative to which one asks about local evolution and entanglement.

How Interactions Create Entanglement

An interaction can turn a product state into a nonproduct state because the time-evolution operator need not factor:

$$U(t) = \exp\left(-\frac{i}{\hbar}Ht\right) \ne U_A(t)\otimes U_B(t).$$

For a concrete two-qubit example, let

$$H_{\mathrm{int}} = J\,Z\otimes Z,$$

and start from

$$\lvert++\rangle = \frac12 \bigl( \lvert00\rangle+\lvert01\rangle +\lvert10\rangle+\lvert11\rangle \bigr).$$

With $\theta=Jt/\hbar$,

$$\begin{aligned} e^{-i\theta Z\otimes Z}\lvert++\rangle = \frac12\bigl[ &e^{-i\theta} (\lvert00\rangle+\lvert11\rangle)\\ &+ e^{i\theta} (\lvert01\rangle+\lvert10\rangle) \bigr]. \end{aligned}$$

The coefficient matrix has determinant

$$\frac14 \left( e^{-2i\theta}-e^{2i\theta} \right) = -\frac{i}{2}\sin(2\theta).$$

For $\sin(2\theta)\ne0$, the state is entangled. For special times, such as $\theta=0$ or $\theta=\pi/2$, this particular state returns to product form up to phases.

This example illustrates a general caution: an interaction term does not entangle every product state at every time. If the initial product state is an eigenstate of the interaction, or if symmetry restricts the accessible subspace, the state may remain product.

Conservation Laws in Composite Systems

Composite Hamiltonians often come with additive conserved quantities. If $Q_A$ acts on $A$ and $Q_B$ acts on $B$, the total quantity is

$$Q_{\mathrm{tot}} = Q_A\otimes I_B + I_A\otimes Q_B.$$

It is conserved when

$$[H,Q_{\mathrm{tot}}]=0.$$

For the noninteracting Hamiltonian, conservation follows if $[H_A,Q_A]=0$ and $[H_B,Q_B]=0$. With interactions, the condition becomes a constraint on $H_{\mathrm{int}}$:

$$[H_{\mathrm{int}},Q_{\mathrm{tot}}]=0.$$

This allows the interaction to exchange the quantity between subsystems while preserving the total. For example, an excitation-exchange coupling may not conserve the excitation number of subsystem $A$ or $B$ separately, but it can conserve their sum.

Conserved quantities often block diagonalize the Hamiltonian into sectors. This is useful for diagonalization and for interpreting dynamics. It is also a warning: if dynamics is restricted to a sector, entanglement and correlations should be analyzed within the accessible subspace, not by ignoring the constraint.

Example: Two Spins

A common two-spin Hamiltonian is

$$H = \frac{\hbar\omega_A}{2}Z\otimes I + \frac{\hbar\omega_B}{2}I\otimes Z + J\,Z\otimes Z.$$

All three terms are diagonal in the computational basis, so the basis states $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$ are energy eigenstates. The interaction shifts the energies depending on whether the two $Z$ eigenvalues are aligned or anti-aligned.

A different coupling,

$$H_{\mathrm{int}} = J\,X\otimes X,$$

mixes computational-basis states:

$$X\otimes X\,\lvert00\rangle = \lvert11\rangle, \qquad X\otimes X\,\lvert01\rangle = \lvert10\rangle.$$

Diagonalizing such a Hamiltonian can produce entangled eigenstates.

Example: Coupled Oscillators

For two distinguishable one-dimensional oscillators, a standard Hamiltonian is

$$H = \frac{p_1^2}{2m_1} + \frac12m_1\omega_1^2x_1^2 + \frac{p_2^2}{2m_2} + \frac12m_2\omega_2^2x_2^2 + \frac{k}{2}(x_1-x_2)^2.$$

The first two terms are $H_1\otimes I_2$, the next two are $I_1\otimes H_2$, and the last term is an interaction. Expanding the coupling gives

$$\frac{k}{2}(x_1-x_2)^2 = \frac{k}{2}x_1^2 + \frac{k}{2}x_2^2 - kx_1x_2.$$

The $x_1^2$ and $x_2^2$ pieces can be absorbed into shifted local oscillator frequencies. The cross term $-kx_1x_2$ couples the factors. This example shows why the split between local and interaction terms may depend on convention, while the full Hamiltonian is unambiguous.

Example: Two Particles with a Potential

For two distinguishable particles in three dimensions,

$$H = \frac{\mathbf p_1^2}{2m_1} + \frac{\mathbf p_2^2}{2m_2} + V(\mathbf x_1-\mathbf x_2).$$

The kinetic terms are local with respect to the two-particle tensor product. The potential couples the particle coordinates. In position representation the wavefunction $\Psi(\mathbf x_1,\mathbf x_2)$ may fail to factor even when the potential is simple.

For identical particles, the same symbolic form must be restricted to the symmetric or antisymmetric subspace. That exchange-symmetry issue belongs to the identical-particle chapter.

Common Mistakes

• Dropping identity factors before the subsystem support of each term is clear.
• Thinking that $H_A+H_B$ is literal addition of operators on different Hilbert spaces rather than shorthand for $H_A\otimes I_B+I_A\otimes H_B$.
• Assuming every interaction entangles every product state.
• Forgetting that a noninteracting Hamiltonian can have entangled eigenvectors if one chooses an entangled basis inside a degenerate eigenspace.
• Treating an expansion $\sum_\mu A_\mu\otimes B_\mu$ as proof that the interaction is local.
• Ignoring conserved sectors when interpreting spectra or entanglement.
• Applying distinguishable-particle tensor-product Hamiltonians to identical particles without imposing exchange symmetry.

Cross-Links

• Operators on Composite Systems
• Product Bases
• Tensor Product Ordering
• Local and Global Observables
• Interactions and Coupling Terms
• Product States
• Local Unitary Equivalence
• Marginals and Correlations
• Many-Particle Hamiltonians
• Tensor Product Exercises
• Hamiltonians
• Time-Evolution Operator
• Propagators in Multiple Dimensions

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• 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.
• A. Messiah, Quantum Mechanics, Dover, 1999.

Exercises

1. Let $H_A\lvert0\rangle_A=E_0\lvert0\rangle_A$, $H_A\lvert1\rangle_A=E_1\lvert1\rangle_A$, and $H_B\lvert b\rangle_B=\epsilon_b\lvert b\rangle_B$. What is the energy of $\lvert1\rangle_A\otimes\lvert b\rangle_B$ under $H_A\otimes I_B+I_A\otimes H_B$?

Solution

The noninteracting Hamiltonian gives additive energies:

$$(H_A\otimes I_B+I_A\otimes H_B) \bigl( \lvert1\rangle_A\otimes\lvert b\rangle_B \bigr) = (E_1+\epsilon_b) \lvert1\rangle_A\otimes\lvert b\rangle_B.$$

The energy is $E_1+\epsilon_b$.

2. Show that a noninteracting time-evolution operator preserves product states.

Solution

For

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

the two terms commute, so

$$U_0(t) = e^{-iH_A t/\hbar} \otimes e^{-iH_B t/\hbar}.$$

Applying this to a product vector gives

$$U_0(t) (\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B) = (U_A(t)\lvert\psi\rangle_A) \otimes (U_B(t)\lvert\phi\rangle_B).$$

The result is still a product vector.

3. For $H_{\mathrm{int}}=JZ\otimes Z$, use the determinant of the coefficient matrix to decide when $e^{-iH_{\mathrm{int}}t/\hbar}\lvert++\rangle$ is entangled.

Solution

With $\theta=Jt/\hbar$, the evolved state has coefficient matrix

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

Its determinant is

$$\det C = \frac14 \left( e^{-2i\theta}-e^{2i\theta} \right) = -\frac{i}{2}\sin(2\theta).$$

A two-qubit pure state is product exactly when this coefficient matrix has rank one, which here means $\det C=0$. Therefore the state is entangled when $\sin(2\theta)\ne0$.

4. Suppose $Q_{\mathrm{tot}}=Q_A\otimes I_B+I_A\otimes Q_B$. What condition must an interaction satisfy to conserve $Q_{\mathrm{tot}}$?

Solution

The interaction must commute with the total quantity:

$$[H_{\mathrm{int}},Q_{\mathrm{tot}}]=0.$$

It may still fail to commute with $Q_A\otimes I_B$ or $I_A\otimes Q_B$ separately. In that case the interaction exchanges the quantity between subsystems while preserving the total.