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.

Helpful background. Operators on composite systems explains identity factors and product operators; product bases explains paired energy labels.

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:

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

The factorization also holds for self-adjoint, time-independent local Hamiltonians in infinite dimensions: their spectral projections act on different factors and commute. The joint spectral calculus defines the self-adjoint sum and its unitary group. Adding an arbitrary unbounded interaction requires a separate self-adjointness argument; a formal sum alone does not establish well-defined evolution.

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

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

If both local Hamiltonians have complete orthonormal eigenbases, their product eigenbasis has additive energies. Its vectors are

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

Continuous spectra instead require spectral measures or generalized eigenvectors, rather than a countable basis of normalizable energy eigenstates.

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, every interaction 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 $\theta=Jt/\hbar$,

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

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. Under the interaction alone, a product eigenstate only acquires a phase. With local terms present, it remains product only if those terms do not drive it out of a nonentangling family of states.

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

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

For time-independent finite-dimensional operators, 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 $H_{\mathrm{int}}$:

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

Here the local Hamiltonians are assumed to conserve their respective quantities already. Otherwise conservation constrains the full commutator, and different contributions can cancel. For unbounded operators, conservation is expressed by commuting spectral projections or an invariant domain argument, not solely a formal commutator.

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. A fixed sector need not itself factor into subsystem spaces. For entanglement, retain the original subsystem split and impose the sector constraint on the allowed joint states.

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 with $m_i>0$, $\omega_i>0$, and a nonnegative spring constant $k$, a stable 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, equal masses and an exchange-symmetric potential, $V(\mathbf r)=V(-\mathbf r)$, allow restriction to the appropriate symmetric or antisymmetric subspace. See many-particle Hamiltonians for this additional constraint.

Common pitfalls

A noninteracting eigenstate can be entangled. Degenerate product energies allow entangled superpositions within an eigenspace. Independent dynamics nevertheless preserves every initially product state.

An interaction need not entangle the chosen input. A coupling can leave its product eigenvectors unchanged while entangling other inputs. Its action and the preparation must both be specified.

A conserved sector is not a new subsystem split. Restricting the accessible states can simplify diagonalization without changing the tensor factors used to define entanglement.

References

• Brian C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chapter on systems and subsystems. Publisher.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapter 2.
• J. J. Sakurai and Jim Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020. Publisher.

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 the evolved $|++\rangle$ state above, find the reduced state of $A$ and its purity. When is the entanglement maximal?

Solution

Multiplying the coefficient matrix by its adjoint gives

$$\rho_A=\frac12\begin{pmatrix}1&\cos(2\theta)\\\cos(2\theta)&1\end{pmatrix}, \qquad \operatorname{Tr}\rho_A^2=\frac{1+\cos^2(2\theta)}2.$$

The global state is pure, so a maximally mixed qubit marginal identifies maximal bipartite entanglement. This occurs at $\theta=\pi/4+n\pi/2$, when the purity is $1/2$. At $\theta=n\pi/2$ the purity is $1$, agreeing with the product-state determinant test.

4. In finite dimensions, suppose $[H_A,Q_A]=[H_B,Q_B]=0$ and all operators are time independent. Let $Q_{\mathrm{tot}}=Q_A\otimes I_B+I_A\otimes Q_B$. 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.