Interactions and Coupling Terms

An interaction term is the part of a composite Hamiltonian that couples tensor factors. For a bipartite system,

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

the interaction term $H_{\mathrm{int}}$ is responsible for dynamics that cannot be reduced to independent evolution on $A$ and $B$.

This page focuses on the mechanism: how couplings convert product evolution into entangling evolution, how coupling constants set scales, and how the same pattern appears in spin systems, oscillators, light-matter models, Coulomb interactions, and controlled-phase gates. The broader Hamiltonian structure is developed in Composite Hamiltonians.

Local Evolution versus Coupled Evolution

If the Hamiltonian is noninteracting,

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

then the time-evolution operator factors:

$$U_0(t) = U_A(t)\otimes U_B(t).$$

Product states remain product. Local unitaries can rotate local bases, add phases, and change local superpositions, but they cannot create entanglement from an initially product state.

With an interaction term,

$$U(t) = \exp\left(-\frac{i}{\hbar}Ht\right)$$

usually does not factor. The resulting time evolution may map some product states to entangled states.

At very short times, starting from $\lvert\psi\rangle_A\lvert\phi\rangle_B$,

$$\lvert\Psi(t)\rangle = \lvert\psi\rangle_A\lvert\phi\rangle_B - \frac{i t}{\hbar} H_{\mathrm{int}} \lvert\psi\rangle_A\lvert\phi\rangle_B + O(t^2),$$

after ignoring purely local terms. Entanglement starts to grow when the interaction pushes the state outside the tangent directions that correspond only to changing $\lvert\psi\rangle_A$ or $\lvert\phi\rangle_B$ separately.

Coupling Terms as Product-Operator Sums

In finite dimensions, a coupling is often written as

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

The operators $A_\mu$ and $B_\mu$ act on the individual factors, while the product $A_\mu\otimes B_\mu$ acts jointly. The constants $g_\mu$ are coupling strengths. Depending on convention, they may have units of energy, angular frequency multiplied by $\hbar$, or an effective low-energy parameter.

Do not read this product-operator expansion as locality. A term such as $Z\otimes Z$ is a joint operator: it assigns phases or energies according to both subsystem labels. It is not a sum of one-subsystem Hamiltonians.

The same physical interaction can be represented differently after basis changes, rotating-frame transformations, normal-mode transformations, or effective-Hamiltonian approximations. The key question is not whether the formula contains tensor-product symbols, but whether the full evolution factors as local dynamics for the chosen subsystem split.

When an Interaction Does Not Entangle

Interactions are often entangling, but they are not magic entanglement dispensers. A product state can remain product under a coupling for several reasons.

If

$$H_{\mathrm{int}} = g\,A\otimes B$$

and $\lvert\psi\rangle_A$ is an eigenstate of $A$ with eigenvalue $a$, then on that input the interaction acts as an effective local Hamiltonian on $B$:

$$H_{\mathrm{int}} \bigl( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \bigr) = \lvert\psi\rangle_A\otimes \bigl(g a B\lvert\phi\rangle_B\bigr).$$

The same is true if $\lvert\phi\rangle_B$ is an eigenstate of $B$. Basis product states can therefore remain product even under a nonlocal coupling. Superpositions are usually more sensitive because the interaction can attach different phases or transitions to different joint labels.

Spin-Spin Couplings

Two-spin models provide the cleanest finite-dimensional examples. An Ising-type coupling is

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

In the computational basis it is diagonal. It assigns one energy to aligned $Z$ eigenvalues and another to anti-aligned eigenvalues. Acting on $\lvert++\rangle$, it produces relative phases that can generate entanglement:

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

Extending this coupling across a chain and adding a noncommuting one-site field gives the Transverse-Field Ising Model.

A transverse exchange-like coupling is

$$H_{\mathrm{int}} = J_x X\otimes X + J_y Y\otimes Y.$$

It can move population between product basis states. For example,

$$X\otimes X\,\lvert01\rangle = \lvert10\rangle.$$

The rotationally symmetric spin-spin coupling is

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

This coupling is diagonal in total-spin sectors rather than in the individual-spin product basis. It separates singlet and triplet energies, making it a standard bridge between tensor products, angular-momentum addition, and exchange physics.

Coupled Harmonic Oscillators

For two distinguishable oscillators, a position-position coupling may arise from

$$V_{\mathrm{int}} = \frac{k}{2}(x_1-x_2)^2.$$

After expanding,

$$V_{\mathrm{int}} = \frac{k}{2}x_1^2 + \frac{k}{2}x_2^2 - kx_1x_2.$$

The first two terms can be absorbed into local oscillator frequencies. The cross term

$$-kx_1x_2$$

is the coupling between the original oscillator factors.

Normal-mode coordinates can diagonalize the quadratic Hamiltonian. That does not mean the original oscillators were never coupled. It means a new set of collective degrees of freedom has been chosen. The ground state may be simple in the normal-mode variables and entangled relative to the original oscillator split.

Jaynes-Cummings Interaction Preview

The Jaynes-Cummings model describes a two-level system coupled to a single oscillator or cavity mode after a rotating-wave approximation. Its interaction has the form

$$H_{\mathrm{JC,int}} = \hbar g \left( a\,\sigma_+ + a^\dagger\sigma_- \right).$$

Here $a$ annihilates a mode excitation, $a^\dagger$ creates one, and $\sigma_+,\sigma_-$ raise and lower the two-level system. The coupling exchanges one excitation between the mode and the two-level system:

$$a^\dagger\sigma_-\lvert e,n\rangle = \sqrt{n+1}\,\lvert g,n+1\rangle,$$

while

$$a\sigma_+\lvert g,n+1\rangle = \sqrt{n+1}\,\lvert e,n\rangle.$$

The total excitation number is conserved by the interaction:

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

This is only a preview. The approximation assumes a near-resonant weak coupling and discards counter-rotating terms; see the Rotating-Wave Approximation for the method behind that simplification.

Coulomb Interaction Between Particles

For two distinguishable charged particles, the nonrelativistic Coulomb interaction is

$$V_C(\mathbf x_1,\mathbf x_2) = \frac{q_1q_2} {4\pi\epsilon_0\lvert\mathbf x_1-\mathbf x_2\rvert}.$$

It is not a sum of a function of $\mathbf x_1$ and a function of $\mathbf x_2$. It couples the coordinates through their separation. The full Hamiltonian

$$H = \frac{\mathbf p_1^2}{2m_1} + \frac{\mathbf p_2^2}{2m_2} + V_C(\mathbf x_1,\mathbf x_2)$$

can be transformed to center-of-mass and relative coordinates for an isolated two-body problem. That transformation simplifies the problem, but it does not make the Coulomb potential a local one-particle term in the original particle split.

In many-particle language the same interaction becomes a two-body operator in field or mode variables. The detailed second-quantized form belongs to Two-Body Operators, and scattering from a long-range Coulomb potential belongs to Coulomb Scattering.

Controlled-Phase Gate Preview

In quantum-circuit language, a controlled-phase gate is a unitary that applies a phase only to one joint basis state. A simple Hamiltonian that generates such a phase is

$$H_{\mathrm{CP}} = \hbar\chi\, \lvert11\rangle\langle11\rvert.$$

Its time evolution is

$$U(t) = \operatorname{diag} \left( 1,1,1,e^{-i\chi t} \right)$$

in the basis $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$. For $\chi t=\pi$, this is the controlled-$Z$ gate:

$$CZ = \operatorname{diag}(1,1,1,-1).$$

Applied to $\lvert++\rangle$, it produces

$$CZ\lvert++\rangle = \frac12 \bigl( \lvert00\rangle+\lvert01\rangle +\lvert10\rangle-\lvert11\rangle \bigr),$$

which is entangled. This is the same algebra behind the edge operation in Graph States.

Controlled-phase gates can also be generated from $Z\otimes Z$ interactions together with local $Z$ rotations and global phases. Hardware implementations differ, but the mathematical point is stable: a phase that depends on a joint label is a coupling operation.

Coupling Strengths and Timescales

A coupling strength is not meaningful without its units and frame. One often sees:

• $J$ as an energy scale in spin Hamiltonians;
• $g$ as a field-theory, oscillator, or cavity coupling;
• $\chi$ as an angular-frequency scale for phase accumulation;
• $U$ as an on-site Hubbard interaction energy;
• $q_1q_2/(4\pi\epsilon_0)$ as the Coulomb strength before the distance dependence is included.

The relevant dimensionless quantity is often a phase or ratio:

$$\theta = \frac{Jt}{\hbar}, \qquad \frac{g}{\Delta}, \qquad \frac{V}{\Delta E}.$$

Small couplings can dominate near resonance or over long times. Large couplings can sometimes be simplified by diagonalizing a different zeroth-order Hamiltonian. The words “weak” and “strong” are therefore relative to the timescale, detuning, temperature, bandwidth, or energy gap in the problem.

Common Mistakes

• Treating every interaction as automatically entangling every product state.
• Calling $A\otimes B$ local merely because it is a product operator.
• Forgetting that a coupling may conserve a total quantity while changing each subsystem’s share.
• Confusing a basis change that diagonalizes a quadratic Hamiltonian with the absence of coupling in the original variables.
• Treating Jaynes-Cummings coupling as exact outside the rotating-wave and near-resonant regime.
• Ignoring the long-range character of the Coulomb interaction.
• Forgetting local phases when comparing $Z\otimes Z$ time evolution with a controlled-phase gate.
• Treating a fitted effective coupling constant as a universal microscopic constant.

Cross-Links

• Composite Hamiltonians
• Operators on Composite Systems
• Product States
• Local Unitary Equivalence
• Entanglement in Quantum Information
• Graph States
• Two-Body Operators
• Many-Particle Hamiltonians
• Transverse-Field Ising Model
• Rotating-Wave Approximation
• Coulomb Scattering
• Time-Evolution Operator

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. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.

Exercises

1. Let $H_{\mathrm{int}}=gA\otimes B$ and suppose $A\lvert a\rangle=a\lvert a\rangle$. Show that this interaction alone cannot entangle an initial state $\lvert a\rangle\otimes\lvert\phi\rangle$.

Solution

On this input,

$$H_{\mathrm{int}} (\lvert a\rangle\otimes\lvert\phi\rangle) = \lvert a\rangle\otimes (g a B\lvert\phi\rangle).$$

The time-evolution operator generated by this interaction acts as

$$e^{-igt A\otimes B/\hbar} (\lvert a\rangle\otimes\lvert\phi\rangle) = \lvert a\rangle\otimes e^{-i g a B t/\hbar}\lvert\phi\rangle.$$

The result remains a product state.

2. Show that $H_{\mathrm{CP}}=\hbar\chi\lvert11\rangle\langle11\rvert$ generates a controlled-$Z$ gate when $\chi t=\pi$.

Solution

The projector $\lvert11\rangle\langle11\rvert$ has eigenvalue $1$ on $\lvert11\rangle$ and eigenvalue $0$ on the other three computational-basis states. Therefore

$$e^{-iH_{\mathrm{CP}}t/\hbar} = \operatorname{diag} \left( 1,1,1,e^{-i\chi t} \right).$$

If $\chi t=\pi$, the last entry is $-1$, so the unitary is

$$\operatorname{diag}(1,1,1,-1) = CZ.$$

3. In the Jaynes-Cummings interaction, verify that $a^\dagger\sigma_-$ maps $\lvert e,n\rangle$ to a state with the same total excitation number.

Solution

The state $\lvert e,n\rangle$ has one excitation in the two-level system and $n$ mode excitations, so its total excitation number is $n+1$.

The operator $\sigma_-$ lowers the two-level system:

$$\sigma_-\lvert e\rangle = \lvert g\rangle.$$

The operator $a^\dagger$ creates one mode excitation:

$$a^\dagger\lvert n\rangle = \sqrt{n+1}\lvert n+1\rangle.$$

Thus

$$a^\dagger\sigma_-\lvert e,n\rangle = \sqrt{n+1}\lvert g,n+1\rangle.$$

The output has zero two-level excitation and $n+1$ mode excitations, again totaling $n+1$.

4. For two oscillators with coupling $\frac{k}{2}(x_1-x_2)^2$, identify the part that couples the original oscillator factors.

Solution

Expand the potential:

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

The $x_1^2$ and $x_2^2$ pieces can be treated as local frequency shifts. The cross term

$$-kx_1x_2$$

depends on both coordinates and couples the original oscillator factors.