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.

Helpful background. Composite Hamiltonians develops local and interaction terms; product states gives the factorization criterion used below.

Local evolution and coupled evolution

Noninteracting evolution factors as $U_A(t)\otimes U_B(t)$, so local unitaries preserve product states. An interaction can instead entangle a product input. The useful test asks how it changes that particular state; failure of the full unitary to factor does not imply that every input becomes entangled.

For a finite-dimensional system, move to the interaction picture to remove the local evolution. At very short times, starting from $\lvert\psi\rangle_A\lvert\phi\rangle_B$,

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

Here $H_{\mathrm{int}}$ is evaluated at the initial time. Terms that change only one factor describe changes of local states. An entangling first-order component is detected by

$$\bigl(I-|\psi\rangle\langle\psi|\bigr)\otimes\bigl(I-|\phi\rangle\langle\phi|\bigr) H_{\mathrm{int}}|\psi,\phi\rangle\ne0.$$

This projection removes both purely local changes. If it vanishes, entanglement can still appear at higher order. For unbounded Hamiltonians, a norm expansion with an $O(t^2)$ remainder requires an appropriate domain, such as $D(H^2)$.

From an entangling amplitude to reduced purity

For a normalized product input, take a finite-dimensional, time-independent Hamiltonian and define

$$Q_A=I_A-|\psi\rangle\langle\psi|,\qquad Q_B=I_B-|\phi\rangle\langle\phi|,\qquad |\chi\rangle=(Q_A\otimes Q_B)H_{\mathrm{int}}|\psi,\phi\rangle.$$

The norm of this component sets the initial loss of reduced-state purity:

$$1-\operatorname{Tr}\rho_A(t)^2 =\frac{2t^2}{\hbar^2}\lVert\chi\rVert^2+O(t^3).$$

To see the coefficient, choose local bases beginning with $|\psi\rangle$ and $|\phi\rangle$. First-order amplitudes that change only one index can be removed by local basis rotations, which preserve purity. The remaining block has entries $-it\chi_{ij}/\hbar$, with both indices orthogonal to the initial factors. Its squared norm is $\epsilon=t^2\lVert\chi\rVert^2/\hbar^2$ to leading order. The largest reduced-state eigenvalue is then $1-\epsilon+O(t^3)$, while the others have total weight $\epsilon+O(t^3)$. Squaring the eigenvalues gives $\operatorname{Tr}\rho_A^2=1-2\epsilon+O(t^3)$. Equivalently, differentiating the reduced density operator gives zero first derivative of purity and second derivative $-4\lVert\chi\rVert^2/\hbar^2$; see Yang, Eqs. (14)–(19).

For a single coupling $H_{\mathrm{int}}=gA\otimes B$, with Hermitian $A,B$ and real $g$,

$$|\chi\rangle =g(A-\langle A\rangle)|\psi\rangle\otimes(B-\langle B\rangle)|\phi\rangle, \qquad \lVert\chi\rVert^2=g^2\operatorname{Var}_{\psi}(A)\operatorname{Var}_{\phi}(B).$$

Both input fluctuations therefore matter. A zero variance makes this quadratic coefficient vanish; local Hamiltonians can still rotate an input and allow entanglement at higher order. For several coupling terms, their projected amplitudes add before the norm is squared, so interference between terms matters. This is a local expansion near $t=0$, not a prediction of indefinite purity loss or an entangling time for every input.

Coupling Terms as Product-Operator Sums

The product-operator expansion writes a finite-dimensional coupling as $H_{\mathrm{int}}=\sum_\mu g_\mu A_\mu\otimes B_\mu$. Each term acts jointly on the two factors. The units of $g_\mu$ depend on the normalization and units of $A_\mu,B_\mu$: dimensionless Pauli matrices take an energy coefficient, while angular-momentum operators require the corresponding powers of $\hbar$.

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

Whether a coupling entangles a state depends on the input. A particularly useful exception occurs when one factor has a definite value of the coupling observable.

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 $B$:

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

Under this interaction alone, the full exponential preserves the product form. The same conclusion holds with local terms if the local evolution preserves the relevant eigenspace. Otherwise a local term can rotate the state away from that eigenspace, and later evolution can entangle it. The corresponding statement applies when $B$ has a definite value in the initial state.

Spin couplings and excitation exchange

An Ising interaction $JZ\otimes Z$, with $J$ in energy units, changes phases according to the two $Z$ labels. Its entangling action on $|++\rangle$ is calculated in composite Hamiltonians. A chain of such couplings with local transverse fields gives the transverse-field Ising model.

Pair flips from an XX coupling

Consider two qubits with

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

where $X,Y,Z$ are Pauli matrices and the real coupling $J$ has energy units. Set $\omega_A=\omega_B=0$ to isolate the interaction. Since $(X\otimes X)^2=I_{AB}$,

$$e^{-iJtX\otimes X/\hbar} =\cos(Jt/\hbar)I_{AB} -i\sin(Jt/\hbar)X\otimes X.$$

Starting from $|00\rangle$, the state therefore becomes

$$|\Psi(t)\rangle =\cos(Jt/\hbar)|00\rangle -i\sin(Jt/\hbar)|11\rangle.$$

When both coefficients are nonzero, the state has Schmidt rank two. Their squared magnitudes are the reduced-state eigenvalues, so equal magnitudes give maximal entanglement. For $J\ne0$, the first positive times are

$$\begin{aligned} t_{\mathrm{ent}}&=\frac{\pi\hbar}{4|J|}, &|\Psi(t_{\mathrm{ent}})\rangle &=\frac{|00\rangle-i\,\operatorname{sgn}(J)|11\rangle}{\sqrt2},\\ t_{\mathrm{prod}}&=\frac{\pi\hbar}{2|J|}, &|\Psi(t_{\mathrm{prod}})\rangle &=-i\,\operatorname{sgn}(J)|11\rangle,\\ t_{\mathrm{return}}&=\frac{\pi\hbar}{|J|}, &|\Psi(t_{\mathrm{return}})\rangle &=-|00\rangle. \end{aligned}$$

The first return to a product state is a transfer to $|11\rangle$ up to phase. Returning to the initial ray takes twice as long. The $J=0$ state remains $|00\rangle$. Local $Z$ terms generally change this evolution and cannot be discarded without the stated zero-frequency assumption.

Excitation exchange from equal XX and YY couplings

A different mechanism preserves the total excitation number. Set

$$H_{\mathrm{ex}}=J(X\otimes X+Y\otimes Y),\qquad J\in\mathbb R.$$

In the one-excitation subspace,

$$H_{\mathrm{ex}}|01\rangle=2J|10\rangle,\qquad H_{\mathrm{ex}}|10\rangle=2J|01\rangle.$$

The terms cancel on $|00\rangle$ and $|11\rangle$. Exponentiating the resulting two-by-two matrix gives

$$e^{-iH_{\mathrm{ex}}t/\hbar}|01\rangle =\cos(2Jt/\hbar)|01\rangle-i\sin(2Jt/\hbar)|10\rangle.$$

The state is maximally entangled when both amplitudes have magnitude $1/\sqrt2$. For $J\ne0$, the first such time is $\pi\hbar/(8|J|)$; complete excitation transfer occurs at $\pi\hbar/(4|J|)$, when the state is product again. Thus entanglement grows and then decreases during a coherent exchange.

The equality of the $XX$ and $YY$ coefficients matters. With coefficients $J_x,J_y$, the matrix element between $|00\rangle$ and $|11\rangle$ is $J_x-J_y$, so excitation number is generally no longer conserved. In the one-excitation subspace the equal $XX$ and $YY$ contributions instead add to $2J$. This doubles the oscillation rate relative to the XX-only pair flip at the same $J$ and halves the first entangling and product-transfer times.

For spin-$1/2$ angular-momentum operators $s_\alpha=\hbar\sigma_\alpha/2$, an isotropic coupling can be written

$$H_{\mathrm{iso}}=\frac{J}{\hbar^2}\sum_{\alpha=x,y,z}s_{1,\alpha}\otimes s_{2,\alpha} =\frac J4(X\otimes X+Y\otimes Y+Z\otimes Z).$$

Here $s_{1,\alpha}$ and $s_{2,\alpha}$ act on their individual spaces and $J$ still has energy units. The singlet and triplet energies are $-3J/4$ and $J/4$, respectively.

Coupled Harmonic Oscillators

For two stable distinguishable oscillators with $k\ge0$, the spring potential $k(x_1-x_2)^2/2$ contains the joint coupling $-kx_1x_2$ and two local frequency shifts. The coupled-oscillator example displays this decomposition.

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

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

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

Choose real $g$ with units of angular frequency and order the factors as atom, then mode. The products abbreviate $a\sigma_+=\sigma_+\otimes a$ and $a^\dagger\sigma_-=\sigma_-\otimes a^\dagger$. 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:

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

The approximation assumes near resonance and coupling matrix elements $|g|\sqrt{n+1}$ small compared with the rapidly oscillating counter-rotating frequency scale; 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

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

For real $\chi$ in angular-frequency units, 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.

The identity

$$|11\rangle\langle11|=\frac14(I-Z)\otimes(I-Z)$$

shows exactly how a controlled phase can be generated from a $Z\otimes Z$ interaction together with local $Z$ rotations and a global phase. These extra phases matter when comparing matrices.

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:

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

In the second ratio, $g$ and the detuning $\Delta$ must both be angular frequencies; in the third, $V$ denotes an energy scale. 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 pitfalls

A coupling constant needs units. Replacing Pauli matrices by spin operators introduces factors of $\hbar/2$. Writing an explicit normalization prevents a wrong energy or timescale.

A product operator is not a product unitary. Although $A\otimes B$ maps a product vector to another product vector, its exponential is generally a sum of such actions and can entangle a superposition.

A change of subsystem variables changes the entanglement question. Normal-mode coordinates can simplify a coupled Hamiltonian while its state remains entangled relative to the original oscillator factors.

References

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapters 2 and 4.
• J. J. Sakurai and Jim Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, angular momentum and time evolution. Publisher.
• Marlan O. Scully and M. Suhail Zubairy, Quantum Optics, Cambridge University Press, 1997, atom–field interaction. Publisher.
• I-Sheng Yang, “The Entanglement Timescale,” Physical Review D 97, 066008 (2018), doi:10.1103/PhysRevD.97.066008, arXiv:1707.05792, Sections II–III, especially Eqs. (14)–(19) — initial purity loss and fluctuations of the coupling.

Exercises

1. Let $A,B$ be finite-dimensional Hermitian operators, $g$ real, $H_{\mathrm{int}}=gA\otimes B$, 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. Factor the controlled-phase unitary into a $ZZ$ evolution, two local $Z$ rotations, and a global phase. Write $\phi=\chi t$.

Solution

All four terms in the projector expansion commute, so

$$U_{\mathrm{CP}}(\phi)=e^{-i\phi/4} \bigl(e^{i\phi Z/4}\otimes e^{i\phi Z/4}\bigr) e^{-i\phi Z\otimes Z/4}.$$

For the three basis states other than $|11\rangle$, the phases cancel. For $|11\rangle$, they multiply to $e^{-i\phi}$. At $\phi=\pi$ this gives $CZ$ exactly, including the displayed global phase.

3. Let $H=JZ\otimes Z$ and start from $(\sqrt3|0\rangle+|1\rangle)_A/2\otimes|+\rangle_B$. Find the leading loss of $A$‘s purity and check it against the exact reduced state. What changes if $A$ instead starts in $|0\rangle$?

Solution

Initially $\langle Z_A\rangle=1/2$, so $\operatorname{Var}(Z_A)=3/4$, while $\operatorname{Var}(Z_B)=1$. With $\theta=Jt/\hbar$, the leading purity loss is therefore $3\theta^2/2$.

Evolving each computational-basis amplitude and tracing out $B$ gives

$$\rho_A(t)=\begin{pmatrix} 3/4&\sqrt3\cos(2\theta)/4\\ \sqrt3\cos(2\theta)/4&1/4 \end{pmatrix},\qquad 1-\operatorname{Tr}\rho_A(t)^2 =\frac38\sin^2(2\theta) =\frac32\theta^2+O(\theta^4).$$

The exact answer also shows recoherence at $\theta=n\pi/2$; the short-time term alone cannot describe that return. For the input $|0\rangle_A|+\rangle_B$, the $A$ variance is zero and this interaction alone leaves the state product for every time: $|0\rangle_A\otimes e^{-i\theta Z}|+\rangle_B$.

4. Let $H=J_xX\otimes X+J_yY\otimes Y$. Find its action on $|00\rangle$ and state when total excitation number is conserved.

Solution

Since $X|0\rangle=|1\rangle$ and $Y|0\rangle=i|1\rangle$,

$$H|00\rangle=(J_x-J_y)|11\rangle.$$

Likewise, $H|11\rangle=(J_x-J_y)|00\rangle$. These transitions change excitation number by two. They vanish exactly when $J_x=J_y$, leaving only transitions within the one-excitation subspace (including the zero-coupling case).