Time-Dependent Hamiltonians

This is the canonical treatment of explicitly time-dependent Hamiltonians, from two-time propagators and time ordering through pulses, driven systems, energy balance, and periodic evolution. The postulate-level entry point is Time-Dependent Hamiltonians: Core First Encounter.

A time-dependent Hamiltonian $H(t)$ has explicit dependence on the time parameter. The state still obeys

$$i\hbar\frac{d}{dt}\lvert\psi(t)\rangle = H(t)\lvert\psi(t)\rangle,$$

but the solution is not generally

$$\exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right].$$

The obstacle is operator ordering: $H(t_1)$ and $H(t_2)$ may not commute. Evolution remains unitary for a well-posed closed system with self-adjoint $H(t)$, but its propagator must remember which generator acted first.

Explicit Time Dependence

Writing $H(t)$ means that the rule generating evolution changes explicitly with the clock parameter. Typical models include

$$H(t) = H_0+\lambda(t)V,$$

where an externally prescribed protocol $\lambda(t)$ changes a field, trap position, boundary condition, coupling, or applied force.

Examples include:

• a spin in a magnetic field whose magnitude or direction is controlled in time;
• a particle in a potential $V(x,t)$;
• a qubit driven by a microwave pulse;
• a trap frequency or lattice depth that is ramped or quenched;
• a piecewise pulse sequence used for control;
• an effective Hamiltonian obtained after eliminating other degrees of freedom.

Explicit time dependence of $H(t)$ is different from the ordinary time dependence of $\lvert\psi(t)\rangle$. A state generally evolves even when $H$ is constant.

Treating $\lambda(t)$ as a prescribed classical function makes the modeled quantum system driven but still closed at the level of its state evolution. Energy may enter or leave through the external protocol. If the source of the drive is instead included as a quantum degree of freedom, a larger time-independent Hamiltonian may describe the combined system and source.

At each time, $H(t)$ should be self-adjoint on suitable domains if it is to generate unitary evolution. In infinite dimensions, pointwise self-adjointness alone is not always enough: the domains and time dependence must be regular enough for a propagator to exist.

Two-Time Propagator

The evolution operator is defined by

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi(t_0)\rangle.$$

It satisfies

$$\begin{aligned} i\hbar \frac{\partial U(t,t_0)}{\partial t} &= H(t)U(t,t_0),\\ U(t_0,t_0)&=I. \end{aligned}$$

Composition and reversal retain their usual forms:

$$U(t_2,t_0) = U(t_2,t_1)U(t_1,t_0),$$

and, for unitary evolution,

$$U(t_0,t) = U^\dagger(t,t_0).$$

The adjoint equation gives

$$\frac{d}{dt} \left( U^\dagger U \right) = \frac{i}{\hbar} U^\dagger \left( H^\dagger-H \right) U.$$

Thus a self-adjoint $H(t)$ preserves $U^\dagger U=I$. Time dependence does not by itself imply dissipation or nonunitarity.

The practical challenge is constructing $U(t,t_0)$.

Integral Equation and Chronological Products

Integrating the operator Schrödinger equation gives

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t} H(s)U(s,t_0)\,ds.$$

This Volterra integral equation already contains the ordering information because the unknown propagator appears to the right of $H(s)$.

Divide the interval into short steps

$$t_0<t_1<\cdots<t_N=t.$$

For a sufficiently regular Hamiltonian,

$$U(t_{j+1},t_j) = I-\frac{i}{\hbar}H(t_j)\Delta t +O(\Delta t^2).$$

Composition produces

$$\begin{aligned} U(t_N,t_0) &\approx e^{-iH(t_{N-1})\Delta t/\hbar} \cdots\\ &\quad\times e^{-iH(t_1)\Delta t/\hbar} e^{-iH(t_0)\Delta t/\hbar}. \end{aligned}$$

The earliest factor acts first and therefore appears on the right. If different factors fail to commute, they cannot be collapsed into an ordinary exponential of the integrated Hamiltonian.

When the Ordinary Exponential Works

A sufficient condition is

$$[H(t_1),H(t_2)]=0$$

for every pair of times in the interval. Then

$$U(t,t_0) = \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right].$$

A useful family is

$$H(t) = f(t)A+g(t)I,$$

where $A$ is one fixed self-adjoint operator. All terms commute, so

$$\begin{aligned} U(t,t_0) &= e^{-iG(t,t_0)/\hbar} e^{-iF(t,t_0)A/\hbar},\\ F(t,t_0) &= \int_{t_0}^{t}f(s)\,ds,\\ G(t,t_0) &= \int_{t_0}^{t}g(s)\,ds. \end{aligned}$$

The $g(t)I$ term contributes a global phase. The $f(t)A$ term changes only the amount of rotation generated by the fixed operator $A$.

Pairwise commutation is sufficient, not the only route to an exact solution. Symmetries, rotating frames, Lie-algebra closure, and specially shaped protocols can also make a driven problem solvable without making all $H(t)$ commute.

Time Ordering and the Dyson Series

Iterating the integral equation gives

$$U(t,t_0) = I+U^{(1)}+U^{(2)}+\cdots,$$

where

$$U^{(1)} = -\frac{i}{\hbar} \int_{t_0}^{t}dt_1\,H(t_1),$$

and

$$\begin{aligned} U^{(2)} &= \left(-\frac{i}{\hbar}\right)^2 \int_{t_0}^{t}dt_1\\ &\quad\times \int_{t_0}^{t_1}dt_2\, H(t_1)H(t_2). \end{aligned}$$

The nested limits enforce $t_1\ge t_2$, so the later-time Hamiltonian appears to the left. The compact notation is

$$U(t,t_0) = \mathcal T \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right].$$

The time-ordering symbol $\mathcal T$ is bookkeeping for noncommuting products, not an additional physical interaction.

A truncated Dyson series need not be exactly unitary. Other expansions, including the Magnus expansion, organize the approximation as an exponential and can preserve unitarity order by order when used consistently. Those methods belong to the advanced approximation pages.

Driven Two-Level Systems

Any traceless two-level Hamiltonian can be written

$$H(t) = \frac{\hbar}{2} \boldsymbol\Omega(t) \mathbin{\boldsymbol\cdot} \boldsymbol\sigma,$$

up to a multiple of the identity. Here $\boldsymbol\Omega(t)$ acts as an effective field in Bloch-vector space.

Using the Pauli identity

$$\left[ \boldsymbol a\mathbin{\boldsymbol\cdot}\boldsymbol\sigma, \boldsymbol b\mathbin{\boldsymbol\cdot}\boldsymbol\sigma \right] = 2i \left( \boldsymbol a\times\boldsymbol b \right) \mathbin{\boldsymbol\cdot}\boldsymbol\sigma,$$

one obtains

$$\begin{aligned} [H(t_1),H(t_2)] &= \frac{i\hbar^2}{2} \left[ \boldsymbol\Omega(t_1) \times \boldsymbol\Omega(t_2) \right]\\ &\quad\mathbin{\boldsymbol\cdot} \boldsymbol\sigma. \end{aligned}$$

Hamiltonians at different times commute when the effective-field directions are parallel or antiparallel. Changing only the magnitude of a fixed-direction field is the commuting case; changing its direction generally creates time-ordering effects.

This geometric criterion underlies driven spins, nuclear magnetic resonance, coherent qubit control, and rotating-frame methods. Exact Rabi solutions and rotating-wave approximations require additional assumptions and live in the later Dynamics and Approximation volumes.

Instantaneous Eigenstates

At each time, one may solve

$$H(t)\lvert n(t)\rangle = E_n(t)\lvert n(t)\rangle.$$

It is tempting to use

$$\lvert\psi(t)\rangle \stackrel{?}{=} \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}E_n(s)\,ds \right] \lvert n(t)\rangle.$$

Differentiation exposes the missing term:

$$\begin{aligned} i\hbar\frac{d}{dt}\lvert\psi(t)\rangle &= E_n(t)\lvert\psi(t)\rangle\\ &\quad+ i\hbar e^{-(i/\hbar)\int_{t_0}^{t}E_n(s)\,ds} \lvert\dot n(t)\rangle. \end{aligned}$$

The second line is not generally parallel to $\lvert n(t)\rangle$. Its components along other instantaneous eigenspaces produce transitions.

In a complete instantaneous basis, define the connection matrix

$$\mathcal A_{mn}(t) = i\langle m(t)\vert\dot n(t)\rangle.$$

Off-diagonal elements encode basis-change couplings. Slow variation can suppress them under the hypotheses of the adiabatic approximation, while the diagonal part contributes a geometric phase. Neither result follows merely from diagonalizing $H(t)$ at each instant.

See Adiabatic Approximation for the controlled approximation.

Sudden Changes and Pulse Sequences

An idealized sudden quench changes the Hamiltonian from $H_i$ to $H_f$ over a time too short for substantial state evolution. The state vector is taken as continuous across the quench:

$$\lvert\psi(0^+)\rangle \approx \lvert\psi(0^-)\rangle.$$

What changes immediately is the energy basis used to analyze that state. If the pre-quench state is $\lvert E_i\rangle$, the probability of post-quench energy $E_n^{(f)}$ is

$$p_n^{(f)} = \left| \langle E_n^{(f)}\vert E_i\rangle \right|^2.$$

For $t>0$, the state evolves under $H_f$. The sudden approximation is a limiting model with a timescale criterion, not an assertion that arbitrary rapid changes are physically harmless.

For a sequence of piecewise-constant pulses,

$$H(t)=H_j \quad \text{on interval }j,$$

the total propagator is an ordered product:

$$U = e^{-iH_N\tau_N/\hbar} \cdots e^{-iH_2\tau_2/\hbar} e^{-iH_1\tau_1/\hbar}.$$

Pulse order matters whenever the generators do not commute. This is both a source of control and a common source of algebraic mistakes.

The full approximation is Sudden Approximation.

Energy and Work

For a closed system evolving under $H(t)$,

$$\frac{d}{dt} \langle H(t)\rangle = \left\langle \frac{\partial H(t)}{\partial t} \right\rangle$$

when the derivatives and domains are well defined. The commutator term vanishes because $[H(t),H(t)]=0$ at the same time.

For a parameterized drive $H(\lambda(t))$,

$$\frac{d}{dt}\langle H\rangle = \dot\lambda \left\langle \frac{\partial H}{\partial\lambda} \right\rangle.$$

The right side is the instantaneous power delivered by the prescribed drive in this closed driven-system model. A changing energy expectation is compatible with unitary evolution because the modeled system is not isolated from the external work source.

If the drive source is included quantum mechanically, energy accounting must be performed for the enlarged system. The explicit time dependence can then emerge from an approximation in which source depletion and backreaction are neglected.

The general expectation-value identity is derived in Conservation Laws.

Periodic Driving

If

$$H(t+T)=H(t),$$

the one-period propagator

$$U_F=U(t_0+T,t_0)$$

is called the Floquet operator. Evolution over an integer number of periods satisfies

$$U(t_0+nT,t_0)=U_F^n.$$

Its eigenphases define quasienergies modulo $2\pi\hbar/T$. This is a preview only: quasienergies are not ordinary energy eigenvalues, and micromotion within each period still matters. The canonical entry point is Periodic Hamiltonians.

Closed and Open Descriptions

A self-adjoint $H(t)$ can generate perfectly unitary dynamics despite explicit driving. This is distinct from an open-system master equation, a stochastic Hamiltonian after averaging, or an effective non-self-adjoint Hamiltonian describing loss or postselection.

If classical noise is represented by a random $H_\xi(t)$, each realization may evolve unitarily:

$$\rho_\xi(t) = U_\xi(t)\rho(0)U_\xi^\dagger(t).$$

The ensemble-averaged state

$$\overline{\rho(t)}$$

need not be related to $\rho(0)$ by one unitary. Averaging over unknown controls creates a quantum channel, not a single closed-system trajectory.

A Calculation Workflow

1. Identify the protocol. State which parameters depend on time and what physical source controls them.
2. Check self-adjointness and domains. In finite dimensions, Hermiticity is straightforward; differential Hamiltonians need boundary and domain care.
3. Test pairwise commutation. If $[H(t_1),H(t_2)]=0$ throughout the interval, use the ordinary integral exponential.
4. Exploit structure. Look for fixed generators, piecewise-constant intervals, symmetries, or a useful rotating frame.
5. Preserve order. For pulses and numerical steps, put later-time factors on the left.
6. Choose an approximation deliberately. Dyson, Magnus, adiabatic, sudden, rotating-wave, and numerical methods have different hypotheses and error structures.
7. Check invariants. Verify $U^\dagger U=I$ for closed evolution and monitor norm or trace in calculations.
8. Track energy accounting. Explicit driving can change $\langle H(t)\rangle$ even when the dynamics is unitary.

Common Mistakes

• Writing an ordinary exponential of $\int H(t)\,dt$ without checking operator order.
• Assuming time dependence automatically means nonunitary evolution.
• Confusing explicit dependence of $H(t)$ with the evolution of the state.
• Reversing pulse factors so that the earliest operation appears on the left.
• Treating instantaneous energy eigenvectors as exact dynamical solutions.
• Calling a sudden quench adiabatic, or applying either approximation without a timescale criterion.
• Assuming energy must be conserved in the presence of an external drive.
• Forgetting that a truncated Dyson series is not exactly unitary.
• Treating quasienergy as an ordinary unique energy.
• Averaging random unitary trajectories and assuming the average is still one unitary trajectory.
• Ignoring time-dependent domains for unbounded Hamiltonians.

Cross-Links

• Hamiltonians
• Time-Evolution Operator
• Unitary Time Evolution
• Energy Eigenstates
• Conservation Laws
• Time Ordering
• Interaction Picture
• Adiabatic Approximation
• Sudden Approximation
• Periodic Hamiltonians

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977, vol. 2, chs. 12–13.
• A. Messiah, Quantum Mechanics, Dover, 1999, vol. 2, chs. 16–17.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chs. 2 and 5.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977, secs. 40–53.
• S. Blanes, F. Casas, J. A. Oteo, and J. Ros, “The Magnus expansion and some of its applications,” Physics Reports 470, 151–238 (2009).
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer (1994).

Exercises

1. Let

$$H(t)=f(t)A+g(t)I$$

for a fixed self-adjoint operator $A$. Derive the exact propagator.

Solution

Every Hamiltonian is a linear combination of the same two commuting operators $A$ and $I$, so

$$[H(t_1),H(t_2)]=0.$$

Define

$$\begin{aligned} F(t,t_0) &= \int_{t_0}^{t}f(s)\,ds,\\ G(t,t_0) &= \int_{t_0}^{t}g(s)\,ds. \end{aligned}$$

The ordinary integral exponential is valid:

$$\begin{aligned} U(t,t_0) &= \exp\!\left[ -\frac{i}{\hbar} \left( FA+GI \right) \right]\\ &= e^{-iG/\hbar}e^{-iFA/\hbar}. \end{aligned}$$

2. For

$$H(t) = \frac{\hbar}{2} \boldsymbol\Omega(t) \mathbin{\boldsymbol\cdot} \boldsymbol\sigma,$$

derive $[H(t_1),H(t_2)]$ and state the geometric commutation condition.

Solution

Using the Pauli-vector commutator,

$$\left[ \boldsymbol a\mathbin{\boldsymbol\cdot}\boldsymbol\sigma, \boldsymbol b\mathbin{\boldsymbol\cdot}\boldsymbol\sigma \right] = 2i \left( \boldsymbol a\times\boldsymbol b \right) \mathbin{\boldsymbol\cdot}\boldsymbol\sigma,$$

gives

$$\begin{aligned} [H(t_1),H(t_2)] &= \frac{i\hbar^2}{2} \left[ \boldsymbol\Omega(t_1) \times \boldsymbol\Omega(t_2) \right]\\ &\quad\mathbin{\boldsymbol\cdot} \boldsymbol\sigma. \end{aligned}$$

The commutator vanishes when the two effective-field vectors are parallel or antiparallel, including the case in which either vector is zero.

3. A system evolves under $H_x=(\hbar\Omega/2)\sigma_x$ for time $\tau$, followed by $H_z=(\hbar\Omega/2)\sigma_z$ for the same duration. Write the total propagator. Why can the order not generally be reversed?

Solution

The $x$ pulse acts first, so it appears on the right:

$$U = e^{-i\Omega\tau\sigma_z/2} e^{-i\Omega\tau\sigma_x/2}.$$

The generators do not commute:

$$[\sigma_z,\sigma_x]=2i\sigma_y.$$

Therefore the two exponential factors generally fail to commute. Reversing them describes a $z$ pulse followed by an $x$ pulse, which is a different rotation sequence on the Bloch sphere.

4. Show directly that a self-adjoint $H(t)$ generates unitary evolution whenever the propagator exists.

Solution

The propagator equations imply

$$\dot U = -\frac{i}{\hbar}H(t)U, \qquad \dot U^\dagger = \frac{i}{\hbar}U^\dagger H(t).$$

Hence

$$\begin{aligned} \frac{d}{dt} \left(U^\dagger U\right) &= \dot U^\dagger U + U^\dagger\dot U\\ &= \frac{i}{\hbar}U^\dagger HU - \frac{i}{\hbar}U^\dagger HU\\ &=0. \end{aligned}$$

Since $U(t_0,t_0)=I$, one has $U^\dagger U=I$ throughout the interval.

5. Derive

$$\frac{d}{dt}\langle H(t)\rangle = \left\langle \frac{\partial H}{\partial t} \right\rangle$$

for a normalized state obeying the time-dependent Schrödinger equation.

Solution

Differentiate the expectation value:

$$\begin{aligned} \frac{d}{dt}\langle H\rangle &= \langle\dot\psi\vert H\vert\psi\rangle + \left\langle \frac{\partial H}{\partial t} \right\rangle\\ &\quad+ \langle\psi\vert H\vert\dot\psi\rangle. \end{aligned}$$

Using

$$\lvert\dot\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle$$

and its adjoint, the first and third terms cancel. The remaining term is the explicit derivative of the Hamiltonian.

6. A qubit is initially in the $+1$ eigenstate of $\sigma_z$. At $t=0$, the Hamiltonian is suddenly changed from

$$H_i=\frac{\hbar\Omega}{2}\sigma_z$$

to

$$H_f=\frac{\hbar\Omega}{2}\sigma_x.$$

Find the two post-quench energy probabilities and the subsequent state.

Solution

The state is continuous through the ideal sudden quench:

$$\lvert\psi(0^+)\rangle=\lvert0\rangle.$$

The eigenstates of $H_f$ are $\lvert+x\rangle$ and $\lvert-x\rangle$, and

$$\lvert0\rangle = \frac{ \lvert+x\rangle+\lvert-x\rangle }{\sqrt2}.$$

Thus the post-quench energies $\pm\hbar\Omega/2$ each occur with probability $1/2$.

For $t>0$,

$$\begin{aligned} \lvert\psi(t)\rangle &= e^{-i\Omega t\sigma_x/2}\lvert0\rangle\\ &= \cos\!\left(\frac{\Omega t}{2}\right)\lvert0\rangle - i\sin\!\left(\frac{\Omega t}{2}\right)\lvert1\rangle. \end{aligned}$$

7. Insert the instantaneous-eigenstate ansatz

$$\lvert\psi(t)\rangle = e^{-(i/\hbar)\int_{t_0}^{t}E_n(s)\,ds} \lvert n(t)\rangle$$

into the Schrödinger equation. What condition would make this ansatz exact up to an additional phase?

Solution

Differentiation gives

$$\begin{aligned} i\hbar\lvert\dot\psi\rangle &= E_n(t)\lvert\psi(t)\rangle\\ &\quad+ i\hbar e^{-(i/\hbar)\int E_n(s)\,ds} \lvert\dot n(t)\rangle. \end{aligned}$$

The first term equals $H(t)\lvert\psi(t)\rangle$. The ansatz fails because of the second term. It can be absorbed into an additional phase only when $\lvert\dot n(t)\rangle$ has no component orthogonal to $\lvert n(t)\rangle$:

$$\langle m(t)\vert\dot n(t)\rangle=0 \qquad \text{for }m\ne n.$$

Adiabatic evolution makes these off-diagonal effects small under additional gap and slowness assumptions; it does not set them identically to zero in general.

8. Let $H(t+T)=H(t)$. Show that evolution over $n$ complete periods is

$$\begin{aligned} U(t_0+nT,t_0) &= U_F^n, \\ U_F &= U(t_0+T,t_0). \end{aligned}$$

Solution

Periodicity implies that the propagator over each corresponding period is the same:

$$U(t_0+(k+1)T,t_0+kT)=U_F.$$

Repeated use of the composition law gives $n$ identical factors:

$$\begin{aligned} U(t_0+nT,t_0) &= \underbrace{ U_F\cdots U_F }_{n\text{ factors}}\\ &= U_F^n. \end{aligned}$$

The product is chronologically ordered, but periodicity makes every factor the same one-period operator.

Additional exercises retained from the earlier canonical treatment

1. Why does the order of factors matter for a piecewise-constant Hamiltonian?

Solution

The state first evolves under $H_0$, then under $H_1$, and so on. Therefore the total operator is

$$U=e^{-iH_{N-1}\Delta t/\hbar}\cdots e^{-iH_1\Delta t/\hbar}e^{-iH_0\Delta t/\hbar}.$$

If the $H_j$ do not commute, changing the order changes the product and therefore changes the final state.