Operators with Explicit Time Dependence

An operator has explicit time dependence when its definition contains the time parameter before any picture-induced evolution is applied. The standard warning sign is an observable written as $A_S(t)$ in the Schrödinger picture.

For such an observable,

$$\frac{d}{dt}\langle A(t)\rangle = \frac{i}{\hbar}\langle[H(t),A(t)]\rangle + \left\langle \frac{\partial A(t)}{\partial t} \right\rangle.$$

The partial-derivative term is not optional. It records that the question being asked of the system can itself change with time.

Three Sources of Time Dependence

It is useful to separate three different effects:

Source	Where it appears	Example
State evolution	$\lvert\psi_S(t)\rangle$	wave packet moving under $H$
Picture-induced operator evolution	$A_H(t)=U^\dagger A_SU$	Heisenberg position $X_H(t)$
Explicit operator time dependence	$A_S(t)$	rotating spin-measurement axis

Only the third source is the partial derivative $\partial A/\partial t$. The first two are generated by the Hamiltonian or by a chosen picture transformation.

Partial Versus Total Time Dependence

The partial derivative means: change the time label in the operator definition while holding the state and canonical operators fixed. For example,

$$A_S(t)=S_x\cos\omega t+S_y\sin\omega t$$

has

$$\frac{\partial A_S}{\partial t} = -\omega S_x\sin\omega t + \omega S_y\cos\omega t.$$

This derivative is not the full time derivative of an expectation value. The full derivative also includes the state evolution generated by $H(t)$.

Heisenberg Equation with Explicit Dependence

In the Heisenberg picture,

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

Differentiating gives

$$\frac{dA_H}{dt} = \frac{i}{\hbar}[H_H,A_H] + \left( \frac{\partial A_S}{\partial t} \right)_H.$$

The final term means:

$$\left( \frac{\partial A_S}{\partial t} \right)_H = U^\dagger(t,t_0) \frac{\partial A_S(t)}{\partial t} U(t,t_0).$$

It is the Schrödinger-picture explicit derivative transformed into the Heisenberg picture.

Rotating Measurement Axes

Suppose a spin is measured along a time-dependent unit vector in the $x$-$y$ plane:

$$A_S(t)=S_x\cos\omega t+S_y\sin\omega t.$$

Even if the state were fixed, this observable would change because the measurement axis rotates. Under a Hamiltonian

$$H=\omega_0S_z,$$

the commutator term is

$$\frac{i}{\hbar}[H,A_S(t)] = -\omega_0S_y\cos\omega t + \omega_0S_x\sin\omega t.$$

The explicit term is

$$\frac{\partial A_S}{\partial t} = -\omega S_x\sin\omega t + \omega S_y\cos\omega t.$$

The two terms cancel when $\omega=\omega_0$. In that case the observable is rotating with the spin precession in just the right way to be a constant of motion. This example shows why conservation tests must include explicit time dependence.

Explicit Constants of Motion

For a free particle,

$$H=\frac{P^2}{2m}.$$

The position operator is not conserved because

$$\frac{i}{\hbar}[H,X]=\frac{P}{m}.$$

However,

$$A_S(t)=X-\frac{t}{m}P$$

is conserved:

$$\frac{\partial A_S}{\partial t} = -\frac{P}{m}, \qquad \frac{i}{\hbar}[H,A_S] = \frac{P}{m}.$$

The explicit derivative cancels the dynamical derivative. The operator is time dependent in the Schrödinger picture but constant along the dynamics.

Time-Dependent Hamiltonians

The Hamiltonian itself may be an explicitly time-dependent operator:

$$H(t)=H_0+\lambda(t)B.$$

For closed evolution under $H(t)$,

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

The commutator term vanishes only because $[H(t),H(t)]=0$ at equal times. The expectation value of the Hamiltonian can still change because the operator itself is changing.

This is not a failure of unitarity. It usually represents work done by or on the external control represented by $\lambda(t)$.

Interaction-Picture Operators

In the interaction picture,

$$A_I(t)=U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0).$$

Even if $A_S$ has no explicit time dependence, $A_I(t)$ usually changes because $U_0(t,t_0)$ changes. If $A_S(t)$ is explicitly time dependent as well, both effects appear:

$$\frac{dA_I}{dt} = \frac{i}{\hbar}[H_0,A_I] + \left( \frac{\partial A_S}{\partial t} \right)_I.$$

This distinction matters in time-dependent perturbation theory. A time-dependent perturbation $V_S(t)$ becomes

$$V_I(t)=U_0^\dagger(t,t_0)V_S(t)U_0(t,t_0),$$

where the time dependence may come both from the drive and from the interaction-picture transformation.

Conservation Criterion

The correct operator-level conservation test is

$$\frac{\partial A_S}{\partial t} + \frac{i}{\hbar}[H,A_S] =0.$$

If $A_S$ has no explicit time dependence, this reduces to $[H,A_S]=0$. But when $A_S(t)$ is explicitly time dependent, commuting with $H$ is not the whole criterion, and failing to commute with $H$ is not the end of the story.

The practical constants-of-motion checklist is Constants of Motion.

Common Mistakes

• Dropping $\partial A/\partial t$ because the operator is being used in the Heisenberg picture.
• Treating every time dependence of an operator as Heisenberg evolution.
• Assuming $[H,A]\ne0$ means $A(t)$ cannot be conserved.
• Assuming $[H,A]=0$ is enough when $A$ itself depends on time.
• Confusing a rotating measurement setting with a rotating quantum state.
• Forgetting that interaction-picture operators can have both transformation-induced and explicit time dependence.
• Calling $\langle\partial H/\partial t\rangle$ nonunitary energy loss rather than external work in a driven closed model.

Cross-Links

• Pictures of Quantum Mechanics
• Heisenberg Equations of Motion
• Picture Transformations
• Interaction Picture
• Constants of Motion
• Time-Dependent Hamiltonians
• Conservation Laws
• Ehrenfest Theorem
• Spin in Magnetic Fields

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.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.

Exercises

1. For $A_S(t)=S_x\cos\omega t+S_y\sin\omega t$, compute $\partial A_S/\partial t$.

Solution

Differentiate the explicit time functions while holding $S_x$ and $S_y$ fixed:

$$\frac{\partial A_S}{\partial t} = -\omega S_x\sin\omega t + \omega S_y\cos\omega t.$$

2. With $H=\omega_0S_z$, show that the rotating spin observable in the previous exercise is conserved when $\omega=\omega_0$.

Solution

Use

$$[S_z,S_x]=i\hbar S_y, \qquad [S_z,S_y]=-i\hbar S_x.$$

Then

$$\frac{i}{\hbar}[H,A_S(t)] = -\omega_0S_y\cos\omega t + \omega_0S_x\sin\omega t.$$

Adding the explicit derivative gives

$$\frac{\partial A_S}{\partial t} + \frac{i}{\hbar}[H,A_S] = (\omega_0-\omega)S_x\sin\omega t + (\omega-\omega_0)S_y\cos\omega t.$$

This vanishes when $\omega=\omega_0$.

3. For a free particle, verify that $A_S(t)=X-tP/m$ is conserved.

Solution

For $H=P^2/(2m)$,

$$\frac{\partial A_S}{\partial t} = -\frac{P}{m}.$$

Using $[X,P]=i\hbar$,

$$\frac{i}{\hbar}[H,X]=\frac{P}{m}, \qquad [H,P]=0.$$

Thus

$$\frac{i}{\hbar}[H,A_S]=\frac{P}{m},$$

and the two terms cancel.

4. Let $A_S(t)$ have explicit time dependence. Write its interaction-picture derivative under a time-independent $H_0$.

Solution

With

$$A_I(t)=U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0),$$

and $i\hbar\,\dot U_0=H_0U_0$, differentiation gives

$$\frac{dA_I}{dt} = \frac{i}{\hbar}[H_0,A_I] + \left( \frac{\partial A_S}{\partial t} \right)_I.$$

The first term is induced by the interaction-picture transformation; the second comes from explicit time dependence already present in $A_S(t)$.