Time Dependence of Quantum Mechanical Operators

In quantum mechanics, the time dependence of operators depends on which representation (picture) we use — primarily the Schrödinger picture or the Heisenberg picture.
Both are equivalent, but they treat the time evolution of states and operators differently.

1. Schrödinger Picture

In the Schrödinger picture:

The operators are typically time-independent (unless they explicitly depend on time, like a time-varying potential).
The state vectors evolve with time according to the Schrödinger equation.

$i\hbar \frac{\partial}{\partial t}|\psi_S(t)\rangle = \hat{H} |\psi_S(t)\rangle$

If an operator itself depends explicitly on time (e.g., an external driving field), then its time dependence is just that explicit one.

$\frac{d}{dt}\langle \hat{A} \rangle = \frac{1}{i\hbar}\langle [\hat{A},\hat{H}] \rangle + \left\langle \frac{\partial \hat{A}}{\partial t} \right\rangle$

2. Heisenberg Picture

In the Heisenberg picture, the situation is reversed:

The state vectors are constant in time (frozen at their initial value).
The operators carry all the time dependence.

The operator evolution is given by the Heisenberg equation of motion:

$\frac{d\hat{A}_H}{dt} = \frac{1}{i\hbar}[\hat{A}_H, \hat{H}] + \left(\frac{\partial \hat{A}_H}{\partial t}\right)$

The solution can also be expressed as a similarity transformation using the time-evolution operator $\hat{U}(t) = e^{-\frac{i}{\hbar}\hat{H}t}$ :

$\hat{A}_H(t) = \hat{U}^\dagger(t)\,\hat{A}_S\,\hat{U}(t)$

3. Examples

(a) Free Particle Momentum and Position

For a free particle with Hamiltonian $\hat{H} = \frac{\hat{p}^2}{2m}$ :

$\frac{d\hat{p}_H}{dt} = \frac{1}{i\hbar}[\hat{p},\hat{H}] = 0 \quad \Rightarrow \quad \hat{p}_H(t) = \hat{p}(0)$

$\frac{d\hat{x}_H}{dt} = \frac{1}{i\hbar}[\hat{x},\hat{H}] = \frac{\hat{p}}{m} \quad \Rightarrow \quad \hat{x}_H(t) = \hat{x}(0) + \frac{\hat{p}(0)}{m}t$

This is a quantum analogue of classical motion with constant momentum and linearly increasing position.

(b) Harmonic Oscillator

For a 1D harmonic oscillator with $\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2$ :

$\frac{d\hat{x}_H}{dt} = \frac{\hat{p}_H}{m}, \quad \frac{d\hat{p}_H}{dt} = -m\omega^2 \hat{x}_H$

Combining gives:

$\frac{d^2\hat{x}_H}{dt^2} + \omega^2 \hat{x}_H = 0$

Solution:

$\hat{x}_H(t) = \hat{x}(0)\cos(\omega t) + \frac{\hat{p}(0)}{m\omega}\sin(\omega t)$

$\hat{p}_H(t) = \hat{p}(0)\cos(\omega t) - m\omega \hat{x}(0)\sin(\omega t)$

This again mirrors classical oscillatory motion but with operator-valued amplitudes.

(c) Spin Precession in a Magnetic Field

For a spin- $\tfrac{1}{2}$ particle in a magnetic field $\vec{B} = B\hat{z}$ , the Hamiltonian is

$\hat{H} = -\gamma B \hat{S}_z$

Then:

$\frac{d\hat{S}_x}{dt} = \frac{1}{i\hbar}[\hat{S}_x, \hat{H}] = \gamma B \hat{S}_y$

$\frac{d\hat{S}_y}{dt} = -\gamma B \hat{S}_x$

So the spin components precess:

$\hat{S}_x(t) = \hat{S}_x(0)\cos(\omega_L t) + \hat{S}_y(0)\sin(\omega_L t)$

$\hat{S}_y(t) = \hat{S}_y(0)\cos(\omega_L t) - \hat{S}_x(0)\sin(\omega_L t)$

where $\omega_L = \gamma B$ is the Larmor frequency.

4. Summary

Picture	State	Operator	Equation of Motion
Schrödinger	$i\hbar \frac{\partial}{\partial t}\|\psi\rangle = \hat{H}\|\psi\rangle$	Usually time-independent	$\frac{d}{dt}\langle \hat{A} \rangle = \frac{1}{i\hbar}\langle [\hat{A},\hat{H}] \rangle + \langle \partial_t \hat{A} \rangle$
Heisenberg	Time-independent	$\hat{A}_H(t)=e^{\frac{i}{\hbar}\hat{H}t}\hat{A}e^{-\frac{i}{\hbar}\hat{H}t}$	$\frac{d\hat{A}_H}{dt}=\frac{1}{i\hbar}[\hat{A}_H,\hat{H}] + \partial_t \hat{A}_H$

Thus, in quantum mechanics, operators evolve in time via commutators with the Hamiltonian, reflecting the deep correspondence between quantum and classical dynamics (where $[A,H]/(i\hbar)$ parallels the classical Poisson bracket $\{A,H\}$ ).