Ehrenfest Theorem

Ehrenfest theorem relates the time evolution of expectation values to classical-looking equations of motion. For

$$H=\frac{p^2}{2m}+V(x),$$

it gives

$$\frac{d}{dt}\langle x\rangle =\frac{\langle p\rangle}{m}, \qquad \frac{d}{dt}\langle p\rangle =-\langle V'(x)\rangle.$$

Derivation

For an operator that may depend explicitly on time, the general identity is

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

It assumes that the state and the relevant operator products lie in domains for which the differentiations and integrations by parts are legitimate. This qualification matters for unbounded operators such as $x$ and $p$.

From the Heisenberg equation,

$$\frac{dA_H}{dt} =\frac{i}{\hbar}[H,A_H]$$

for operators with no explicit time dependence. Taking expectation values gives

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\langle[H,A]\rangle.$$

For $A=x$ and $H=p^2/(2m)+V(x)$,

$$\frac{d}{dt}\langle x\rangle =\frac{\langle p\rangle}{m}.$$

For $A=p$,

$$\frac{d}{dt}\langle p\rangle =-\langle V'(x)\rangle.$$

Relation to Newton’s Law

Classical Newtonian motion would use

$$m\frac{d^2}{dt^2}x_{\rm cl} =-V'(x_{\rm cl}).$$

Ehrenfest theorem gives instead

$$m\frac{d^2}{dt^2}\langle x\rangle =-\langle V'(x)\rangle.$$

These agree with the classical equation for $\langle x\rangle$ only when

$$\langle V'(x)\rangle \approx V'(\langle x\rangle).$$

This approximation is exact for potentials at most quadratic in $x$, and often useful for narrow wave packets in slowly varying potentials.

To quantify the approximation, put

$$q=\langle x\rangle, \qquad \delta x=x-q, \qquad \sigma_x^2=\langle(\delta x)^2\rangle.$$

Expanding the force about $q$ gives

$$\langle V'(x)\rangle =V'(q) +\frac12V'''(q)\sigma_x^2 +\frac16V^{(4)}(q)\langle(\delta x)^3\rangle +\cdots.$$

The first correction is controlled by the packet width and the third derivative of the potential. For a quartic potential $V(x)=\lambda x^4/4$,

$$\langle V'(x)\rangle =\lambda\left(q^3+3q\sigma_x^2 +\langle(\delta x)^3\rangle\right).$$

Even a symmetric packet therefore acquires the finite-width correction $3\lambda q\sigma_x^2$.

Exact Closures and Moment Hierarchy

• For a free particle, $d\langle p\rangle/dt=0$ and the center moves uniformly.
• For a uniform force, $V'(x)$ is constant, so the mean obeys the exact classical acceleration law.
• For a harmonic oscillator, $V'(x)=m\omega^2x$, so the first moments close exactly.
• For generic anharmonic potentials, first moments couple to variances and higher central moments.

The width already has its own equation. If

$$C_{xp}=\frac12\langle\delta x\,\delta p+\delta p\,\delta x\rangle,$$

then

$$\frac{d\sigma_x^2}{dt}=\frac{2}{m}C_{xp}.$$

The covariance in turn couples to momentum variance and force-position correlations. This hierarchy explains why correct center motion does not imply a rigid, classical packet.

For $A=H$, the same general identity gives

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

Energy is conserved for a time-independent Hamiltonian, provided the domain assumptions needed for the commutator argument hold.

What It Does Not Prove

Ehrenfest theorem does not by itself explain the full classical limit. It says something about expectation values, not about definite trajectories, measurement records, decoherence, or the suppression of interference between macroscopically different histories. For the broader conceptual bridge, see Correspondence Principle.

Wave-Packet Spreading

Even when $\langle x\rangle$ follows a classical-looking trajectory, the wave packet can spread. A free Gaussian packet is the standard example: its center moves uniformly, while its width changes in time.

Common Mistakes

• Replacing $\langle V'(x)\rangle$ by $V'(\langle x\rangle)$ without justification.
• Saying Ehrenfest theorem proves that particles follow classical paths.
• Forgetting that broad or split wave packets can have expectation values that are poor summaries of the state.
• Ignoring wave-packet spreading.
• Treating the theorem as a substitute for semiclassical analysis or decoherence.

Cross-Links

• Heisenberg Equations of Motion
• Pictures of Quantum Mechanics
• Conservation Laws
• Expectation Values
• Ehrenfest Theorem Overview
• Ehrenfest Theorem Revisited
• Gaussian Wave Packets
• Free Particle
• Correspondence Principle

References

• P. Ehrenfest, “Bemerkung über die angenäherte Gültigkeit der klassischen Mechanik innerhalb der Quantenmechanik,” Zeitschrift für Physik 45, 455-457 (1927).
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. For the harmonic oscillator $V(x)=m\omega^2x^2/2$, show that $\langle x\rangle$ obeys the classical oscillator equation.

Solution

Here $V'(x)=m\omega^2x$, so

$$\langle V'(x)\rangle =m\omega^2\langle x\rangle.$$

Ehrenfest theorem gives

$$m\frac{d^2}{dt^2}\langle x\rangle =-m\omega^2\langle x\rangle,$$

or

$$\frac{d^2}{dt^2}\langle x\rangle+\omega^2\langle x\rangle=0.$$

2. Show that expectation-value motion closes exactly for every state if $V(x)$ is at most quadratic.

Solution

For $V(x)=ax^2+bx+c$, the force is affine:

$$V'(x)=2ax+b.$$

Linearity of expectation values gives

$$\langle V'(x)\rangle=2a\langle x\rangle+b =V'(\langle x\rangle).$$

Thus the equations for $\langle x\rangle$ and $\langle p\rangle$ form a closed classical system for every state, even though higher moments may still evolve.

3. For $V(x)=\lambda x^4/4$ and a packet symmetric about its mean, compute the leading finite-width correction to the force on the mean.

Solution

Symmetry about the mean gives $\langle(\delta x)^3\rangle=0$. Since

$$\langle x^3\rangle=q^3+3q\sigma_x^2,$$

Ehrenfest’s equation becomes

$$m\ddot q=-\lambda(q^3+3q\sigma_x^2).$$

The correction to the point-particle force is $-3\lambda q\sigma_x^2$.