Heisenberg Equations of Motion

The Heisenberg equation of motion gives the time derivative of an operator in the Heisenberg picture:

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

It is the operator analogue of Hamiltonian time evolution. The dynamics-side discussion of commutators as generators is Commutator Dynamics.

The Core expectation-value consequence, including the conservation-law test, is Conservation Laws. For a practical symmetry-oriented checklist, see Constants of Motion.

Differentiating the Heisenberg operator

Assume a differentiable unitary propagator. For unbounded operators, use states in an invariant domain on which the products and derivatives below exist. Write $H_H(t)=U^\dagger(t,t_0)H_S(t)U(t,t_0)$. For an operator with no explicit time dependence,

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

Using

$$i\hbar\frac{\partial U}{\partial t}=HU,$$

and its adjoint, differentiation gives

$$\frac{dA_H}{dt} = \frac{i}{\hbar}H_HA_H - \frac{i}{\hbar}A_HH_H = \frac{i}{\hbar}[H_H,A_H].$$

If $A_S$ has explicit time dependence, the extra term $\left(\partial A/\partial t\right)_H$ must be included.

Free Particle

For

$$H=\frac{p^2}{2m},$$

the canonical commutator gives

$$\frac{dx_H}{dt} =\frac{p_H}{m}, \qquad \frac{dp_H}{dt}=0.$$

Thus

$$x_H(t)=x_H(0)+\frac{p_H(0)}{m}t.$$

Particle in a Potential

For

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

one obtains

$$\frac{dx_H}{dt}=\frac{p_H}{m}, \qquad \frac{dp_H}{dt}=-V'(x_H).$$

These are operator equations on the appropriate domain. Their expectation values give Ehrenfest theorem, where $\langle V'(x)\rangle$ cannot generally be replaced by $V'(\langle x\rangle)$.

Harmonic Oscillator

For

$$H=\hbar\omega\left(a^\dagger a+\frac12\right),$$

the ladder operator obeys

$$\frac{da_H}{dt}=-i\omega a_H,$$

so

$$a_H(t)=a_H(0)e^{-i\omega t}.$$

Spin in a Magnetic Field

With the magnetic-moment convention $H=-\gamma\mathbf B\cdot\mathbf S$,

$$\frac{d\mathbf S_H}{dt}=\gamma\mathbf S_H\times\mathbf B.$$

The sign of the gyromagnetic ratio $\gamma$ sets the sense of precession. For $\mathbf B=B\hat z$, $\dot S_x=\gamma B S_y$ and $\dot S_y=-\gamma B S_x$.

The spin-specific derivation is Larmor Precession.

Common Mistakes

• Forgetting the explicit time-derivative term.
• Reversing the commutator sign.
• Treating operator equations as ordinary commuting equations.
• Assuming Heisenberg equations automatically solve the whole classical-limit problem.
• Dropping domain and boundary-condition issues for unbounded operators.

References

• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Use $[x,p]=i\hbar$ to derive $dx_H/dt=p_H/m$ for the free particle.

Solution

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

$$\frac{dx_H}{dt} =\frac{i}{\hbar}[H,x_H] =\frac{i}{2m\hbar}[p_H^2,x_H].$$

Using $[p^2,x]=p[p,x]+[p,x]p=-2i\hbar p$ gives

$$\frac{dx_H}{dt} =\frac{i}{2m\hbar}(-2i\hbar p_H) =\frac{p_H}{m}.$$