Constants of Motion

A constant of motion is an observable whose value is preserved by the dynamics in a specified sense. The most useful operator criterion is:

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

If $A$ has no explicit time dependence, this reduces to the familiar test

$$[H,A]=0.$$

This page is a practical checklist for using that test correctly. The Core Formalism derivation is Conservation Laws; the symmetry interpretation is Commutators and Conservation Laws and Quantum Noether Principle.

Several Strengths of Conservation

The phrase “conserved” can mean several different things.

Statement	Meaning	Strength
$\frac{d}{dt}\langle A\rangle=0$ in one state	one expectation value is stationary	weak
$\frac{d}{dt}\langle A\rangle=0$ in all states	expectation is conserved dynamically	stronger
$\frac{dA_H}{dt}=0$	the Heisenberg operator is time independent	operator constant
spectral projectors of $A$ are preserved	the full measurement distribution is fixed	strong observable conservation

The strongest everyday meaning is that the full distribution of $A$ is preserved. A state need not be an eigenstate of $A$ for $A$ to be conserved.

Schrodinger-Picture Criterion

Let the state obey

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

and let $A(t)$ be a possibly time-dependent observable in the Schrodinger picture. Then

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

If the operator identity

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

holds, then $\langle A\rangle$ is conserved for every state evolving under $H(t)$.

For a time-independent observable and Hamiltonian, this becomes

$$[H,A]=0.$$

This is the most common constants-of-motion test.

Heisenberg-Picture Criterion

In the Heisenberg picture,

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

The equation of motion is

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

Thus $A$ is an operator constant of motion when

$$\frac{dA_H}{dt}=0.$$

This is often the cleanest definition because it says that the observable itself, not only one of its expectation values, is unchanged by time evolution.

Explicit Time Dependence

A quantity can be conserved even when the Schrodinger-picture operator contains explicit time dependence. The explicit derivative can cancel the commutator term.

For a free particle,

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

Momentum is conserved because $[H,P]=0$. Position is not conserved because

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

However, the explicitly time-dependent operator

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

is a constant of motion:

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

Therefore

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

This example is a useful warning: commuting with $H$ is sufficient only when the observable has no explicit time dependence.

Energy and Time-Dependent Hamiltonians

For $A=H(t)$,

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

because $[H(t),H(t)]=0$ at equal times. Thus energy is conserved for a closed system with a time-independent Hamiltonian, but not generally for a driven system.

If $H(t)$ depends on an externally controlled parameter, the changing expectation value of $H$ usually represents work done by or on that external drive. It is not a failure of unitary quantum mechanics.

Constants from Symmetry

The most common source of constants of motion is symmetry. If

$$U(\alpha)=e^{-i\alpha G/\hbar}$$

is a continuous unitary symmetry of a time-independent Hamiltonian, then

$$[G,H]=0.$$

When $G$ has no explicit time dependence, $G$ is a constant of motion. This is the ordinary quantum-mechanical Noether pattern:

• translation symmetry gives momentum conservation;
• rotational symmetry gives angular-momentum conservation;
• time-translation symmetry gives energy conservation;
• global phase symmetry gives conservation of the associated charge or number.

Not every constant of motion is obviously tied to a manifest geometric symmetry. Some are hidden or accidental, such as the additional conserved structure behind the Coulomb problem.

Compatible Constants

Several quantities may each commute with the Hamiltonian without commuting with one another. If

$$[H,A]=0, \qquad [H,B]=0,$$

it does not follow that $[A,B]=0$.

This matters because only mutually commuting observables can generally be used simultaneously to label stationary states. In a rotationally invariant system,

$$[H,J_x]=[H,J_y]=[H,J_z]=0,$$

but

$$[J_x,J_y]=i\hbar J_z.$$

One usually chooses a compatible set such as $H$, $J^2$, and $J_z$, not all three components $J_x,J_y,J_z$.

Complete Sets and Integrability Preview

A complete set of commuting conserved quantities gives enough labels to distinguish states up to the remaining degeneracies. In ordinary central-potential problems, the labels $E$, $l$, and $m$ come from a commuting set involving $H$, $L^2$, and $L_z$.

In classical mechanics, integrability is tied to having enough independent constants of motion in involution. In quantum mechanics, the analogous phrase usually means a sufficiently large commuting family of conserved operators. In many-body physics, “integrable” often means the presence of an extensive set of commuting charges.

This is only a preview. The practical state-labeling story is developed in Simultaneous Eigenstates and Good Quantum Numbers. The important practical point here is modest: conserved operators are most useful as labels when they are mutually compatible.

Examples

For the free particle,

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

both $P$ and $H$ are constants of motion. The position $X$ is not, but $X-tP/m$ is.

For a central potential,

$$H=\frac{\mathbf P^2}{2m}+V(R),$$

the constants include $H$, $L^2$, and a chosen component such as $L_z$. The three components of $\mathbf L$ are each conserved but not mutually commuting.

For a spin in a constant magnetic field along $z$,

$$H=-\gamma B S_z.$$

Then $S_z$ and $S^2$ are constants of motion. The transverse components $S_x$ and $S_y$ are not constant; they precess.

Common Mistakes

• Treating a conserved expectation value in one state as an operator conservation law.
• Forgetting the explicit $\partial A/\partial t$ term.
• Assuming energy is conserved whenever evolution is unitary.
• Assuming all conserved quantities commute with one another.
• Confusing a conserved distribution with a sharp value in each state.
• Treating approximate conservation as exact after adding a small symmetry-breaking term.
• Ignoring boundary conditions and domains for unbounded operators.

Cross-Links

• Conservation Laws
• Time-Dependent Hamiltonians
• Heisenberg Equations of Motion
• Ehrenfest Theorem
• Commutators and Conservation Laws
• Quantum Noether Principle
• Simultaneous Eigenstates and Good Quantum Numbers
• Symmetry Constraints on Hamiltonians
• Accidental Symmetry
• Free Particle
• Central Potentials
• Larmor Precession

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.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. For a free particle, show that $A(t)=X-tP/m$ is a constant of motion.

Solution

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

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

Also,

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

Since $[H,P]=0$ and $(i/\hbar)[H,X]=P/m$,

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

The two terms cancel, so

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

2. Suppose $[H,A]=0$ and $[H,B]=0$. Must $A$ and $B$ commute?

Solution

No. Angular momentum in a rotationally invariant Hamiltonian is the standard example:

$$[H,J_x]=[H,J_y]=[H,J_z]=0,$$

but

$$[J_x,J_y]=i\hbar J_z.$$

Each component is conserved, but the components cannot all be diagonalized simultaneously.

3. Let $H(t)=H_0+\lambda(t)B$. Compute $d\langle H(t)\rangle/dt$ for closed evolution.

Solution

Use the expectation-value equation with $A=H(t)$:

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

The commutator is zero, and

$$\frac{\partial H}{\partial t} = \dot\lambda(t)B.$$

Therefore

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

4. In a central potential, why can one label states by $L^2$ and $L_z$ but not by all three components $L_x,L_y,L_z$?

Solution

For a central potential, $H$ commutes with each component of $\mathbf L$ and with $L^2$. However, the angular momentum components do not commute with one another:

$$[L_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k.$$

Thus one can choose a mutually commuting set such as $H$, $L^2$, and $L_z$, but not $H$, $L_x$, $L_y$, and $L_z$ all at once.