Constants of Motion

A constant of motion is a quantity preserved by a specified time evolution. The basic test is dynamical: compare the observable with the Hamiltonian that generates the motion, including any explicit time dependence of the observable itself.

For an observable $A(t)$ in the Schrödinger picture,

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

Thus an operator identity

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

implies that $\langle A(t)\rangle$ is conserved for every state evolving under $H(t)$. When $A$ has no explicit time dependence, this reduces to the familiar condition $[H,A]=0$ for a time-independent Hamiltonian, or $[H(t),A]=0$ for all relevant times in a driven problem.

This page focuses on the time-evolution meaning of conservation. The Core Formalism page Conservation Laws gives the direct derivation, and the symmetry-volume page Constants of Motion gives practical symmetry diagnostics.

Conservation of Expectation Values

The weakest useful statement is that one expectation value is constant along one solution:

$$\frac{d}{dt} \langle\psi(t)\rvert A(t)\lvert\psi(t)\rangle =0.$$

This can happen for special states even when $A$ is not a conserved observable. For example, $\langle X\rangle$ may be momentarily stationary in a particular wave packet even though position is not a constant of motion for a free particle.

A stronger statement is conservation for every state evolving under the same Hamiltonian. That requires the operator combination

$$D_t A \equiv \frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A]$$

to vanish as an operator, at least on the domain of states being considered. Then no special property of the initial state is being used.

Operators Commuting with the Hamiltonian

Suppose $H$ is time independent and $A$ has no explicit time dependence. If

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

then $A$ is conserved in the Heisenberg sense:

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

Equivalently, $A$ commutes with the evolution operator

$$U(t,t_0)=e^{-iH(t-t_0)/\hbar}.$$

For a discrete spectral decomposition

$$A=\sum_a aP_a,$$

the projectors $P_a$ are preserved:

$$U^\dagger(t,t_0)P_aU(t,t_0)=P_a.$$

Therefore the full measurement distribution of $A$ is conserved:

$$\Pr_t(a) = \langle\psi(t)\rvert P_a\lvert\psi(t)\rangle = \Pr_{t_0}(a).$$

This is stronger than saying that the mean value is constant. A conserved observable need not have a sharp value; a superposition of $A$ eigenstates can keep the same probabilities for all $A$ outcomes while still evolving by phases or by dynamics inside degenerate subspaces.

Explicit Time Dependence

Commuting with $H$ is not the whole story when the observable itself depends on time. 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:

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

But the explicitly time-dependent operator

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

is conserved, because

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

The two terms cancel. This example is conceptually important: a constant of motion is a constant along the dynamics, not necessarily a time-independent formula written in the Schrödinger picture.

Time-Dependent Hamiltonians

If the Hamiltonian depends explicitly on time, the same expectation-value formula applies:

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

For a time-independent observable $A$, conservation for all states follows if

$$[H(t),A]=0$$

for every time in the interval. It is not enough to check the commutator at one instant.

Energy illustrates the difference between a Hamiltonian as generator and a Hamiltonian as conserved quantity. Taking $A=H(t)$ gives

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

because $[H(t),H(t)]=0$ at equal times. A driven closed system can evolve unitarily while its system energy changes because the external drive is represented by the time dependence of $H(t)$.

Symmetries and Conservation Laws

Continuous symmetries are the most common source of constants of motion. If

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

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

$$U(\alpha)HU^\dagger(\alpha)=H.$$

For a continuous symmetry generated by $G$, differentiating at $\alpha=0$ gives

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

If $G$ has no explicit time dependence, then $G$ is conserved. Translation symmetry gives momentum conservation, rotational symmetry gives angular-momentum conservation, and time-translation symmetry gives energy conservation.

This page uses the dynamical consequence. The symmetry-volume treatment of the generator logic and Noether pattern begins with Commutators and Conservation Laws and Quantum Noether Principle.

Degeneracies and Simultaneous Eigenstates

If $[H,A]=0$, then $H$ preserves the eigenspaces of $A$. In a nondegenerate eigenspace this is very restrictive: an $A$ eigenstate remains in the same one-dimensional subspace up to a phase. In a degenerate eigenspace, the state may still evolve nontrivially inside that subspace while the value of $A$ remains fixed.

This is why conserved quantities become state labels only after one chooses a mutually commuting family. If

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

it does not follow that $[A,B]=0$. Conserved observables may fail to be simultaneously measurable.

For rotationally invariant systems,

$$[H,L_x]=[H,L_y]=[H,L_z]=0,$$

but

$$[L_x,L_y]=i\hbar L_z.$$

One usually labels states by a compatible set such as $H$, $L^2$, and one component $L_z$, not by all three components of $\mathbf L$. The broader state-labeling language is developed in Simultaneous Eigenstates and Good Quantum Numbers.

Standard Examples

For a time-independent Hamiltonian, energy is conserved because $[H,H]=0$. In a time-dependent Hamiltonian, $\langle H(t)\rangle$ changes according to $\langle\partial H/\partial t\rangle$.

For a free particle,

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

each component of momentum is conserved:

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

For a particle in a potential $V(\mathbf X)$, momentum in direction $i$ is conserved when the potential is invariant under translations in that direction. The commutator form is

$$\frac{d}{dt}\langle P_i\rangle = -\left\langle \frac{\partial V}{\partial X_i} \right\rangle.$$

For a central potential,

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

angular momentum is conserved. A compatible stationary-state labeling set is built from $H$, $L^2$, and $L_z$.

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

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

The operators $S_z$ and $S^2$ are conserved, while $S_x$ and $S_y$ precess. Conservation of one component does not mean all components are fixed.

Common Mistakes

• Checking $\langle[H,A]\rangle=0$ in one state and concluding that $A$ is conserved as an observable.
• Forgetting the explicit $\partial A/\partial t$ term.
• Saying that unitary evolution always conserves energy, even when $H(t)$ is explicitly time dependent.
• Assuming a conserved observable must have a sharp value in the state.
• Assuming separately conserved observables commute with one another.
• Labeling degenerate states without checking whether the proposed labels form a mutually commuting set.
• Ignoring domain and boundary-condition subtleties for unbounded operators.

Cross-Links

• Conservation Laws
• Symmetries and Dynamical Automorphisms
• Unitarity and Conservation of Probability
• Time-Independent Hamiltonians
• Time-Dependent Hamiltonians
• Heisenberg Equations of Motion
• Operators with Explicit Time Dependence
• Ehrenfest Theorem
• Commutators and Conservation Laws
• Quantum Noether Principle
• Simultaneous Eigenstates and Good Quantum Numbers
• Translation-Invariant Hamiltonians
• Central Potentials

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. Let $A$ have no explicit time dependence. Starting from the Schrödinger equation, show that $[H,A]=0$ implies $d\langle A\rangle/dt=0$ for every state.

Solution

For no explicit time dependence,

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

If $[H,A]=0$ as an operator, then its expectation value vanishes in every state:

$$\frac{d}{dt}\langle A\rangle=0.$$

The condition is state-independent; it is not merely a cancellation in one chosen state.

2. For a free particle with $H=P^2/(2m)$, verify that $A(t)=X-tP/m$ is a constant of motion.

Solution

The explicit derivative is

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

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

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

Therefore

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

and

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

3. A particle moves in a central potential. Explain why one may label stationary states by $L^2$ and $L_z$, but not by $L_x$, $L_y$, and $L_z$ simultaneously.

Solution

For a central potential, $H$ commutes with $\mathbf L$ and with $L^2$, so angular momentum is conserved. However, the components of angular momentum do not commute:

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

A simultaneous eigenbasis requires mutually commuting labels. The standard compatible set is $H$, $L^2$, and one component such as $L_z$.

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

Solution

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

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

The equal-time commutator vanishes, and

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

Thus

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