Commutators and Conservation Laws

The basic quantum conservation test is a commutator test. For the Core Formalism derivation of the expectation-value formula, see Conservation Laws. For a closed system with Hamiltonian $H$, an observable $A$ with no explicit time dependence is conserved when

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

The essential distinction is between an operator conserved in every state and an expectation that happens to be constant in one special state. The commutator tests the former.

The classical counterpart is the Poisson-bracket test $\{f,H\}=0$, reviewed in Poisson Brackets.

Expectation-Value Equation

For finite matrices, or states in suitable differentiable common domains for unbounded operators, closed-system dynamics gives

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

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

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

Therefore $[H,A]=0$ implies $\langle A\rangle$ is constant in every state evolving under $H$.

Operator Equation

In the Heisenberg picture,

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

If $A$ has no explicit time dependence and commutes with the Hamiltonian, then

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

This is a stronger statement than conservation of one expectation value in one state: the observable is a constant of motion.

Symmetry Generator Implies Conservation

Suppose a continuous unitary symmetry is generated by $G$:

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

If the Hamiltonian is invariant under this symmetry, then

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

for all $\alpha$. Differentiating at $\alpha=0$ gives

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

Equivalently, $[H,G]=0$, so $G$ is conserved when it has no explicit time dependence.

Free Particle Momentum Conservation

For a free particle in one dimension,

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

Since $P$ commutes with any function of $P$,

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

Momentum is conserved. This is the operator statement behind translation invariance.

Central Potential Angular Momentum Conservation

For a three-dimensional central potential,

$$H=\frac{\mathbf p^2}{2m}+V(r), \qquad r=\lvert\mathbf r\rvert,$$

with rotationally invariant boundary conditions and domain, the Hamiltonian is rotationally invariant. Consequently,

$$[H,L_i]=0$$

for each component of orbital angular momentum. It follows that $L^2$ and a chosen component, usually $L_z$, can be used to label energy eigenstates.

Energy Conservation

If the Hamiltonian itself has no explicit time dependence, then

$$\frac{d}{dt}\langle H\rangle = \frac{i}{\hbar}\langle[H,H]\rangle = 0.$$

Energy conservation in closed quantum mechanics is therefore the conservation law associated with time-translation symmetry.

What Conservation Does Not Mean

Conservation of $A$ does not mean every state has a sharp value of $A$. It means the distribution of $A$ is constant under time evolution when $A$ commutes with $H$. A superposition of $A$ eigenstates can remain a superposition while the probabilities for measuring each $A$ eigenvalue remain fixed.

A stationary expectation is not enough

Take $H=\hbar\omega\sigma_z/2$ and $A=\sigma_x$, with $\omega\ne0$. In the stationary state $|+_z\rangle$, $\langle\sigma_x\rangle=0$ at all times. Yet $[H,\sigma_x]=i\hbar\omega\sigma_y\ne0$: starting in $|+_x\rangle$ gives $\langle\sigma_x\rangle=\cos\omega t$. The first state hides the dynamics of this observable.

For unbounded observables, preserve domains and use the spectral criterion: time evolution must commute with every spectral projection of $A$. This guarantees that the entire distribution, including continuous outcomes, is unchanged. See Constants of Motion for explicitly time-dependent conserved operators.

Exercises

1. On the full line with compatible domains and a differentiable potential, let $H=p^2/(2m)+V(x)$. Use $[p,V(x)]=-i\hbar V'(x)$ to determine when momentum is conserved.

Solution

Since $[p,p^2/(2m)]=0$,

$$[p,H]=[p,V(x)]=-i\hbar V'(x).$$

Momentum is conserved when this vanishes as an operator, which requires $V'(x)=0$ on the region considered. Thus the potential is constant there.

2. If $[H,A]=0$, what can be said about the probabilities of measuring the eigenvalues of $A$?

Solution

When $A$ has no explicit time dependence and commutes with $H$, the projectors onto eigenspaces of $A$ also commute with $H$ under the usual spectral assumptions. The probabilities for the corresponding outcomes are constant under time evolution.

3. A spin one half evolves under $H=\omega S_z$, where $S_i=\hbar\sigma_i/2$ and $\omega\ne0$ is real. Derive the equations for the transverse means $\langle S_x\rangle$ and $\langle S_y\rangle$. Does conservation of $S_z$ imply a stationary state?

Solution

The angular-momentum commutators give $[H,S_x]=i\hbar\omega S_y$ and $[H,S_y]=-i\hbar\omega S_x$. Therefore

$$\begin{aligned} \frac{d}{dt}\langle S_x\rangle&=-\omega\langle S_y\rangle,\\ \frac{d}{dt}\langle S_y\rangle&=\omega\langle S_x\rangle,\\ \frac{d}{dt}\langle S_z\rangle&=0. \end{aligned}$$

For an initial $|+_x\rangle$ state, the transverse means are $\hbar\cos(\omega t)/2$ and $\hbar\sin(\omega t)/2$. The two $S_z$ components acquire a changing relative phase, although their measurement probabilities remain fixed. Thus conservation of one spin component does not make the state stationary. The Larmor-precession calculation relates the rotation direction to the magnetic-moment convention.

References

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