Conserved Quantities

An observable is conserved by an autonomous evolution when its full measurement distribution is unchanged for every initial state. For self-adjoint observables, the exact criterion is strong commutation with the Hamiltonian. This spectral statement remains meaningful when means or variances are infinite and when products of unbounded operators are unavailable. The physical introduction and examples of symmetry diagnostics belong to Constants of Motion.

Helpful background. Compatible Observables defines strong commutation. Born Rule from Spectral Measures identifies spectral projections with outcome probabilities. Stone’s Theorem supplies the autonomous unitary group.

Conservation is a statement about every spectral event

Let $H$ and $A$ be self-adjoint and $U(t)=e^{-itH/\hbar}$. The following statements are equivalent:

1. For every normalized $\psi\in\mathcal H$, every Borel set $B\subset\mathbb R$, and every $t$, $\langle U(t)\psi,E_A(B)U(t)\psi\rangle=\langle\psi,E_A(B)\psi\rangle$.
2. Every spectral projection $E_A(B)$ commutes with every $U(t)$.
3. $A$ and $H$ strongly commute: all their spectral projections commute.

To prove the first equivalence, fix $B,t$ and set

$$K=U(t)^*E_A(B)U(t)-E_A(B).$$

This is a bounded self-adjoint operator. Statement 1 says $\langle\psi,K\psi\rangle=0$ for every vector, after rescaling. Polarization recovers every matrix element and gives $K=0$. Multiplication by $U(t)$ then gives statement 2; the converse is immediate.

The equivalence with statement 3 uses the bounded functional-calculus criterion developed in Compatible Observables. Commuting with the whole group is equivalent to commuting with the Hamiltonian’s spectral projections. A commutator calculation on a convenient test space is not an alternative proof unless its domain hypotheses justify this promotion.

Domains and all existing moments are preserved

Under these equivalent conditions, spectral calculus gives

$$U(t)D(A)=D(A),\qquad AU(t)\psi=U(t)A\psi \quad(\psi\in D(A)).$$

More generally, $U(t)$ preserves $D(f(A))$ for every finite Borel function $f$ and intertwines $f(A)$ on its domain. In the domain formula, the integral of $|f|^2$ is unchanged because the entire scalar spectral measure is unchanged.

Every initially finite absolute moment of $A$ is therefore conserved. In particular, a finite first moment requires $\int|\lambda|\,d\mu_\psi^A<\infty$, while a finite second moment is equivalent to $\psi\in D(A)$. A conserved probability law may have neither. Conservation does not make an undefined expectation meaningful.

The same event identity immediately gives conservation for every density operator:

$$\operatorname{Tr}\!\left(U(t)\rho U(t)^*E_A(B)\right) =\operatorname{Tr}\!\left(\rho E_A(B)\right).$$

All factors other than the trace-class state are bounded here, so the trace manipulation is justified without unbounded cyclicity assumptions.

State-specific constancy is much weaker

Take $H=(\hbar\omega/2)\sigma_z$ with $\omega\ne0$ and $A=\sigma_x$. An initial $\sigma_z$ eigenvector changes only by phase. Its $A$ distribution remains equally split between the two outcomes for all time, although $A$ is not conserved as an observable.

Starting instead from the positive $\sigma_x$ eigenvector gives

$$\Pr_t(A=+1)=\cos^2(\omega t/2),$$

so the failure is exposed by another preparation. Even constancy of the whole distribution in one state does not establish an operator-level conservation law.

There is a useful qualification to the phrase “the mean is weaker than the distribution.” For bounded self-adjoint $A$, constancy of its mean for every initial vector implies $U(t)^*AU(t)=A$ by polarization, and consequently conservation of its full spectral measure. The weakness is in a restricted set of preparations or insufficient domain information, not in an absolute hierarchy that ignores quantifiers.

Conserved labels allow motion within degenerate sectors

On $\mathbb C^3$, take real $E,\omega$ and define

$$A=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}, \qquad H=\begin{pmatrix}0&\hbar\omega/2&0\\ \hbar\omega/2&0&0\\0&0&E\end{pmatrix}.$$

The operators commute. An initial $e_1$ evolves into

$$U(t)e_1=\cos(\omega t/2)e_1-i\sin(\omega t/2)e_2,$$

while $A=0$ remains certain. Conservation fixes the spectral subspace, not a preferred vector inside it. This distinction is especially important when a conserved charge labels a large degeneracy.

The boundary counterexample in Compatible Observables shows a different obstruction: formal commuting differential actions can belong to incompatible self-adjoint realizations. Their apparent conserved labels cannot be inferred from the local differential expressions alone.

Moving invariants for driven evolution

For a unitary propagator $U(t,s)$ and a self-adjoint $A_0$, define

$$A(t)=U(t,s)A_0U(t,s)^*,\qquad D(A(t))=U(t,s)D(A_0).$$

Its spectral projections are the same conjugates of $E_{A_0}(B)$. Consequently its distribution in $\psi(t)=U(t,s)\psi_0$ equals the initial $A_0$ distribution. This exact statement needs no formal derivative of an unbounded $A(t)$.

For bounded norm-differentiable objects, it reduces to

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

The plus sign combines the explicit change of the chosen observable with the change of the state. A driven Hamiltonian need not itself be such an invariant. The regularity needed for other Heisenberg equations is treated in Heisenberg Dynamics.

Exercises

1. Show that parity is conserved for the free Hamiltonian on $L^2(\mathbb R)$.

Solution

Parity is the bounded self-adjoint unitary $(R\psi)(x)=\psi(-x)$. Fourier transformation makes the free Hamiltonian multiplication by $\hbar^2k^2/(2m)$ and parity reversal of $k$. The even multiplier commutes with reversal, as do all its bounded spectral functions. Thus $R$ commutes with $U(t)$, and its two outcome projections $(I\pm R)/2$ are conserved.

2. If $A$ is conserved and $f$ is bounded Borel, prove that the expectation of $f(A)$ is conserved in every density state.

Solution

Spectral commutation gives $U^*f(A)U=f(A)$. Trace-class cyclicity with bounded factors gives $\operatorname{Tr}(U\rho U^*f(A))=\operatorname{Tr}(\rho f(A))$. Indicator functions recover the individual spectral probabilities.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, publisher edition — strong commutation and unbounded functional calculus.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Sections 4.2 and 5.1, author’s text — commuting spectral functions and conserved energy.