Continuous Symmetries and Their Generators

A continuous symmetry has a self-adjoint generator only after its ray action has been lifted to a strongly continuous unitary one-parameter group. Stone’s Theorem supplies the exact equivalence and proof. This page applies it to symmetry questions: which phase choices shift the generator, what periodicity requires of its spectrum, and when the generator represents a conserved quantity.

Required background. Stone’s Theorem supplies the generator and its domain. Projective Representations and Multipliers separates ray composition from unitary composition.

First establish an exact one-parameter group

Suppose $U(s)$ is unitary, $U(0)=I$, and

$$U(s+t)=U(s)U(t), \qquad \|U(s)\psi-\psi\|\longrightarrow0 \quad(s\to0)$$

for every $\psi$. In the convention

$$U(s)=e^{-isA/\hbar},$$

Stone’s theorem gives a unique self-adjoint $A$. The units of $A$ are $\hbar$ divided by the units of $s$; for a dimensionless angle, $A$ has angular-momentum units.

The norm derivative is defined exactly on $D(A)$:

$$A\psi=i\hbar\lim_{s\to0} \frac{U(s)\psi-\psi}{s}.$$

An arbitrary phase choice for the lift of a continuous ray action need not obey the exact group law. The lifting problem comes first. Standard continuous projective representations of $\mathbb R$ admit an appropriate unitary lift, but this is a lifting result, not an extra conclusion of Stone’s theorem. Several one-parameter lifts need not assemble into one ordinary representation of a larger group.

Two exact lifts differ by a character

Assume two strongly continuous unitary groups $U,V$ induce the same ray action in dimension at least two. Phase uniqueness gives $V(s)=\beta(s)U(s)$. Strong continuity makes $\beta$ continuous, and the two group laws imply

$$\beta(s+t)=\beta(s)\beta(t).$$

A continuous character of the additive real line is $\beta(s)=e^{-ics/\hbar}$ for a real constant $c$. Hence

$$V(s)=e^{-is(A+cI)/\hbar}.$$

The generator of the ray action is therefore fixed only up to an additive scalar. The operator domain remains the same. Spectral differences and commutators with other operators are unaffected, while absolute eigenvalue labels shift.

A phase function nonlinear in $s$ generally breaks the exact one-parameter group law and cannot be interpreted as a constant shift of a Stone generator.

Circle symmetry imposes a spectral condition

Now let $s=\theta$ be an angle. If the unitary representation descends to the circle, then $U(\theta+2\pi)=U(\theta)$, so

$$e^{-2\pi iA/\hbar}=I.$$

Spectral calculus implies that the spectral measure of $A/\hbar$ is supported on $\mathbb Z$. One way to see this is to apply the positive function $|e^{-2\pi ix}-1|^2$: its spectral integral vanishes, so all spectral weight lies on its zero set. Thus the generator has integer-spaced spectral support, with arbitrary allowed multiplicities and possibly missing integers.

If only the ray action is periodic, the weaker condition is $U(2\pi)=e^{i\alpha}I$. The same argument gives support on $\mathbb Z-\alpha/(2\pi)$. A scalar shift can remove this phase for the circle subgroup. It need not remove the projective obstruction for all spatial rotations simultaneously.

This is why a periodic ray orbit and an ordinary representation of the full rotation group are different questions. Spin as a Projective Representation treats the global covering-group constraint.

A symmetry generator is not automatically conserved

The existence of $A$ says nothing yet about a chosen Hamiltonian $H$. A robust invariance condition is

$$U(s)e^{-itH/\hbar} =e^{-itH/\hbar}U(s) \quad\text{for every }s,t.$$

For self-adjoint generators, this is equivalent to strong commutation of their spectral measures. It entails the required domain invariance and expresses conservation without relying on an unspecified product commutator.

A calculation $[A,H]\psi=0$ on a dense test domain can be useful but is not, without further hypotheses, equivalent to strong commutation. Domain and integration theorems must connect the infinitesimal calculation to the bounded unitary groups.

For several Lie-group generators, differentiating on an appropriate common smooth-vector domain produces Lie-algebra relations. The global representation can still contain topology or central phases not visible in those local commutators.

Exercise: a harmless phase with a nonconstant rate

Let $U(s)$ be a strongly continuous unitary group and define $V(s)=e^{is^2}U(s)$ for dimensionless $s$. It induces the same ray map at every $s$. Does it define another one-parameter group?

Solution

Its product has phase $e^{i(s^2+t^2)}$, while $V(s+t)$ has phase $e^{i(s^2+t^2+2st)}$. These differ for generic $s,t$, so the exact group law fails. Equality of ray actions at each parameter value does not make every phase-chosen lift suitable for Stone’s theorem.

References

• V. Bargmann, “On Unitary Ray Representations of Continuous Groups”, Annals of Mathematics 59, 1–46, 1954 — continuity and lifting of ray representations.
• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — self-adjoint generators and quantum symmetry.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapters 4–5 — commuting operators and unitary dynamics.