Generators

A fixed generator is the self-adjoint observable that produces a strongly continuous one-parameter unitary group. The smooth symmetry groups behind such transformations are Lie groups, and their infinitesimal commutator structure is a Lie algebra. Translations are generated by momentum, rotations by angular momentum, and time translations by the Hamiltonian.

The classical counterpart is an infinitesimal canonical transformation generated by a phase-space function through the Poisson bracket.

The standard form is

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

where $\alpha$ is the transformation parameter and $G$ is the generator.

Generator as a Derivative

For a strongly continuous one-parameter group, the following expansion holds in operator norm when $G$ is bounded:

$$U(\alpha) = I-\frac{i}{\hbar}\alpha G+O(\alpha^2).$$

Therefore

$$G = i\hbar \left.\frac{dU}{d\alpha}\right|_{\alpha=0}.$$

For an unbounded generator, interpret the derivative on $D(G)$: $[U(\alpha)\psi-\psi]/\alpha\to-iG\psi/\hbar$. The remainder is $o(\alpha)$ there and $O(\alpha^2)$ on $D(G^2)$. An arbitrary continuous curve of unitaries need not have a derivative or be generated by one fixed $G$.

Self-Adjointness and Observables

Stone’s theorem gives a unique self-adjoint generator for a strongly continuous unitary group. In finite dimensions its Hermiticity can be checked directly:

For small $\alpha$,

$$U^\dagger(\alpha)U(\alpha) = I + \frac{i\alpha}{\hbar}(G^\dagger-G) + O(\alpha^2).$$

For matrices, first-order unitarity requires $G^\dagger=G$. For an unbounded operator, a formal first-order calculation establishes symmetry on the chosen domain, not self-adjointness by itself; the group theorem supplies the stronger conclusion.

Infinitesimal State Transformation

An active transformation of the state is

$$\lvert\psi\rangle \longmapsto U(\alpha)\lvert\psi\rangle.$$

For small $\alpha$,

$$\delta\lvert\psi\rangle = -\frac{i}{\hbar}\alpha G\lvert\psi\rangle.$$

This is the state-space version of an infinitesimal symmetry transformation.

Infinitesimal Operator Transformation

With the active convention used in this volume,

$$A\longmapsto U(\alpha)AU^\dagger(\alpha).$$

To first order,

$$\delta A = -\frac{i}{\hbar}\alpha [G,A].$$

If instead one uses the passive or Heisenberg-style convention $A\mapsto U^\dagger A U$, the sign flips:

$$\delta A = \frac{i}{\hbar}\alpha [G,A].$$

Many sign disagreements in books come from switching between these two conventions without saying so.

Translation Generator

For spatial translations along $x$,

$$T(a)=e^{-iaP/\hbar}.$$

The generator $P$ is the momentum operator. In the position representation on the line,

$$P=-i\hbar\frac{d}{dx}.$$

Indeed, for a differentiable wavefunction,

$$\psi(x-a)=\psi(x)-a\psi'(x)+o(a).$$

Comparing with $T(a)\psi=\psi-iaP\psi/\hbar+o(a)$ gives $P\psi=-i\hbar\psi'$. On the full line its self-adjoint domain consists of square-integrable functions with square-integrable weak first derivative. Boundaries require a separate domain choice.

Rotation Generator

For a rotation by angle $\theta$ about unit vector $\hat{\mathbf n}$,

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J \right).$$

The generator is $\hat{\mathbf n}\cdot\mathbf J$. The components of $\mathbf J$ satisfy the angular momentum algebra

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}J_k.$$

Time-Translation Generator

For time-independent closed-system evolution,

$$U(t)=e^{-iHt/\hbar}.$$

The generator is the Hamiltonian $H$. This is why the Hamiltonian is more than an energy function: it is the operator that generates time evolution.

Sign and Parameter Conventions

The parameter $\alpha$ may have dimensions. For translations, $\alpha=a$ has dimensions of length and $G=P$ has dimensions of momentum. For rotations, $\alpha=\theta$ is dimensionless and $G=\hat{\mathbf n}\cdot\mathbf J$ has dimensions of angular momentum. The combination $\alpha G/\hbar$ must be dimensionless.

Exercises

1. In a finite-dimensional Hilbert space, starting from $U(\alpha)=I-i\alpha G/\hbar+O(\alpha^2)$, show that first-order unitarity implies $G^\dagger=G$.

Solution

Compute

$$U^\dagger U = \left(I+\frac{i\alpha G^\dagger}{\hbar}\right) \left(I-\frac{i\alpha G}{\hbar}\right) +O(\alpha^2).$$

This equals

$$I+\frac{i\alpha}{\hbar}(G^\dagger-G)+O(\alpha^2).$$

For this to equal $I$ to first order for arbitrary $\alpha$, one needs $G^\dagger=G$.

2. What is the generator of a rotation by angle $\theta$ about the $z$ axis?

Solution

With the convention

$$U(\hat z,\theta) = e^{-i\theta J_z/\hbar},$$

the generator is $J_z$.

3. Does the generator of a transformation automatically commute with the Hamiltonian? Use $G=P$ and the oscillator Hamiltonian $H=P^2/(2m)+m\omega^2X^2/2$.

Solution

On a common invariant domain, $[P,H]=-i\hbar m\omega^2X$, which is nonzero. Momentum still generates translations, but those translations change the oscillator’s center and are not symmetries of the fixed Hamiltonian. A generator and a conserved generator are different notions.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapters 1–4.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.