In Classical Mechanics to describe symmetries like translations and rotations we use diffeomorphisms on the configuration manifold. In Quantum Mechanics we use unitary operators in state space. We impose unitarity because we don't want the action of the symmetry to interfere with the normalization of the states.

Now, examples of that are the cases of translations and rotations. Those are defined directly in terms of the classical ones. In the case of rotations if $\mathcal{E}$ is the state space of a spinless particle in three dimensions and if $R\in SO(3)$ is a rotation in three dimensions we have the induce rotation $\hat{R}\in \mathcal{L}(\mathcal{E})$ by

$$\langle \mathbf{r} | \hat{R}|\psi\rangle = \langle R^{-1}\mathbf{r}|\psi\rangle.$$

Now in Classical Mechanics we often want to talk about the infinitesimal version of a symmetry which is known its generator. In Quantum Mechanics the same idea is quite important. The generators of rotations, for example, are the angular momentum operators.

The whole point with generators is that

  1. They can be interpreted as the infinitesimal version of a symmetry.
  2. In analogy with Lie groups if $A$ and if $\xi$ is its generator we should be able to write

$$A_\alpha = \exp(i\alpha \xi),$$

where $\alpha$ is a parameter characterizing the extent to which we are applying the symmetry. The operators generators $\xi$ are then hermitian operators.

These are facts that I know in a totally informal and non rigorous manner. What I want is to make the idea of generators of a symmetry in QM precise.

One problem that we already have is that the exponential might converge since there are operators which are unbounded. In any case: how do we precisely define the generators of an operator, how do we show they exist and how do we write the operator as an exponential in terms of its generators in a rigorous manner?


1 Answer 1


The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$.

The definition of the exponential for an unbounded self-adjoint operator requires theorems from Borel functional calculus that say that for every measurable function $f$ on the reals the expression $f(O)$ for $O$ a self-adjoint operator defines a unique operator with the property that $f(O)v_\lambda = f(\lambda) v_\lambda$ for every eigenstate $v_\lambda$ with eigenvalue $\lambda$. Naively, you might even take this as the definition of $f(O)$.

You can find the proofs of these assertions for example in chapter VIII of "Methods of Modern Mathematical Physics" by Reed and Simon.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.