Operators in quantum mechanics are basically related to each other through their Lie algebra i.e. the commutator $\times \frac{1}{i\hbar}$. This is then connected to the state space i.e. the Hilbert space through whatever relations between the vectors and operators in the form of matrix elements, or their eigenvectors, etc.

Usually, when we talk about symmetries, we talk about transformations. These transformations can either act on the space or the operators. We then observe how this transformation change relationships between the two (the space and the operators) for symmetries. When we act on the space, in order to preserve the relations between the states, we look at unitary transformation i.e. Hilbert space automorphisms. If we want to look at how to change the operators, we would consider matrix Lie algebra automorphisms, some of which can be written

$A\rightarrow U^{-1}AU$ for $U$ unitary

In fact, you can show that the unitary matrix you would use for the space transformation and the operator transformation are the same. However, it doesn't encompass all the automorphisms, only the inner automorphisms for the Lie algebra. I understand, however, that in the theory of Lie algebras, there exist outer automorphisms which preserve the Lie algebra but corresponds to no unitary matrix conjugation and so I am unclear on how this affects the states due to the strong correspondence earlier. I am not sure if these do or can come up in the study of quantum mechanics. Please let me know if you know about any other additional structure that must be preserved for the operator relations if you want to reduce them all to unitary conjugations.


1 Answer 1


What you write is not correct, because it crucially depends on what your algebra of observables is. Even in the first quantized formalism, your algebra (typically a $C^*$- or von Neumann algebra) is small enough so that there are plenty of automorphisms which are not inner. The simplest example is when your algebra is not unital. Then unitaries cannot be elements of the algebra (because then $U \, U^* = \mathbf{1}$ would also be an element of the algebra, but that would mean the algebra is unital), but conjugating an element with some unitaries could still give you an element of the algebra. Another example are the algebra of compact operators on an infinite-dimensional Hilbert space. Because the compact operators are a two-sided ideal, for any compact operator $K$ and unitary $U$ the product $U \, K , U^*$ is necessarily a compact operator.

So when these are represented as bounded operators on a Hilbert space, and then in that potentially larger algebra of bounded operators, they are inner. In the previous example, you can picture the compact operators being a proper subset of the bounded operators (still assuming that the Hilbert space is infinite-dimensional). If the relevant algebra are the compact operators, then not all automorphisms are inner. If you instead consider the algebra of bounded operators, conjugating by unitaries is necessarily an inner automorphism.

  • $\begingroup$ Hmm, sounds convincing. Maybe I’m not really sure when outer automorphisms actually manifest because I haven’t really studied Lie algebras enough. Is it the case that expanding the space sometimes makes all outer automorphisms inner? What would be an outer automorphism for compact but not for bounded? I’m really looking for those transformations that can’t be expressed as conjugating by unitaries, not so much determining if all unitary conjugations yield a valid automorphism at this point. $\endgroup$ Commented Jan 28, 2019 at 4:53
  • $\begingroup$ Yes, there are cases where outer automorphisms become inner. The example with the compact operators is one such case. There are also examples where automorphisms have nothing to do with conjugations by unitaries. As an example of an automorphism that is not written in terms of unitaries, consider translations on the $C^*$-algebra $\mathcal{C}_{\infty}(\mathbb{R}^d)$, the continuous functions on $\mathbb{R}^d$, which vanish at infinity. $\theta_x(f) := f(\cdot - x)$ is clearly an automorphism, but cannot be written as a conjugation with a unitary. $\endgroup$
    – Max Lein
    Commented Jan 28, 2019 at 5:32

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