# Why are all transformations of quantum operators inner automorphisms?

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.

Quantum field theory in Minkowski spacetime has CPT symmetry, which is antiunitary (section II.5 in [3]), so it's not an inner automorphism. But it's not an outer autmorphism, either; it's not a (linear) automorphism at all. Since the question refers specifically to outer automorphisms, I'll try to address that. This is pushing the limits of things I've only barely begun to understand, so I'll lean heavily on the literature.

In the context of quantum field theory, in any given model, we can consider the von Neumann algebra $$A$$ generated by the model's observables. Any region $$R$$ of spacetime is associated with a subalgebra $$A(R)\subset A$$, whose observables are interpreted as being localized within that region. The algebra $$A(R)$$ is not necessarily type I. In other words, it's not necessarily isomorphic to an algebra of all bounded linear operators on any Hilbert space. In fact, according to section 6.5 in [1], there are at least some examples in which $$A(R)$$ is a type III factor. (A factor is a von Neumann algebra with trivial center.) According to exercise 14.4.12 in [2], a type III factor on a separable Hilbert space has an outer automorphism.

Even if $$A(R)$$ has an outer automorphism, it might not represent any symmetry of the model. However, if my shaky inferences are correct, then sometimes it does. One example is described in section V.4 in [3], and it is also nicely reviewed in [4]. (I don't necessarily endorse the speculations proposed in [4], but it does include a nice review.) It involves the modular group (that is, the group of modular automorphisms), which is also reviewed in [1]. Page 7 in [4] says "In general, the modular group... is not a group of inner automorphisms." The modular group depends on a choice of vector in the Hilbert space. (Interestingly, as explained on pages 7-8 in [4], the modular automorphism groups defined by different vectors are all inner-equivalent, but I think that's beside the point here.) If $$R$$ is a wedge region in Minkowski space, such as the wedge covered by a Rindler coordinate system, then the modular group for $$A(R)$$ associated with the vacuum state acts geometrically in $$R$$, namely as Lorentz boosts (page 248 in [3]). I didn't find a direct statement confirming that this particular modular group consists of outer automorphisms, but the context strongly suggests that it does, especially the context in [4]. If that inference is correct, then this seems to be an example of a symmetry implemented by outer automorphisms of the von Neumann algebra.

I'm not sure how to reconcile this with the fact that the Lorentz group can be implemented by unitary (that is, inner) automorphisms in $$A$$. I suspect that the aforementioned outer automorphisms are only outer in $$A(R)$$, and that they can be implemented as inner automorphisms in $$A$$, but I'm not sure about that. Really pushing beyond the limits of my understanding here.

A couple of extra notes:

• According to section V.4.2 in [3], "In general the wedge regions are the only ones for which the modular automorphisms (induced by the vacuum state) correspond to point transformations in Minkowski space. However, if the theory is conformally invariant there are wider classes of regions for which the modular automorphisms act geometrically. They include, as the most important case, the diamonds...''

• Section 6.2 in [5] describes an idea relating the modular group (of presumably outer automorphisms) to the AdS/CFT correspondence, and this section together with appendix D also constitutes another nice review of the mathematical background.

References:

[1] Witten (2018), "Notes on Some Entanglement Properties of Quantum Field Theory," http://arxiv.org/abs/1803.04993

[2] Kadison and Ringrose (1997), Fundamentals of the Theory of Operator Algebras, Volume II: Advanced Theory (American Mathematical Society)

[3] Haag (1996), Local Quantum Physics (Springer)

[4] Connes and Rovelli (1994), "Von Neumann algebra automorphisms and time-thermodynamics relation in general covariant quantum theories," http://www.alainconnes.org/docs/carlotime.pdf

[5] Papadodimas (2013), "State-Dependent Bulk-Boundary Maps and Black Hole Complementarity," https://arxiv.org/abs/1310.6335

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.

• 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. – Aakash Lakshmanan Jan 28 at 4:53
• 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. – Max Lein Jan 28 at 5:32