Hermitian (or anti-Hermitian) operators are of central importance in quantum mechanics in at least two different incarnations:

  • Observables are represented by Hermitian operators on the quantum mechanical state space.
  • Transformations of the state space have to preserve the Hilbert space structure: they are unitary. (Anti-)Hermitian operators are the infinitesimal generators of unitary transformations.

It follows that every observable generates a transformation of the state space, and conversely, that to a transformation of the state space corresponds an observable (or at least a Hermitian operator).

My question: is this generally a meaningful correspondence?

I learned elsewhere on this site (I cannot find the question anymore) that not all Hermitian operators correspond to observables (superselection rules), therefore a first part of the question would be if among Hermitian operators the observables correspond to a physically identifiable subset of all unitary operators.

I know of some cases in which the correspondence does seem to be meaningful:

  • when the observable is the Hamiltonian, the generated transformation (one-parameter group of transformations) is the time evolution.
  • when the transformation is a symmetry of the system, the associated generator is a conserved quantity.
  • $\begingroup$ "every observable generates a transformation of the state space" -- not really. A Hermitian operator is not anti-Hermitian. To obtain an anti-Hermitian operator from a Hermitian $A$, there are 2 equally natural ways: to multiply $A$ by $i$ or to divide $A$ by $i$. Thus, to every Hermitian operator we cannot associate just one anti-Hermitian without making an arbitrary choice, we have to associate two. $\endgroup$
    – Alexey
    Commented Feb 21 at 19:19
  • $\begingroup$ I would argue that you can only associate a one-dimensional space of Hermitian operators without making an arbitrary choice, and this corresponds to a one-parameter group of unitary transformations. Also note that in physics it is very common to speak about Hermitian (rather than anti-Hermitian) operators as generators of unitary transformations, and they write the exponential map as $X\mapsto e^{iX}$, in that case the choice is made for you. $\endgroup$
    – doetoe
    Commented Feb 23 at 21:38
  • $\begingroup$ How would you associate an anti-Hermitian operator to a Hermitian one without making an arbitrary choice? The choice between $X\mapsto e^{iX}$ and $X\mapsto e^{-iX}$ is arbitrary. $\endgroup$
    – Alexey
    Commented Feb 23 at 21:45
  • $\begingroup$ I didn't claim it was not. I claimed you can only avoid such a choice when you consider the correspondence between the whole one dimensional space $\mathbb{R}X$ and $i\mathbb{R}X$. $\endgroup$
    – doetoe
    Commented Feb 24 at 7:28
  • $\begingroup$ But there are two choice of correspondence between these spaces. None is intrinsically better than the other. So I do not see how "every observable generates a transformation of the state space", because it generates two. We are talking about physics, not about psychological or cultural phenomenon, aren't we? $\endgroup$
    – Alexey
    Commented Feb 24 at 12:47

1 Answer 1


A first remark: the term "Hermitian", even if very popular in physics is in my opinion quite misleading (because someone uses it for symmetric operators, others for self-adjoint ones).

A second remark: the self-adjoint operators of a given Hilbert space $\mathscr{H}$ are in one-to-one correspondence with the strongly continuous groups of unitary operators; not with any group of unitary operators. So it is not possible to associate observables with "unitary operators", but it is possible to associate them with strongly continuous (abelian, locally compact) groups of unitary operators.

These distinctions, even if in some sense subtle, may be important. In fact there are representations of unitary groups that does not admit a self-adjoint generator; for example the canonical commutation relations (in the exponentiated Weyl form) have such "non-regular" representations for fields, and are physically related to infrared problems (see e.g this link).

Concerning observables, the point is that it is quite difficult to give a satisfactory algebraic setting in order to collect together observables that are unbounded (as they actually are the majority of physically relevant quantities: e.g. energy, momentum...). One option is to construct an algebra of unbounded operators, but there are all kinds of domain "nightmares" to be taken into account. Another is to consider an algebra of bounded operators (a $C^*$ or von Neumann algebra), and "affiliate" unbounded self-adjoint operators to it in a suitable fashion. Both procedures are not, in my opinion, completely satisfactory; anyways the algebraic approach gives a very nice framework to understand some of the aspects of quantum theories, especially representations of groups of operators.

My personal point of view is to consider any self-adjoint operator on a given Hilbert space (usually a suitable representation of a $C^*$ algebra, or its bicommutant) as an observable. This choice is justified from the fact that any real-valued physically measurable quantity that is actually measured by physicists behaves like a self-adjoint operator (and not a symmetric one); in particular it has (mathematically speaking) an associated spectral family, as it is the case for self-adjoint operators but not for symmetric ones.

A last mathematical comment: Of course you can associate to a given self-adjoint operator $A$ a strongly continuous unitary group $e^{itA}$; and for example construct the $C^*$ algebra $\{e^{itA},t\in\mathbb{R}\}\overline{\phantom{ii}}$; where the bar stands for the closure (in the operator norm). That algebra may be very interesting to study, and be related to a certain symmetry group of transformations and so on. However, there are other algebras that could be even more interesting, for example the resolvent algebra $\{(A-i\lambda)^{-1}, \lambda\in\mathbb{R}\}\overline{\phantom{ii}}$.

In the case of CCR, the resolvent algebra has a "richer" structure of affiliated self-adjoint operators, and more importantly of automorphisms. That means that more types of quantum dynamics can be defined on the resolvent algebra, preserving it, than for the Weyl (unitary exponential) algebra. In the viewpoint of observables being only the operators affiliated to a given algebra, this means that the resolvent algebra contains more observables, and a less trivial structure of possible evolutions than the Weyl algebra.

  • $\begingroup$ Thanks for your answer. I am (to some extent) aware of the technical inaccuracies, but I didn't want to focus on those and deliberately kept the formulation loose. Let's say that my point of departure is that loosely speaking, observables correspond to transformations, and my question is whether this correspondence has a general physical significance. $\endgroup$
    – doetoe
    Commented Jul 21, 2015 at 8:56
  • 1
    $\begingroup$ @doetoe The same correspondence can be made, loosely speaking, also in classical mechanics. Flows (transformations) are generated by phase space functionals, i.e. observables. The difference is only that the classical transformations may be non-linear and act on a commutative space, while quantum transformations are restricted to be linear and act on a non-commutative space. However, there are both at the classical and quantum level, transformations that are not generated by observables (even if from observables you can always "artificially" construct a transformation). $\endgroup$
    – yuggib
    Commented Jul 21, 2015 at 9:12
  • $\begingroup$ Might be worth noting that the strongly continuous assumption in the one parameter unitary group $\leftrightarrow$ self-adjoint operator theorem can be relaxed to weakly measurable for separable Hilbert spaces Wikipedia's Article and the von Neumann reference cited there. $\endgroup$ Commented Jul 21, 2015 at 11:49
  • 1
    $\begingroup$ Actually, you can associate self-adjoint operators to unitary ones in a natural way through the Cayley Transform. This is a one-one to one correspondence between self-adjoint operators and unitary operators which do not have $1$ as an eigenvalue. This transformation is a useful one for the study of self-adjoint extensions of symmetric operators and can be used to prove the spectral theorem for unbounded self-adjoint operators from the bounded normal form. $\endgroup$ Commented Jul 22, 2015 at 18:32

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