# Symmetric, (essentially) self-adjoint operators and the spectral theorem

At the moment I got a bit confused about the notation in some QM textbooks. Some say the operators should be symmetric, some say they should be essentially self-adjoint, others that they has to be self-adjoint (or in many cases hermitian what maybe means symmetric or maybe self-adjoint). Which condition do we need for our observables (cause they are not the same in the case of an infinite-dimensional Hilbert space)?

If symmetric or essentially self-adjoint is enough why can we find an othonormal basis of eigenvectors (since the spectral theorem holds only for self-adjoint operators)?

## 2 Answers

The reason why we need self-adjointness for observables is mainly that we require their spectrum to be real.

However characterizing the exact domain of self-adjointness is often difficult, while it is easier to characterize some of its cores. Therefore essential self-adjointness on a given (usually easily chracterizable) core is helpful since the self-adjoint extension is then unique.

Symmetric but not self-adjoint operators may not have a real spectrum, and in addition they do not generate unitary groups of operators that physically implement symmetries as self-adjoint operators do instead. Therefore they're not a good candidate to be physically relevant.

• Thanks, I got the point. But in this case there is one further problem left: If we assume that all the operators are self-adjoint and not defined everywhere (since they are unbounded) how can we make sure that the products of operators are well-definied, i.e. why is the image of the first in the domain of the second and so on? – StrangeField Jan 21 '17 at 22:30
• There is no general strategy, and often it is not possible to define the product on a given domain. There are results in that sense for concrete operators of physical importance (e.g. position and momentum of quantum mechanics), and suitable cores of essential self-adjointness where the product is defined. For example, both $x$ and $-i\nabla$ map the Schwartz space $\mathscr{S}(\mathbb{R}^d)$ of rapidly decreasing functions into itself, and the latter is a core for both. Therefore their product and the commutator are well-defined on that space. – yuggib Jan 22 '17 at 10:02

First of all, from the practical / Physics point of view you will see that it actually makes no difference, it's just mathematical details. But I do understand the desire to have a mathematically precisely defined theory.

Observables are in fact required to be self-adjoint, and the reason is - like you guessed - that we need the spectral theorem (see e.g. here).
Since an essentially self-adjoint operator has a unique self-adjoint extension, it does not really matter whether we write down a self-adjoint operator or "just" an essentially self-adjoint one.

In a Physics lecture, the professor will usually only write down the momentum operator as $p = -\mathrm i\, \partial_x$ without specifying a domain, and prove that it is symmetric while implicitly assuming that the wave functions are continuously differentiable or something similar. Explaining the issue of domains of unbounded operators and introducing Sobolev spaces etc would take a lot of time for arguably little benefit.
The property will then be called "hermitian" which seems to be used in the meaning of "self-adjoint but we don't really care about the details". (As far as I know, hermitian is originally supposed to mean bounded self-adjoint. See also here.)