Symmetric, self-adjoint operators and the spectral theorem

[Neutrinos02]

Hi Guys,

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 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 Hilbertspace)?

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

Thanks for your help

 

[dextercioby]

The observables should be self-adjoint operators, but essential self-adjointness would do. There's no guarrantee for a purely real spectrum for a symmetric operator.

 

[vanhees71]

A very nice paper about the trouble you can get in, if you don't take care that the operators representing observables should be (essentially) self-adjoint is

https://arxiv.org/abs/quant-ph/9907069

Another one, treating self-adjointness properly

https://arxiv.org/abs/quant-ph/0103153

 

[Neutrinos02]

Thanks for your answers. The fact that the operator should be self-adjoint makes sense but there is one 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?

 

[dextercioby]

This is a very good question. Self-adjoint operators won't typically have the same domain, but it may happen that the maximal common domain of them is an essential self-adjointness domain and moreover this domain is also invariant for the polynomial algebra.
Example: the Schwartz test function space in R is a common dense everywhere invariant domain for the x, p and p^2 (this is the free particle Hamiltonian) operators. All 3 of them are esa when restricted to this space.

 

[A. Neumaier]

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-defined, i.e. why is the image of the first in the domain of the second and so on?

In general this is not the case and the product need not exist. However, in most quantum theories of interest, there is an algebra of operators of interest with a fixed common dense domain (for $N$-particle QM, the Schwartz space in $3N$ dimensions).