# Superselection rules and non-observable Hermitian operators

Gold Member
Its usually said(like https://en.wikipedia.org/wiki/Superselectiond [Broken]) that superselection rules imply that not all Hermitian operators can be considered to be physical observables. But I don't understand how that follows. Can someone explain?
Thanks

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## Answers and Replies

Superselection means that some form of physical constraint is imposed on the general structure of observables on the Hilbert space.

As an example, the geometry of spacetime implies that rotating a system by one full cycle must not change any properties of the system. The double cover representations in quantum theory however may introduce a phase shift upon such an operation.

Fermions and Bosons will therefore be out of phase after such a rotation has been performed. Since nothing observable (in the future or the past) may change, the Hamiltonian, and with it all measurable quantities, may not depend on any phase difference between a Fermionic and a Bosonic component of the state. The two realisations of the permutation group are therefore said to live in separate superselection sectors. The observable operators then separate into a direct sum of operators on each sector.

Hoper this helps!

Jazz

ShayanJ
Gold Member
The observable operators then separate into a direct sum of operators on each sector.
So it implies that any operator which can't be decomposed that way, should not represent an observable. Right?

So it implies that any operator which can't be decomposed that way, should not represent an observable. Right?
Yes.

vanhees71
Science Advisor
Gold Member
You can put it also a bit differently and say a superselection rule constrains the possible superpositions of states. One example is the "spin selection rule". In usual quantum theory you cannot make superpositions of a state with half-integer and integer spin. Suppose you do so, i.e., having, e.g., ##s_1=1/2## and ##s_2=0## and consider the state
$$|\psi \rangle=|1/2,-1/2 \rangle+|1,-1 \rangle,$$
then the rotation around the ##z## axis by ##2 \pi## does not give ##\exp(\mathrm{i} \varphi) |\psi \rangle## for any real ##\varphi##.

Then of course you have a restriction on the possible operators, representing observables, particularly the Hamiltonian: Such an operator must not mix any half-integer spin state with an integer-spin state and vice versa.

ShayanJ