# The operator X for the position in QFT

#### DrDu

Can you name one hermitian operator in non-relativistic QM which is not an observable?
Let c be the anihilation operator of an electron. Then $c+c^\dagger$ is hermitian but unobservable due to charge or univalence superselection. Based on the latter, you can show that any fermionic operator is not observable.

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#### Demystifier

2018 Award
Let c be the anihilation operator of an electron. Then $c+c^\dagger$ is hermitian but unobservable due to charge or univalence superselection. Based on the latter, you can show that any fermionic operator is not observable.
I said an operator in non-relativitic QM. By that I meant an operator acting in the space of a fixed number of particles.

#### DrDu

The number of particles being fixed is a consequence of mass superselection, which can be shown to be a consequence of Galilean symmetry. To prove it, you have to work in Fock space. Besides this, working in Fock space is also very convenient in non relativistic QM and is almost always done in solid state physics, as you can use a unified approach for all particle numbers.

Besides your argumentation makes it impossible to give a counter example. You can always argue that a non-observable quantity is superfluous even in a relativistic context and that you can live in principle without it. However, it is mostly very convenient, if not mandatory, mathematically to work with non-observable quantities.

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