Exact value for every observable in QM?

[asimov42]

Folks - I'm asking a lot of questions lately (hopefully useful not just for me).

By chance, reading about quantum states, I referenced Wikipedia (dubious I know), and came across the following phrase (with a citation, that I will check): "Even in quantum theory, however, for every observable there are some states that have an exact and determined value for that observable."

If I consider, say, the position operator, then I can certainly see the above being true (e.g., a state that result in a delta function for the position) - but this state cannot be physically realizable, can it? (i.e., they are non-normalizable, correct?).

 

[andrewkirk]

I expect it is talking about observables that have a finite or countable set of measurable values, such as spin (finite) or energy (infinite but countable). The exact states (eigenstates) of those observables are physically realisable, whereas those of observables whose potential values comprise a continuum (such as location) are not.

 

[vanhees71]

Folks - I'm asking a lot of questions lately (hopefully useful not just for me).

By chance, reading about quantum states, I referenced Wikipedia (dubious I know), and came across the following phrase (with a citation, that I will check): "Even in quantum theory, however, for every observable there are some states that have an exact and determined value for that observable."

If I consider, say, the position operator, then I can certainly see the above being true (e.g., a state that result in a delta function for the position) - but this state cannot be physically realizable, can it? (i.e., they are non-normalizable, correct?).

Ironically for the position operator it's not true. Since the corresponding self-adjoint operator has an entirely continuous spectrum, there is no proper eigenvector (in the wave-function language a square-integrable eigenfunction) for any of its spectral values, and thus there is no state for which the position has a determined value.

 

[Demystifier]

Ironically for the position operator it's not true. Since the corresponding self-adjoint operator has an entirely continuous spectrum, there is no proper eigenvector (in the wave-function language a square-integrable eigenfunction) for any of its spectral values, and thus there is no state for which the position has a determined value.

Position operator is not an exception. When particle is not in a bound state, momentum and energy also have a continuous spectrum.

 

[vanhees71]

That's true. The argument holds for all values of an observable in the continuous part of the representing operator's spectrum.