Recent content by hexiphysics

Yes or the distance from the point charge to the field point in question.

Sorry I confused myself, in Griffiths it is stated as an axiom that all observables with Hermitian operators have a complete set of eigenfunctions. If xp+px does not have a complete set, it is not observable.

Yes I understand this. I am saying that I do not believe that xp+px is measurable because it does not have a complete set of eigenstates. I used the uncertainty principle to offer an explanation as to why it may not be able to be measured.

I do not believe xp+px is considered an observable because you would not be able to measure an exact eigenstate due to the uncertainty principle.

In order for an operator to represent an observable it must be Hermitian. Hermitian operators always have eigenfunctions that form a complete set, so in order for something to be an observable it must be Hermitian and that implies that its eigenfunctions are complete.

Think of distances in different dimensions. What is the length of a radial vector in 1,2, and 3 dimensions?

Think about what you know about solutions to the time independent Schrödinger equation. What does integral (-inf, inf) Psi1(x)^2 equal? What do you know about integrating stationary states of different energy levels from (-inf, inf). Hope this helps! (sorry I don't know how to make the...