phrygian said:
I am talking about the energy eigenstates of the infinite square well, how are these not also momentum eigenstates since p = Sqrt[(2*m*KE)]
It's not that simple in quantum mechanics. The formula you have really only works well for regular numbers. But when you're dealing with operators (as you are in QM), you run into problems with multiple operators squaring to the same thing. It's like how you can have two different numbers that have the same square, one positive and one negative, but more complex, because there can be a much larger number of operators with the same square, and only one of them can really be considered the "square root" of the original operator.
You can actually check this for yourself: take the square well energy eigenfunctions
\psi(x) = A \sin(k_n x)
and first apply the momentum operator to them:
\hat{p}\psi = -i\hbar\frac{d}{d x}\psi = \cdots
Notice that what you get is
not a multiple of the original wavefunction, so it's not an eigenfunction. But if you apply the momentum operator again (try it!), you'll get something that
is a multiple of the original eigenfunction. So the square well energy eigenstates are eigenstates of \hat{p}^2 (because that's what is in the Hamiltonian), but not of \hat{p}. This can happen because \hat{p} is not the operator that you'd call the "square root" of \hat{p}^2:
\hat{p}\neq \sqrt{\hat{p}^2}
kind of like how -2 \neq \sqrt{(-2)^2}.
