Why Are Infinite Square Well Eigenstates Not Energy or Momentum Eigenstates?

[phrygian]

Homework Statement

The eigenstates of the infinite square well are not energy eigenstates and are not momentum eigenstates.

Homework Equations

The Attempt at a Solution

I don't understand how this can be? If the eigenstates of the infinite square well are energy eigenstates, and thus have a definite energy, how can that not have a definite momentum given that momentum in the well is just Sqrt[(2*m*KE)]?

 

[diazona]

What eigenstates are you talking about? i.e. the eigenstates of which operator? There's no such thing as "eigenstates of the infinite square well", there are only eigenstates of each individual operator in the infinite square well potential.

 

[phrygian]

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)]

 

[diazona]

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 $-2 \neq \sqrt{(-2)^2}$.