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**Is there a "magnitude of momentum" quantum operator?**

## Homework Statement

Is the ground state of the infinite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not? What can you say about the magnitude of the momentum?

## Homework Equations

Ground state wavefunction of the infinite square well: [itex]\psi(x) = \sqrt(\frac{2}{a})sin(\frac{\pi}{a}x)[/itex] (0 ≤ x ≤ a).

Momentum operator: [itex]\hat{p} = -i\hbar\frac{d}{dx}[/itex].

## The Attempt at a Solution

Since the wavefunction is normalizable, if it were an eigenfunction then we would have a state with a definite momentum, violating the uncertainty principle.

We can show this directly by plugging it in:

[itex]\hat{p}\psi(x) = -i \hbar \frac{d}{dx}(\sqrt(\frac{2}{a}) sin(\frac{\pi}{a}x))[/itex]

[itex] = -i \hbar \sqrt(\frac{2}{a}) \frac{\pi}{a} cos(\frac{\pi}{a}x) [/itex]

[itex] = \frac{-i \hbar \pi}{a} cot(\frac{\pi}{a} x) \psi(x)[/itex].

So we can see that it is not an eigenfunction.

The problem I am having is with the magnitude of the momentum. Now I must confess here that I have the solutions manual to the book, so I cheated a bit and looked in there. It says "the magnitude of the momentum [itex]\sqrt(2mE_{1}) = \pi\hbar/a[/itex] is determinate, but the particle is just as likely to be found travelling to the left (negative momentum) as to the right (positive momentum)". I don't understand how we can see this from the above equation. Is there an operator for the magnitude of the momentum, for which ψ is an eigenfunction?

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