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Consider a particle confined in an infinite square well potential of width L,

$$V(x)=\left\{ \begin{array}{ll}\infty, &{\rm for}\ (x \le 0)\vee (x \ge L) \\0, &{\rm for} \ 0 < x < L \end{array}\right. $$

The ground state solution of the time independent Schrödinger equation is given by $\psi$(x)=Asin(kx), where k = $\frac{\pi}{L}$ and A = $\sqrt{\frac2L}$.

Determine whether the ground state is an eigenfunction of $[p]$ and of $[p^2]$. Discuss the implication of the results.

So far I have guessed the expectation value to be $\frac{L}{2}$ and confirmed it by evaluating the integral (from an earlier part of the question). I'm stuck on how to do the last part though. From a previous second year quantum mechanics exam, any help would be greatly appreciated.

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closed as too localized by David Z Jun 25 '13 at 17:15

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I have fixed your formatting of the fraction that wasn't a fraction, including better fonts and characters. –  Luboš Motl Jun 25 '13 at 9:24
    
Welcome to Physics Stack Exchange! This is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. –  David Z Jun 25 '13 at 17:16

2 Answers 2

The momentum operator, $p=-i\frac{\partial}{\partial x}$. Apply that to the wavefunction and see what you get. For it to be an eigenfunction, it must satisfy the equation $\hat{p}\psi \propto \psi$.

Then do the same for $\hat{p}^2$.

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Forget everything and remember math.

your sine can be decomposed in two exponential functions with their arguments being opposite. You know that momentum eigenvectors are exponential functions, so you see that your energy state does not define a momentum state so nope, your ground state is not a momentum eigenfunction.

But it is pretty damn close. Those two momentum states your ground state can be decomposed into happen to have the same modulus, so yup, it is a $p²$ eigenstate.

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