Question

Is the ground state of the infinite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not?

Short answer

The wavefunction ${\psi }_n$is not a momentum eigenfunction.

The magnitude of the momentum is a constant, and it is given by,

$\left|p\right|=\sqrt{2mE}=\frac{n\pi \hslash }{a}$

Step-by-step solution

Concept used

The wave function of the infinite square well's stationary states is calculated as follows:

${\mathit{\psi }}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{n\pi x}}{\mathbf{a}}\mathbf{\right)}$

Where $\mathit{a}$ is the width of the well.

Calculate the momentum

The wave function of the infinite square well's stationary states is calculated as follows:

$\hat{\mathbf{p}}=-i\hslash \frac{d}{dx}$

We can check that directly as:

$$\begin{gathered} \hat{p}{\psi }_n=-i\hslash \frac{d}{dx}\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right) \\ =-i\hslash \sqrt{\frac{2}{a}}\frac{n\pi }{a}\cos \left(\frac{n\pi x}{a}\right) \\ =-i\frac{\hslash n\pi }{a}\cot \left(\frac{n\pi x}{a}\right){\psi }_n\left(x\right) \end{gathered}$$

Since the momentum operator doesn't yield the original wave function multiplied by a constant, then the wavefunction ${\psi }_n$is not an eigenfunction of momentum. The mean momentum is:

$$\begin{gathered} ⟨p⟩={\int }_0^a{\psi }_n\hat{p}{\psi }_{n}dx \\ =-i\frac{2n\pi \hslash }{a^2}{\int }_0^a\sin \left(\frac{n\pi x}{a}\right)\cos \left(\frac{n\pi x}{a}\right)dx \\ =0 \end{gathered}$$

Which means the particle is just as likely to be found traveling to the left as to the right. The magnitude of momentum is a constant that is given by,

$\left|p\right|=\sqrt{2mE}=\frac{n\pi \hslash }{a}$