91影视

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

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

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

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

$\hat{\mathbf{p}}=-i\mathrm{鈩�}\frac{d}{dx}$

We can check that directly as:

$$\begin{gathered} \hat{p}{蠄}_n=-i\mathrm{鈩�}\frac{d}{dx}\sqrt{\frac{2}{a}}\sin \left(\frac{n蟺x}{a}\right) \\ =-i\mathrm{鈩�}\sqrt{\frac{2}{a}}\frac{n蟺}{a}\cos \left(\frac{n蟺x}{a}\right) \\ =-i\frac{\mathrm{鈩�}n蟺}{a}\cot \left(\frac{n蟺x}{a}\right){蠄}_n\left(x\right) \end{gathered}$$

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

$$\begin{gathered} 鉄�p鉄�={鈭�}_0^a{蠄}_n\hat{p}{蠄}_{n}dx \\ =-i\frac{2n蟺\mathrm{鈩�}}{a^2}{鈭�}_0^a\sin \left(\frac{n蟺x}{a}\right)\cos \left(\frac{n蟺x}{a}\right)dx \\ =0 \end{gathered}$$

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,

$\left|p\right|=\sqrt{2mE}=\frac{n蟺\mathrm{鈩�}}{a}$