# 3.8: The Average Momentum of a Particle in a Box is Zero

- Page ID
- 142969

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Alternative SolutionAn alternative, albeit more complicated, approach is to recognize that the uncertainty of \(p\) must be zero if the wavefunction is an eigenstate of the momentum operator. Hence

\[\sqrt{\langle p^{2}\rangle - \langle p \rangle ^{2}}=0 \nonumber \]

This requires calculating the \(\langle p^{2}\rangle\) and \(\langle p \rangle\) expectation values:

\[\begin{align*} \langle p \rangle &=\int_{0}^{L} \psi^{*}\left[-i \hbar \dfrac{d}{d x}\right] \psi d x \\[4pt] &=-i \hbar \int_{0}^{L} \dfrac{2}{L} \sin \left(\dfrac{n \pi x}{L}\right) \dfrac{d}{dx} x \sin \left(\dfrac{n \pi x}{L}\right) dx \\[4pt] &=-i \dfrac{\mathrm{h}^2}{L} \int_{0}^{L} \sin \left(\dfrac{n \pi x}{L}\right) \cos \left(\dfrac{n \pi x}{L}\right) dx \\[4pt] &= 0 \end{align*} \nonumber \]

This integral is zero via orthonormality of the sine and cosine functions (although you can expand the integrand and confirm this).

\[ \begin{align*} \langle p^{2} \rangle &=\int_{0}^{L} \psi^{*} \left[-i \hbar \dfrac{d}{d x}\right]^{2} \psi d x \\[4pt] &=\dfrac{\mathrm{h}^2}{L} \int_{0}^{L} \sin \left(\dfrac{n \pi x}{L}\right) \dfrac{d^{2} }{d x^{2}} \sin \left(\dfrac{n \pi x}{L}\right) d x \\[4pt] &=\dfrac{-\mathrm{h}^2}{L} \int_{0}^{L} \sin \left(\dfrac{n \pi x}{L}\right) \sin \left(\dfrac{n \pi x}{L}\right) d x \\[4pt] &= -\dfrac{\mathrm{h}^2}{L} \end{align*} \nonumber \]

Now the integral above is 1 using orthonormality (although you can expand the integrand and confirm this). Now we calculate the uncertainty in momentum in the PIB wavefunctions:

\[ \begin{align*} \sqrt{\langle p^{2} \rangle - \langle p \rangle^2} &=\sqrt{\dfrac{-\mathrm{h}^2}{L} -0^{2}} \\[4pt] &= \sqrt{\dfrac{-\mathrm{h}^2}{L}} \\[4pt] &\neq 0 \rightarrow \end{align*} \nonumber \]

Since the uncertainty is not 0, different measurements (experiments) will results in different values of momentum being quantified. Hence, the PIB wavefunctions are

noteigenfunctions of the momentum operator.