Question

The expectation value of \(p_x\) of the momentum of a particle trapped in a one-dimensional box is:

• A. Zero
• B. \(\dfrac{p_x}{2}\)
• C. \(\dfrac{p_x}{4}\)
• D. \(\dfrac{p_x}{\sqrt{2}}\)

Hint:

The box eigenstate is a real standing wave (equal left- and right-moving parts), so the mean momentum cancels to zero.

Answer 1

The Correct Option is A

Step 1: Use a symmetry argument. The box eigenfunction $\psi_n(x) = \sqrt{2/L}\,\sin(n\pi x/L)$ is a real function.

Step 2: For any real, normalisable wavefunction the expectation value of the momentum operator $\hat{p}_x = -i\hbar\, d/dx$ is purely imaginary from the $-i\hbar$ factor, yet $\langle p_x \rangle$ must be real because momentum is an observable (Hermitian operator). The only number that is both real and purely imaginary is zero.

Step 3: Equivalently, decompose the standing wave: $\sin(n\pi x/L) = \dfrac{1}{2i}\left(e^{i n\pi x/L} - e^{-i n\pi x/L}\right)$. This is an equal-weight mixture of a right-moving component with momentum $+n\pi\hbar/L$ and a left-moving component with momentum $-n\pi\hbar/L$.

Step 4: The two opposite momenta carry equal probability, so they cancel on average.
\[\boxed{\langle p_x \rangle = 0}\]