Question

Quantum-mechanical stationary states are of the general form $\mathit{\Psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\omega }\mathbf{t}}$. For the basic plane wave (Chapter 4), this is $\mathit{\Psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\omega }\mathbf{t}}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}$, and for a particle in a box it is $\mathbf{A}\left(\mathbf{sinkx}\right){\mathbf{e}}^{\mathbf{-}\mathbf{i\omega t}}$. Although both are sinusoidal, we claim that the plane wave alone is the prototype function whose momentum is pure-a well-defined value in one direction. Reinforcing the claim is the fact that the plane wave alone lacks features that we expect to see only when, effectively, waves are moving in both directions. What features are these, and, considering the probability densities, are they indeed present for a particle in a box and absent for a plane wave?

Short answer

Feature for a particle in a box: the wave function is equal to zero at any point where $kx=n\pi$.

Feature for a plane wave: the function could never be equal to zero, hence has pure momentum and well-defined value in one direction.

Step-by-step solution

Given data

For a basic plane wave, the wave function is given by $\Psi (x,t)=A{e}^{i(kx-\omega t)}$ and for a particle in a box the wave function is given by $\Psi (x,t)=A\sin \left(kx\right)e^{-i\omega t}.$

Concept of quantum mechanics

In quantum mechanics, a particle does not have a fixed boundary or a definite momentum. Here, particles are considered to be a superposition of a large number of waves forming a wave packet.

Time-dependent wave

If a wave is moving in both directions, then it will form a standing wave, and hence nodes will be formed. For a particle in a box, the solution has the term $\left(\sin \left(kx\right)\right)$, which is definitely zero when $kx=n\pi$but for a plane wave, the function $\exp \left(i\left(kx-\omega t\right)\right)$is not zero, though its complex square is a constant.

Hence, we see that a wave function for a plane wave cannot be zero.