Question

A study of classical waves tells us that a standing wave can be expressed as a sum of two travelling waves. Quantum-Mechanical travelling waves, discussed in Chapter 4, is of the form $\mathit{\psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{=}\mathbf{\omega }\mathbf{t}\mathbf{)}}$. Show that the infinite well’s standing wave function can be expressed as a sum of two traveling waves.

Short answer

$\psi (x,t)=2Ai\sin \left(kx\right)e^{-iex}$

What we ended up with is the indefinite-well’s standing-wave function, thus proving it is just the sum of two waves moving in opposite directions.

Step-by-step solution

Step (1): Introduction

To show that the infinite well’s standing-wave function can be expressed as the sum of two travelling waves, we add two waves moving in opposite directions.

Step (2): Finding sum of two waves

The equation for each wave is

$\psi (x,t)=A{e}^{i(kx-\omega t)}$

Since our two waves are moving in opposite directions, they have opposite momentum(so our second wave is negative) and an opposite sign for K. This gives us equation for the sum of the travelling waves:

$\psi (x,t)=A{e}^{(kx-\omega t)}$

So,role="math" localid="1658425779738" $\begin{array}{rcl}\psi (x,t)& =& A\left(e^{i(kx-\omega t)}-A{e}^{i(kx-\omega t)}\right)\\ & =& A{e}^{i(kx-\omega t)}\left(e^{ikx}-e^{-ikx}\right)\\ & =& 2Ai\left(\frac{e^{ikx}-e^{-ikx}}{2i}\right)e^{iex}\end{array}$

We know that,

$\begin{array}{rcl}& & \left(\frac{e^{ikx}-e^{-ikx}}{2i}=\sin kx\right)\end{array}$

Thus,

$\psi (x,t)=2Ai\sin \left(kx\right)e^{-iex}$

What we ended up with is the indefinite-well’s standing-wave function, thus proving it is just the sum of two waves moving in opposite directions.