Energies for Simple Quantum Well

Classical Mechanics Level 3

A particle in a one-dimensional quantum well is governed by a variant of the time-independent Schrodinger equation, expressed in terms of wave function $\Psi (x)$ .

The quantities $V$ and $E$ are the potential energy and total energy, respectively. $E$ is a positive constant.

$-\frac{d^2}{d x^2} \, \Psi (x) + V(x) \, \Psi (x) = E \, \Psi (x)$

The potential energy varies as follows:

$V(x)= \begin{cases} \infty, &\,\, x < 0 \\ 0, &\,\, 0 \leq x \leq \pi \\ \infty, &\,\, x > \pi \\ \end{cases}$

The boundary conditions on $\Psi (x)$ are:

$\Psi(x) = \begin{cases} 0, \,\, x \leq 0 \\ 0, \,\, x \geq \pi \\ \end{cases}$

Determine the sum of the five smallest non-zero allowable values of $E$ .

Note: This problem is easily solvable by hand

Solution Strategy:
1) Solve the differential equation within the well $(V = 0)$ to derive a general solution for $\Psi(x)$ within the well
2) Apply the boundary conditions $\Psi(0) = 0$ and $\Psi(\pi) = 0$ to the solution
3) Energy quantization arises naturally as a result of the prior two steps

The answer is 55.

1 solution

Karan Chatrath
Jan 1, 2021

We are asked to solve:

$\frac{d^2\Psi}{dx^2} + E\Psi = 0 \implies \Psi(x) = A \cos(x\sqrt{E}) + B \sin(x\sqrt{E})$

Applying boundary conditions: $\Psi(0) = 0$ $\implies A = 0$ $\Psi(\pi) = 0 \implies B \sin(\pi \sqrt{E}) = 0$ $\pi\sqrt{E} = n\pi$ Where $n = 1,2,3 \dots$ .

Based on this:

$E_n = n^2$ $\implies \sum_{n=1}^{5} E_n = \sum_{n=1}^{5} n^2 = 55$

I have always wanted to explore the quantum harmonic oscillator (QHM) problem. Never done so before. This exercise has proved to be a neat gateway into QHM. I will read about it at leisure.

Karan Chatrath - 5 months, 1 week ago

Energies for Simple Quantum Well

The answer is 55.

1 solution

0 pending reports