Quantum Qual Problems

1.) Given a quantum mechanical particle of mass [math]M[/math] confined inside a box of sides [math]a,b,c[/math]. The particle is allowed to move freely between [math]0 \lt x \lt a, 0\lt y\lt b [/math] and [math]0\lt z\lt c[/math].

• Use the time-independent Schrodinger equation for this problem to obtain the general form for the eigenfunctions of the particle

• Now apply boundary conditions to obtain the specific eigenfunctions and eigenenergies for this specific problem.

• Assume [math]a=b=c[/math] and find the first 6 eigenenergies of the problem in terms of the box side length ([math]a[/math]), the particle mass ([math]M[/math]) and standard constants. What are their quantum number? Make a sketch of the eigenvalue spectrum, a table listing these eigenenergies and the quantum numbers of all the states that correspond to them.

Solution:

2.)

• [math] [- \frac{\hbar^2}{2M}\Delta^2 + V]W(x,y,z)=E W(x,y,z) [/math]
  

In our case, using separation of variables, we will get 3 differential equations for x, y and z. W(x,y,z)=w(x)w(y)w(z)

[math]\frac{d^2 w(x)}{dx^2} - m^2 w(x) = 0[/math](1)
The same will be for y and z.

Solution of equation (1) is following
[math]w(x) = A\sin(mx)+B\cos(mx)[/math]
[math]w(y) = C\sin(ky)+D\cos(ky)[/math]
[math]w(z) = E\sin(qz)+F\cos(qz)[/math]

• Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have
  

[math]w(x) = A\sin(mx) [/math]
[math]w(y) = C\sin(ky) [/math]
[math]w(z) = E\sin(qz) [/math]

Also, w(a)=0 which gives [math]A\sin(ma)=0, m=\frac{\pi n_x}{a}[/math]. For y component [math]C\sin(kb)=0, k=\frac{\pi n_y}{b}[/math] and for z [math]E\sin(qc)=0, q=\frac{\pi n_z}{c}[/math]

A, C and E are normalization constants

[math]\frac{1}{A^2}=\int\sin^2 (\pi nx/a) dx = \frac{a}{2} [/math], limits are from 0 to a.

The eigenfunction for each component will be

[math]w(x) = \sqrt{\frac{2}{a}} \sin(\pi n_x x/a)[/math]
[math]w(y) = \sqrt{\frac{2}{b}} \sin(\pi n_y y/b)[/math]
[math]w(z) = \sqrt{\frac{2}{c}} \sin(\pi n_z z/c)[/math]

The eigenenergies

[math]E_{n_x} = \frac{\pi^2 \hbar^2 {n_x}^2}{2Ma^2}[/math], [math]E_{n_y} = \frac{\pi^2 \hbar^2 {n_y}^2}{2Mb^2}[/math], [math]E_{n_z} = \frac{\pi^2 \hbar^2 {n_z}^2}{2Mc^2}[/math]
Total energy is sum of these energies.

• [math]E = \frac{\pi^2 \hbar^2 n^2}{2M^2 a^2}[/math], where [math]n^2=n_x ^2[/math]