Question

Consider a cubic 3D infinite well.

(a) How many different wave functions have the same energy as the one for which $\left(n_x,n_y,n_z\right)\mathbf{=}\left(5,1,1\right)$?

(b) Into how many different energy levels would this level split if the length of one side were increased by $5\%$ ?

(c) Make a scale diagram, similar to Figure 3, illustrating the energy splitting of the previously degenerate wave functions.

(d) Is there any degeneracy left? If so, how might it be “destroyed”?

Short answer

(a) There are two other wave functions having the same energy as the one for which $\left(n_x,n_y,n_z\right)=\left(5,1,1\right)$.

(b) This energy level would split into two new energy levels if the length of one side is increased by $5\%$.

(c) The scale diagram is as follows

(d) There will still be two wave functions having the same energy. This can be removed by changing any of the side lengths by a small amount.

Step-by-step solution

Given data

There is a 3D infinite cubic well.

Energy in an infinite well

The energy of a particle of mass m in a 3D infinite well of sides $L_x$, $L_y$and $L_z$ is

$E=\frac{{\pi }^2{\hslash }^2}{2m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\right)$ ..... (I)

Here, $\hslash$ is the reduced Planck's constant and $n_x$, $n_y$ and $n_z$ are the quantum numbers

Step 3:Determining the degeneracy of energy in a cubic well

(a)

For a cubic well, the energy in equation (I) reduces to

$E=\frac{{\pi }^2{\hslash }^2}{2m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$ ..... (II)

Thus the energy when any one of $\left(n_x,n_y,n_z\right)$ is 5 and the rest two are 1 are the same. Hence the wave functions corresponding to $\left(n_x,n_y,n_z\right)=\left(5,1,1\right)$, $\left(n_x,n_y,n_z\right)=\left(1,5,1\right)$ and $\left(n_x,n_y,n_z\right)=\left(1,1,5\right)$ have the same energies. There are three wave functions having the same energy.

Determining the splitting in energy level

(b)

Let the length along X axis, that is$L_x$ be increased by $5\%$. The new length is then $\$1.05L\$$. The energy in equation (II) becomes

$E=\frac{{\pi }^2{\hslash }^2}{2m{L}^2}\left(\frac{n_x^2}{{1.05}^2}+n_y^2+n_z^2\right)$

Thus the wave function corresponding to$\left(n_x,n_y,n_z\right)=\left(5,1,1\right)$will now have lower energy than the wave functions corresponding to $\left(n_x,n_y,n_z\right)=\left(1,5,1\right)$and$\left(n_x,n_y,n_z\right)=\left(1,1,5\right)$. The previous energy level will thus split into two separate energy levels.

Plotting the energy levels

(c)

The plot of the initial and final energy levels are as follows

Determining the residual degeneracy

(d)

There are still two wave functions corresponding to $\left(n_x,n_y,n_z\right)=\left(1,5,1\right)$ and $\left(n_x,n_y,n_z\right)=\left(1,1,5\right)$ that are in the same energy level. Thus a degeneracy of 2 is left. This can be removed if one of the lengths along x or y axis is slightly changed making the energy equation equal to equation (I).