Question

Use separation of variables in Cartesian coordinates to solve infinite cubical well

_$V(x,y,z)=0$if x,y,z are all between 0 to a;

_{$V(x,y,z)=\infty$}Otherwise

a) Find the stationary states and the corresponding energies

b) Call the distinct energies $E_1,E_2,E_3,..$in the order of increasing energy. Find_{localid="1658127758806" $E_1,E_2,E_3,E_4,E_5,E_6$}determine their degeneracies (that is, the number of different states that share the same energy). Comment: In one dimension degenerate bound states do not occur but in three dimensions they are very common.

c) What is the degeneracy of E₁₄ and why is this case interesting?

Short answer

(a) _{$$\begin{gathered} \Psi (x,y,z)={\left(\frac{2}{a}\right)}^{\frac{3}{2}}\sin \left(\frac{n_x\mathrm{\pi }}{a}X\right)s\mathrm{in}\left(\frac{n_y\mathrm{\pi }}{a}y\right)s\mathrm{in}\left(\frac{n_z\mathrm{\pi }}{a}z\right) \\ E(x,y,z)=\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2}(n_x^2+n_y^2+n_z^2)n_x,n_y,n_z=1,2,3,....... \end{gathered}$$}

(b) Distinct energies calculated in the step 3.

(c) $E_{14}=4$

Step-by-step solution

Define the Schrödinger equation

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

Determine the corresponding energies

_{$-\frac{h^2}{2m}(\frac{{\partial }^2\Psi }{\partial x^2}+\frac{{\partial }^2\Psi }{\partial y^2}+\frac{{\partial }^2\Psi }{\partial z^2})=E\Psi$}

Separable solutions:_{$\Psi (x,y,z)=X\left(x\right)Y\left(y\right)Z\left(z\right)$}

Put this in and divide by $XYZ$

_{role="math" localid="1658128747853" $$\begin{gathered} -\frac{h^2}{2m}{\nabla }^2\Psi +V\Psi =E\Psi \\ \frac{1}{X}\frac{d^{2}x}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}+\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}=-\frac{2m}{h^2}E \end{gathered}$$}

The three terms on the left are functions of $x,y,z$, so each must be a constant, call the separation constants _{$K_x^2,K_y^2,K_z^2$}

$$\begin{gathered} \frac{d^{2}X}{d{x}^2}=-K_x^{2}X;\frac{d^{2}Y}{d{y}^2}=-K_y^{2}Y;\frac{d^{2}Z}{d{z}^2}=-K_y^{2}Z \\ X\left(x\right)=A_x\sin k_{x}x+B_x\cos k_{x}x \\ Y\left(y\right)=A_y\sin k_{y}Y+B_y\cos k_{y}Y \\ Z\left(z\right)=A_z\sin k_{z}Z+B_z\cos k_{z}Z \\ X\left(0\right)=0,B_x=0; \\ Y\left(0\right)=0,B_y=0 \\ Z\left(0\right)=0,B_z=0 \\ X\left(a\right)=0 \\ \sin \left(k_{x}a\right)=0k_{x}a=n\tau \tau \\ {k}_x=\frac{n_x\tau \tau }{a},k_y=\frac{n_y\tau \tau }{a},k_z=\frac{n_z\tau \tau }{a} \\ \psi (x,y,z)=A_x{A}_y{A}_z\sin \left(\frac{n_x\tau \tau }{a}X\right)\sin \left(\frac{n_y\tau \tau }{a}Y\right)\sin \left(\frac{n_z\tau \tau }{a}Z\right) \\ {A}_x=A_Y=A_z=\sqrt{\frac{2}{a}} \\ \psi (x,y,z)={\left(\frac{2}{a}\right)}^{\frac{3}{2}}\sin \left(\frac{n_x\tau \tau }{a}X\right)\sin \left(\frac{n_y\tau \tau }{a}Y\right)\sin \left(\frac{n_z\tau \tau }{a}Z\right) \\ E(x,y,z)=\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2a}(n_x^2+n_y^2+n_z^2) \end{gathered}$$

Determine the distinct energies

$$\begin{gathered} d=1;E_1=E_{111}=\frac{3{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \\ d=2;E_2=E_{211}=E_{121}=E_{112}=\frac{6{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \\ d=3;E_3=E_{221}=E_{212}=E_{122}=\frac{9{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \\ d=3;E_4=E_{311}=E_{131}=E_{113}=\frac{11{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \\ d=1;E_5=E_{222}=\frac{12{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \\ d=1;E_6=E_{123}=E_{231}=E_{312}=E_{132}=E_{213}=E_{321}=\frac{14{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{a}^2} \end{gathered}$$

Step 4:Determine the degeneracy

After calculating$E_7,E_8,.....,E_{13}$we find that:

$E_{14}=E_{333}=E_{115}$, so the degeneracy is 4.

This is called "accidental degeneracy" since $(3,3,3)$and $(1,1,5)$conspired to have the same energy eigenvalue 27.