Question

Question: Do Problem 6 in polar coordinates to find the eigenfunctions and energy eigenvalues of a particle in a circular box$\mathit{r}\mathbf{<}\mathit{a}$.

Short answer

The solution is${\mathrm{\Psi }}_{\mathrm{nm}}={\mathrm{J}}_{\mathrm{n}}\left({\mathrm{k}}_{\mathrm{nm}}\mathrm{r}\right)\left\{\begin{array}{l}\sin \left(n\theta \right)\\ \cos \left(n\theta \right)\end{array}\right.{\mathrm{e}}^{-\mathrm{i}\frac{{\mathrm{E}}_{\mathrm{nn}}}{h}}$

Step-by-step solution

Given Information.

An expression has been given as$0⟨x⟨1,\text{0}⟨\text{y}⟨\text{1}$ .

Definition of Schrodinger equation.

A linear partial differential equation that determines the wave function of a quantum-mechanical system is known as the Schrödinger equation.

The Schrodinger equation is,

$\mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}{\mathbf{\nabla }}^{\mathbf{2}}\mathit{\Psi }\mathbf{+}\mathit{V}\mathit{\Psi }\mathbf{=}\mathit{i}\mathit{\hslash }\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}\mathit{\Psi }$.

Use the Schrodinger equation.

Use the Schrodinger equation.

$-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi =i\hslash \frac{\partial }{\partial t}\Psi$

Assume a solution of the form mentioned below.

$\Psi \left(x,y,t\right)=\psi \left(x,y\right)T\left(t\right)$

The equation can be separated into a time equation with the solution given below.

$T\left(t\right)=e^{-\frac{iEt}{\hslash }}$

It can be separated into a space equation (with V = 0 )

$-\frac{{\hslash }^2}{2m}{\nabla }^2\psi =E\psi$

Write the polar coordinate of the equation.
$-\frac{{\hslash }^2}{2m}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \psi }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2\psi }{\partial {\theta }^2}\right)=E\psi$

Put a solution of the form given below.

$\psi =R\left(r\right)\Theta \left(\theta \right)$

Rewrite the above equation.
$-\frac{{\hslash }^2}{2m}\left(\Theta \frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)+R\frac{1}{r^2}\frac{{\partial }^2\Theta }{\partial {\theta }^2}\right)=E·R·\Theta \mathit{0}$

Then divide by and then multiplying by $-r^2\frac{2m}{{\hslash }^2}$.

$r\frac{1}{R}\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)+\frac{1}{\Theta }\frac{{\partial }^2\Theta }{\partial {\theta }^2}+\frac{2mE}{{\hslash }^2}=0$

The second term a function of only$\frac{1}{\Theta }\frac{{\partial }^2\Theta }{\partial {\theta }^2}=-n^2$ .

Solve further.

Write the solutions.
$\mathrm{\Theta }\left(\mathrm{\theta }\right)=\left\{\begin{array}{l}\sin \left(n\theta \right)\\ \cos \left(n\theta \right)\end{array}\right.$

It is known that the function is a periodic function must be a natural number.

Rewrite the full equation.
$r\frac{1}{R}\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)-n^2+\frac{2mE}{{\hslash }^2}=0$

Multiply by R.
$r\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)+\left(\frac{2mE}{{\hslash }^2}-n^2\right)R=0$

The solution must be as mentioned below.
$R\left(r\right)=J_n\left(Kr\right)$00

As in the text, the value on the boundary$r=a$ is 0.
$$\begin{gathered} R\left(a\right)=0 \\ =J_n\left(Ka\right) \end{gathered}$$

Define $K=\frac{k}{a}$where are zeros of the Bessel function.

Combining the space and the time solutions.
width="228">Ψnm=Jn(knmr)sinnθcosnθe-iEnnh

Here,$E_{nm}=K_{nm}^2\frac{{\hslash }^2}{2m}$and$k_{nm}$is the$m^{th}$ root of J_n