91影视

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

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

The Schrodinger equation is,

.$\mathbf{-}\frac{{\mathbf{鈩�}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{唯}\mathbf{+}\mathbf{痴唯}\mathbf{=}\mathbf{i}\mathbf{鈩�}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{t}}\mathbf{唯}$

Use the Schrodinger equation.
$-\frac{{\mathrm{鈩�}}^2}{2m}{鈭�}^2唯+V唯=i\mathrm{鈩�}\frac{鈭�}{鈭�t}唯$

Assume a solution of the form mentioned below.
$唯\left(x,y,t\right)=蠄\left(x,y\right)T\left(t\right)$

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

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

It can be separated into a space equation (with V = 0)
$-\frac{{\mathrm{鈩�}}^2}{2m}{鈭�}^2蠄=E蠄$

Write the polar coordinate of the equation.
$-\frac{{\mathrm{鈩�}}^2}{2m}\left(\frac{1}{r}\frac{鈭�}{鈭�r}\left(r\frac{鈭�蠄}{鈭�r}\right)+\frac{1}{r^2}\frac{{鈭�}^2蠄}{鈭�{胃}^2}\right)=E蠄$

Put a solution of the form given below.

$蠄=R\left(r\right)螛\left(胃\right)$

Rewrite the above equation.
$-\frac{{\mathrm{鈩�}}^2}{2m}\left(螛\frac{1}{r}\frac{鈭�}{鈭�r}\left(r\frac{鈭�R}{鈭�r}\right)+R\frac{1}{r^2}\frac{{鈭�}^2螛}{鈭�{胃}^2}\right)=E路R路螛$

Then divide by and then multiplying by$-r^2\frac{2m}{{\mathrm{鈩�}}^2}$.
$r\frac{1}{R}\frac{鈭�}{鈭�r}\left(r\frac{鈭�R}{鈭�r}\right)+\frac{1}{螛}\frac{{鈭�}^2螛}{鈭�{胃}^2}+\frac{2mE}{{\mathrm{鈩�}}^2}=0$

The second term a function of only.
$\frac{1}{螛}\frac{{鈭�}^2螛}{鈭�{胃}^2}=-n^2$

Write the solutions.
$螛\left(胃\right)\mathit{=}\{\begin{array}{l}\sin \left(n胃\right)\\ \cos \left(n胃\right)\end{array}$

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

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

Multiply by R.
$r\frac{鈭�}{鈭�r}\left(r\frac{鈭�R}{鈭�r}\right)+\left(\frac{2mE}{{\mathrm{鈩�}}^2}-n^2\right)R=0$

The solution must be as mentioned below.

$R\left(r\right)=J_n\left(Kr\right)$

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 k are zeros of the Bessel function.

Combining the space and the time solutions.

${\mathrm{唯}}_{\mathrm{nm}}={\mathrm{J}}_{\mathrm{n}}\left({\mathrm{k}}_{\mathrm{nm}}\mathrm{r}\right)\left\{\begin{array}{l}\sin \left(n胃\right)\\ \cos \left(n胃\right)\end{array}{\mathrm{e}}^{-\frac{{\mathrm{E}}_{\mathrm{mn}}}{\mathrm{h}}}\right.$

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