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,

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

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.
$-\frac{{\mathrm{鈩�}}^2}{2m}\left(螛\frac{1}{r}\frac{鈭�}{鈭�r}\left(r\frac{鈭�R}{鈭�r}\right)+R\frac{1}{r^2}\frac{{鈭�}^2螛}{鈭�{胃}^2}\right)=E路R路螛\mathit{0}$

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.
$\mathrm{螛}\left(\mathrm{胃}\right)=\left\{\begin{array}{l}\sin \left(n胃\right)\\ \cos \left(n胃\right)\end{array}\right.$

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

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

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)$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">唯nm=Jn(knmr)sinn胃cosn胃e-iEnnh

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