91影视

The Schrodinger equation in two dimensionsis

$\left(\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}\right)蠄\left(x,y\right)=\frac{-2m\left(E-U\right)}{h^2}蠄\left(x,y\right)路路路路路路\left(1\right)$$\left(\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}\right)蠄\left(x,y\right)=\frac{-2m\left(E-U\right)}{h^2}蠄\left(x,y\right)路路路路路路\left(1\right)$

The potential of a 2D infinite well is

$$\begin{gathered} U=\left\{00<x<L,0<y<L \\ 鈭�\text{otherwise} \\ \right. \end{gathered}$$

The general solution to the differential equation

$\frac{d^{2}X}{d{x}^2}=-k^{2}X路路路路路路路\left(2\right)$

is of the form

$X=A\sin \left(kx\right)+B\cos \left(kx\right)路路路路路路\left(3\right)$ $X=A\sin \left(kx\right)+B\cos \left(kx\right)路路路路路路\left(3\right)$

where A and B are constants

Let

$蠄\left(x,y\right)=f\left(x\right)g\left(y\right)$

Substitute this in equation (1) to get

$$\begin{gathered} \left(\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \\ g\left(y\right)\frac{{鈭�}^{2}f\left(x\right)}{鈭�x^2}+f\left(x\right)\frac{{鈭�}^{2}g\left(y\right)}{鈭�y^2}=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \end{gathered}$$$$\begin{gathered} \left(\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \\ g\left(y\right)\frac{{鈭�}^{2}f\left(x\right)}{鈭�x^2}+f\left(x\right)\frac{{鈭�}^{2}g\left(y\right)}{鈭�y^2}=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \end{gathered}$$

Divide throughout by f(x) g(y) to get

$\frac{1}{f\left(x\right)}\frac{{鈭�}^{2}f\left(x\right)}{鈭�x^2}+\frac{1}{g\left(y\right)}\frac{{鈭�}^{2}g\left(y\right)}{鈭�y^2}=\frac{-2m\left(E-U\right)}{h^2}$$\frac{1}{f\left(x\right)}\frac{{鈭�}^{2}f\left(x\right)}{鈭�x^2}=-C_x路路路路路路\left(4\right)$

Call

$\frac{1}{f\left(x\right)}\frac{{鈭�}^{2}f\left(x\right)}{鈭�x^2}=-C_x路路路路路路\left(4\right)$

and

$\frac{1}{g\left(y\right)}\frac{{鈭�}^{2}g\left(y\right)}{鈭�y^2}=-C_y路路路路路路\left(5\right)$

Finally, $C_x+C_y=\frac{2m\left(E-U\right)}{{\mathrm{鈩�}}^2}路路路路路路\left(6\right)$

Hence, the solution is $C_x+C_y=\frac{2m\left(E-U\right)}{h^2}$

Since the well is infinite there should be no probability of any particle of staying outside it. Hence f(x) and g(y) should be 0 outside the wall.

The equations (IV) and (V) are of the form (II). Their general solutions inside the well are of the form of equation (III) as follows

$$\begin{gathered} f\left(x\right)=A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right) \\ g\left(y\right)=C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right) \end{gathered}$$

Here A and B are functions of x and C and D are functions of y. In general C_x and C_Y can be positive, negative or zero.

The complete wave function becomes

$蠄\left(x,y\right)=\left\{A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right)\right\}\left\{C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right)\right\}$

The first set of boundary conditions are

$$\begin{gathered} 蠄\left(x,0\right)=0 \\ 蠄\left(0,y\right)=0 \end{gathered}$$

Substitute these to get

$$\begin{gathered} 蠄\left(x,0\right)=\left\{A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right)\right\}D \\ =0 \\ D=0 \end{gathered}$$

and

$$\begin{gathered} 蠄\left(0,y\right)=B\left\{C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right)\right\} \\ =0 \\ B=0 \end{gathered}$$

Combine C to A and get

$蠄\left(x,y\right)=A\sin \left(\sqrt{C_x}x\right)\sin \left(\sqrt{C_y}y\right)$

The second set of boundary conditions are

$$\begin{gathered} 蠄\left(x,L\right)=0 \\ 蠄\left(L,y\right)=0 \end{gathered}$$

The first one gives

$$\begin{gathered} A\sin \left(\sqrt{C_x}x\right)\sin \left(\sqrt{C_y}L\right)=0 \\ \sin \left(\sqrt{C_y}L\right)=0 \\ \sqrt{C_y}L=n_y蟺n_y=1,2.... \\ \sqrt{C_y}=\frac{n_y蟺}{L} \\ {C}_y={\left(\frac{n_y蟺}{L}\right)}^2 \end{gathered}$$

The second one gives

$$\begin{gathered} A\sin \left(\sqrt{C_x}L\right)\sin \left(\sqrt{C_y}y\right)=0 \\ \sin \left(\sqrt{C_x}L\right)=0 \\ \sqrt{C_x}L=n_x蟺n_x=1,2.... \\ \sqrt{C_x}=\frac{n_x蟺}{L} \\ {C}_x={\left(\frac{n_x蟺}{L}\right)}^2 \end{gathered}$$

The final wave function becomes

$蠄\left(x,y\right)=A\sin \left(\frac{n_x蟺x}{L}\right)\sin \left(\frac{n_y蟺y}{L}\right)$

Hence, The functions andare zero outside the wall. Inside the wall the general solutions are

$$\begin{gathered} f\left(x\right)=A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right) \\ g\left(y\right)=C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right) \end{gathered}$$

Allowed values of the constants are

$$\begin{gathered} C_x={\left(\frac{n_x蟺}{L}\right)}^2{n}_x=1,2,3,.. \\ {C}_y={\left(\frac{n_y蟺}{L}\right)}^2{n}_y=1,2,3,.. \end{gathered}$$

From equation (VI) and the value of potential inside the well

$$\begin{gathered} C_x+C_y=\frac{2mE}{h^2} \\ \frac{n_x^2{蟺}^2}{L^2}+\frac{n_y^2{蟺}^2}{L^2}=\frac{2mE}{h^2} \\ E=\frac{h^2{蟺}^2}{2m{L}^2}\left(n_x^2+n_y^2\right) \end{gathered}$$

Thus the allowed values of energy are $\frac{h^2{蟺}^2}{2m{L}^2}\left(n_x^2+n_y^2\right)$ with two independent quantum numbers n_x and n_y .

The energies for which n_x and n_y are not equal have no unique corresponding wave functions.

Separate set of quantum numbers that have the same $n_x^2+n_y^2$ have the same energy. For example the wave functions $蠄\left(x,y\right)=A\sin \left(\frac{蟺x}{L}\right)\sin \left(\frac{2蟺y}{L}\right)$ and $蠄\left(x,y\right)=A\sin \left(\frac{2蟺x}{L}\right)\sin \left(\frac{蟺y}{L}\right)$ have quantum numbers (1 , 2) and (2 ,1) but they both have energy $\frac{{\mathrm{鈩�}}^2{蟺}^2}{2m{L}^2}\left(1^2+2^2\right)=\frac{5{\mathrm{鈩�}}^2{蟺}^2}{2m{L}^2}$.