Question

Question:Let $\mathit{V}\mathbf{=}\mathbf{0}$in the Schrodinger equation (3.22) and separate variables in 2-dimensional rectangular coordinates. Solve the problem of a particle in a 2-dimensional square box, $\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{l}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{y}\mathbf{<}\mathbf{l}$This means to find solutions of the Schrodinger equation which are 0 for $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{I}\mathbf{,}\mathbf{}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{I}$, that is, on the boundary of the box, and to find the corresponding energy eigenvalues. Comments: If we extend the idea of a “particle in a box” (see Section 3, Example 3) to two or three dimensions, the box in 2D might be a square (as in this problem) or a circle (Problem 8); in 3D it might be a cube (Problem 7.17) or a sphere (Problem 7.19). In all cases, the mathematical problem is to find solutions of the Schrodinger equation with $\mathbf{V}\mathbf{=}\mathbf{0}$inside the box and $\mathbf{\psi }\mathbf{=}\mathbf{0}$on the boundary of the box, and to find the corresponding energy eigenvalues. In quantum mechanics, $\mathbf{\psi }$describes a particle trapped inside the box and the energy eigenvalues are the possible values of the energy of the particle.

Short answer

The solution is $\mathrm{\psi }(\mathrm{x},\mathrm{y},\mathrm{t})=\sum _{\mathrm{n}=1}^{\infty }\sum _{\mathrm{m}=1}^{\infty }\sin \left(\frac{\mathrm{n\pi }}{\mathrm{I}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{I}}\mathrm{y}\right){\mathrm{e}}^{-{\mathrm{ih\pi }}^2}\frac{({\mathrm{n}}^2+{\mathrm{m}}^2)}{2{\mathrm{ml}}^2}\mathrm{t}$ .

Step-by-step solution

Given Information.

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

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

Use Schrodinger equation:

Use the Schrodinger equation.

$\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{\psi }\mathbf{}\mathbf{+}\mathbf{V\psi }\mathbf{=}\mathbf{i}\mathit{h}\frac{\mathit{\partial }}{\mathit{\partial }\mathit{t}}\mathit{}\mathit{\psi }$

Assume a solution of the form mentioned below.

$\mathrm{\psi }\left(\mathrm{x},\mathrm{y},\mathrm{t}\right)=\mathrm{\psi }(\mathrm{x},\mathrm{y})\mathrm{T}\left(\mathrm{t}\right)$

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

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

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

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

Put a solution of the form given below.

$\mathrm{\psi }=\mathrm{X}\left(\mathrm{x}\right)\mathrm{Y}\left(\mathrm{y}\right)$

Then divide by$\mathrm{X}\left(\mathrm{x}\right)\mathrm{Y}\left(\mathrm{y}\right)$ .

$$\begin{gathered} \frac{1}{\mathrm{x}}\frac{d^{2}X}{d{x}^2}+\frac{1}{\mathrm{Y}}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}=-\frac{2mE}{h^2}=-k_x^2-k_y^2 \\ \left(\frac{1}{x}\frac{d^{2}X}{d{x}^2}+k_x^2\right)+\left(\frac{1}{Y}\frac{{\mathrm{d}}^2}{{\mathrm{dy}}^2}+k_y^2\right)=0 \end{gathered}$$

The solutions are a trigonometric solutions.
$$\begin{gathered} \mathrm{X}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\\ \sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\end{array}\right. \\ \mathrm{Y}\left(\mathrm{y}\right)=\left\{\begin{array}{l}\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\\ \sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\end{array}\right. \end{gathered}$$

Use boundary condition:

$\mathrm{Y}\left(0\right)=0$

On the boundary$\psi =0$ .
$$\begin{gathered} \mathrm{\psi }(0,\mathrm{y},\mathrm{t})=0 \\ \mathrm{\psi }(\mathrm{l},\mathrm{y},\mathrm{t})=0 \\ \mathrm{\psi }(\mathrm{x},0,\mathrm{t})=0 \\ \mathrm{\psi }(\mathrm{x},\mathrm{I},\mathrm{t})=0 \end{gathered}$$

$$\begin{gathered} \mathrm{x}\left(0\right)=0 \\ \Rightarrow 0=\mathrm{Csin}\left({\mathrm{k}}_{\mathrm{x}}0\right)+\mathrm{D}\cos \left({\mathrm{k}}_{\mathrm{x}}0\right) \end{gathered}$$
$\mathrm{D}=0$
$$\begin{gathered} \mathrm{X}\left(\mathrm{I}\right)=0 \\ \Rightarrow 0=\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{I}\right) \end{gathered}$$
$$\begin{gathered} {\mathrm{k}}_{\mathrm{x}}\mathrm{I}=\mathrm{n\pi } \\ {\mathrm{k}}_{\mathrm{x}}=\frac{\mathrm{n\pi }}{\mathrm{I}} \end{gathered}$$

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$\Rightarrow 0=\mathrm{E}\sin \left({\mathrm{k}}_{\mathrm{y}}0\right)+\mathrm{F}\cos \left({\mathrm{k}}_{\mathrm{y}}0\right)$

$\mathrm{F}=0$

$\mathrm{Y}\left(\mathrm{I}\right)=0$
$0=\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{l}\right)$
${\mathrm{k}}_{\mathrm{y}}\mathrm{l}=\mathrm{m\pi }$
${\mathrm{k}}_{\mathrm{y}}=\frac{\mathrm{m\pi }}{\mathrm{l}}$

Calculate Energy:

Calculate the energy.

$$\begin{gathered} {\mathrm{n}}_{\mathrm{x}}^2+{\mathrm{n}}_{\mathrm{y}}^2=\frac{2\mathrm{mE}}{{\mathrm{h}}^2} \\ ={\left(\frac{\mathrm{n\pi }}{\mathrm{I}}\right)}^2+{\left(\frac{\mathrm{m\pi }}{\mathrm{l}}\right)}^2 \\ =\frac{{\mathrm{\pi }}^2}{{\mathrm{l}}^2}({\mathrm{n}}^2+{\mathrm{m}}^2) \end{gathered}$$

$\mathrm{E}={\mathrm{h}}^2{\mathrm{\pi }}^2\frac{({\mathrm{n}}^2+{\mathrm{m}}^2)}{2{\mathrm{ml}}^2}$

It can be seen that the energies are a degenerate. n and m can vary to get the same value.

Write the solution.
$$\begin{gathered} {\mathrm{\psi }}_{\mathrm{N}}=\sin \left(\frac{\mathrm{n\pi }}{\mathrm{l}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right){\mathrm{e}}^{-\frac{{\mathrm{iE}}_{{\mathrm{N}}^{\mathrm{t}}}}{h}} \\ {\mathrm{E}}_{\mathrm{N}}=h^{\mathit{2}}{\pi }^{\mathit{2}}\frac{\mathit{(}{\mathit{n}}^{\mathit{2}}\mathit{+}{\mathit{m}}^{\mathit{2}}\mathit{)}}{\mathit{2}{\mathit{ml}}^{\mathit{2}}} \end{gathered}$$
$\mathrm{\psi }(\mathrm{x},\mathrm{y},\mathrm{t})=\sum _{\mathrm{n}=1}^{\infty }\sum _{\mathrm{m}=1}^{\infty }\sin \left(\frac{\mathrm{n\pi }}{\mathrm{I}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right){\mathrm{e}}^{-\mathrm{i}h{\pi }^{\mathit{2}}\frac{\mathit{(}{\mathit{n}}^{\mathit{2}}\mathit{+}{\mathit{m}}^{\mathit{2}}\mathit{)}}{\mathit{2}{\mathit{ml}}^{\mathit{2}}}\mathrm{t}}$
Hence, the final solution is$\mathrm{\psi }(\mathrm{x},\mathrm{y},\mathrm{t})=\sum _{\mathrm{n}=1}^{\infty }\sum _{\mathrm{m}=1}^{\infty }\sin \left(\frac{\mathrm{n\pi }}{\mathrm{I}}\mathrm{x}\right)\sin \left(\frac{\mathrm{m\pi }}{\mathrm{l}}\mathrm{y}\right){\mathrm{e}}^{-\mathrm{i}h{\pi }^{\mathit{2}}\frac{\mathit{(}{n}^{\mathit{2}}\mathit{+}{m}^{\mathit{2}}\mathit{)}}{\mathit{2}m{l}^{\mathit{2}}}\mathrm{t}}$