Chapter 35, Problem 21P

(a)

To determine

The wave-function of the particle.

Answer to Problem 21P

Thewave function of the particles is ${\text{ψ}}_{n_x,n_y}\left(x,y\right)=\left(\frac{2}{L}\right)\sin \frac{n_x\pi x}{L}\sin \frac{n_y\pi y}{L}$ .

Explanation of Solution

Given:

The particle is constrained to move in the two dimensional region defined by,

  $0\le x\le L\text{and}0\le y\le L$

Formula used:

The expression for Schrodinger equation in two dimensionis given by,

  $\frac{{\hslash }^2}{2m}\left[\frac{{\partial }^2}{{\partial }^{2}x}+\frac{{\partial }^2}{{\partial }^{2}y}\right]\text{ψ}\left(x,y\right)=E\text{ψ}\left(x,y\right)$

Calculation:

Theexpression for Schrödinger's equation with separation of variables for the term $x$ is written as,

  $\begin{array}{c}\text{ψ}\left(x,y\right)=X\left(x\right)Y\left(y\right)\\ \frac{d^{2}X\left(x\right)}{x^2}+k_y^{2}X\left(x\right)=0\\ k_x^2=\frac{2m{E}_x}{{\hslash }^2}\end{array}$

Theexpression for Schrödinger's equation with separation of variables for the term $y$ is written as,

  $\begin{array}{c}\frac{d^{2}Y\left(y\right)}{y^2}+k_y^{2}Y\left(y\right)=0\\ k_y^2=\frac{2m{E}_y}{{\hslash }^2}\end{array}$

The above familiar differential equations have the usual solution,

  $\begin{array}{l}X\left(x\right)=A_x\sin \left(k_{x}x\right)+B_x\cos \left(k_{x}x\right)\\ Y\left(y\right)=A_y\sin \left(k_{y}y\right)+B_y\cos \left(k_{y}y\right)\end{array}$

Applying the boundary conditions,

  $\begin{array}{l}\text{ψ}\left(0,y,z\right)=\text{ψ}\left(L,y,z\right)\\ \text{ψ}\left(x,0,z\right)=\text{ψ}\left(x,L,z\right)\end{array}$

Which implies it will have solutions,

  $\begin{array}{l}X\left(x\right)=\sqrt{\frac{2}{L}}\sin \left(k_{x}x\right)\\ Y\left(y\right)=\sqrt{\frac{2}{L}}\sin \left(k_{y}y\right)\end{array}$

Here, $k_x=\frac{n_x\pi }{L}$ and $k_y=\frac{n_y\pi }{L}$ .

Therefore the wave function corresponds for the state ${\text{ψ}}_{n_x,n_y}\left(x,y\right)=\left(\frac{2}{L}\right)\sin \frac{n_x\pi x}{L}\sin \frac{n_y\pi y}{L}$

Conclusion:

Therefore,the wave function of the particle is ${\text{ψ}}_{n_x,n_y}\left(x,y\right)=\left(\frac{2}{L}\right)\sin \frac{n_x\pi x}{L}\sin \frac{n_y\pi y}{L}$ .

(b)

To determine

The energy corresponding to the wave function.

Answer to Problem 21P

The energy corresponding to the wave function are $\frac{{\hslash }^2{n}_x^2{\pi }^2}{2m{L}^2}$ and $\frac{{\hslash }^2{n}_y^2{\pi }^2}{2m{L}^2}$ .

Explanation of Solution

Calculation:

The $x$ component of wave function is given as,

  $-\frac{{\hslash }^2}{2m}\left[\frac{{\partial }^{2}X\left(x\right)}{{\partial }^{2}x}\right]=E_{x}X\left(x\right)$

Solving with the corresponding wave function,

  $\begin{array}{l}\frac{{\hslash }^2}{2m}\frac{n_x^2{\pi }^2}{L^2}X\left(x\right)=E_{x}X\left(x\right)\\ E_x=\frac{{\hslash }^2{n}_x^2{\pi }^2}{2m{L}^2}\end{array}$

The $y$ component of wave function is given as,

  $\begin{array}{l}\frac{{\hslash }^2}{2m}\frac{n_y^2{\pi }^2}{L^2}Y\left(y\right)=E_{y}Y\left(y\right)\\ E_y=\frac{{\hslash }^2{n}_y^2{\pi }^2}{2m{L}^2}\end{array}$

Conclusion:

Therefore,the energy corresponding to the wave function are $\frac{{\hslash }^2{n}_x^2{\pi }^2}{2m{L}^2}$ and $\frac{{\hslash }^2{n}_y^2{\pi }^2}{2m{L}^2}$ .

(c)

To determine

The quantum numbers of the two lowest states that has degeneracy.

Answer to Problem 21P

The quantum numbers of the two lowest states that has degeneracy are $n_x=1$ , $n_y=2$ and $n_x=2$ , $n_y=1$ .

Explanation of Solution

Calculation:

The energy for the state $n_x=1$ and $n_y=2$ is calculated as,

  $\begin{array}{c}{E}_{1,2}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_x^2\right)\\ =\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(1^2+2^2\right)\\ =\frac{5{\hslash }^2{\pi }^2}{2m{L}^2}\end{array}$

The energy for the state $n_x=2$ and $n_y=1$ is calculated as,

  $\begin{array}{c}{E}_{2,1}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_x^2\right)\\ =\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(2^2+1^2\right)\\ =\frac{5{\hslash }^2{\pi }^2}{2m{L}^2}\end{array}$

Conclusion:

Therefore,the quantum numbers of the two lowest states that has degeneracy are $n_x=1$ , $n_y=2$ and $n_x=2$ , $n_y=1$ .

(d)

To determine

The quantum numbers of the three lowest states that has degeneracy.

Answer to Problem 21P

The quantum numbers of the three lowest states that has degeneracy are $n_x=5,n_y=5$ , $n_x=1,n_y=7$ and $\text{and}{n}_x=7,n_y=1$ .

Explanation of Solution

Calculation:

The state will have three fold degeneracy are $n_x=5,n_y=5$ , $n_x=1,n_y=7$ and $n_x=7,n_y=1$ .

The energy for the state $n_x=5,n_y=5$ is calculated as,

  $\begin{array}{c}{E}_{5,5}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_x^2\right)\\ =\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(5^2+5^2\right)\\ =\frac{50{\hslash }^2{\pi }^2}{2m{L}^2}\end{array}$

The energy for the state $n_x=1,n_y=7$ is calculated as,

  $\begin{array}{c}{E}_{1,7}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_x^2\right)\\ =\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(1^2+7^2\right)\\ =\frac{50{\hslash }^2{\pi }^2}{2m{L}^2}\end{array}$

The energy for the state $n_x=7,n_y=1$ is calculated as,

  $\begin{array}{c}{E}_{1,7}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_x^2\right)\\ =\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(7^2+1^2\right)\\ =\frac{50{\hslash }^2{\pi }^2}{2m{L}^2}\end{array}$

Conclusion:

Therefore,the quantum numbers of the three lowest states that has degeneracy are $n_x=5,n_y=5$ , $n_x=1,n_y=7$ and $n_x=7,n_y=1$ .