Question

Consider an 2D infinite square well that extends from 0 to a in the x direction, and from 0 to b in the y direction. a) Write down the time independent wave function Psi(x, y) for an electron inside this 2D infinite square well in terms of n_x and n_y, the quantum numbers corresponding to the x and y part of the wave function, respectively. b) Calculate the energy of the first excited state. Assume that a = 0.5 nm and b = 1 nm. Express your answer in eV. c) Calculate the lowest degenerate energy level, and give all combinations of n_x and n_y that yield this energy level. Use the values of a and b given in the part b). Express your answer in eV. d) Draw a contour plot of the probability density function P(x, y) for n_x = 1, n_y = 2. Be sure to label your axis and use the correct dimensions. Also, label the peak(s) of P(x, y) with the actual value in S.I. units.

Answer 1

(a) The time independent wave function will be

$\psi(x,y) = A \sin\left(\frac{n_x\pi x}{a} \right ) \sin\left(\frac{n_y\pi y}{b} \right )$

The constant A can be determined by the normalisation condition:

$\int_{0}^{a}\int_{0}^{b}\left | \psi(x,y) \right |^2dxdy = 1$

$A^2\int_{0}^{a}\sin^2\left(\frac{n_x\pi x}{a} \right )dx\int_{0}^{b}\sin^2\left(\frac{n_y\pi y}{b} \right )dy = 1$

$A^2 \frac{a}{2}\frac{b}{2} = 1$

$A = \frac{2}{ab}$

So

$\psi(x,y) = \frac{2}{ab} \sin\left(\frac{n_x\pi x}{a} \right ) \sin\left(\frac{n_y\pi y}{b} \right )$

(b) Energy of a state is given by

$E = \frac{\hbar^2\pi^2}{2m}\left(\frac{n_x^2}{a^2}+\frac{n_y^2}{b^2} \right )$

Substituting the following values in the above equation:

m = 9.1*10^-31 kg

hbar = 1.054*10^-34m² kg/s

a = 0.5*10^-9m, b = 1*10^-9m

For first excited state, n_x=1, n_y=2

So energy of first excited state is

$E = 4.8\times 10^{-19}\text{ J} = \frac{4.8\times 10^{-19} }{1.6\times 10^{-19}}\text{ eV} = 3.01 eV$

(c) Putting in values of a and b in the expression for energy:

$E = \frac{\hbar^2\pi^2}{2m}\left(\frac{n_x^2}{(0.5\times 10^{-9})^2}+\frac{n_y^2}{(1\times 10^{-9})^2} \right )$

$E = \frac{\hbar^2\pi^2}{2m}\left( 4n_x^2 + n_y^2 \right)\left(\frac{1}{(10^-9)^2} \right )$

By taking different combinations of n_x and n_y we can see that the lowest degenerate states are obtained for

n_x=1, n_y=4   and    n_x=2, n_y=2

The energy of lowest degenerate state can be calculated by substituting the above values:

$E = 1.2\times 10^{-18}\text{ J} = 1.2\times 10^{-18}/1.6\times 10^{-19} \text{ eV} = 7.5 \text{ eV}$

(d) The wavefunction for n_x=1, n_y=2 is

$\psi(x,y) = 4 \sin\left(\frac{\pi x}{0.5} \right ) \sin\left(\frac{2\pi y}{1} \right )$

I've plotted it in ROOT. You can use any other software you like:

4 sin(3.14*2*x) sin(2 3.14 y) 0.8 0.6 0.4 y (nm) 0.2 0 80 0.1 0.2 0.0.4 0.5 x (nm)