Determine the density of states$\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}$for a 2D infinite well (ignoring spin) in which${\mathit{E}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}}\mathbf{=}\mathbf{(}{\mathbf{n}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{)}\frac{{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}$

The base equation for the density of states is,

$D\left(E\right)=\frac{\text{differentialnumberofstatesinrangedE}}{dE}.$

Given equation

$E_{n_x,n_y}=(n_x^2+n_y^2)\frac{{\pi }^2{\hslash }^2}{2m{L}^2}$ ……. (1)

$n^2=n_x^2+n_y^2$

Substitute$n^2$ for i$n_x^2+n_y^2$n equation (1).

$E_{n_x,n_y}=(n_x^2+n_y^2)\frac{{\pi }^2{\hslash }^2}{2m{L}^2}$

Solve that for $n$,

$\begin{array}{c}{E}_n=n^2\frac{{\pi }^2{\hslash }^2}{2m{L}^2}\\ \frac{2mL{E}_n}{{\pi }^2{\hslash }^2}=n^2\\ n={\left(\frac{2mL{E}_n}{{\pi }^2{\hslash }^2}\right)}^{1/2}\end{array}$

Differentiate on both sides with respective to energy$E$ .

$\frac{dn}{dE}=\frac{1}{2}{\left(\frac{2m{L}^2}{{\pi }^2{\hslash }^2{E}_n}\right)}^{1/2}$

Differential number of states in range$dE=\frac{1}{4}2\pi ndn$

Here,$\frac{1}{4}$ is because we are just using positive values from $n_x$and ,$n_y$ meaning that only one quadrant is being looked at, or just $1/4$of the total circumference.

Divide with$dE$ on both sides,

$\begin{array}{c}\frac{\text{DiferntialnumberofstatesinrangedE}}{dE}=\frac{1}{4}2\pi n\frac{dn}{dE}\\ D\left(E\right)=\frac{1}{4}2\pi n\frac{dn}{dE}\end{array}$

Substitute${\left(\frac{2mL{E}_n}{{\pi }^2{\hslash }^2}\right)}^{1/2}$for $n$and$\frac{1}{2}{\left(\frac{2m{L}^2}{{\pi }^2{\hslash }^2{E}_n}\right)}^{1/2}$for$\frac{dn}{dE}$,

$\begin{array}{c}D\left(E\right)=\frac{1}{4}2\pi n\frac{dn}{dE}\\ =\frac{1}{4}2\pi \left[{\left(\frac{2m{L}^2{E}_n}{{\pi }^2{\hslash }^2}\right)}^{1/2}\right]\left[\frac{1}{2}{\left(\frac{2m{L}^2}{{\pi }^2{\hslash }^2{E}_n}\right)}^{1/2}\right]\\ =\frac{\pi }{4}\left(\frac{2m{L}^2}{{\pi }^2{\hslash }^2}\right)\\ =\frac{m{L}^2}{2\pi {\hslash }^2}\end{array}$

The density of the given energy state is$\frac{m{L}^2}{2{\pi }^2}$ .