# Density of States for a quantum well: Derivation?

Consider a quantum well, where we have:

$$E_{k_x,k_y,n_z}=\frac{\hslash^2k_x^2}{2m}+\frac{\hslash^2k_y^2}{2m}+f(n_z)$$

with $$k_x$$ and $$k_y$$ having widths of $$\frac{2\pi}{L}$$ and $$n_z$$ varing in integers,

the density of states is a staircase with the steps occuring at steps of $$\frac{m}{\pi\hslash^2}$$ with the steps happening at the quantised values of $$f(n_z)$$. I understand that this increase is due to contribution of multiple energy levels coming into play but was not satisfied with that. I tried to find a derivation elsewhere but could not find it.

Can anybody rigourously derive the staircase structure?

You can use the general formula for denisty-of-states $$N(E)=2\sum_\alpha\delta(E-E_\alpha)$$ Now using our energy dispersion relation we get \begin{align*} N(E) &= 2\sum_{\alpha=n,k_x,k_y}\delta(E-E_\alpha)\\ &\approx 2\frac{L_xL_y}{(2\pi)^2}\int_{-\infty}^\infty dk_x \int_{-\infty}^\infty dk_y\sum_n\delta(E-E_n(k))\\ &= 2\frac{L_xL_y}{(2\pi)^2}\int_0^\infty dk 2\pi k\sum_n\delta(E-E_n(k))\\ &= 2\frac{L_xL_y}{(2\pi)^2}\sum_n \int_{\epsilon_n}^\infty dE'2\pi\frac{m}{\hbar^2}\delta(E-E')\\ &=2\frac{L_xL_y}{(2\pi)^2}\sum_n\int_{-\infty}^\infty \delta(E-E')\Theta(E'-\epsilon_n) \\ &=\frac{mL_x L_y}{\pi\hbar^2}\sum_n \Theta(E-\epsilon_n) \end{align*} where I put $$f(n_z)=\epsilon_n$$. In this way we obtain the typical DOS of a 2D electron gas: $$n(E)=\frac{m}{\pi\hbar^2}\sum_n \Theta(E-\epsilon_n)$$ where $$\Theta$$ is the stepfunction.