### Density of states in nanostructures

19May15

Some notes from my classes and worth seeing (if not remembering) results.

We’d like to calculate the function $D(E)$ called density of states, for systems with partially restricted dimensions. In fact all of the considered cases are 3D, they’re justÂ confined spatially in one (2D) or two directions (1D).

Starting from the general term for density of states

$D(E)=\sum _{\mu } \delta \left(E-E_{\mu }\right)$

where $\mu$ is the collection of all quantum numbers, we can calculate the terms for density of states for 3D, 2D and 1D cases.

#### 3D case

In the 3D case we don’t have any restrictions, therefore changing the sums over $\mu$ to integrals over all possible states enumerated by $k$ vector we get:

$D(E)= 2 \frac{V}{(2\pi )^3} \underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }} \delta \left(E-\frac{\hbar ^2}{2m}\left(k_x^2+k_y^2+k_z^2\right)\right)dk_xdk_ydk_z$

Where the term $\frac{V}{ (2 \pi )^3 }$ is the phase space occupied by a single state and the integer $2$ comes from two possible spins for each state.

$D(E)=2 \frac{V}{(2\pi )^3}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar^2k^2}{2m}\right)k^2dk \sin \theta d\theta d\varphi =$
$2\frac{V}{(2\pi )^3}4\pi \underset{0}{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar ^2k^2}{2m}\right)k^2dk= \begin{vmatrix}{c} \frac{2m E_{\zeta }}{\hbar ^2}=k^2 \\ \frac{1}{2}\frac{\sqrt{2m} }{\hbar }\frac{dE_{\zeta }}{\sqrt{E_{\zeta }}}= dk \end{vmatrix} =$
$2\frac{V}{(2\pi )^3}4\pi \frac{1}{2}\frac{\sqrt{2m} }{\hbar }\frac{2m}{\hbar ^2}\underset{0}{\overset{\infty }{\int }}\sqrt{ E_{\zeta }}\delta \left(E-\sqrt{E_{\zeta }}\right)dE_{\zeta }=\frac{V}{2\pi ^2}\frac{(2m)^{3/2}}{\hbar ^3}\sqrt{E}$

#### 2D case

In 2D case the values $E_z^n$ are enumerated by integer $n$, therefore we don’t convert one of the sums.

$D(E) =2 \frac{S}{(2\pi )^2}\sum _n \underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar ^2}{2m}\left(k_x^2+k_y^2\right)-E_z^n\right)dk_xdk_y=$
$2\frac{S}{(2\pi )^2}2\pi \sum _n \underset{0}{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar ^2k^2}{2m}-E_z^n\right)kdk= \begin{vmatrix}{c} \frac{\hbar ^2}{m}kdk=dE' \\ k dk = \frac{m}{\hbar ^2}dE' \\ \end{vmatrix}$

$\frac{S}{\pi }\frac{m}{\hbar ^2}\sum _n \underset{0}{\overset{\infty }{\int }}\delta \left(E-E_n-E'\right)dE'=\frac{S}{\pi }\frac{m}{\hbar ^2}\sum _n \theta \left(E-E_n\right)$

#### 1D case

In 1D case the values $E_z^n$ $E_y^m$ are enumerated by integers $n$ and $m$, therefore we don’t convert two of three sums.

$D(E)= 2 \frac{L}{2\pi }\sum _n \sum _m \underset{-\infty }{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar ^2}{2m}k_x^2-E_z^n-E_y^m\right)dk_x=$
$\frac{L}{\pi }\sum _n \sum _m \underset{-\infty }{\overset{\infty }{\int }}\delta \left(E-\frac{\hbar ^2}{2m}k_x^2-E_z^n-E_y^m\right)dk_x= \begin{vmatrix}{c} \frac{\hbar ^2}{m}kdk=dE' \\ dk = \frac{m}{\hbar }\frac{dE'}{\sqrt{2m E'}} \\ \end{vmatrix}=$
$\frac{L}{\pi }\frac{m}{\hbar }\frac{2}{\sqrt{2m}}\sum _n \sum _m \underset{0 }{\overset{\infty }{\int }}\delta \left(E-E_z^n-E_y^m-E'\right)\frac{1}{\sqrt{E'}}dE'=$
$\frac{L}{\pi \hbar }\sqrt{2 m}\sum _n \sum _m \frac{1}{\sqrt{E-E_z^n-E_y^m}}\theta \left(E-E_z^n-E_y^m\right)$