The density of states can be calculated using the semi-classical Thomas-Fermi-Equation: $$ \rho(E) = \frac{1}{(2\pi\hbar)^n} \int \int \delta(E-H(\mathbf{p},\mathbf{q})) \; d^n p \; d^n q $$

Calculate the semi-classical density of states of a particle with mass $m$ in a n-dimensional infinite well $$ V(x_1,\dots,x_n) = \begin{cases}0 \quad \mathrm{if}\quad 0\leq, x_1,\dots,x_n \leq L \quad \\ \infty \;\;\mathrm{else}\end{cases} $$

$H(\mathbf{p},\mathbf{q})$ is the classical Hamilton function. In this case it should be $H(\mathbf{p},\mathbf{q})=\frac{\mathbf{p}^2}{2m}$ inside the well. $$ \Rightarrow \rho(E) = \frac{1}{(2\pi\hbar)^n} \int \int \delta\left(E-\frac{\mathbf{p}^2}{2m}\right) \; d^n p \; d^n q $$

Now the delta function only depends on $\mathbf{p}$, so the p integral should just be equal to 1, shouldn't it?

So the equation becomes

$$\rho(E) = \frac{1}{(2\pi\hbar)^n} \int 1\; d^n q $$

Am I correct so far? If so, what limits should I use to solve the integral? I'd think I should integrate from 0 to L. This would give me the solution: $$ \rho(E) = \frac{L^n}{(2\pi\hbar)^n} $$ I doubt this is correct, as we should also plot our result as a function of E. I think my error lies in eliminating the delta function. It would be great if someone could help me find the right solution.


1 Answer 1


You are right, your integral over the delta function is wrong. The "problem" is that the argument in the delta function is squared. To evaluate the integral it is convenient to use the identity $$\delta(x^2-a^2)=\frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]$$

Then one gets $$\delta\left(E-\frac{p^2}{2m}\right) = \frac{1}{2 \sqrt{E}} \left[\delta\left(\sqrt{E}-\frac{p}{\sqrt{2m}}\right) + \delta\left(\sqrt{E}+\frac{p}{\sqrt{2m}}\right)\right]$$

Integrating this gives $$ \int \delta\left(E-\frac{\mathbf{p}^2}{2m}\right) d^np = (2m)^{n/2} E^{n/2-1} $$ Thus the density is a function of the energy.

Edit: Fixed problem with dimensions.

  • $\begingroup$ Thank you very much. I suspected the error to be in the way I handled the delta function. Now the q integral should just add a L^n, shouldn't it? $\endgroup$
    – kaos
    Nov 28, 2016 at 0:05
  • $\begingroup$ Yes, the integral over q^n just gives the volume L^n $\endgroup$
    – Virft
    Nov 28, 2016 at 17:03

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