Density of state

1. Dec 18, 2008

KFC

Consider electrons in metal as a quantum ideal gas, the quantized energy would be

$$E(n_x, n_y, n_z) = \frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)$$

Now, letting

$$\alpha = \frac{n_x}{L_x}, \qquad \beta = \frac{n_y}{L_y}, \qquad \gamma = \frac{n_z}{L_z}$$

and

$$\rho^2 = \alpha^2 + \beta^2 + \gamma^2$$

and setting up an coordinate with $$\alpha, \beta, \gamma$$. One could see that the volume b/w E-E+dE is proportional to the volume b/w $$\rho - \rho + d\rho$$

Note that the volume within $$\rho - \rho + d\rho$$ is just one-eighth of the volume of the spherical shell with thickness $$d\rho$$, namely,

$$V_\rho = \frac{4\pi \rho^2 d\rho}{8}$$

Hence, the density of states within $$\rho - \rho + d\rho$$ becomes

$$N(\rho)d\rho = \frac{V_\rho}{\textnormal{volume per state}} = \frac{4\pi\rho^2d\rho /8}{1/V} = \frac{\pi V \rho^2 d\rho}{2}$$

Well, we know that

$$E = \frac{h^2}{8m}\rho^2$$

$$N(\rho)d\rho = N(E)dE$$

which gives

$$N(E) = \frac{1}{2}\frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}$$

But the correct answer for density of state in energy should be (two times of the above reuslt), i.e.

$$N(E) = \frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}$$

Where is the '2' come from?

Someone suggests

$$N(\rho)d\rho = 2\times\frac{\pi V \rho^2 d\rho}{2}$$

This will give the correct answer, but where is "2" come form?

2. Dec 18, 2008

Avodyne

Each electron has two possible spin states.

3. Dec 18, 2008

KFC

Oh, how can I forget that! Thanks a lot :)

4. Dec 9, 2009

nsundar

Can someone help me calculating the density of states from phonon dispersion relations.
I have computed the phonon dispersion relations from considering the vibrations of crystals. Now as a next step, i want to compute the density of states. not sure how.