Density of States (Condensed Matter)

[roam]

Homework Statement

I need some help with the following problem:

Homework Equations

$\rho(k) dk = \frac{L}{\pi} dk$

$L=Na$

$\omega^2= \omega_m^2 \ sin^2 (qa/2)$

The Attempt at a Solution

The density of states is given by:

$g(\omega)= \rho (k) / \frac{dw}{dk}$​

Where

$\frac{d\omega}{dk} = \frac{a}{2} \ cos \frac{qa}{2}$

$g(\omega) = \frac{L}{\pi} \frac{2}{a} \frac{1}{\omega_m \ cos (qa/2)}$

Using the identity

$sin^2 x + cos^2 x =1 \implies cos x = \sqrt{1-sin^2 x} , \ cos(qa/2)=\sqrt{1-sin^2(qa/2)}$

We get

$g(\omega)=\frac{L}{\pi} \frac{2}{a \omega_m \sqrt{1-sin^2(qa/2)}} = \frac{2Na}{a \pi \omega_m \sqrt{1-sin^2(qa/2)}}$

$\therefore g(\omega)= \frac{N}{\pi} \frac{2}{\sqrt{\omega_m^2 -\omega^2}}$​

But this is not the correct answer. Why is there a "2" on the numerator, and how can we get rid of this factor of 2?

Did I make a mistake, or is this a typo in the question?

Any help is greatly appreciated.

 

[TSny]

I believe your derivation is correct. The factor of 2 should be there as you have shown. See equation (35) here.

Keeping track of factors of 2 can get confusing when comparing periodic boundary conditions and fixed boundary conditions. But the density of states should end up the same for either boundary condition.