# State density in one dimension

For a phonon we took in our lectures the state density for a 3D crystal and in order to find the number of states with an energy value between $$[0,E)$$ we did the division between the volume of the sphere with radius k and the volume of a single state which is $$(2\pi/L)^3$$.

So how would it look for one dimension? I thought of being the division between the length of the crystal and the length of one state. Is that a correct assumption or no?

Density of state may be evaluated by imposing a periodic boundary condition: $$\Psi(x + L) = \Psi(x)$$

For $$\Psi(x) = \sqrt{\frac{1}{L}} \exp( i k x )$$ , the boundary condition implys:

$$\exp ikL = 1; \text{ therefore, } kL = 2n\pi;$$

$$k= k_n = \frac{2n\pi}{L}; \text{ for } n = ...-3, -2, -1, 0, 1, 2, 3, ...$$ each integer corresponding to a allowed state. The separation between two states is $$\frac{2\pi}{L}$$ . Therefore, the density of state is $$1 / \frac{2\pi}{L} = \frac{L}{2\pi}$$. The total number of state between $$k$$ and $$k+dk$$ has states $$dN$$:

$$dN = \frac{L}{2\pi} \times dk$$

But typical density of states is defined referring to the energy spacing - the number of states in unit energy specing bewteen $$E$$ and $$E+DE$$

$$dN = \frac{L}{2 \pi} d k = \frac{L}{2 \pi} \frac{d k}{d E} d E = D(E) d E.$$

Thus, the density of state in energy is

$$D(E) = \frac{L}{2 \pi} \frac{dk}{d E}.$$

For typical energy dispersion $$E = \frac{\hbar^2 k^2}{2 m}$$ and $$k = \frac{\sqrt{2mE}}{\hbar}$$,

$$D(E) = \frac{L}{2 \pi} \frac{\sqrt{2m}}{\hbar} \frac{1}{2 \sqrt{E}} = \frac{L}{4 \pi} \sqrt{ \frac{2m}{\hbar^2}} \frac{1}{\sqrt{E}}$$