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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?

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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 $k = \frac{\sqrt{2mE}}{\hbar}$,

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

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