density of state can be computed by 2 ways. i will demon a usual way on the k-space. then i will show the phase-space method.

in the k-space, k is the wave vector. the wave can be matter wave from Schrödinger equation or EM wave from Maxwell equation. In volume with length a, with boundary condition that the wave must be zero at boundary. the solution is:

\Phi (x) = sin( k x )

k = n \frac {\pi} {a}

so, the element or basic unit of the k-vector is

\pi / a = u

the number of state on between k+dk is

dn = \Omega_d(k) dk \times m / u^d

where \Omega_d(k) is the volume of  (d-1) dimension and m is multiplicity. then, we have to relate the k-vector with Energy. it is different from matter wave and EM wave.

E_m = \frac { \hbar^2 k^2 }{2 m}

E_{\gamma} = \hbar k

Thus, the density of state per unit volume is

g(E) = \frac {1}{a^d} \frac {dn}{dE}

_______________example______________________

For matter wave,

\frac {dk}{dE} = \frac { \sqrt{2m} } { \hbar 2 \sqrt{E} }

\Omega_3 (E) = 4 \pi \frac { 2 m E } {\hbar^2 }

for spin half

m = 2

g(E) = \frac { 8 m \sqrt {2m } } { \pi^2 \hbar^3 } \sqrt{E}

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in phase space, a unit volume is ( by the uncertainly principle )

\Delta x \Delta p = 2 \pi \hbar

thus the total number of state is

n = m \frac {1}{ (2 \pi \hbar )^d } \int {dx^d} \int {dp^d}

the density of state in unit energy is

\frac{dn}{dE} = \frac {1}{ (2 \pi \hbar )^d } \frac{d}{dE} \int {dx^d} \int {dp^d}

then using

E^2 = p^2 c^2 + m_0^2 c^4

the density of state per unit volume is

g(E) = m \frac {1} { (2 \pi \hbar )^d } \frac {d} {dE} \int {dp^d}

 

 

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