The density of states

In classical physics the states available to a particle in a volume $V$ and with some maximum value for the momentum $p_{max}$ are infinite, because the space is continuous both for the particle's spacial degrees of freedom ${\bf r}$ and for its momentum ones p. By contrast, in quantum mechanics they are finite, because of the requirements of the particle's wavefunction to be zero on the edges of the box, which means that the momentum of the particle can only assume discrete values. In particular, the time independent Schrödinger equation for a free particle is:

$$\hat{K} \psi = E \psi,$$ (11.1)

where $\hat K = \frac{\hat p^2}{2m} = -\frac{\hbar^2}{2m} \nabla^2 = -\frac{\hbar^2}{2... ...\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})$ is the kinetic energy operator, with $m$ the mass of the particle, $\hbar = h/2\pi$ the reduced Plank's constant and $\hat p$ the momentum operator. The three components $x, y$ and $z$ can be treated independently, and writing $E = \frac{p^2}{2m}$, with $p^2 = p_x^2 + p_y^2 + p_z^2$, we have for the $x$ component:

$$-\hbar^2\frac{\partial^2 \psi}{\partial x^2} = p_x^2 \psi,$$ (11.2)

from which we get:

$$\psi(x)= A \sin\left(\frac{p_x x}{\hbar}\right),$$ (11.3)

where $A$ is an unimportant normalisation constant. The three-dimensional solution is therefore of the type:

$$\psi({\bf r})= A \sin\left(\frac{p_x x}{\hbar}\right)\sin\left(\frac{p_y y}{\hbar}\right)\sin\left(\frac{p_z z}{\hbar}\right),$$ (11.4)

where ${\bf r} = (x,y,z)$ and $A$ is a different, still unimportant, normalisation constant.

If the particle is enclosed in a cubic box of side $L$ then the wavefunction  must be equal to zero on the faces of the box. Choosing a set of cartesian coordinates with the origin on one vertex of the box, we must have $\psi([0/L,0/L,0/L]) = 0$ and therefore $\sin\left(\frac{p_x L}{\hbar} \right) = \sin\left(\frac{p_y L}{\hbar} \right) = \sin\left(\frac{p_z L} {\hbar} \right) = 0$, which implies

$$\frac{p_x}{\hbar} L = \pi n_x; \quad \frac{p_y}{\hbar} L = \pi n_y; \quad \frac{p_z}{\hbar} L = \pi n_z,$$ (11.5)

or

$$p_x = \frac{n_x h}{2 L}; \quad p_y = \frac{n_y h}{2 L}; \quad p_z = \frac{n_z h}{2 L},$$ (11.6)

with $n_x, n_y, n_z$ integer numbers. Eq.  tells us that states in momentum space are separated by units of $\frac{h}{2L}$ in each direction and therefore the volume of momentum space occupied by each state would appear to be $\frac{h^3}{8 L^3} = \frac{h^3}{8 V}$, which means that the number of states per unit volume of momentum space would be $\frac{8V}{h^3}$. However, note that the point in momentum space ${\bf p} = \frac{h}{2 L} (-n_x,n_y,n_z)$ would correspond to the same solution of the Schrödinger equation as ${\bf p} = \frac{h}{2 L} (n_x,n_y,n_z)$, because $\sin(-\frac{n_x h}{2 L}) = - \sin(\frac{n_x h}{2 L})$, and a simple change of sign does not make the two wavefunctions linearly independent. It follows that, out of the eight possible combinations obtained by inverting the sings of the three cartesian coordinates of ${\bf p}$, we only have to count one state, or, which is the same, the number of states per unit volume of momentum space is not $\frac{8V}{h^3}$ but actually $\frac{V}{h^3}$. We now ask how many states are available to a particle that has the modulus of its momentum between $p$ and $p+dp$. This is equal to the number of states in the volume of the shell delimited by the two spheres with radius $p$ and $p+dp$, times the number of states per unit volume in momentum space. The volume of this shell is equal to $4\pi p^2 dp$ and therefore the number of states in such an element of volume is:

$$f(p) dp = \frac{V }{h^3}4\pi p^2 dp.$$ (11.7)

Consider now the phase space, made by the points in configuration space and momentum space, $({\bf r},{\bf p})$. A volume of phase space is obtained by the product of a volume of configuration space and a volume of momentum space. For example, an infinitesimal element of volume of phase space can be expressed as $d\Gamma = d^3{\bf r} d^3{\bf p}$. Eq.  can be interpreted as the number of state in a volume of phase space given by $V 4\pi p^2 dp$ ($V$ in configuration space and $4\pi p^2 dp$ in momentum space) and therefore the number of states per unit volume of phase space is simply equal to $1/h^3$. In other words, we can divide the phase space in little cells of volume $h^3$, and each of these cells will contain just one state. It follows that the number of states in an element of volume $d\Gamma$ is:

$$\frac{1}{h^3} d^3 {\bf r}d^3 {\bf p}.$$ (11.8)