Counting Standing-Wave States

Consider a three-dimensional standing wave confined in a cubic box that extends from $x=0$ to $x=a$, from $y=0$ to $y=a$, and from $z=0$ to $z=a$. (See Section 4.4.2.) The wavefunction, $\psi(x,y,z)$, must satisfy the boundary conditions

$$\psi(0,y,z)=\psi(a,y,z) = 0,$$ (5.439)

$$\psi(x,0,z)=\psi(x,a,z) = 0,$$ (5.440)

$$\psi(x,y,0)=\psi(x,y,a) = 0.$$ (5.441)

Thus, standing-wave solutions of the form

$$\psi(x,y,z)=\psi_0\,\sin(k_x\,x)\,\sin (k_y\,y)\,\sin(k_z\,z)$$ (5.442)

are only acceptable if

$$k_x = n_x\,\frac{\pi}{a},$$ (5.443)

$$k_y =n_y\,\frac{\pi}{a},$$ (5.444)

$$k_z = n_z\,\frac{\pi}{a},$$ (5.445)

where $n_x$, $n_y$, and $n_z$ are positive integers. (Note that negative values of $n_x$ do not give rise to wave states that are physically distinct from the corresponding positive values, et cetera.) It follows that $k_x$, $k_y$, and $k_z$ are all quantized in units of $\pi/a$.

Now,

$$\frac{\partial n_x}{\partial k_x} = \frac{a}{\pi},$$ (5.446)

$$\frac{\partial n_y}{\partial k_y} = \frac{a}{\pi},$$ (5.447)

$$\frac{\partial n_z}{\partial k_z} = \frac{a}{\pi}.$$ (5.448)

Thus, the number of translational wave states that are such that $k_x$ lies between $k_x$ and $k_x+dk_x$, $k_y$ lies between $k_y$ and $k_y+dk_y$, and $k_z$ lies between $k_z$ and $k_z+dk_z$, is

$$N({\bf k})\,dk_x\,dk_y\,dk_z = \left(\frac{\partial n_x}{\partial... ...n_z}{\partial k_z}\,dk_z \right)= \left(\frac{a}{\pi}\right)^3dk_x\,dk_y\,dk_z.$$ (5.449)

Note that

$$N({\bf k})= \left(\frac{a}{\pi}\right)^3$$ (5.450)

is independent of the wavevector ${\bf k} = (k_x$, $k_y$, $k_z)$, because the allowed wave states are uniformly distributed in ${\bf k}$-space.

The wavenumber is defined $k = \vert{\bf k}\vert$. The number of translational wave states such that $k$ lies between $k$ and $k+dk$ is denoted $\rho(k)\,dk$, where $\rho(k)$ is termed the density of states. Now, $\rho(k)\,dk$ is the number of wave states that lie in an octant of a spherical annulus in ${\bf k}$-space whose inner radius is $k$, and whose outer radius is $k+dk$. We have to take an octant of the annulus because only wave states characterized by positive values of $n_x$, $n_y$, and $n_z$ have physical significance. (See Section 4.4.3.) The volume of the octant in ${\bf k}$-space is

$${\cal V} = \frac{1}{8}\,4\pi\,k^2\,dk.$$ (5.451)

Hence,

$$\rho(k)\,dk = N({\bf k})\,{\cal V} = \left(\frac{a}{\pi}\right)^3\frac{1}{8}\,4\pi\,k^2\,dk,$$ (5.452)

which implies that

$$\rho(k) = \frac{V\,k^2}{2\pi^2},$$ (5.453)

where $V=a^3$ is the volume of the box. Although the previous expression was derived for the special case of a cubic box, we shall assume that it is valid for a macroscopic box of any shape. This assumption is reasonable provided that the wavelengths of most of the standing waves confined in the box are much smaller than the dimensions of the box (i.e., provided that $n_x$, $n_y$, and $n_z$ are all typically much greater than unity).

Consider electromagnetic waves confined in a box. Such waves satisfy the dispersion relation

$$\omega = k\,c,$$ (5.454)

where $c$ is the speed of light in vacuum. (See Section 2.4.4.) Note that $c$ is not a function of $\omega$. Let $\rho(\omega)\,d\omega$ be the number of translational electromagnetic wave states for which $\omega$ lies between $\omega$ and $\omega+d\omega$. It follows that

$$\rho(\omega)\,d\omega = 2\,\rho(k)\,\frac{dk}{d\omega}\,d\omega = 2\,\rho(k)\,\frac{1}{c}\,d\omega,$$ (5.455)

which yields

$$\rho(\omega) = 2\,\frac{\rho(k)}{c}= 2\,\frac{V\,k^2}{2\pi^2\,c},$$ (5.456)

giving

$$\rho(\omega) = \frac{V}{\pi^2}\,\frac{\omega^2}{c^3},$$ (5.457)

where use has been made of Equations (5.453) and (5.454). Here, the factor of 2 in Equation (5.455) is required because electromagnetic waves are transverse waves, so there are two independent polarization states for each allowed translational state. (See Section 2.4.4.)

Consider sound waves propagating through a solid. Such waves satisfy the dispersion relation

$$\omega= v_s\,k,$$ (5.458)

where $v_s$ is the sound speed. Note that $v_s$ is not (usually) a function of $\omega$. However, solids suppose both transverse and longitudinal sound waves (unlike gases, which only support longitudinal waves). Of course, for transverse waves, there are two independent polarization states for each allowed translational state. However, for longitudinal waves, there is only one polarization state for each allowed translational state. Thus, by analogy with electromagnetic waves, the density of transverse sound wave states is

$$\rho_t(\omega) = \frac{V}{\pi^2}\,\frac{\omega^2}{v_{s\,t}^{\,3}},$$ (5.459)

where $v_{t\,s}$ is the characteristic phase velocity of transverse waves, whereas the density of longitudinal sound wave states is

$$\rho_l(\omega) = \frac{V}{2\pi^2}\,\frac{\omega^2}{v_{s\,l}^{\,3}},$$ (5.460)

where $v_{t\,s}$ is the characteristic phase velocity of longitudinal waves. The total density of sound wave states, irrespective of the wave polarization, is

$$\rho(\omega) = \frac{3\,V}{2\,\pi^2}\,\frac{\omega^2}{v_{s}^{\,3}},$$ (5.461)

where

$$\frac{1}{v_s^{\,3}}= \frac{2}{3}\,\frac{1}{v_{s\,t}^{\,3}}+\frac{1}{3}\,\frac{1}{v_{l\,t}^{\,3}}.$$ (5.462)

Here, $v_s$ is the average sound speed.

Finally, consider electrons of mass $m_e$ confined in a box. According to quantum mechanics, electrons have wavelike properties such that the electron energy, $\epsilon$, is related to the wavenumber, $k$, according to the dispersion relation

$$\epsilon = \frac{\hbar^2\,k^2}{2\,m_e}.$$ (5.463)

(See Section 4.4.2.) Let $\rho(\epsilon)\,d\epsilon$ be the number of translational electron states for which $\epsilon$ lies between $\epsilon$ and $\epsilon+d\epsilon$. It follows that

$$\rho(\epsilon)\,d\epsilon = \rho(k)\,\frac{dk}{d\epsilon}\,d\epsilon= \rho(k) \,\frac{(2\,m_e)^{1/2}}{2\,\hbar\,\epsilon^{1/2}}\,d\epsilon.$$ (5.464)

However, according to the Pauli exclusion principle  (see Section 4.4.3), only two electrons (corresponding to a spin-up electron, and a spin-down electron) can be put into each translational state. Hence, reinterpreting $\rho(\epsilon)\,d\epsilon$ as the number of electrons whose energies lies between $\epsilon$ and $\epsilon+d\epsilon$, we get

$$\rho(\epsilon) = 2\,\frac{V\,k^2}{2\pi^2} \,\frac{(2\,m_e)^{1/2}}{2\,\hbar\,\epsilon^{1/2}},$$ (5.465)

which gives

$$\rho(\epsilon)= \frac{\sqrt{2}\,V\,m_e^{\,3/2}\,\epsilon^{1/2}}{\pi^2\,\hbar^3},$$ (5.466)

where use has been made of Equations (5.453) and (5.463).