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Sommerfeld Expansion

Let us examine the conduction electrons in a metal in slightly more detail. In particular, let us try to obtain a more exact expression for the electronic specific heat. We saw in the previous section that the Fermi energy, $ \mu$ , is determined by the equation

$\displaystyle N= \sum_r \frac{1}{{\rm e}^{ \beta (\epsilon_r-\mu)}+1}.$ (8.154)

Likewise, the mean electron energy is

$\displaystyle \overline{E} =\sum_r\frac{\epsilon_r}{{\rm e}^{ \beta (\epsilon_r-\mu)}+1}.$ (8.155)

Because, in general, the energies of the quantum states are very closely spaced, the sums in the previous two expressions can be replaced by integrals. Now, according to Section 8.12, the number of quantum states per unit volume with wavenumbers in the range $ k$ to $ k+dk$ is

$\displaystyle \rho_k(k) dk = \frac{k^{ 2}}{2\pi^{ 2}} dk.$ (8.156)

However, the energy of a state with wavenumber $ k$ is

$\displaystyle \epsilon = \frac{\hbar^{ 2} k^{ 2}}{2 m},$ (8.157)

where $ m$ is the electron mass. Let $ \rho(\epsilon) d\epsilon$ be the number of electrons whose energies lies in the range $ \epsilon$ to $ \epsilon+d\epsilon$ . It follows that

$\displaystyle \rho(\epsilon) d\epsilon = 2 V \rho_k(k) \frac{dk}{d\epsilon} d\epsilon,$ (8.158)

where the factor of $ 2$ is to take into account the two possible spin states which exist for each translational state. Hence,

$\displaystyle \rho(\epsilon) = \frac{V}{2\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \epsilon^{ 1/2}.$ (8.159)

Moreover, Equations (8.154) and (8.155) become

$\displaystyle N$ $\displaystyle =\int_0^\infty F(\epsilon) \rho(\epsilon) d\epsilon,$ (8.160)
$\displaystyle \overline{E}$ $\displaystyle = \int_0^\infty F(\epsilon) \epsilon \rho(\epsilon) d\epsilon,$ (8.161)


$\displaystyle F(\epsilon)= \frac{1}{{\rm e}^{ \beta (\epsilon-\mu)}+1}$ (8.162)

is the Fermi function.

The integrals on the right-hand sides of Equations (8.160) and (8.161) are both of the general form

$\displaystyle \int_0^\infty F(\epsilon) \varphi(\epsilon) d\epsilon,$ (8.163)

where $ \varphi(\epsilon)$ is a smoothly varying function of $ \epsilon$ . Let

$\displaystyle \psi(\epsilon)= \int_0^\epsilon \varphi(\epsilon') d\epsilon'.$ (8.164)

We can integrate Equation (8.163) by parts to give

$\displaystyle \int_0^\infty F(\epsilon) \varphi(\epsilon) d\epsilon = \left[F...
...psilon)\right]_0^\infty -\int_0^\infty F'(\epsilon) \psi(\epsilon) d\epsilon,$ (8.165)

which reduces to

$\displaystyle \int_0^\infty F(\epsilon) \varphi(\epsilon) d\epsilon = -\int_0^\infty F'(\epsilon) \psi(\epsilon) d\epsilon,$ (8.166)

because $ \varphi(0)=\varphi(\infty) F(\infty)=0$ . Here, $ F'(\epsilon)\equiv dF/d\epsilon$ .

Now, if $ k T\ll \mu$ then $ F(\epsilon)$ is a constant everywhere, apart from a thin region of thickness $ k T$ , centered on $ \epsilon=\mu$ . (See Figure 8.4.) It follows that $ F'(\epsilon)$ is approximately zero everywhere, apart from in this region. Hence, the relatively slowly-varying function $ \psi(\epsilon)$ can be Taylor expanded about $ \epsilon=\mu$ :

$\displaystyle \psi(\epsilon) = \sum_{m=0,\infty}\frac{1}{m!}\left[\frac{d^{ m}\psi}{d\epsilon^{ m}}\right]_\mu (\epsilon-\mu)^m.$ (8.167)

Thus, Equation (8.166) becomes

$\displaystyle \int_0^\infty F(\epsilon) \varphi(\epsilon) d\epsilon =-\sum_{m...
...silon^{ m}}\right]_\mu\int_0^\infty F'(\epsilon) (\epsilon-\mu)^m d\epsilon.$ (8.168)

From Equation (8.162), we have

$\displaystyle \int_0^\infty F'(\epsilon) (\epsilon-\mu)^m d\epsilon=-\int_0^\...
...u)}}{[{\rm e}^{ \beta (\epsilon-\mu)}+1]^{ 2}} (\epsilon-\mu)^m d\epsilon,$ (8.169)

which becomes

$\displaystyle \int_0^\infty F'(\epsilon) (\epsilon-\mu)^m d\epsilon=-\beta^{\...
...-\beta \mu}^\infty \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}} x^{ m} dx,$ (8.170)

where $ x=\beta (\epsilon-\mu)$ . However, because the integrand has a sharp maximum at $ x=0$ , and because $ \beta \mu\gg 1$ , we can replace the lower limit of integration by $ -\infty$ with negligible error. Thus, we obtain

$\displaystyle \int_0^\infty F'(\epsilon) (\epsilon-\mu)^m d\epsilon=-(k T)^{ m} I_m,$ (8.171)


$\displaystyle I_m =\int_{-\infty}^\infty \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}} x^{ m} dx.$ (8.172)

Note that

$\displaystyle \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}}\equiv \frac{1}{({\rm e}^{ x}+1) ({\rm e}^{-x}+1)}$ (8.173)

is an even function of $ x$ . It follows, by symmetry, that $ I_m$ is zero when $ m$ is odd. Moreover,

$\displaystyle I_0 = \int_{-\infty}^\infty \frac{{\rm e}^{ x}}{({\rm e}^{ x}+1)^{ 2}} dx= -\left[\frac{1}{{\rm e}^{ x}+1}\right]_{-\infty}^{\infty} = 1.$ (8.174)

Finally, it can be demonstrated that

$\displaystyle I_2 = \frac{\pi^{ 2}}{3}.$ (8.175)

(See Exercise 3.) Hence, we deduce that

$\displaystyle \int_0^\infty F(\epsilon) \varphi(\epsilon) d\epsilon = \int_0^...
...{ 2}}{6} (k T)^{ 2} \left(\frac {d\varphi}{d\epsilon}\right)_\mu + \cdots.$ (8.176)

This expansion is known as the Sommerfeld expansion, after its inventor, Arnold Sommerfeld.

Equation (8.160) yields

$\displaystyle N = \int_0^\mu \rho(\epsilon) d\epsilon + \frac{\pi^{ 2}}{6} (k T)^{ 2}\left(\frac{d\rho}{d\epsilon}\right)_\mu + \cdots.$ (8.177)

However, it follows from Equation (8.159) that

$\displaystyle \int_0^\mu\rho(\epsilon) d\epsilon$ $\displaystyle = \frac{V}{3\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \mu^{ 3/2},$ (8.178)
$\displaystyle \left(\frac{d\rho}{d\epsilon}\right)_\mu$ $\displaystyle = \frac{V}{4\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \mu^{-1/2}.$ (8.179)


$\displaystyle N\simeq \frac{V}{3\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \...
...3/2}+ \frac{V}{24} \frac{(2 m)^{3/2}}{\hbar^{ 3}} (k T)^{ 2} \mu^{-1/2},$ (8.180)

which can also be written

$\displaystyle \mu_F^{ 3/2} \simeq \mu^{ 3/2} + \frac{\pi^{ 2}}{8} (k T)^{ 2} \mu^{-1/2},$ (8.181)

where $ \mu_F$ is the Fermi energy at $ T=0$ . [See Equation (8.144).] The previous equation can be rearranged to give

$\displaystyle \mu\simeq \mu_F\left[1+\frac{\pi^{ 2}}{8}\left(\frac{k T}{\mu}\right)^2\right]^{-2/3},$ (8.182)

which reduces to

$\displaystyle \mu\simeq \mu_F\left[1-\frac{\pi^{ 2}}{12}\left(\frac{k T}{\mu_F}\right)^2\right],$ (8.183)

assuming that $ k T/\mu_F\ll 1$ . Figure 8.6 shows a comparison between the previous approximate expression for the temperature variation of the Fermi energy of a degenerate electron gas and the numerically-calculated exact value. It can be seen that our approximate expression is surprisingly accurate (at least, for $ T\sim \mu_F/k$ ).

Figure: The solid curve shows the numerically-calculated exact Fermi energy of a degenerate electron gas as a function of the temperature. The dashed curve shows the analytic approximation $ \mu/\mu_F\simeq 1-(\pi^{ 2}/12) ) (T/T_F)^{ 2}$ . Here, $ T_F=\mu _F/k$ is known as the Fermi temperature.
\epsfysize =3in

Equation (8.161) yields

$\displaystyle \overline{E} = \int_0^\mu \epsilon \rho(\epsilon) d\epsilon + \...
...{6} (k T)^{ 2}\left[\frac{d(\epsilon \rho)}{d\epsilon}\right]_\mu + \cdots.$ (8.184)

However, it follows from Equation (8.159) that

$\displaystyle \int_0^\mu\epsilon \rho(\epsilon) d\epsilon$ $\displaystyle = \frac{V}{5\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \mu^{ 5/2},$ (8.185)
$\displaystyle \left[\frac{d(\epsilon \rho)}{d\epsilon}\right]_\mu$ $\displaystyle = \frac{3 V}{4\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \mu^{ 1/2}.$ (8.186)


$\displaystyle \overline{E}\simeq \frac{V}{5\pi^{ 2}} \frac{(2 m)^{3/2}}{\hba...
...5/2}+ \frac{V}{8} \frac{(2 m)^{3/2}}{\hbar^{ 3}} (k T)^{ 2} \mu^{ 1/2},$ (8.187)

which can also be written

$\displaystyle \overline{E} \simeq \frac{3}{5} N \mu_F\left[\left(\frac{\mu}{\...
...4}\left(\frac{k T}{\mu_F}\right)^2\left(\frac{\mu}{\mu_F}\right)^{1/2}\right].$ (8.188)

Making use of Equation (8.183), and only retaining terms up to second order in $ k T/\mu_F$ , we obtain

$\displaystyle \overline{E}\simeq \frac{3}{5} N \mu_F + \frac{\pi^{ 2}}{4} N \mu_F\left(\frac{k T}{\mu_F}\right)^{ 2}.$ (8.189)

Hence, the specific heat capacity of the conduction electrons becomes

$\displaystyle C_V^{ (e)} = \frac{d\overline{E}}{d T} = \frac{\pi^{ 2}}{2} k N \frac{k T}{\mu_F},$ (8.190)

and the molar specific heat is written

$\displaystyle c_V^{(e)} = \frac{\pi^{ 2}}{2} \left(\frac{k T}{\mu_F}\right) R.$ (8.191)

Of course, because $ k T/\mu_F\ll 1$ , this value is much less than the classical estimate, $ (3/2) R$ .

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Next: White-Dwarf Stars Up: Quantum Statistics Previous: Conduction Electrons in Metal
Richard Fitzpatrick 2016-01-25