Example 4: Electrons in Metals

The goal of this section is to show that even at room temperature, the conducting electrons in metals can be modeled as an ideal gas of fermions contained in a box of volume $V=L^3$, where $L$ defines the dimensions of the piece of metal. Such a goal is accomplished by comparing the kinetic energy of conducting electrons, modeled as an ideal gas of fermions, with typical energy fluctuations due to thermal motion.
 

The average number of electrons occupying the j-th energy state is

$$\overline{n_j}=\frac{1}{e^{\beta(\epsilon_j -\mu)}+1},$$ (202)

where

$$\epsilon_j =\frac{\hbar^2 K_j^2}{2m},$$ (203)

and

$$K_j = (n_x(j),n_y(j),n_z(j))\pi/L$$ (204)

with $n_x(j),n_y(j),n_z(j)=1,2,...$ Therefore, the average number of electrons is

$$\overline{N}= \sum_j \overline{n_j},$$ (205)

$$\overline{N} = 2 \sum_{n_x} \sum_{n_y} \sum_{n_z} \frac{1}{e^{\beta(\epsilon(n_x,n_y,n_z) -\mu)}+1},$$ (206)

or,

$$\overline{N} = 2 \int_{-\infty}^{\infty} d{k_x} \int_{-\infty}^{\......( \frac{L}{2 \pi} \Bigg)^3 \frac{1}{e^{\beta(\frac{\hbar^2 K^2}{2m} -\mu)} +1}.$$ (207)

In particular, at $T=0$,

$$\frac{1}{1+e^{\beta(\frac{\hbar^2 K^2}{2m} -\mu)}} = \begin{cas......2}{2m} < \mu,\\ 0, & \hspace{.2cm} \frac{\hbar^2 K^2}{2m} > \mu, \end{cases}$$ (208)

therefore,

$$\overline{N} = 8 \pi \int_0^{K_f} d K K^2 \Bigg ( \frac{L}{2 \pi}......{2 \pi} \Bigg)^3 \frac{K_f^3}{3}= \frac{2 V }{(2 \pi)^3} \frac{4}{3} \pi K_f^3,$$ (209)

where $K_f$ is the Fermi momentum defined as follows

$$\frac{\hbar^2 K_f^2}{2m} =\mu.$$ (210)

The value of $K_f$ for a specific metal can be found, according to Eq. (209) and using the values of the density and atomic weight of the corresponding metal, assuming that each atom in the metal donates an electron to the conducting electron gas. Such value of $K_f$, can be used to compute the chemical potential according to Eq. (210). The calculation for Cu, with a density of 9 g/cm$^3$ and atomic weight of 63.5 g/mol gives

$$\mu/k \approx 80,000 K,$$ (211)

which indicates that even at room temperature the ideal gas approximation is accurate.
 
 

Thermal Energy and Heat Capacity

The remaining of this section proves that at low temperature $T$ the heat capacity $C_v$ of electrons in metals is proportional T. The thermal energy $E$ of electrons in metals is

$$E = \sum_j 2 \overline{n_j} \epsilon_j,$$ (212)

where index $j$ specifies a one-electron quantum state with energy $\epsilon_j$ and $\overline{n_j}$ is the average number of electrons with one kind of spin in such state. The factor 2, introduced by Eq. (212) counts for the spin degeneracy. Substituting $\overline{n_j}$ according to Eq. (202) and changing the sum over $j$ by a sum over energy levels we obtain

$$E = 2 \int_0^{\infty} d\epsilon \frac{\rho(\epsilon) \epsilon}{e^{\beta(\epsilon - \mu)}+1},$$ (213)

where $\rho(\epsilon)$ is the degeneracy of the energy level. Eq. (213) can be integrated by parts, according to

$$\int_a^b u d{\phi} = u*{\phi} \Bigg \vert _a^b - \int_a^b \phi du,$$ (214)

defining $d{\phi} = \epsilon \rho(\epsilon) d \epsilon$ and $u(\epsilon) = 1/(exp(\beta(\epsilon-\mu))+1)$. Note that according to this choice of variables ${\phi}(\epsilon)=\int_0^{\epsilon} d \bar{\epsilon} \rho(\bar{\epsilon}) \bar{\epsilon}$. We obtain

$$E=\lim_{\epsilon \rightarrow \infty} \frac{2\int_0^{\epsilon} d \......' ] \frac{(-e^{\beta(\epsilon -\mu)} \beta)}{(e^{\beta(\epsilon -\mu)} +1 )^2}.$$ (215)

Note that the first term, introduced by Eq. (215) is equal to 0 since in the limit when $\epsilon \rightarrow \infty$ the denominator becomes extremely large. The second term introduced by Eq. (215) is also equal 0 since in the limit when $\epsilon \rightarrow 0$ the numerator is equal to 0. Therefore, introducing the definition

$$F(\epsilon) \equiv \frac{1}{e^{\beta(\epsilon -\mu)} +1},$$ (216)

we obtain that Eq. (215) can be rewritten as follows,

$$E=-2 \int_0^{\infty} d \epsilon \phi(\epsilon) \frac{\partial F}{\partial \epsilon}.$$ (217)

At this point, it is important to note that $\partial F/\partial \epsilon$ is a function peaked at $\epsilon = \mu$, as represented by the following diagram,

since $F(\epsilon)$ is a step function represented by the following diagram,

Therefore we can approximate $\phi(\epsilon)$, in Eq. (217), by its expansion around $\epsilon = \mu$,

$$\phi (\epsilon) = \phi (\mu + x) \approx \sum_0^{\infty} \frac{\p......{\partial \epsilon ^n} \Bigg \vert_{\epsilon=\mu} \frac{(\epsilon -\mu)^n}{n!},$$ (218)

where $x \equiv \epsilon -\mu$. Substituting Eq. (218) into Eq. (217) we obtain,

$$E = -\int_0^{\infty } d \epsilon \sum_{n=0}^{\infty} \frac{\parti......epsilon=\mu} \frac{(\epsilon -\mu)^n}{n!} \frac{\partial F}{\partial \epsilon}.$$ (219)

Considering the first few terms of Eq. (219) we obtain,

$$\begin{array}{ll}E &= \phi(\mu) \Bigg ( - \int_0^{\infty} d{......on-\mu)}+2 e^{\beta(\epsilon-\mu)}+1} \Bigg) + ...,\end{array}$$ (220)

where the first term is equal to $\phi(\mu)$ because $F(\infty)=0$ and $F(0)=1$. In order to show that the second term in Eq. (220) is equal to 0 we rewrite Eq. (220) in terms of the variable $\bar{x}=\beta x$,

$$\begin{array}{ll}E & = \phi(\mu) + \frac{\partial \phi}{\par......x}^2}{e^{\bar{x}} + e^{-\bar{x}} + 2} \Bigg) + ...,\end{array}$$ (221)

where the lower integration limit has been changed from 0 to $-\infty$ since the integrand is approximately equal to zero whenever abs$(\epsilon-\mu)$ is large. Note that the integral introduced by the second term of Eq. (221) is equal to 0 because the integrand is odd. In addition,

$$\int_{-\infty}^{\infty} d{\bar{x}} \frac{\bar{x}^2}{e^{\bar{x}} + e^{-\bar{x}} + 2} = \frac{\pi^2}{3},$$ (222)

therefore,

$$E = \phi(\mu) + \frac{(k T)^2}{2} \frac{\partial^2 \phi}{\partial {\epsilon^2}} \Bigg )_{\epsilon=\mu} \Bigg (- \frac{\pi^2}{3} \Bigg ) + ...,$$ (223)

At sufficiently low T, higher order terms in Eq. (223) are negligible. Therefore, at low T

$$C_v \propto T,$$ (224)

as observed in experiments.