6.20 Intro to Electrical Conduction

Some of the basic physics of electrical conduction in metals can be understood using a very simple model. That model is a free-electron gas, i.e. noninteracting electrons, in a periodic box.

The classical definition of electric current is moving charges. That can readily be converted to quantum terms for noninteracting electrons in a periodic box. The single-particle energy states for these electrons have definite velocity. That velocity is given by the linear momentum divided by the mass.

Consider the possibility of an electric current in a chosen $x$-​direction. Figure 6.18 shows a plot of the single-particle energy ${\vphantom' E}^{\rm p}$ against the single-particle velocity $v^{\rm {p}}_x$ in the $x$-​direction. The states that are occupied by electrons are shown in red. The parabolic outer boundary reflects the classical expression ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12{m_{\rm e}}{v^{\rm {p}}}^2$ for the kinetic energy: for the single-particle states on the outer boundary, the velocity is purely in the $x$-​direction.

Figure 6.18: Conduction in the free-electron gas model.
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In the system ground state, shown to the left in figure 6.18, no current will flow, because there are just as many electrons that move toward negative $x$ as there are that move towards positive $x$. To get net electron motion in the $x$-​direction, electrons must be moved from states that have negative velocity in the $x$-​direction to states that have positive velocity. That is indicated to the right in figure 6.18. The asymmetric occupation of states now produces net electron motion in the positive $x$-​direction. That produces a current in the negative $x$-​direction because of the fact that the charge $\vphantom0\raisebox{1.5pt}{$-$}$$e$ of electrons is negative.

Note that the electrons must pick up a bit of additional energy when they are moved from states with negative velocity to states with positive velocity. That is because the Pauli exclusion principle forbids the electrons from entering the lower energy states of positive velocity that are already filled with electrons.

However, the required energy is small. You might just briefly turn on an external voltage source to produce an electric field that gets the electrons moving. Then you can turn off the voltage source again, because once set into motion, the noninteracting electrons will keep moving forever.

In physical terms, it is not really that just a few electrons make a big velocity change from negative to positive due to the applied voltage. In quantum mechanics electrons are completely indistinguishable, and all the electrons are involved equally in the changes of state. It is better to say that all electrons acquire a small additional drift velocity $\Delta{v}^{\rm {p}}_x$ in the positive $x$-​direction. In terms of the wave number space figure 6.17, this shifts the entire sphere of occupied states a bit towards the right, because velocity is proportional to wave number for a free-electron gas.

The net result is still the energy versus velocity distribution shown to the right in figure 6.18. Electrons at the highest energy levels with positive velocities go up a bit in energy. Electrons at the highest energy levels with negative velocities go down a bit in energy. The electrons at lower energy levels move along to ensure that there is no more than one electron in each quantum state. The fact remains that the system of electrons picks up a bit of additional energy. (The last subsection of derivation {D.45} discusses the effect of the applied voltage in more detail.)

Conduction electrons in an actual metal wire behave similar to free electrons. However, they must move around the metal atoms, which are normally arranged in some periodic pattern called the crystal structure. The conduction electrons will periodically get scattered by thermal vibrations of the crystal structure, (in quantum terms, by phonons), and by crystal structure imperfections and impurities. That kills off their organized drift velocity $\Delta{v}^{\rm {p}}_x$, and a small permanent electric field is required to replenish it. In other words, there is resistance. But it is not a large effect. For one, in macroscopic terms the conduction electrons in a metal carry quite a lot of charge per unit volume. So they do not have to go fast. Furthermore, conduction electrons in copper or similar good metal conductors may move for thousands of Ångstroms before getting scattered, slipping past thousands of atoms. Electrons in extremely pure copper at liquid helium temperatures may even move millimeters or more before getting scattered. The average distance between scattering events, or collisions, is called the “free path” length $\ell$. It is very large on an atomic scale.

Of course, that does not make much sense from a classical point of view. Common sense says that a point-size classical electron in a solid should pretty much bounce off every atom it encounters. Therefore the free path of the electrons should be of the order of a single atomic spacing, not thousands of atoms or much more still. However, in quantum mechanics electrons are not particles with a definite position. Electrons are described by a wave function. It turns out that electron waves can propagate through perfect crystals without scattering, much like electromagnetic waves can. The free-electron gas wave functions adapt to the crystal structure, allowing the electrons to flow past the atoms without reflection.

It is of some interest to compare the quantum picture of conduction to that of a classical, nonquantum, description. In the classical picture, all conduction electrons would have a random thermal motion. The average velocity $v$ of that motion would be proportional to $\sqrt{{k_{\rm B}}T/m_{\rm e}}$, with $k_{\rm B}$ the Boltzmann constant, $T$ the absolute temperature, and $m_{\rm e}$ the electron mass. In addition to this random thermal motion in all directions, the electrons would also have a small organized drift velocity $\Delta{v}^{\rm {p}}_x$ in the positive $x$-​direction that produces the net current. This organized motion would be created by the applied electric field in between collisions. Whenever the electrons collide with atoms, they lose much of their organized motion, and the electric field has to start over again from scratch.

Based on this picture, a ballpark expression for the classical conductivity can be written down. First, by definition the current density $j_x$ equals the number of conduction electrons per unit volume $i_{\rm {e}}$, times the electric charge $\vphantom0\raisebox{1.5pt}{$-$}$$e$ that each carries, times the small organized drift velocity $\Delta{v}^{\rm {p}}_x$ in the $x$-​direction that each has:

\begin{displaymath}
j_x = - i_{\rm {e}} e \Delta v^{\rm {p}}_x
\end{displaymath} (6.30)

The drift velocity $\Delta{v}^{\rm {p}}_x$ produced by the electric field between collisions can be found from Newton’s second law as the force on an electron times the time interval between collisions during which this force acts and divided by the electron mass. The average drift velocity would be half that, assuming for simplicity that the drift is totally lost in collisions, but the half can be ignored in the ballpark anyway. The force on an electron equals $-e{\cal E}_x$ where ${\cal E}_x$ is the electric field due to the applied voltage. The time between collisions can be computed as the distance between collisions, which is the free path length $\ell$, divided by the velocity of motion $v$. Since the drift velocity is small compared to the random thermal motion, $v$ can be taken to be the thermal velocity. The “conductivity” $\sigma$ is the current density per unit electric field, so putting it all together,
\begin{displaymath}
\sigma \sim \frac{i_{\rm e} e^2 \ell}{m_{\rm e}v}
\end{displaymath} (6.31)

Neither the thermal velocity $v$ nor the free path $\ell$ will be the same for all electrons, so suitable averages have to be used in more detailed expressions. The “resistivity” is defined as the reciprocal of the conductivity, so as 1/$\sigma$. It is the resistance of a unit cube of material.

For metals, things are a bit different because of quantum effects. In metals random collisions are restricted to a small fraction of electrons at the highest energy levels. These energy levels are characterized by the Fermi energy, the highest occupied energy level in the spectrum to the left in figure 6.18. Electrons of lower energies do not have empty states nearby to be randomly scattered into. The velocity of electrons near the Fermi energy is much larger than the thermal value $\sqrt{{k_{\rm B}}T/m_{\rm e}}$, because there are much too few states with thermal-level energies to hold all conduction electrons, section 6.10. The bottom line is that for metals, in the ballpark for the conductivity the free path length $\ell$ and velocity $v$ of the Fermi-level electrons must be used. In addition, the electron mass $m_{\rm e}$ may need to be changed into an effective one to account for the forces exerted by the crystal structure on the electrons. That will be discussed in more detail in section 6.22.3.

The classical picture works much better for semiconductors, since these have much less conduction electrons than would be needed to fill all the quantum states available at thermal energies. The mass correction remains required.


Key Points
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The free-electron gas can be used to understand conduction in metals in simple terms.

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In the absence of a net current the electrons are in states with velocities in all directions. The net electron motion therefore averages out to zero.

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A net current is achieved by giving the electrons an additional small organized motion.

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The energy needed to do this is small.

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In real metals, the electrons lose their organized motion due to collisions with phonons and crystal imperfections. Therefore a small permanent voltage must be applied to maintain the net motion. That means that there is electrical resistance. However, it is very small for typical metals.