# Band Structure in metals

I have some confusion about the band structure of metals and was wondering if anyone could help me understand.

One can approximate a metal as being equivalent to a "nearly free electron" system. When doing this, one finds that the energy eigenvalues of the system are given by:

\begin{equation} \epsilon = E_{\mathbf{k}}\pm |U_{\mathbf{k}}| \end{equation} where $E_{\mathbf{k}}$ is the kinetic energy of the electron of mode $\mathbf{k}$ and $U_{\mathbf{k}}$ is a shift in energy which defines the "band gap."

I have a couple questions:

1. Does this result tell me anything about what the specific energy levels are for the electrons in the atoms?
2. Building from question 1, I have read that metals have a continuum of energy levels. Does this mean electrons can be excited by any frequency? Or is this continuum limited to a certain band? Does the above result shed any light on this?
• (1) No, the Bloch solutions in the metal are not solutions to the atoms. (2) The bands have a continuum of states in them, but space between them. The band structure is mapped out by various experimental techniques to excite electrons from one band to another, determining the distribution of allowable states. – Jon Custer Apr 19 '17 at 16:15
• Would you happen to know where I can get information on the band structure for different materials? Say for example, steel? – user41178 Apr 20 '17 at 13:29
• Steel is not a single phase material - it is composed of multiple phases, compositions, and crystal structures. There is no single band structure for 'steel'. – Jon Custer Apr 20 '17 at 13:35
• So would it be correct of me to go as far as saying, for any kind of macroscopic solid structure, the electron band will be a continuum and span all frequencies? This must be true as there is always scattering by EM waves regardless of the frequency. – user41178 Apr 20 '17 at 13:37
• No, that would not be correct. The variations in color or reflectivity vs wavelength or absorption vs wavelength are all rooted in the band structure of either the entire thing, or the 'effective' response of a composite solid. While there is always an interaction, it isn't the same for all solids. – Jon Custer Apr 20 '17 at 13:48

## 2 Answers

(1) No. The premise is that what you call a metal is a many body system, a dense one. So the hamiltonian for the valence electrons is no longer the atomic one, but the one proper of the total system. By using the "nearly free electron" approximation you also assume that the valence electrons of the atoms are no longer "confined" in the surroundings of the single atom, but are required to stay inside the volume of the metal. Thus it doesn't have sense, in this approximation, to talk about atomic energy levels for valence electrons.

(2) The energy levels of the system are arranged in a band structure which is a discrete set of energy intervals, each of which is separated from the next one by a band gap, which is a region of forbidden energies for your electrons. The size of the band gap, in the case of the "nearly free electron" approximation, is given by twice the value of the perturbation: $\Delta E_{gap}=2|U_{\vec k}|$ . Now, you have to imagine that a metal has a band which is half filled with electrons. Roughly speaking, this means that you can excite one electron by giving to him just a tiny amout of energy (more precisely, this is the case of the electrons that are at the Fermi surface, but this is not important at the moment). But you cant excite electrons with any energy (so frequency, if you think of photons) you want, because if the sum of the energy you give plus the energy of the electron is a value that is comprised in the band gap, that electron will never be excited.

EDIT

No, I don't know a place with lots of bad structures, but you just can google it, or look in a ghood solid physics book, like the Ashcroft & Mermin.

And no, your assumption is not correct; it's not true that metal will reflect every EM frequency. Just think about gamma rays being able to penetrate a thin layer of lead, or about the photoelectric effect. The key for this understanding is the fact that electrons, being fermions, obey Pauli exclusion principle: they populate the band structures occupying all the states from the lowest to the maximum energy. The surface (in the reciprocal space) that containins all the occupied states with maximum energy (called Fermi energy) is called the Fermi surface. So, when a photon of very low energy hits the metal, it won't be absorbed by electrons with energy far smaller than the Fermi energy, because then it would imply that it's promoted to an already occupied level. Insted, it will be absorbed by an electron which has a Fermi energy, because it has no upper levels occupied. Reasoning in a similar fashion, a metal can absorb photons of any frequency (up to a certain value) because there will be a suitable electron which has an energy that grants that it won't be promoted in an already occupied level nor it will end in the band gap.

Not all the frequency can be reflected or absorbed, it is true for low frequency, but it's not true for high frequencies. The rough but intuitive model you can use to understand this is to think about electrons as a collective gas, that responds to an incoming EM wave oscillating as an harmonic oscillator with the frequency of the wave. For low ones, the elecron gas is able to oscillate properly with the wave, thus reflecting it. If the EM frequency gets higher than a certain critical value (called the plasmon frequency) tipical of the material, the gas is no longer able to oscillate as fast as the wave, and so stops the motion, thus enabling the wave to get through the metal.

• Thanks for the reply. Would you happen to know where I can get more information about what exactly these bands are for different materials? For example, steel. Also, the other answer makes a good point, a metal will always reflect photons, no matter the frequency (it seems). So to me this says that the continuum of energy of the electron energy levels is such that it spans all frequencies. Is this not correct?? – user41178 Apr 20 '17 at 12:49
• I've edited the reply, hope it can be useful – MattiaBenini Apr 20 '17 at 15:46

A useful way to visualize the difference between conductors, insulators and semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands. In a sense, the atoms/molecules in a solid's lattice allow for overall solutions for the electrons , which may be considered as shared by all the lattice, one quantum mechanical solution.

Within a band the energy differences are small, and in the case of metals the electrons can easily exist in the conduction band and be considered practically free.

Does this mean electrons can be excited by any frequency?

An antenna is made of metal and its electrons can be excited practically by any frequency bar x-rays and higher.