Is it an accurate statement to say that free electrons in a metal experience NO restoring force when they interact with electromagnetic waves? I understand that the electrons exist in a space filled with ions, and doesn't the cumulative potential that is present due to the presence of the ions exert an electric field on the electrons. Even in the case of simple metals, where you can say that the nucleus is shielded by the valence electrons, so called Coulomb shielding, how significant is the shielding. Naively, it seems to me that because the charge in the nucleus is not balanced by the charge in the bound electrons, there should be some net potential that the free electron sees.
Electromagnetism – Are Free Electrons in a Metal Really Free?
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The details depend on exactly what level of theory you want to work at, but the basic principle is that electrons feel forces (and accumulate energies from them) whenever there are electromagnetic fields. A photon is essentially a quantum of excitation in the electromagnetic field, so when one is present there will be fields and these will affect the electron.
Anything after that depends on how you want to work: you can do a QFT type of treatment, or you can also quantize the field more naively and then heuristically give the interaction energies in analogy with the classical theory, or you can also work in a semiclassical treatment where the electron is quantized but the EM field is some fixed classical wave.
It's important to note, though, that even though the Coulomb fields of charged particles to induce electromagnetic fields, not all electromagnetic fields can (or at least should) be understood as the result of Coulomb interactions. Indeed, electromagnetic waves are fundamentally different to the Coulomb attraction fields; they are what is known as "null fields". Most importantly, trying to ascribe an origin to the field is misguided: fields create forces at different points in space based on what the value of the field is there, regardless of how that field was actually generated.
One nice way to put the problem is if you have an atom (single-electron for simplicity) interacting with photons at some wavelength below X-rays. The electric field of a given mode of the radiation can be written, classically, in the form $$ \mathbf{E}(\mathbf{r},t)=\mathbf{E}_0a(t)e^{i\mathbf{k}\cdot\mathbf{r}}+\text{c.c.}, $$ where $a(t)$ is the interesting dynamical variable, whose real and imaginary parts follow simple harmonic motion (and will therefore vary as $a(t)=e^{-i\omega t}$ classically) and $\text{c.c.}$ denotes the complex conjugate. The corresponding quantized version of this quantity is given by $$ \hat{\mathbf{E}}(\mathbf{r},t)=\mathbf{E}_0\, \hat{a} \, e^{i\mathbf{k} \cdot \mathbf{r}} +\text{h.c.}, $$ where $\text{h.c.}$ denotes the hermitian conjugate, and now $\hat{\mathbf{E}}$ is a position-dependent operator (or operator-valued function of position, however you want to see it). The 'core' dynamical variable $a(t)$ has been replaced by the operator $\hat a$, which is the annihilation operator of the corresponding quantum harmonic oscillator.
So what's a photon in this formalism? By the term "photon", we refer to an elementary excitation of the field, which is the state with one quantum of energy, i.e. the first Fock state, $$|1\rangle=\hat a^\dagger|0\rangle.$$ You can of course have more than one photon (e.g. a higher number state like $|7\rangle$), or superpositions of different numbers of photons (like $\frac1{\sqrt{2}}(|0\rangle+|1\rangle)$), or more complicated states such as classical-like coherent states.
OK, so that's how the photons look. (Or at least, they can be modelled like this.) Now for the interaction.
It turns out that atoms are much smaller than most of the wavelengths they interact with. That is, if $R$ is the largest distance an electron might get from the nucleus - a few Bohr or so - and $k=2p/\lambda$ is the wavenumber of the radiation, then the only way not to have $kR\ll 1$ is to be using fairly hard X-rays, in the sub-nanometre regime.
This means that as far as any electron will be concerned, the electric field is constant, and the $e^{i\mathbf{k}\cdot \mathbf{r}}$ term can be completely neglected. Taking this a bit further, you can use the classical formula for the corresponding potential energy, the dipole coupling $$ V_\text{int}=e\mathbf{r}\cdot\mathbf{E}, $$ and simply upgrade all the quantities in it to operators to get the corresponding coupling operator: $$ \hat{V}_\text{int}=e\hat{\mathbf{r}}\cdot\left(\mathbf{E}_0\hat a+\mathbf{E}_0^\ast\hat a^\dagger\right). $$
This gives you, finally, your hamiltonian. You need to combine it with whatever the atomic hamiltonian, $\hat H_0$ of the electron is (including its kinetic energy and its Coulomb interaction with the nucleus) as well as the energy contained in the field, which is given by the good old harmonic oscillator formula. That means that the total energy, and the total hamiltonian, is $$ \hat H =\hat H_0 +e\hat{\mathbf{r}}\cdot\left(\mathbf{E}_0\hat a+\mathbf{E}_0^\ast\hat a^\dagger\right) +\hbar\omega(\hat a^\dagger\hat a+\tfrac12). $$ Note that the hatted position $\hat{\mathbf{r}}$, is of course the electron's position, which is now an operator. The corresponding Schrödinger equation is $i\hbar\tfrac{d}{dt}|\Psi\rangle=\hat H|\Psi\rangle$, and you must be careful to note that the state $|\Psi\rangle$ is now the combined state of both the atom and the field, which includes such superpositions as $$ |\Psi\rangle=\frac1{\sqrt{2}}\left( |\text{ground}\rangle|1\rangle+|\text{excited}\rangle|0\rangle \right). $$
Finally, then, what does one mean, for example, by a phrase such as "the atom was excited by absorbing a photon"? Suppose you start off with your field in the one-photon state $|1\rangle$ and the atom in the ground state $|g\rangle$. This is no longer an eigenstate of the system, because the coupling does not respect it: $$ \hat V_\text{int}|g\rangle|1\rangle=e(\mathbf{E}_0\cdot \hat{\mathbf{r}})|g\rangle|0\rangle. $$ Therefore, the state of your system must evolve. What will it evolve into? To a first approximation, you can take the time-evolution operator to first order, $$\hat U(t)=e^{-i\hat H t}\approx1-i\hat H t;$$ this means, for example, that you will have transitions to any excited state $|e\rangle$ with amplitude $$ \langle e|\langle 0|\hat V_\text{int}|g\rangle|1\rangle=e\mathbf{E}_0\cdot \langle e|\hat{\mathbf{r}}|g\rangle\langle e|0\rangle, $$ which will tend to be nonzero.
Atoms are really distinguished from each other due to their nuclear properties . It is the number of protons in the nucleus that defines the electric potential which will trap electrons and create a neutral atom. Thus metals have a specific type of nucleus that generates the potential which gives rise to loosely bound electrons in the outer shells. It is a complicated interplay between the potentials of electrons and protons that generates the electron shells described by the quantum numbers.
So it is a many body interaction that will keep electrons mobile in metals, and the positive "holes" immobile at the location of each atom.
In a conductor, electric current can flow freely, in an insulator it cannot. Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them. "Conductor" implies that the outer electrons of the atoms are loosely bound and free to move through the material. Most atoms hold on to their electrons tightly and are insulators. In copper, the valence electrons are essentially free and strongly repel each other. Any external influence which moves one of them will cause a repulsion of other electrons which propagates, "domino fashion" through the conductor.
In semiconducting materials there is small mobility of both electrons and the holes left behind, again because of the potential solutions of the many body problem:
A silicon crystal is different from an insulator because at any temperature above absolute zero temperature, there is a finite probability that an electron in the lattice will be knocked loose from its position, leaving behind an electron deficiency called a "hole".
If a voltage is applied, then both the electron and the hole can contribute to a small current flow.
The conductivity of a semiconductor can be modeled in terms of the band theory of solids.
You cannot explain the behavior of matter below the nano scale , classically. Atoms and electrons are already quantum mechanical entities. It is the reason why quantum mechanics was invented and believed to be the underlying level of nature. Quantum numbers make a huge difference in the behavior of "particles" in the microcosm of atoms and molecules.
Edit looking at your figure: It is not talking of motion of physical positive charges, holes, but of hypothetical ones. In a semiconductor, the holes move, i.e.a neutral atoms become positively charged sequentially. In metals , the positive charge is attached to the individual atom, generating the potential that gives the energy levels. For holes to move, atoms at the atomic level have to exchange electrons. In metals the electrons are shared by all atoms and the mobility of holes is zero
The word "hole" is used in semiconductors as the subtraction of an electron from a neutral atom. The word "hole" that you use classically is the "space left from the motion of an electrin in a band" Electrons in bands are associated with the whole crystal, not with individual atoms.
Best Answer
The free electron model is surprisingly good at predicting the properties of electrons in metals, and this implies that the electrons really are nearly free. However when you look more closely there is of course an interaction with the lattice. This is modelled using the (rather predictably named) nearly free electron model.
The conduction electrons are delocalised, so you shouldn't think of them as little balls bouncing off the ion cores. The spatial extent of their wavefunction is typically far greater than the lattice repeat, hence the relatively weak interactions. However interactions with the lattice are responsible for electrical resistance and thermal conductivity, and at very low temperatures for superconductivity. However note that these aren't interactions between a single electron and a single ion core, but rather interactions between electron waves and lattice waves (phonons).