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The solutions to the time independent Schrodinger equation for a periodic potential are Bloch waves of the form $$\psi(r) = u(r)e^{ik.r}$$ where $u(r)$ is a periodic function with the same periodicity as the potential, and $k$ is the crystal wave vector.

At a given temperature, the crystal is vibrating due to its thermal energy (a superposition of its various vibrational modes) and therefore at any instant in time the crystal should be quite disordered violating the assumption of periodicity. So why are Bloch waves often the starting point for so many calculations in semiconductor physics?

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It is a quantitative question how much disorder is necessary to break up the basic Bloch structure of the electron wave functions on a lattice. Figuring out the answer was a key piece of the work that garnered Philip Anderson the Nobel Prize in 1977. Thermal noise is one source of randomness, which affects all atoms in the lattice more or less equally. Other disturbances to the lattice structure may be localized around impurities or lattice defects.

Macroscopically, the observable that most easily characterizes the possible disturbance of the Bloch wave solutions of the Schrödinger equation is whether a crystalline material is a conductor or an insulator. Normally, we expect that crystals in which the electron Bloch bands are partially filled will be conductors. They possess unoccupied Bloch states available immediately above the Fermi energy. Electrons in these states can propagate long distances (in a perfect lattice, the Bloch states are completely delocalized, corresponding to an infinite mean free path) and thus conduct electricity. Obviously, however, the presence of sufficient quantities of impurity or disorder will introduce the possibility of dissipative collisions that limit this conductivity; in this case, instead of Bloch waves, the electron wave functions near the Fermi level become localized.

Any disturbance to the the perfect regularity of the lattice will create some possibility for electron-atom collisions. There will be some reduction in the spatial extent of some of the electron orbitals. However, the quantitative question of whether all the Bloch waves are disturbed by a small amount of disorder turns out to depend on the dimension of the crystal lattice. In two dimensions, any amount of disorder will destroy the extended Bloch wave eigenstates, but in three dimensions, there is a threshold below which fluctuations and disorder do not change the Bloch structure too much. Above the disorder threshold, there is what is known as "strong localization" [or "Anderson localization"], in which the conductivity of a sufficiently disordered (e.g. hot) three-dimensional crystal is destroyed—converting the substance into an insulator. Below the threshold, there is only a modest increase in the resistivity, while the material remains, by and large, a good conductor.

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    $\begingroup$ "Electrons in these states can propagate long distances ... and thus conduct electricity" — electrons in all Bloch states can propagate long distances due to delocalization. This alone doesn't let them conduct electricity. $\endgroup$
    – Ruslan
    Commented Oct 25, 2022 at 18:58
  • $\begingroup$ "In two dimensions, any amount of disorder will destroy the extended Bloch wave eigenstates" can you provide some reading reference for this, I have heard this many times and always wanted to read more about details of reasoning $\endgroup$
    – Kutsit
    Commented Oct 26, 2022 at 13:45
  • $\begingroup$ @Ruslan, can you tell describe what more is needed? I am not getting why won't delocalized states in a partially filled band be sufficient to generate current? $\endgroup$
    – Kutsit
    Commented Oct 26, 2022 at 13:48
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    $\begingroup$ @Kutsit Partial filling is a sufficient additional condition. My comment was about the sentence quoted that seems to imply that delocalization alone is a sufficient condition, which it is not. $\endgroup$
    – Ruslan
    Commented Oct 26, 2022 at 21:02
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Well, there are also defects in every real material that break the lattice translational symmetry. But as you probably encountered it many times in quantum mechancis, we can look at an easier version of the problem, diagonalize the Hamiltonian and then after that look at the real problem (with defects, phonons, etc.) and how much these imperfections change the problem. If the effect is small, one can treat them in perturbation theory.

But there are many examples where that is not possible. An example would be the Kondo effect (mininum in electrical resistance at a temperature $T_K>0$, called the Kondo temperature, that is observed in materials with magnetic impurities) which is explained by the Kondo bound state. The derivation is explicitly non-perturbative and the state is not a (superposition of) bloch waves.

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    $\begingroup$ It is easy to imagine why a defect would be a small perturbation. However in the case of phonons, it seems to me that the translational symmetry breaks dramatically. Is there a way to say/estimate that the potential the electron feels is still mostly periodic? $\endgroup$
    – AbelT
    Commented Oct 23, 2022 at 19:44
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    $\begingroup$ A quantitative approach would be to compare the involved enrgy scales. The relevant energy scale for electrons is the Fermi energy (or temperature), which is for metals of the order of $10^4$K. Compared to the typical phonon energy, this is huge, since $\theta_D \sim 10^2$K ($\theta_D$ is called the Debye temperature). $\endgroup$
    – Samuel
    Commented Oct 23, 2022 at 19:58
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    $\begingroup$ @AbelThayil To get a rough intuitive sense of the scale of thermal vibrations, a crystalline solid typically melts when the mean amplitude $A$ of the atomic vibrations about their equilibrium position is about $10\%$ of the interatomic distance $a$. This (or rather, the more quantiatively accurate version of this rule) is called the Lindemann melting criterion. At the same time, the vibrational amplitude scales with $\sqrt{T}$, so we would loosely have that $A \sim 0.1 a \sqrt{T/T_m}$, where $T_m$ is the (absolute) melting temperature. $\endgroup$
    – J. Murray
    Commented Oct 24, 2022 at 2:44
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Actually, even without phonons, Bloch's theorem hypothesis of a periodic potential is badly violated by all real crystals: finite samples have boundaries, then cannot be periodic.

Then, how do Bloch states, although impossible in the real world, are so useful in Solid State Physics?

The answer is Perturbation Theory. The main ideas to justify it in the case of crystal physics are:

  1. the existence of an ideal periodic problem close enough to the real-world non-periodic system, at least locally;
  2. the difference between observable quantities evaluated for the ideal and real system should be negligible.

The two ingredients work differently in different contexts, but both are required.

For example, Bloch's states are useless in describing a glass because the atomic structure of glass has no resemblance with a periodic structure, even at the scale of a few neighbors. They are useful for finite-size crystals, even if the translational symmetry is broken at the boundary because it is possible to show that the difference between properties of a finite part of a perfect (infinite) periodic crystal and of a finite-size crystal is proportional to the surface area of the surface. Therefore, this difference becomes negligible for macroscopic samples (it grows with the volume of the sample as $V^{\frac23}$, while bulk properties grow as $V$).

The same argument can be used to justify the usefulness of Bloch's states in the presence of thermal motion (phonons) or defects. Provided the number of phonons or defects is not too high, in most of cases, the perfect crystal solution is an excellent zeroth-order approximation. Only when thermal disturbance becomes too high (too many phonons, resulting in too large amplitude vibrations$^{(*)}$) or if the number of defects reaches a threshold of concentration, the structure of the real system departs so much from the ideal crystal to make Bloch description inadequate.

$^{(*)}$ The empirical Lindemann's melting criterion predicts melting as soon as the mean square fluctuation around equilibrium positions reaches a given percentage of the first neighbor distance.

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Periodic crystal is the closest exactly solvable case for considering more complex situations. Building a theory as a perturbation around an exact solution is a commonplace approach in physics.

Thus, distortions of the crystal, applied bias, deformation and even interactions can be treated as small perturbations to the perfectly periodic crystal lattice. (Note also that a great part of interactions is already incorporated into the periodic potential for the Bloch waves, which are typically used only for valence and conduction electrons, i.e., for the outer shells of the atoms constituting the crystal.)

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  • $\begingroup$ It is incorrect that Bloch waves are typically used only for valence and conduction electrons. Bloch's theorem is valid for every electronic state in a periodic potential, even for strongly localized states. $\endgroup$ Commented Oct 24, 2022 at 22:54
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    $\begingroup$ @GiorgioP Bloch theorem is valid for any periodic potential, but one does need to have a periodic potential in order to apply it. Classical solid state physics texts do away with the core electrons well before they get to Bloch waves. $\endgroup$
    – Roger V.
    Commented Oct 25, 2022 at 5:12
  • $\begingroup$ Focus on valence electrons has nothing to do with Bloch states. Indeed, even very localized states in tight-binding approximation are written as Bloch's waves. $\endgroup$ Commented Oct 25, 2022 at 5:27
  • $\begingroup$ @GiorgioP do you have an example of any specific physical problem in mind? $\endgroup$
    – Roger V.
    Commented Oct 25, 2022 at 6:03
  • $\begingroup$ @GiorgioP am talking essentially about the Born-Oppenheimer approximation: one assumes that the crystal exists - which means that some core electrons are grouped with nuclei, forming the periodic potential, which is then solved for Bloch states. FYI: I understand that English is the native language for neither of us, but your combative style is rather offensive. $\endgroup$
    – Roger V.
    Commented Oct 25, 2022 at 6:12

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