8
$\begingroup$

In my introductory condensed matter physics class, I learnt that the solutions to the time independent Schrodinger equation in a periodic potential are Bloch states. These wave functions are stationary states that have very high uncertainty in position - they are spread out over the whole crystal.

My question is: are electrons in crystals really delocalised over the whole crystal, with their wave functions static? Or do they move around from place to place with more localised wave functions?

I was specifically wondering about this questions with respect to conductors. I know that the expectation value of position for localised wave functions moves approximately according to F=ma, and I also know that Drude theory (which uses F=ma to model electrons in a conductor) correctly predicts V=IR. Is the reason that Drude theory works decently the fact that electrons in conductors are actually fairly localised?

If the electrons in a conductor were highly delocalised (like in Bloch states) then I would not expect them to be well modelled by F=ma and would hence expect the classical Drude model to fail badly and not correctly predict V=IR.

I'm quite confused, so thanks for any help!

$\endgroup$
2
  • 1
    $\begingroup$ The Bloch theorem supposes a infinite periodic potential, what is a reasonable assumption for a real crystal. But when this crystal is under an external potential, we can not say that the Hamiltonian has a periodic potential. So Bloch waves are not more solutions as far as I can see. $\endgroup$ Mar 2, 2023 at 23:12
  • $\begingroup$ Make sure you aren't conflating solutions of the time-independent Schrodinger equation with solutions of the time-dependent Schrodinger equation. $\endgroup$
    – user196574
    Mar 3, 2023 at 1:13

2 Answers 2

13
$\begingroup$

In a perfect crystal at zero temperature the wave functions are completely delocalized.

Now, as we get more and more realistic, and consider various physical phenomena in conductors, we need to account for imperfections, interactions, non-equilibrium, etc.

Core electrons of atoms, corresponding to deep lying energy bands are pretty much in Bloch states (although strongly peaked at the ions.) Valence and conduction band electrons are rather spread - after all, they still assure the binding between the atoms, which holds the crystal together. However, various interaction/scattering processes, as well as crystal imperfections (impurities, dislocations) result in finite coherence length - the characteristic distance after which the phase of electron is randomized. This coherence length can be considered roughly as the size of electron.

The coherence length is usually much longer than the lattice constant, so that we can still use the Bloch theory (that is the energy bands theory), however compared to the sample size we can consider it very short and hence treat electrons as particles.

Nowadays there are possibilities to create semiconductor nanostructures with very large coherence lengths - of the order of microns, that is thousands or tens of thousands of lattice spacings. This allows studying various quantum effects, where conductance deviates from Drude description - such as weak localization, conductance quantization, Coulomb blockade, etc. This is often referred to as mesoscopic physics, mesoscopic structures, meaning that the structures in question are not microscopic (they are much bigger than atoms), but they cannot be treated macroscopically either, due to the essential quantum effects.

Update

  • At not too low temperatures we can actually estimate the conduction electron size from a simple argument: in effective mass approximation, the density matrix of free electrons is $\rho(\mathbf{k})\propto \exp\left(-\frac{\mathbf{p}^2}{2m^*k_BT}\right)$, that is the electron wave packet has momentum uncertainty $\sigma_p^2=m^*k_B T$. The position uncertainty then can be estimated as $\sigma_x^2\approx\frac{\hbar^2}{4\sigma_p^2}=\frac{\hbar^2}{4m^*k_BT}$. Electrons become more and more localized at higher temperatures.

  • Related question is Why do Drude/Sommerfeld models even work? - this question does not specifically focus on localization, but addresses a similar problem of why a particle description is appropriate for discussing properties of semiconductors and metals.

$\endgroup$
2
  • 1
    $\begingroup$ When you say core electrons do you mean electrons that are, for example, in 1s states in atoms with multiple shells? When you say they are in Bloch states does that mean these core electrons are "delocalized" in the sense that their wavefunction repeats at every atom? I somehow always assumed that these core electrons were delocalized, in the sense that the wavefunction would look like it was centered at a single atom, so this would be interesting to hear. $\endgroup$ Mar 2, 2023 at 13:13
  • 3
    $\begingroup$ @AccidentalTaylorExpansion I don't have special expertise on this subject, but I assume that due to their indistinguishability we must consider them as Bloch states. In fact, even if electrons do not tunnel between atoms at all, their wave functions can be written as Bloch states - here it is simply a basis transformation. So yes, I think of 1s electrons as Bloch states, but localized on the atoms, which simply means that the Bloch amplitude $u(\mathbf{r})$ is very sharply peaked on the atoms, and almost zero between them. $\endgroup$
    – Roger V.
    Mar 2, 2023 at 13:19
2
$\begingroup$

Note that Bloch states in a crystal are just a generalization of the plane waves in a vacuum (the momentum eigenstates). Just as it doesn't make much sense to speak about Newton's laws for momentum eigenstates, it also doesn't make sense to apply them to Bloch states.

But, as soon as you consider a more realistic kind of wavefunctions, like wave packets, you can immediately see that these lumps of probability will indeed move according to the Newton's laws as described by the Ehrenfest theorem (with the usual caveat of how the derivative of the potential is taken and, in case of the crystal, the use of the effective mass and quasimomentum).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.