# why do periodic lattices conduct better

by Gobil
Tags: conduct, lattices, periodic
 P: 59 Hi all, In a periodic array of ions, e.g. a metal crystal, the conduction electrons are free to move around. I have read that distortions to the periodic array can cause a decrease in conductivity of the crystal. These can be crystal impurities, phonons etc. My question is, why should the conductivity change? For example, if we had a perfect gold crystal, and measure the conductivity, it should give a finite value (?). If we then melt this crystal and keep the density more or less the same, we will have less conductivity. This I dont get, the arrangement of the atoms is different, but on average the distance between them is more or less the same, so the average potentials in the two cases are equal, and the electron feels the same forces throughout the crystal. why then are the conductivities different?
 P: 59 it is, but from a quick read it does not really answer the physics of the question. It is calculating the structure factor (which Iīm not entirely clear on) for various systems. Iīm thinking in terms of a 1D string of ions, an electron that overcomes the potential between the ions is conducting. If these ions are not periodic, why should this make a difference to the average potential felt?
P: 1,757

## why do periodic lattices conduct better

The conductivity (and resistivity) of metals is better described by considering the electrons as waves scattered by the lattice. It's not so much a matter of electrons interacting with individual ions but rather with the lattice as a whole.
P: 59
 Quote by nasu The conductivity (and resistivity) of metals is better described by considering the electrons as waves scattered by the lattice. It's not so much a matter of electrons interacting with individual ions but rather with the lattice as a whole.
in these terms, could it then be explained why the crytalline structure gives good conduction in metals? if the electrons are being scattered by a regular array, there will be interference effects, if the array is amorphous, like in a liquid metal, there will be no intereference effects. I still donīt understand how this should effect the conductivity...
 Sci Advisor PF Gold P: 2,175 It is a quantum mechanical effect, and I don't think it can be understood at the level of individual "classical" electrons. However, the QM is reasonably straightforward (solving the SE for a periodic potential) and the result are so-called Bloch waves. Have a look at the wiki http://en.wikipedia.org/wiki/Bloch_wave especially the bit about the Bloch wave vector is a conserved quantity. i.e. there is no scattering in an ideal periodic potential.
P: 59
 Quote by f95toli i.e. there is no scattering in an ideal periodic potential.
this is the part I have trouble with. what is scattering in the QM sense?

NO SCATTERING: when the the electron is translated from one area to another, but has the same momentum (wavevector) in both areas?

SCATTERING:when the the electron is translated from one area to another, and has a different energy?

I just donīt get why they scatter more in a disordered lattice.
 Sci Advisor PF Gold P: 2,175 No, you can have scattering without a change in energy (just as in classical physics), i.e. only the wavevector changes. I think your problem is that you are thinking about the electron as a particle travelling among ions, but the point is that this does not work: in a periodic potential the wave-nature of the electron dominates.
P: 59
 Quote by f95toli in a periodic potential the wave-nature of the electron dominates.

but why does this wave nature change in a disordered system, where the spacing on average is the same as the average spacing in the crystal state. The electrons are still close, so all the QM effects still apply, but for some reason they are not as "free" to move around.
 Sci Advisor PF Gold P: 2,175 The wave nature is still there, but because you no longer have a periodic lattice (e.g. a impurity means that the potential suddenly changes) means that the waves get "distorted"; you no longer get plane waves solutions.
 P: 59 so if the solutions are no longer plane wave solutions, what does this mean for the conduction? In my head if we have some distorted wavefunction, it means the probability distribution will not be regular anymore. I.e. there will be places where it is more likely to find an electron, and places where it is less likely. Is this just the same as saying we now have some potentials which the electron can not overcome (or is less likely to overcome)?
 Sci Advisor P: 3,258 You are right, it is not necessary to have plane wave solutions to get conductivity. It is sufficient that the expectation of the momentum operator does not vanish in the ground state which is also possible for wavefunctions in an aperiodic substance. Electronic properties of aperiodic substances are very complex and many interesting phenomena like Anderson localization occur. In less dimensions than 3d, electrons and holes may scatter repeatedly and coherently from the same impurity and resistivity diverges with the sample size. However, as far as I understand, no one has ever questioned that an aperiodic substance in 3d may be conductive. But the temperature dependence is different. At least at low temperatures and sufficiently pure samples, resistivity is due to electron phonon umklapp scattering. In a crystal, the combined electronic and phonon crystal momentum changes by 2K where K is a wavevector of the reciprocal lattice. In a crystal, the difference of any two electronic states at the Fermi surface is different from 2K (if not, the crystal would show some Peierls instability), so that necessarily a phonon with finite k has to be absorbed. At sufficiently low temperatures, these phonons are absent and resisitivity goes to 0. In an aperiodic crystal, Umklapp scattering may take place for any k value so that electron phonon-scattering leads to a finite value of resistivity also at low temperatures.
P: 59
 Quote by DrDu In an aperiodic crystal, Umklapp scattering may take place for any k value so that electron phonon-scattering leads to a finite value of resistivity also at low temperatures.
so in the Unmklapp process there is actually a momentum exchange between the phonon and electron. This happens when the lattice does not exhibit perfect periodicity, because in going from A to B in the lattice, the electron no longer has the same energy. If it has less energy at the end, it has scattered and given some energy to the lattice. Is this correct?
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 Quote by Gobil so in the Unmklapp process there is actually a momentum exchange between the phonon and electron. This happens when the lattice does not exhibit perfect periodicity, because in going from A to B in the lattice, the electron no longer has the same energy. If it has less energy at the end, it has scattered and given some energy to the lattice. Is this correct?
Yes, in electron phonon scattering, the perfect symmetry of the lattice is broken due to the phonon which is equivalent to a lattice deformation. More important than the change of energy is the change of momentum. Namely if the electron has momentum k before and k' after scattering and the phonon the momentum q, then k+q-k'=K, where K is some reciprocal lattice vector.
In a crystal, k,k' and K and hence also q are of the order of the Fermi momentum as k-k' cannot be too near to K as this would induce an instability of the lattice. However E(q), the energy of the phonon, is then finite and the Boltzmann factor for the presence of a phonon would be very small at low temperatures. That's how resistance increases at very low temperatures and clean crystalline samples. At higher temperatures, you can assume that there are many phonons with a Maxwell Boltzmann distribution which is not considerably disturbed by the current. Hence the electrons scatter simply from the random distortions of the phonons which act at the same time as a heat bath and take up the energy.
 P: 59 ok itīs much clearer now thanks. So if we consider a cold ~0K solid with some crystal defects, there will also be scattering due to the distortions to the wavefunctions from the aperiodic potential. and hence we have some resistance. right?