Quasiparticles

Spastik_Relativity

I dont have a great understanding of quantum mechanics but i have been learning about superconductors at school. We've learnt about cooper pairs and how they are attracted and about their boson-like behaviour. But theres one question that i cant completely answer and whilst looking on the internet i got a couple of contrasting and undefined answers.

What exactly happens to the unpaired electrons in superconductors?
i know a couple of things such as they are influenced by the electric field and have no resistance. I have also read they are to do with the bose-einstien fluids in superconductors which i dont quite understand. Also i read that they have their certain properties due to their interaction or relationship with positive holes created by cooper pairs (this i dont understand at all).

Any information or help on this topic is much appreciated.

Thanks

ZapperZ

The unpaired electrons in the conduction band behaves like normal conduction band electrons in a metal. ONLY the electrons that have condensed in the Cooper pairs can transmit electrical current without resistance. The unpaired electrons behaves normally. The reason why you do not see their resistivity when you apply a DC field is because the supercurrent "short-circuits" them out since the condensed cooper pair are better able to conduct current.

However, when you do an AC field, you will notice that the resistance is NOT zero for a superconductor. In such a case, depending on the AC freq., the field causes both the cooper pairs and the normal electrons to oscillate back and forth. So now the resistance is a result of both the cooper paired electrons and the normal electrons. Thus, you detect AC resistivity.

The rest of your question, I don't understand, since it appears that you associated these unpaired electrons with a list of behaviors attributed to superconductivity, which they don't participate in.

Zz.

Gokul43201

I wonder if the OP is not talking about (Cooper) paired electrons after all ? The second paragraph seems to contradict some of the things in the first. Besides, there doesn't seem to be a specific question about quasiparticles, which is the thread title.

ZapperZ

That certainly is a possibility, because the rest of the question didn't make much sense.

Zz.

Spastik_Relativity

ZapperZ, your first post answered most of my question about the resistivity of the unpaired electrons. so, THANKS!

the second bit confuses even me.
basically i read that the reason they have no resistance is an interactin between the positive holes left other electrons and the unpaired electrons (which really confused me).

i guess what im trying to ask is exactly what are quasiparticles and why do the free electrons become quasiparticles.

like i said before, i dont have a great understanding of quantum mechanics and dont quite get the bose-einstien fluids in superconductors so an easy explination would be helpful.

thanks

ZapperZ

The reason why there is no DC resistivity in a superconductor is because when the Cooper pairs condenses in a BE state, there is what is known as "long range coherence". In such a state, naively we can think of the paired electrons being described as "plane waves". If you have done a bit of quantum physics, you would know that plane waves are "everywhere" simultaneously. It doesn't scatter (i.e. no dispersion), etc. So this is what we describe the supercurrent as. It is the origin of the zero resistivity.

The word "quasiparticles" came out of Landau's Fermi Liquid theory of how particles move in matter when there is a lot of interactions. As opposed to free particles, when electrons for example, move in a conductor, it not only see the potential well from the lattice ions, but also coulombic potential from OTHER electrons. Now it is VERY difficult (impossible?) to exactly solve such a many-body problem when you have a gazillion electrons to take care of.

What Landau did was to show that, in the case where the interactions between electrons are "weak", one can "renormalize" such interactions and dump it into the electron's mass, turning it into an effective mass. When one does this, one gets back a "non-interacting" particle back, but with a different mass. So he has turned one many-body problem (difficult) to a many one-body problem (easy). So you can use all you know about a "free electron" case, except the mass of the electron is now different. This different electron is called a "quasiparticle". It is a particle arising out of a single-particle excitation that has been renormalized to take into account the many-body interactions.

Zz.

Gokul43201

To the OP : It would help us to know what your level of education/learning is, so we can address your questions at the appropriate level.

1. If you are recreating this from your text/class notes, it would help to either quote the entire paragraph of interest and/or provide the name of the text.

2. I can only guess that you (your text or your notes) are talking about the mechanism for Cooper pairing - through the interaction of an electron with a lattice phonon (mode of oscillation of the positive ions).

http://www.ornl.gov/info/reports/m/ornlm3063r1/pt3.html

To stress the key points of what Zz just covered :

1. One talks of quasiparticles in strongly interacting systems (where the interaction between electrons can not be treated as a small perturbation to the energy)

2. A Quasiparticle is a mathematical trick that allows us to calculate the properties of such systems by making them look like non-interacting systems of different particles (these different particles are the quasiparticles).

ZapperZ

This is slightly off-topic from the original question of this thread, but I want to address this issue since I spent practically all of my graduate and postdoc years on this.

The Fermi Liquid picture is valid within the "weak coupling" regime. We know that from the model. The question is whether it works outside of this approximation.

Now, in the "strongly interacting" regime, there are many evidence to indicate that the "quasiparticle" picture may no longer be valid. For example, the normal state of optimally doped, and underdoped cuprates, ARPES measurements show NO quasiparticle peak (which are observed in the superconducting state). What this means is that the normal state of these cuprates are not the normal Fermi Liquid metals. This is very different than the conventional superconductor case where the superconductivity arises out of the already present quasiparticles.

The self-energy dependence as a function of temperature and the Matsubara frequency are also different than the predicted Fermi liquid model. There is more of a linear dependence than a quadratic depends that Fermi liquid predicted. Chandra Varma came up with a phenomenological model to describe this, calling it a "marginal fermi liquid".

So in the strongly interacting regime, there are many indications that the "quasiparticle" concept either does not exist, or not very well-defined.

Zz.

Gokul43201

Here's a nice summary written by Varma on the underdoped Cuprates : http://physicsweb.org/articles/world/13/2/8

ZapperZ

What, you didn't believe what I just told you?

:)

Zz.

Spastik_Relativity

very interesting

Im attending school in Australia and am currently at high school.
We've breifly done a topic about superconductivity at school but the majority of the explianations that i have read in textbooks from school try to use a classical theory to descirbe superconductivity ( and it just doesnt make sense).
Like i said i dont have a great understand of QM but i have a reasonable understanding (ive still got a alot to learn).

I am very interested in physics etc. so i wanted to learn about how superconductors "actually" work.

Spastik_Relativity

Furthermore, I have another question to do with supercondeuctors etc.

If the classical model of ion lattices and electrons in fixed orbits etc is incorrect and infact electron waves and potential wells and what not is correct. Then why does the BSC theory use lattices to describe how to electrons are attracted to eaqch otehr with the phonon being created?

Or is the distortion of the lattice similar to the distortion of the potenial well of the ions?

ZapperZ

The classical model of electrons "orbiting" the nucleus similar to the planetary model is incorrect. However, in a solid, one can picture the atoms relatively fixed in rigid locations. This doesn't contradict the first statement. In a metal, the valence electrons are no longer localized to just one atom - it can meander throughout the crystal. Thus, in a solid, the isolated model of the atom is no longer completely valid. The valence state has been modified due to the overlapping between other atoms nearby. This is what forms the conduction bands in metals.

So in a metal, one can still picture ions in rigid locations. Such a description is perfectly valid.

Zz.

Gokul43201

Hey !! Don't you recall this thread ?

ZapperZ

I did, but you should also look at what I said in that final posting by me on that thread - that the "Fermi-Liquid like" characteristics was on the OVERDOPED regime - something that I studied extensively. This is the only doping range in which the QP peak DOES persists in the normal state. The optimally and underdoped regime have no such thing.

This is getting to be quite interesting and I'm having a deja vu feeling. :)

Zz.

Spastik_Relativity

so basically u can picture the lattice, only now one will apply band theory instead of the fixed orbits view.

Gokul43201

Note that even in the case of isolated atoms/molecules, you do not have the "orbits" that Bohr first proposed.