# A Can the electron-electron interaction have an impact on the conductivity?

#### fluidistic

Gold Member
Can the electron-electron interaction have an impact on the conductivity? Google seems to point out that yes for some 2d materials such as graphene. But what about 3d materials?

Intuitively, any lost momentum by an electron would be gained by another electron, so in average, this e-e interaction should have no impact on the current and thus on the conductivity. So why would 2d materials behave differently in that aspect?

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#### ZapperZ

Staff Emeritus
2018 Award
Can the electron-electron interaction have an impact on the conductivity? Google seems to point out that yes for some 2d materials such as graphene. But what about 3d materials?

Intuitively, any lost momentum by an electron would be gained by another electron, so in average, this e-e interaction should have no impact on the current and thus on the conductivity. So why would 2d materials behave differently in that aspect?
It does. Why shouldn't it, since e-e scattering contributes to the scattering rate and quasiparticle lifetime, just like any other mechanism?

The problem is that this is such a small effect that you need (i) very clean material (very few defects and impurities), and (ii) at very low temperatures to be able to detect the impact of e-e interaction in the metal's resistivity. See, for example, https://journals.aps.org/prb/abstract/10.1103/PhysRevB.19.6172.

Zz.

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#### fluidistic

Gold Member
It does. Why shouldn't it, since e-e scattering contributes to the scattering rate and quasiparticle lifetime, just like any other mechanism?
Well, as I tried to explain, the e-e interaction is different in that there's no momentum transfer to the lattice, unlike the electron-phonon interaction. So there's no degradation of the electron's (quasi)momentum in the e-e interaction. At least, that's what I thought.

It looks like you have the Matthiesen rule in mind ($\rho = \rho_\text{e-p} +\rho_\text{e-e}$, using the nomenclature of the paper you link to) when you mention the scattering times, where there's a term from the e-e interaction. If that's the case, then my question is why is that term ($\rho_\text{e-e}$) non zero? Your answer in your first paragraph is "because you have that term".

This is what I'm trying to understand. From a quick search, https://arxiv.org/abs/1805.03650 mentions
paper 1 said:
Kohn’s theorem states that interactions do not change the cyclotron resonance frequencies in single-band metals with parabolic band dispersion relations strictly obeying translational invari-ance or, if applied in the absence of a magnetic field, that electron-electron interactions cannot degrade the current in the system and therefore cannot affect the dc long-wavelength electrical conductivity. As a result, in a clean material with a parabolic dispersion relation, the con-ductivity cannot be changed except by umklapp scatter-ing (which, arising from an underlying lattice, explicitly breaks translational invariance and momentum conser-vation) or Baber scattering (which involves multiband systems also manifesting a breaking of translational in-variance)
Emphasis mine.

ZapperZ said:
The problem is that this is such a small effect that you need (i) very clean material (very few defects and impurities), and (ii) at very low temperatures to be able to detect the impact of e-e interaction in the metal's resistivity. See, for example, https://journals.aps.org/prb/abstract/10.1103/PhysRevB.19.6172.

Zz.
Now that's something more tangible. From what I can understand from the paper you link, in compensated systems, the electron-electron interaction (more precisely, the electron-hole scattering) has no impact whatsoever on the total resistivity. It is only in uncompensated systems where the e-h scattering can cause a resistivity impact. That's where Matthiesen rule "fails". But the article makes a clear distinction between e-e scattering and the e-h one, so I'm a bit confused.
As far as I understand, Baber scattering is the process responsible for the $\rho \sim T^2$ behavior, which is hinted by the paper https://arxiv.org/abs/1206.0623 as

paper2 said:
Therefore the suggestion [3] that the T2-dependence of the DC resistivity in SrTi1−xNbxO3 can be provided by the Baber scattering mechanism is supported by the present calculation.
and from the paper https://epjb.epj.org/articles/epjb/abs/2017/08/b170367/b170367.html I get the impression that an Umklapp process is required in order to affect the resistivity (so the e-e interaction is discarded). But then it points out that electrons with different effective mass can lead to an impact on $\rho$ (for example with (quasi)electrons from s and d bands).

To add another paper to the pile, https://arxiv.org/abs/1204.3591 claims
another paper said:
While it is well-known that the electron-electron (ee) interaction cannot affect the resistivity of a Galilean-invariant Fermi liquid (FL), the reverse statement is not necessarily true (...)
Overall, from the little I understand, it is only under very specific conditions that the e-e interaction can have an impact on $\rho$. And that a scattering time does not necessarily correlates with an impact on $\rho$. Thus, in general, it is not clear at all to me why the e-e interaction leads to a contribution to $\rho$. If someone could clarify things, I would be glad.

Edit: From another (well cited) recent paper: https://arxiv.org/abs/1508.07812
the last paper said:
However, electron-electron scattering alone does not generate a finite contribution to resistivity, because such a scattering event would conserve momentum with no decay in the charge current. The presence of an underlying lattice is required in any scenario for generating T2 resistivity from electron-electron scattering.

#### ZapperZ

Staff Emeritus
2018 Award
Well, as I tried to explain, the e-e interaction is different in that there's no momentum transfer to the lattice, unlike the electron-phonon interaction. So there's no degradation of the electron's (quasi)momentum in the e-e interaction. At least, that's what I thought.
I don't understand how this would affect charge transport. e-e interaction is the surest way for an electron to lose its momentum and energy (like masses collision). In fact, e-lattice collision quite elastic and the electron doesn't tend to lose that much energy at all, if that is what we care about.

You need to look at the Boltzmann transport equation, and look at whether the mean free path comes into play. If it does, then why shouldn't e-e interaction affects the conductivity? Or do you want me to explain in detail why mean free path is a factor in the first place?

Zz.

#### fluidistic

Gold Member
I don't understand how this would affect charge transport. e-e interaction is the surest way for an electron to lose its momentum and energy (like masses collision). In fact, e-lattice collision quite elastic and the electron doesn't tend to lose that much energy at all, if that is what we care about.

You need to look at the Boltzmann transport equation, and look at whether the mean free path comes into play. If it does, then why shouldn't e-e interaction affects the conductivity? Or do you want me to explain in detail why mean free path is a factor in the first place?

Zz.
The e-lattice "collision" refers to the electron-phonon interaction. It doesn't change much the electron's energy as you say, but it can have a huge impact on the electron's momentum, which matters regarding conduction, and in particular the resistivity. No surprise here, it's well known that the e-p interaction has the major contribution to $\rho$ for metals near room temperature.

When looking at the Boltzmann equation, the e-e interaction does not affect the conductivity, unless the system is "noncompensated" or when the masses between the 2 electrons differ (as well as several other assumptions). At least, that's what I get from the papers so far.

#### ZapperZ

Staff Emeritus
2018 Award
Let's boil this down to the simplest level.

1. Do you think that e-e interaction contributes to the scattering rate and mean-free path, especially at very low temps?
2. Do you see a connection between scattering rate/mean free path, and charge transport?

Zz.

#### fluidistic

Gold Member
Let's boil this down to the simplest level.

1. Do you think that e-e interaction contributes to the scattering rate and mean-free path, especially at very low temps?
2. Do you see a connection between scattering rate/mean free path, and charge transport?

Zz.
1.) Yes I think so.

2) Not so obviously. In the sense that I fail to see that a scattering time necessarily implies an impact on resistivity. For e-p interaction, I do understand that the resistivity will be impacted, but I do not understand why for the e-e case $\rho$ would be impacted.

Let's say 2 quasielectrons interact (not with the lattice, only between themselves). Their energy will not change much, after all there aren't available states far away from $E_F$ where they already are. Their individual momentum can change a lot (by about $2 k_F$ as maximum), but since we are focusing strictly on this e-e system, we're leaving out Umklapp processes. Thus the total momentum of these 2 electrons is unchanged despite the interaction. That's why there should be no change in the current, hence the conductivity and the resistivity, despite the fact that these electrons have interacted and have an associated scattering time.

However there seem to be a Baber scattering mechanism which I wasn't aware, which seems to involve electrons of different effective masses, from which there would be an impact on the resistivity. I haven't figured out whether this involves an interaction with the lattice or only between the electrons.

#### ZapperZ

Staff Emeritus
2018 Award
1.) Yes I think so.

2) Not so obviously. In the sense that I fail to see that a scattering time necessarily implies an impact on resistivity. For e-p interaction, I do understand that the resistivity will be impacted, but I do not understand why for the e-e case $\rho$ would be impacted.

Let's say 2 quasielectrons interact (not with the lattice, only between themselves). Their energy will not change much, after all there aren't available states far away from $E_F$ where they already are. Their individual momentum can change a lot (by about $2 k_F$ as maximum), but since we are focusing strictly on this e-e system, we're leaving out Umklapp processes. Thus the total momentum of these 2 electrons is unchanged despite the interaction. That's why there should be no change in the current, hence the conductivity and the resistivity, despite the fact that these electrons have interacted and have an associated scattering time.

However there seem to be a Baber scattering mechanism which I wasn't aware, which seems to involve electrons of different effective masses, from which there would be an impact on the resistivity. I haven't figured out whether this involves an interaction with the lattice or only between the electrons.
Then this is the source of your problem.

The scattering rate in a metal is the SUM of all the scattering rates via the different mechanicsm, i.e. e-e, e-ph, e-defect, e-impurity. This sum determines the charge transport rate inside the metal.

Now, it doesn't mean that they all contribute equally at all temperatures. But the physics here clearly shows that all the scattering mechanism are there and do contribute to the overall scattering rate.

Zz.

#### fluidistic

Gold Member
Then this is the source of your problem.

The scattering rate in a metal is the SUM of all the scattering rates via the different mechanicsm, i.e. e-e, e-ph, e-defect, e-impurity. This sum determines the charge transport rate inside the metal.

Now, it doesn't mean that they all contribute equally at all temperatures. But the physics here clearly shows that all the scattering mechanism are there and do contribute to the overall scattering rate.

Zz.
I know Matthiesen's rule in terms of scattering time is $1/\tau = 1/\tau_\text{e-p} + 1/\tau_\text{other scattering mechanisms}$, that it fails in many cases. It can also be written in the form I wrote in post #3, but again, the fact that there's a scattering time for the e-e interaction does not imply that there is a resistivity part that comes from the e-e interaction.

Anyway I've read a chapter from Ziman's "Electrons and Phonons" textbook as well as the paper by Baber (1939) and the paper by Xiao Lin et al. (mentioned above). Figure 4 of that paper is really illustrative. Essentially it shows that in order for the e-e interaction to cause resistivity (in particular the $T^2$ dependence mentioned in the paper you referred to in post #2), there must be an interaction with the lattice.

The Baber mechanism or scattering is a bit special, in that a free electron interacts with a bound (!) electron. This is an interaction that leads to the $T^2$ dependence of $\rho$, though it is not the only interaction that leads to such a temperature dependence of the resistivity.

So in the end the answer is that the e-e interaction between 2 free electrons does not cause any perturbation to $\rho$, despite the fact that there exist a scattering time associated to that interaction. However, an interaction between a free electron and a bound electron (also refered to an e-e interaction) does lead to an impact on $\rho$. There are other e-e interactions that also lead to an impact on $\rho$, but they all involve an interaction with the lattice.

Problem solved, to me.

#### Henryk

Gold Member
Well, as I tried to explain, the e-e interaction is different in that there's no momentum transfer to the lattice, unlike the electron-phonon interaction
Actually, that is not exactly true. In crystal, the momentum is conserved to within a reciprocal lattice vector. So, you can have to electrons interactions that result in a net loss of the momentum.
Next thing, even when the total momentum is conserved in a scattering process, there may be a loss of current. The reason is that the electron velocity is not simply ,momentum divided by mass. Mass can, actually, be negative for some carriers. Therefore, you might have a scattering process when one electron gains momentum, the other loses it but the velocity of the first electron decreases and so does the velocity of the second electron.

#### Gigaz

Can the electron-electron interaction have an impact on the conductivity? Google seems to point out that yes for some 2d materials such as graphene. But what about 3d materials?

Intuitively, any lost momentum by an electron would be gained by another electron, so in average, this e-e interaction should have no impact on the current and thus on the conductivity. So why would 2d materials behave differently in that aspect?
As others have said, it is difficult to have an impact on the conductivity from electron-electron scattering. But electron-electron interaction can turn materials which should be metallic into Mott insulators:
In this case, you have an extremely strong impact of electron-electron interaction on conductivity.

#### fluidistic

Gold Member
Actually, that is not exactly true.
I agree I was wrong, as I recapitulate in post #9.

Henryk said:
In crystal, the momentum is conserved to within a reciprocal lattice vector. So, you can have to electrons interactions that result in a net loss of the momentum.
That is true, however in that case it isn't solely an e-e interaction, there is an electron-phonon interaction involved, and there, there is no doubt/question/surprise that the current can be degraded and so the conductivity impacted. This is not related to the e-e interaction discussed in the question.

Henryk said:
Next thing, even when the total momentum is conserved in a scattering process, there may be a loss of current. The reason is that the electron velocity is not simply ,momentum divided by mass. Mass can, actually, be negative for some carriers. Therefore, you might have a scattering process when one electron gains momentum, the other loses it but the velocity of the first electron decreases and so does the velocity of the second electron.
Good point, I must study this in more details. It sounds like electrons of different bands interacting together.

#### DrDu

That is true, however in that case it isn't solely an e-e interaction, there is an electron-phonon interaction involved, and there, there is no doubt/question/surprise that the current can be degraded and so the conductivity impacted. This is not related to the e-e interaction discussed in the question.
That's not true, an Umklapp event in electron scattering does not involve phonons, although there are obviously also Umklapp events for phonons.

#### fluidistic

Gold Member
That's not true, an Umklapp event in electron scattering does not involve phonons, although there are obviously also Umklapp events for phonons.
You are right, I was not aware of this. I am sorry. Someone really needs to answer properly this question as clearly as possible, I have already tried and messed up a few times.

#### DrDu

Best have a look at the masters own words (p. 63ff):

"Can the electron-electron interaction have an impact on the conductivity?"

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