Does electron-electron scattering contribute to resistivity?

Electron-phonon and electron-defect scattering clearly contributes to resistance, but pure electron-electron scattering conserves the total momentum (and energy) of all the electrons. Then, how is it possible for electron-electron interactions to contribute to electrical resistance?

• If it is an elastic scatter you are talking about, then it does not. If inelastic then it does. Inelastic would be to change energy levels in electrons in the molecule for example. Defects are also collective behavior of electrons of the lattice anyway. May 21 '14 at 6:36

According to the Landau theory of Fermi liquids, there is a straight correpondance between the eigenstates of a system of interacting fermions (here electrons) and a system of non-interacting fermions with a renormalized mass $m\rightarrow m^*$, said effective mass (which is the case for electrons under Coulomb repulsion). These "renormalized particles" are called quasi-particles. Such renormalization come up from a collective effect between the particles.

Lets consider a fermion with a given momentum $\,\vec{p}$ in the Fermi sea ; i.e. its energy $E_\vec{p}<E_F$ with $E_F=\mu(T=0)$ is the Fermi energy. Interactions in a scattering process with an other particle can modify its momentum so that

$$\,\vec{p} \rightarrow\vec{p'}$$

which has now a "blurry" defintion reagrding the initial state. But the physical relevant quantity here seems to be the scattering rate $\Gamma_\vec{p}$, which is the rate for the particle to lose its initial momentum $\vec{p}$. The wave-function of this scattered particle can be euristically expressed as :

$$\Psi_\vec{p}\sim\exp\left(-i\frac{E_\vec{p}}{\hbar}t\right)\exp(-\Gamma_\vec{p}\,t)$$

which is nothing but the wave-function of the so-called corresponding quasi-particle. Then, the quantity $\tau_\vec{p}=\Gamma_\vec{p}^{-1}$ can be understood has the life-time of the quasi-particle, or the electron/electron relaxation time between two scattering events.

Moreover, one can show that, for Fermi liquids, $\Gamma_\vec{p}$ goes like :

$$\Gamma_\vec{p}\simeq\frac{(E_\vec{p}-E_F)^2}{\hbar E_\vec{p}}$$

Since a quasi-particle is well defined when $\Gamma_\vec{p}<<\frac{E_\vec{p}}{\hbar}$ ($\Psi_\vec{p}$ has to do enough oscillations before being damped by $\Gamma_\vec{p}$) and given the fact that a gas of electrons is degenerated at all temperature ; i.e. $E_F>>k_B T$, you will have :

$$E_F-E_\vec{p}\sim k_B T$$

Now, considering electrons in solids, Drude model gives you with a very good approximation the contribution to the conductivity of the electron/electron scattering process : $$\sigma_{e^-/e^-}=\frac{ne^2\tau}{m}\;\;\;\text{where}\;\;\;\tau=\frac{\hbar\mu}{(k_B T)^2}$$

with $\forall \,T, \,\mu(T)\sim E_F$ the chemical potential of degenerated electrons gas.

Typically, $\sigma_{e^-/e^-}$ is relevant for $T$ around $10\,K$, for higher temperature, the conductivity is governed by the electron/phonon scattering processes. For smaller $T$, there is a residual conductivity due to static impurities.

• Thank you for the detailed response! I'm still somewhat confused. Even in the presence of electron-electron interactions, we have translational invariance, so the total momentum is a good quantum number. How do we reconcile Fermi liquid theory with the requirement that the total momentum of the system is conserved? May 23 '14 at 0:31
• I don't see why one should discuss the phenomenology of Fermi liquids here. You cannot use the $\tau$ given by Laudau theory for transport, because it does not contribute to the dissipation of momentum (and current). (technically, one has to include vertex correction to the current-current correlation function, that will suppress the effect of dissipation due to electron-electron scattering).
May 23 '14 at 20:20
• @dolan: Yes, so the Landau theory is not the answer of the OP. And Drude theory assume an external source of dissipation (i.e. other than e-e scattering), which is again not the OP's question.
May 24 '14 at 2:51
• @dolan: you are wrong: e-e scattering is not a source of dissipation of momentum, since momentum commutes with the Hamiltonian. Then, if momentum is equivalent to current (for example if there are only electrons and no lattice), this implies that the conductivity is infinite. Have a look at the Mahan where all that is discussed at length.
May 25 '14 at 16:37
• @dolan: The finite life-time of the excitations does not imply a finite conductivity. This is a well-known consequence of gallilean invariance and momentum conservation. That can be showed in a few lines using the equivalence of momentum and current.
May 25 '14 at 18:37

Short answer: they don't. You need that the current operator and the (total) momentum operator are different (and of course if the current operator does not commute with the Hamiltonian) that you can get a finite conductivity. Another possibility is that momentum is not conserved (in presence of a lattice, for instance).

• Pretty sad, the only correct answer gets down voted. Jul 1 '17 at 2:26
• @Thomas I think it is more subtle than this. I have posted a (short) answer, hinting at other e-e interactions leading to a contribution to rho. But I do agree that a downvote is way too harsh here. Jun 10 '19 at 21:13

I agree that e-e scattering does not contribute to resistivity. When the medium is conducting a current the electrons system has a net momentum in the direction of the current. Resistivity occurs when this net momentum is transferred to the lattice. e-e scattering conserves the momentum of the electron system. Only a coupling between the electrons and the lattice can cause resistivity.

• As I posted in my answer, there can be a lattice interaction while still dealing with an e-e interaction (I exclude Umklapp processes). A free electron interacting with a bound electron in an atom. That's called a Baber mechanism, or Baber scattering. Jun 10 '19 at 21:15

The short answer is that yes, the e-e interaction can lead to a non zero contribution to the resistivity, but it is very subtle and could maybe be interpreted as a no, such as my2cents, Thomas and Adam have already considered.

While it is true that an e-e interaction between 2 free electrons conserve momentum (that's the interaction my2cents, Thomas and Adam probably have in mind), there are other kinds of e-e interactions, not all of which conserve (quasi)momentum.

The Baber scattering consists of an e-e interaction between a free electron and a bound one (usually in the s or d level of an atom). Such an interaction does not conserve (quasi)momentum and this causes a resistive term that has a $$T^2$$ dependence. Reference: https://arxiv.org/abs/1508.07812. Please see Figure 4. Note that in this case, one may argue that the bound electron is part of the lattice and may not qualify as a pure e-e interaction, at least not the one one might have in mind at first. But it is commonly considered as an e-e interaction, leading to a resistive contribution. (See Piers Coleman's Condensed Matter textbook and many papers.)

• While I see your point, I think that given the confusion on this issue, including this kind of interaction in the discussion does not help. But you are technically correct, and I should have answered interaction between conducting electrons does not contribute to electrical resistance.