Does electron-electron scattering contribute to resistivity?

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

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.)

Answer 5

Previous discussions were all performed at the quantum mechanical level. However, the same question can obviously be asked about classical systems. For example, can the viscosity of liquids (e.g. water) arise solely from the interactions between the constituents? The answer could be found in Boltzmann's transport theory. Clearly, if only that type of interactions existed, a flowing water would never stop by virtue of conservation of the momentum of center of mass, in contradiction with intuition. Indeed, inter-particle interactions alone can never bring about viscosity (resistivity), because the latter is fundamentally related to irreversible processes (energy dissipation into heat). Microscopic dynamical laws always obey time-reversal symmetry and can therefore not bring about irreversible processes. Irreversible elements have to be put in by hand. In Boltzmann's theory, for instance, such element is introduced by a distribution function, which emphasizes the probabilistic feature of the system rather than the motion of individual particles.

The key message is that, resistivity originates from irreversible processes.