Suppose there's a metal in which electrons interact with themselves and with the phonons. The hamiltonian might look like this

\begin{equation} H= \sum_{k}\epsilon_k c^\dagger_k c_k + \sum_{k}\omega_ka^\dagger_ka_k+H_{e-e}+H_{e-p} \end{equation}

where $\epsilon_k$ is the energy of the electrons, $\omega_k$ is the energy of the phonons and the other two terms are interaction terms. Without going into the details, let's say that I calculate the resistivity only due to e-e scattering and only due to e-p scattering and I get

\begin{equation} \rho_{ee}=a T^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rho_{ep}=b T \end{equation}

where $a$ and $b$ are just constants. The question is: How should I add these resistivities? Is it reasonable to just say $\rho_{total}=\rho_{ee}+\rho_{ep}$? I'm trying to avoid a calculation where I need to treat both scattering processes simultaneously, of course.

Some extra information: The e-e resistivity is calculated in the context of Umklapp scattering at low temperatures when the Fermi surface has a similar size as the Brillouin zone. The e-p resistivity is the tipical calculation at high temperatures when the phonons become classical. There's a temperature range where both effects coincide and that's where I'm trying to get the total resistivity.

P.S.: I read about Matthiesen rule and also, in a different place, about the "parallel resistor" formula but I'm not sure how to apply those concepts here. (Just naming those terms in case they ring a bell).


1 Answer 1


I would recommend reading more about Matthiessen's rule, preferably the section on it in Ashcroft and Mermin's near-indispensable textbook. In short the idea is that scattering rates from distinguishable scattering mechanisms (as you seem to have here) are additive. In the relaxation-time approximation that implies that the total relaxation time $\tau$ satisfies $$\frac{1}{\tau}=\sum_i \frac{1}{\tau_i}.$$ I.e. the relaxation times add like parallel resistances. If none of the $\tau_i$ depends on the momentum $k$ we obtain Matthiessen's rule for the resistivity: $$\rho=\sum_i \rho_i.$$ In other words, physical resistivities add like serial resistors, just like you proposed. If you think about it, this is quite natural. A bunch of electron going from point A to point B in a slab of the material will experience both scattering mechanisms - it can't split up and have a portion of it undergo only electron-phonon scattering.

However, there's a big assumption made above: that $\tau_i$ is $k$-independent for each $i$. When this doesn't hold Matthiessen's rule becomes an inequality $$\rho \ge \sum_i \rho_i.$$

We can obtain also obtain a formal upper bound. There's a variational result due to Kohler [cited as Z. Phys. 126, 495 (1949)], described in Ziman's book Electrons and Phonons (1960). It's an interesting result for metals, but I don't think it's particularly useful for microscopic descriptions/lattice system.

Introduce a function $\Phi_\mathbf{k}$ such that the electron distribution becomes $$f_\mathbf{k}\equiv f_\mathbf{k}^0 - \Phi_\mathbf{k} \frac{\partial f_\mathbf{k}^0}{\partial \epsilon_\mathbf{k}}.$$ We can view $\Phi_\mathbf{k}$ as the deviation from equilibrium, or the extra energy due to the transport process. For two given scattering mechanisms $P_1$, $P_2$ we can associate two exact solutions $\Phi_1$, $\Phi_2$. The resistivity of each mechanism in isolation can be written as an inner product, $\rho_i=\langle \Phi_i | P_1 \Phi_i\rangle$. Forming the linear combination $$\Phi' = \eta \Phi_1 + \left( 1-\eta\right) \Phi_2$$ we have that $$\rho \le \langle \Phi' | \left( P_1 + P_2 \right) \Phi'\rangle.$$

Minimizing with respect to $\eta$ we obtain the bound $$\rho - \left( \rho_1 + \rho_2 \right) \le \beta_1 \beta_2 \frac{\rho_1 \rho_2}{\beta_1 \rho_1 + \beta_2 \rho_2},$$ where $\langle \Phi_j|P_i\Phi_j\rangle = \left( 1+ \beta_i\right)\rho_i$.

  • $\begingroup$ Great Answer! Is there a way to get an upper limit on the resistivity as well? Or adding them up gives a lower boundary on resistivity and that’s as far as we can go? $\endgroup$ Commented Aug 29, 2019 at 13:37
  • $\begingroup$ @P.C.Spaniel I'm not aware of any general results along those lines. The upside is that the inequality actually doesn't rely on the relaxation-time approximation - it can be derived under more general conditions than the equality. If you need a precise value for the resistivity (rather than rough temperature scaling trends) it'd likely be better to use something like Kubo's formula. $\endgroup$
    – Anyon
    Commented Aug 29, 2019 at 13:59
  • $\begingroup$ @P.C.Spaniel I did find an upper bound in Ziman's old book. I think it's mostly useful in analytically tractable cases, but if you have an idea about how far your system is from equilibrium it does let you go slightly further. $\endgroup$
    – Anyon
    Commented Sep 2, 2019 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.