Say I have an ellipsoidal dispersion:

$$E(k) = \frac{\hbar^2}{2}\left[\frac{k_x^2}{m_x}+\frac{k_y^2}{m_y}+\frac{k_z^2}{m_z}\right]$$

If I want to talk about the conductivity in the x-direction, I need to use $m_x$.

But what if I want to talk about the conductivity along the unit vector $(1,1,1)$? More generally, what is the effective mass we use when talking about the conductivity along some non-principal directions?

I'm studying is the application of Drude theory in semiconductors, where the conductivity along a principal direction is given by $ne^2\tau/m_i$, where $m_i$ is the effective mass along the direction.

Say I want to know the conductivity in the direction of the electric field $\vec{E}$, given that it does not point along any of the principal axes.

Here is how I analyze the problem:

Due to the electric field, the electron would be moving in all three of the $x,y,z$ direction, driven by $E_x,E_y,E_z$, respectively.

By this argument, the obvious way is to just write the conductivity $\sigma$ as a tensor, as shown in textbooks.

However, I was wondering if there is a way to express $\sigma$ as a scalar so that the analysis can be reduced to a one-dimensional problem by simply writing the equation as scalars: $j=\sigma E$.

I naively tried to add all the contributions together along with the corresponding cosines and sines of the angles the field makes with the principal axes. What I mean is that, if $\theta$ is the angle made by $\vec{E}$ with the $z$-axis, and $\phi$ is the angle made by the $xy$-projection with the $x$-axis, then

$$\sigma_{\parallel \vec{E}}=ne^2\tau\left[\frac{\sin{\theta}\cos{\phi}}{m_x}+\frac{\sin{\theta}\sin{\phi}}{m_y}+\frac{\cos{\theta}}{m_z}\right]$$

but this is clearly nonsensical.

  • 2
    $\begingroup$ The reciprocal mass turns into a tensor, and you evaluate the tensor on the unit vector and obtain a ``velocity" that is not in the same direction as the momentum, which is fine. Just do it naïvely. $\endgroup$ Apr 27, 2023 at 9:22
  • $\begingroup$ Could you add more details? Do you consider a metal, a semiconductor? Which conductivity model do you use, and from which equations do you find it? $\endgroup$
    – E. Anikin
    Apr 27, 2023 at 10:55

1 Answer 1


For the case of anisotropic mass tensor, the conductivity also becomes a tensor, as noted by naturallyInconsistent: \begin{equation} \sigma = ne^2\tau\, \mathrm{diag}(m_x^{-1}, m_y^{-1}, m_z^{-1}). \end{equation} In general, when $m_x \ne m_y \ne m_z$, the current $\vec{j} = \sigma \vec{E}$ is not parallel to the electric field: it will also contain a component perpendicular to the electric field. Therefore, there is no way to express $\sigma$ as a scalar and to reduce the problem to the one-dimensional case.

However, one can ask about the projection of the current $\vec{j}$ on the unit vector parallel to $\vec{E}$: \begin{equation} j_\parallel = \frac{(\vec{j},\vec{E})}{|\vec{E}|} = \frac{(\sigma\vec{E},\vec{E})}{|\vec{E}|} = ne^2\tau|\vec{E}| \left[\frac{\sin^2{\theta}\cos^2{\phi}}{m_x}+ \frac{\sin^2{\theta}\sin^2{\phi}}{m_y} + \frac{\cos^2{\theta}}{m_z}\right] = \sigma_{\parallel\vec{E}}|\vec{E}|. \end{equation} The proportionality coefficient $\sigma_{\parallel\vec{E}}$ can be called "the conductivity along the direction of $\vec{E}$":


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