The effective mass is a tensor property of rank two. It may be anisotropic if the corresponding electron band is not perfectly isotropic. For silicon, this seems to be the case.

However, there exists an alternative approach, which describes tensors of physical significance for crystals based on their symmetry. This is Neumann's principle, which states that the tensor form must be invariant under all symmetry transformations of the crystal. Since silicon is cubic, I conclude that the effective mass must be isotropic.

What is wrong? Does Neumann's principle not apply here?

As much as HTNW's answer is correct, it is not exactly what I am looking for. A second rank tensor describing a physical quantity in a cubic crystal has the form $\sigma_{ij} = \sigma\delta_{ij}$ where $i$ and $j$ are $\{x,y,z\}$. Is this tensor isotropic or not? It doesn't matter to me what we call it; that is only semantics.

Nevertheless, experiments indicate that the effective mass has one value in the "longitudinal direction" and a different value in the "transversal direction". This is certainly not captured in the tensor given by Neumann's principle. The question still stands: what causes this discrepancy?

Edit: The edited answer is very satisfactory and is now accepted.

  • $\begingroup$ I am not too familiar with the matter but: Cube means Octahedral symmetry. But if you take a rotation of the unitcell by lets say 42 degree around some axis, this is not giving same image... Hence the effective mass in that direction might also differ. Id rather say that every symmetry operation in the Octahedral symmetry group can also be applied on the effective mass tensor. $\endgroup$
    – Martin
    Commented Jun 16, 2021 at 16:18
  • 2
    $\begingroup$ "Neumann" refers to Franz Ernst Neumann, not John von Neumann. From Neumann's principle: Neumann's principle, or principle of symmetry, states that, if a crystal is invariant with respect to certain symmetry operations ... generalized to physical phenomena by Curie laws. ... Franz Neumann's (1795-1898) principle was first stated ... (1873/1874) $\endgroup$ Commented Jun 17, 2021 at 9:58

3 Answers 3


Neumann's rule can actually predict the form of these anisotropic efective mass tensors and the existence of the longitudinal and transverse masses, if you provide it with the right information. Therefore, I will first make explicit some preliminaries, so that the final calculation resulting in this prediction makes sense.

Silicon's crystal structure has the $O_h$ (octahedral) point group, therefore any quantity associated with the crystal itself should transform under this point group. Specifically, the symmetry elements $\mathbf{A},\mathbf{B}\in O_h,$ can be specified by matrices that act on vectors $\mathbf{v}\in\mathbb{R}^3$ by $$(\mathbf{A}\mathbf{v})_i=A_{ij}v_j,$$ act on each other by $$(\mathbf{A}\mathbf{B})_{ij}=A_{ik}B_{kj},$$ and as a consequence must act on rank-2 tensors ${\boldsymbol\sigma}\in\mathbb{R}^3\otimes\mathbb{R}^3$ by $$(\mathbf{A}\boldsymbol\sigma)_{ij}=A_{ii'}A_{jj'}\sigma_{i'j'}.$$

The group $O_h$ can be generated by three reflections, which in our presentation (choosing Cartesian coordinates) can be written $$A_{ij}=\begin{bmatrix}1&&\\&1&\\&&-1\end{bmatrix}\quad B_{ij}=\begin{bmatrix}1&&\\&&1\\&1&\end{bmatrix}\quad C_{ij}=\begin{bmatrix}&1&\\1&&\\&&1\end{bmatrix}.$$

Neumann's rule is that any rank-2 tensor-valued property of the crystal must be invariant under all of $O_h$, and because $\mathbf{A},\mathbf{B},\mathbf{C}$ generate $O_h$, it boils down to equations like $\sigma_{ij}=A_{ii'}A_{jj'}\sigma_{i'j'}$ (resp. for $B_{ij}$ and $C_{ij}$). If you calculate this out, you indeed find $\sigma_{ij}=\sigma\delta_{ij}$—any rank-2 tensor (such as effective mass) should be diagonal and thus isotropic (same in every direction), since it acts like just a scalar $\sigma$.

This means silicon crystals cannot have any "one" (i.e. in some sense preferred or unique) anisotropic effective mass. However, silicon's conduction band actually has, in addition to one minimum centered at wavevector $\mathbf{k}=\mathbf{0}$, 6 lower minima centered at $\mathbf{k}_{+x}=\begin{bmatrix}k\\0\\0\end{bmatrix},\mathbf{k}_{-x}=\begin{bmatrix}-k\\0\\0\end{bmatrix},$ etc. for $\mathbf{k}_{\pm y},\mathbf{k}_{\pm z}.$ Note that each of these "preferred directions" alone is not invariant under the symmetry of the crystal, and so you might think Neumann's rule excludes their existence, but actually they transform into each other, so they can exist when they are all together. Similarly, the effective mass around each of these points can be anisotropic, but together you find that the effective mass around one point transforms under the crystal's symmetries into the effective mass around another, and so the entire system is symmetric enough to exist in the crystal. (Note that the central minimum's effective mass is isotropic as predicted.)

Explicitly, we can define the effective mass as a function $\sigma_{ij}(\mathbf{k})$ of which "well" we're in (where $\mathbf{k}$ may only be the $\mathbf{k}_{\pm x},\mathbf{k}_{\pm y},\mathbf{k}_{\pm z}$ from above, and optionally $\mathbf{0}$). Neumann's rule then predicts that this function must satisfy $$\sigma_{ij}(\mathbf{A}\mathbf{k})=A_{ii'}A_{jj'}\sigma_{i'j'}(\mathbf{k})$$ for all $\mathbf{A}\in O_h$ in order to be allowed. If you compute this out for each of the generators above you will again come to some set of requirements for $\sigma_{ij}(\mathbf{k}).$

Doing the calculation, I get $$\sigma_{ij}(\mathbf{0})=m_0\delta_{ij}\\\sigma_{ij}(\mathbf{k}_{+x})=\sigma_{ij}(\mathbf{k}_{-x})=\begin{bmatrix}m_l&&\\&m_t&\\&&m_t\end{bmatrix}\\\sigma_{ij}(\mathbf{k}_{+y})=\sigma_{ij}(\mathbf{k}_{-y})=\begin{bmatrix}m_t&&\\&m_l&\\&&m_t\end{bmatrix}\\\sigma_{ij}(\mathbf{k}_{+z})=\sigma_{ij}(\mathbf{k}_{-z})=\begin{bmatrix}m_t&&\\&m_t&\\&&m_l\end{bmatrix},$$ where $m_{0,l,t}$ are the remaining degrees of freedom after all equations have been eliminated. They correspond respectively to the isotropic (central minimum) and longitudinal and transverse (anisotropic minima) effective masses. You can see how Neumann's rule is actually rather predictive (42 DOF have become 3!): I only inputted that there was some symmetric set of privileged directions each with an associated effective mass tensor and it suggested the existence of longitudinal and transverse masses along/perpendicular to those directions, while eliminating cross terms. Further, these masses must be the same in every well.

To get really abstract about what's going on here, note that Neumann's principle only applies to those quantities that truly belong to the crystal, without anything external breaking the symmetry. If I just gave you a piece of silicon, you could determine and write down as a 2-tensor the effective electron mass in the central minimum of the conduction band without breaking the crystal's symmetry, but you cannot do that with the effective mass in the anisotropic minima. You'd have to pick one minimum to do the measurement around, and then you've broken the symmetry of the crystal and Neumann's principle can't be blamed if the tensor you get is not symmetric enough. However, you can preserve the symmetry by realizing that "the effective mass around the anisotropic minima" is a quantity on its own right, which contains all 6 anisotropic effective masses in a symmetric container. This symmetric quantity is no longer "contaminated" by your choice of minimum to measure it at, so it indeed "belongs" to the crystal alone and you can apply Neumann's principle to this larger object (which is the function ${\boldsymbol\sigma}(\mathbf{k})$) to inspect the structure of all of the anisotropic effective masses at once. More generally, $\boldsymbol\sigma$ is a function of all of (continuous) $\mathbf{k}$-space anyway, so the principle ${\boldsymbol\sigma}(\mathbf{A}\mathbf{k})=\mathbf{A}{\boldsymbol\sigma}(\mathbf{k})$ comes up that way too.

Partial Reference


A second rank tensor describing a physical quantity in a cubic crystal has the form $\sigma_{ij} = \sigma\delta_{ij}$ where $i$ and $j$ are $\{x,y,z\}$. <...> Nevertheless, experiments indicate that the effective mass has one value in the "longitudinal direction" and a different value in the "transversal direction".

Effective mass tensor describes the behavior of electron in a single valley, of which there are six equivalent instances for the conduction band minimum. If this tensor described the properties of a crystal, then indeed it should be isotropic*. But actually, it describes properties of a single point (its small neighborhood) in the Brillouin zone—and it's not the $\Gamma$ point—so it can be anisotropic without breaking any principles.

*Notice how the average of the 6 equivalent-valley effective masses does become isotropic, unlike each of the masses.

  • $\begingroup$ You still need to explain the anisotropic hole mass. $\endgroup$
    – my2cts
    Commented Jun 17, 2021 at 19:35
  • $\begingroup$ @my2cts why do I need to explain it? $\endgroup$
    – Ruslan
    Commented Jun 17, 2021 at 19:44
  • $\begingroup$ Because your argument relies on the degeneracy of the indirect band minima. The top of the valence band is not degenerate and still the effective mass is anisotropic. $\endgroup$
    – my2cts
    Commented Jun 17, 2021 at 21:36
  • $\begingroup$ @my2cts From the footnote at the bottom of my reference I glean that the (heavy?) hole band is isotropic at $\mathbf{k}=\mathbf{0}$ and quickly becomes anisotropic as you move out, which just means that the approximation that energy is quadratic in $\mathbf{k}$ breaks down. Then this argument is mostly fine: at the one point that transforms to itself under symmetry ($\mathbf{k}=\mathbf{0}$) the effective mass has the required symmetry (isotropy), but any nearby points won't have that symmetry so that is no longer required (but you could still write symmetry relations between the points). $\endgroup$
    – HTNW
    Commented Jun 17, 2021 at 23:05
  • $\begingroup$ @HTNW The effective mass is defined as $m_{ij} = \partial^2 E / \partial k_i \partial k_j$ so there is no meaning to considering only $\bf k =0$. $\endgroup$
    – my2cts
    Commented Jun 17, 2021 at 23:20

This is because the band minima of silicon are not located at the center of the Brillouin Zone, but at some finite $\pm {\bf k^0_i}$. The cubic symmetry of the Silicon crystal means that there are six of these minima and they are the same distance from the center of the zone ($k^0_x = k^0_y = k^0_z$). The bands around these minima, however, are not isotropic, such that this gives rise to anisotropic effective masses for the electrons occupying them.

Note that this does not break the overall symmetry of the system, since rotating the crystalline structure by $90$ degrees (and so rotating the Brillouin zone the same way) would take each minima into another and since they are all at the same energy level they should all remain (in)occupied.

Reference image of the constant energy surfaces around these minima: #/media/File:Silicon_conduction_band_ellipsoids.JPGReference image of the constant energy surfaces around these minima

Brews ohare, CC BY-SA 3.0 https://creativecommons.org/licenses/by-sa/3.0, via Wikimedia Commons


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.