i am trying to understand this.

I know that the effective mass of electrons or holes is calculated as:

$$m^* = \frac{h^2}{(4\pi^2)\frac{d^2E}{dk^2}}$$

Now,if i look at this plot for example:


I can see that for the holes(they should be on the blue part because they belong to the valence band), the absolute value of the curvature $\frac{d^2E}{dk^2}$ is smaller than the one for the electrons(red part). So, because the effective mass is inversely proportional to that curvature, the effective mass of holes should be larger than the electron's one.

I think that is correct, but again, i am considering the absolute value of the curvature, because for the blue part, the curvature is negative, so. Shouldn't the effective mass of holes be negative aswell?

  • 1
    $\begingroup$ See Wikipedia. $\endgroup$ – Trimok Jul 5 '14 at 14:59
  • $\begingroup$ short answer, a hole behaves as an effective particle (thus moves and has momentum etc..) also the effective mass of the hole should be the same as the mass of the particle, this was what made dirac fail to account in one equation for both an electron and a proton $\endgroup$ – Nikos M. Jul 5 '14 at 21:11
  • $\begingroup$ More on effective mass of holes: physics.stackexchange.com/… $\endgroup$ – Qmechanic Aug 1 '14 at 10:17

Your equation is considering the effective mass of electrons.

The holes are lack of electrons. To talk about them, we effectively invert the energy axis, i.e. if we compute electron and hole energies with respect to valence band ceiling $E_V=0$, we have:


Then it's straightforward to see that $m^*_e=-m^*_h$ in the same valley.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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