Physical significance of electrons with negative effective mass. Are they holes or what?

For metals,

• the conduction band is less than fully filled,

• the effective mass $$m^*=\hbar^2\Big(\frac{d^2E}{dk^2}\Big)^{-1}$$ is positive for the interval $$k\in[-\frac{\pi}{2a},+\frac{\pi}{2a}]$$ of the first Brillouin zone, and

• negative for the intervals $$k\in[-\frac{\pi}{2a},-\frac{\pi}{a}]$$ and $$k\in[+\frac{\pi}{2a},+\frac{\pi}{a}]$$ of the first Brillouin zone.

What is the physical significance of negative effective mass for electrons lying in the intervals $$k\in[-\frac{\pi}{2a},-\frac{\pi}{a}]$$ and $$k\in[+\frac{\pi}{2a},+\frac{\pi}{a}]$$? It appears that when the magnitude of $$k$$ increases (with the applied electric field) beyond the value $$\frac{\pi}{2a}$$, the electron starts to move along the applied field behaving like a positive charge. Will it be appropriate to regard these electrons having negative effective mass as holes?

More surprising to me is that the effective mass $$m^*$$ suffers an infinite discontinuity at the points $$\pm\frac{\pi}{2a}$$. What is the meaning of this discontinuity? I'll highly appreciate if someone can explain what is going on here.

What is really effective mass?
Effective mass emerges as a result of expanding the energy dispersion near its minimum/maximum, where it is correspondingly positive/negative.

Energy spectrum of a crystalline solid consists of energy bands of finite width, described by an energy dispersion relation $$\epsilon_n(\mathbf{k})$$, where $$n$$ is the band index and $$\hbar\mathbf{k}$$ is quasi-momentum - this is not the real momentum of an electron, but a quantum number entering the Bloch theorem.

Let us take for simplicity a one-dimensional band with dispersion $$\epsilon(k) = \Delta\cos(ka).$$ This band has minima at $$k=\pm\pi/a$$ and maximum at $$k=0$$, and its width is $$2\Delta$$. If we expand this energy-quasi-momentum relation at $$k=0$$, we obtain $$\epsilon(k)\approx\Delta -\frac{\Delta a^2 k^2}{2} = \Delta + \frac{\hbar^2k^2}{2m^*},$$ where the effective mass is defined as $$m^*=-\frac{\hbar^2}{\Delta a^2}.$$ The effective mass is introduced by analogy with the free-electron dispersion relation $$\epsilon(k) = \frac{p^2}{2m*} = \frac{\hbar^2k^2}{2m},$$ and simplifies calculations, when the electrons are indeed close to the band extrema.

If instead we wanted to expand the dispersion relation near its minimum, we could write $$k=\pm\pi/a + q$$, and obtain $$\epsilon(k) = \Delta\cos(\pm\pi + qa) = -\Delta\cos(qa) \approx\Delta +\frac{\Delta a^2 q^2}{2} = \Delta + \frac{\hbar^2q^2}{2m^*}.$$

For a real semiconductor we are usually interested in the phenomena happening near the maximum of the valence band, which is filled with electrons up to the top, and the bottom of the conduction band, which is empty. Therefore the effective mass in the conduction band is positive, whereas in the valence band it is negative. Since in real materials the energy bands have complicated form, we often have to deal with the effective mass tensor, resulting from the expansion of the three-dimensional energy-momentum relation: $$\epsilon(\mathbf{k}) \approx \epsilon(0) + \frac{1}{2}\sum_{i,j}\frac{\partial^2\epsilon(\mathbf{k})}{\partial k_i\partial k_j}|_{\mathbf{k}=0}k_i k_j = \epsilon(0) + \sum_{i,j}\frac{\hbar^2k_ik_j}{2m_{ij}^*},\\ \frac{1}{m_{ij}^*} = \frac{1}{\hbar^2}\frac{\partial^2\epsilon(\mathbf{k})}{\partial k_i\partial k_j}|_{\mathbf{k}=0}$$ (more precisely, it is the inverse effective mass that has tensor properties.) Moreover, in a real material the bottom of the conduction band and the top of the valence band do not necessarily occur at the same point in k-space.

Effective mass near the edges of the Brillouin zone
Finally, when we go away from the band extremum, the expansion is not valid anymore. However, the effective mass does not diverge at $$k=\pm\frac{\pi}{2a}$$, since it is not a function of $$k$$, but the value of the derivative at a particular point(i.e., a band extremum): $$m^*=\hbar^2\left(\frac{d^2E(k)}{dk^2}\right)|_{k=0},$$ it is NOT $$m^*(k)=\hbar^2\left(\frac{d^2E(k)}{dk^2}\right).$$

Holes vs. electrons with negative effective mass
Holes are vacancies in the valence band, obtained by removing a few electrons at its top. All the electrons at the top of the valence band have negative effective mass, so holes are more than just electrons with a negative effective mass. In fact, holes are rather complex many-particle excitation.

• Actually effective mass is a function of $k$, otherwise we wouldn't have different $m^*$ in different valleys. And Wikipedia appears to agree with me: The effective mass tensor generally varies depending on k, meaning that the mass of the particle actually changes after it is subject to an impulse. The only cases in which it remains constant are those of parabolic bands, described above. – Ruslan Aug 2 at 16:28
• One particular consequence of this is the Bloch oscillation, where an electron subject to external electric field gets accelerated (with $m^*>0$) up to the $k$-point where $1/m^*=0$, and decelerated after that until it gets to $1/m^*=0$ from the negative direction to restart the acceleration cycle. (The stationary states in this case form a ladder of identical-but-shifted localized states bound by the boundaries of the allowed band, which are tilted in the $x$-space due to the electric field.) – Ruslan Aug 2 at 16:28
• @Ruslan one need not define the effective mass in order to describe the Bloch oscillations. Firmally defining it as the inverse of the k-dependent second derivative of energy is possible, but it pretty much negates the whole concept. I think this Wikipedia article is misleading. It is worth however noting that the extrema of energy bands do not necessarily occur at the same $k$, and a band may have multiple minima/maxima - in this sense the effective mass define my way is still k-dependent. – Vadim Aug 2 at 20:26
• @Vadim According to your band dispersion $\epsilon=\Delta\cos(ka)$ with $\Delta>0$, your calculation shows that the effective mass is negative at the bottom of the band and positive at the top. This result will be the opposite if the dispersion were $\epsilon=-\Delta\cos(ka)$ with $\Delta>0$. So electron effective mass does not have a definite sign at the bottom or top of a band; it depends on the band dispersion relation. Is this true? – mithusengupta123 Oct 17 at 17:29
• @Vadim Sorry, my mistake. I was thinking of the tight-binding dispersion relation $\epsilon(k)=\epsilon_0-2t\cos(ka)$ with $t>0$, and for which $k=0$ represents the bottom of the band and $k=\pm\pi/a$ represent the top. However, like Ruslan, I think effective mass is not only defined only at the bottom or top of a band but across the band, and in general, is a function of $k$ and it does become infinity at $\pm\pi/2a$. See page 178 of this Oxford lecture notes by Steven Simon: www-thphys.physics.ox.ac.uk/people/SteveSimon/condmat2012/… – mithusengupta123 Oct 17 at 23:02