Physics Asked on January 30, 2021
For metals,
the conduction band is less than fully filled,
the effective mass $m^*=hbar^2Big(frac{d^2E}{dk^2}Big)^{-1}$ is positive for the interval $kin[-frac{pi}{2a},+frac{pi}{2a}]$ of the first Brillouin zone, and
negative for the intervals $kin[-frac{pi}{2a},-frac{pi}{a}]$ and $kin[+frac{pi}{2a},+frac{pi}{a}]$ of the first Brillouin zone.
What is the physical significance of negative effective mass for electrons lying in the intervals $kin[-frac{pi}{2a},-frac{pi}{a}]$ and $kin[+frac{pi}{2a},+frac{pi}{a}]$? 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 $pmfrac{pi}{2a}$. 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 $hbarmathbf{k}$ 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 $$epsilon(k) = Deltacos(ka).$$ This band has minima at $k=pmpi/a$ and maximum at $k=0$, and its width is $2Delta$. If we expand this energy-quasi-momentum relation at $k=0$, we obtain $$epsilon(k)approxDelta -frac{Delta a^2 k^2}{2} = Delta + frac{hbar^2k^2}{2m^*},$$ 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 $k=pmpi/a + q$, and obtain $$epsilon(k) = Deltacos(pmpi + qa) = -Deltacos(qa) approxDelta +frac{Delta a^2 q^2}{2} = Delta + frac{hbar^2q^2}{2m^*}.$$
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: $$epsilon(mathbf{k}) approx epsilon(0) + frac{1}{2}sum_{i,j}frac{partial^2epsilon(mathbf{k})}{partial k_ipartial k_j}|_{mathbf{k}=0}k_i k_j = epsilon(0) + sum_{i,j}frac{hbar^2k_ik_j}{2m_{ij}^*},\ frac{1}{m_{ij}^*} = frac{1}{hbar^2}frac{partial^2epsilon(mathbf{k})}{partial k_ipartial k_j}|_{mathbf{k}=0} $$ (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 $k=pmfrac{pi}{2a}$, since it is not a function of $k$, but the value of the derivative at a particular point(i.e., a band extremum):
$$m^*=hbar^2left(frac{d^2E(k)}{dk^2}right)|_{k=0},$$
it is NOT
$$m^*(k)=hbar^2left(frac{d^2E(k)}{dk^2}right).$$
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.
Answered by Vadim on January 30, 2021
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