= Solution
Twice the quadratic coefficient from the previous part is the <band curvature> $E_\pm''(G_1/2)$. Thus the <band effective masses> are
$$
\frac1{m_\pm^*}=\frac1{\hbar^2}E_\pm''\left(\frac{G_1}2\right)
=\frac1m\pm\frac{\hbar^2G_1^2}{4m^2U},\qquad
\boxed{m_\pm^*=\frac{m}{1\pm\hbar^2G_1^2/(4mU)}=\frac{m}{1\pm2E_B/U}.}
$$
Here $U=|V(G_1)|$ and $E_B=\hbar^2G_1^2/(8m)$. In the weak-potential regime $U\ll E_B$,
$$
\boxed{m_+^*\simeq+\frac{4m^2U}{\hbar^2G_1^2},\qquad
m_-^*\simeq-\frac{4m^2U}{\hbar^2G_1^2}.}
$$
Both have units of <mass>; their magnitudes are much smaller than $m$ in this asymptotic regime. The upper <energy band> has a local minimum and positive <effective mass>, while the lower <energy band> has a local maximum and negative <effective mass>. The exact lower-branch sign changes at $U=2E_B$ in the two-wave formula; the intended weak-coupling limit lies below that threshold.
The physical meaning follows from the semiclassical equations for an <Electron> <wave packet> in one <energy band>:
$$
v=\frac1\hbar\frac{dE}{dk},\qquad \hbar\dot k=F,\qquad \dot v=\frac F{m^*}.
$$
An upper-band <Electron> near its minimum accelerates in the direction of the applied <force>, whereas a lower-band <Electron> near its maximum accelerates opposite to it. At the extremum itself the <group velocity> is zero. A nearly filled lower <energy band> is conveniently described by <electron holes> with positive <electric charge> $+e$ and positive <effective mass> $m_h^*=-m_-^*$. Negative curvature is a response of a crystal <energy eigenstate>, not a negative bare <electron mass>. These statements assume slow forcing that does not induce transitions across the <band gap>.
Back to article page
