= Solution
For a potential of period $a$, define the <Floquet matrix for a one-dimensional periodic potential> by
$$
\begin{pmatrix}\psi(x+a)\\\psi'(x+a)\end{pmatrix}
=M(E)\begin{pmatrix}\psi(x)\\\psi'(x)\end{pmatrix}.
$$
The Schrodinger equation has no first-derivative term, so the Wronskian is constant and $\det M=1$. Its multipliers are therefore reciprocal. They lie on the unit circle precisely when
$$
\boxed{|\operatorname{tr}M(E)|\leq2}.
$$
Those energies form continuous allowed bands of bounded <Bloch theorem> solutions. Values with $|\operatorname{tr}M|>2$ have a growing and a decaying multiplier and form forbidden gaps. The band edges have $\operatorname{tr}M=\pm2$, as in <Floquet discriminant and energy bands>.
For the delta-comb <Kronig-Penney model>, start immediately to the right of one delta function. Free propagation to the next delta and the derivative jump there are represented by
$$
P=
\begin{pmatrix}
\cos ka&\dfrac{\sin ka}{k}\\
-k\sin ka&\cos ka
\end{pmatrix},
\qquad
D=
\begin{pmatrix}
1&0\\
\dfrac{2mV_0}{\hbar^2}&1
\end{pmatrix}.
$$
Thus
$$
\boxed{
M(E)=DP
=
\begin{pmatrix}
\cos ka&\dfrac{\sin ka}{k}\\
\dfrac{2mV_0}{\hbar^2}\cos ka-k\sin ka&
\cos ka+\dfrac{2mV_0}{\hbar^2k}\sin ka
\end{pmatrix}
}.
$$
Consequently the <Floquet discriminant of the delta-comb Kronig-Penney model> is
$$
\boxed{
\frac12\operatorname{tr}M(E)
=\cos(ka)+\gamma\sin(ka)
},
\qquad
\gamma=\frac{mV_0}{\hbar^2k}.
$$
All band edges are therefore determined by
$$
\cos(ka)+\gamma\sin(ka)=\sigma,
\qquad \sigma=\pm1.
$$
In factorized form, the periodic edges $\sigma=1$ obey
$$
\boxed{
\sin\frac{ka}{2}=0
\quad\text{or}\quad
\tan\frac{ka}{2}=\gamma
},
$$
while the antiperiodic edges $\sigma=-1$ obey
$$
\boxed{
\cos\frac{ka}{2}=0
\quad\text{or}\quad
\tan\frac{ka}{2}=-\frac1\gamma
}.
$$
To express the same edges through the single-barrier scattering data, the <One-dimensional transfer matrix from scattering amplitudes> gives
$$
\operatorname{tr}M
=\frac{(t^2-r^2)e^{ika}+e^{-ika}}t.
$$
At an edge put $z=e^{-ika}$ and $\operatorname{tr}M=2\sigma$. Then
$$
z^2-2\sigma tz+t^2-r^2=0,
$$
so the <scattering-amplitude equations for one-dimensional band edges> are
$$
\boxed{e^{-ika}=\sigma t\pm r}.
$$
Explicitly,
$$
\begin{array}{c|c}
\operatorname{tr}M=2&e^{-ika}=t+r\quad\text{or}\quad t-r\\
\operatorname{tr}M=-2&e^{-ika}=-t+r\quad\text{or}\quad -t-r.
\end{array}
$$
For the delta barrier, $t-r=1$ and
$$
t+r=\frac{1-i\gamma}{1+i\gamma},
$$
which reproduces the four factorized edge equations above.
Back to article page
