= Solution
Between delta barriers, $V=0$ and the <Schrödinger equation> gives $\psi''+k^2\psi=0$, with $k^2=2mE/\hbar^2$ for positive <energy>. Write the state and derivative as a two-component vector $\mathbf v=(\psi,\psi')^T$. Free propagation across one interval of length $a$ is
$$
\mathbf v(a^-)=P\mathbf v(0^+),\qquad
P=\begin{pmatrix}\cos ka&\sin ka/k\\-k\sin ka&\cos ka\end{pmatrix}.
$$
The wavefunction is continuous at a delta barrier, while its derivative jumps by $g\psi$, with $g=2m\alpha/\hbar^2$. Thus the barrier matrix is
$$
D=\begin{pmatrix}1&0\\g&1\end{pmatrix},\qquad
\mathbf v(a^+)=M\mathbf v(0^+),\quad M=DP.
$$
<Bloch's theorem> requires $\mathbf v(a^+)=e^{i\phi a}\mathbf v(0^+)$. Both $D$ and $P$ have determinant one, so the two eigenvalues of $M$ are $e^{i\phi a}$ and $e^{-i\phi a}$. Taking its trace gives
$$
2\cos(\phi a)=\operatorname{tr}M=2\cos ka+\frac gk\sin ka.
$$
Therefore the <Floquet discriminant of the delta-comb Kronig-Penney model> yields
$$
\boxed{\cos(\phi a)=\cos(ka)+\frac{m\alpha}{\hbar^2k}\sin(ka).}
$$
This trace argument is the one-cell transfer method for the <Kronig-Penney model>. The loop imposes the additional discrete values of $\phi$ derived in part (b). Real $\phi$ requires the right-hand side to lie in $[-1,1]$, giving allowed <energy bands>; values outside this range are <band gaps>. The strength $\alpha$ has units of <energy> times length, because a <Dirac delta distribution> has units of inverse length.
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