 Solution

Between delta barriers, $V=0$ and the Schrödinger equation gives ${\psi }^{\prime \prime }+k^2\psi =0$, with $k^2=2mE/{\hslash }^2$ for positive energy. Write the state and derivative as a two-component vector $\mathbf{v}=(\psi ,{\psi }^{\prime }{)}^T$. Free propagation across one interval of length $a$ isThe wavefunction is continuous at a delta barrier, while its derivative jumps by $g\psi$, with $g=2m\alpha /{\hslash }^2$. Thus the barrier matrix isBloch's theorem requires $\mathbf{v}(a^{+})=e^{i\varphi a}\mathbf{v}(0^{+})$. Both $D$ and $P$ have determinant one, so the two eigenvalues of $M$ are $e^{i\varphi a}$ and $e^{-i\varphi a}$. Taking its trace givesTherefore the Floquet discriminant of the delta-comb Kronig-Penney model yieldsThis trace argument is the one-cell transfer method for the Kronig-Penney model. The loop imposes the additional discrete values of $\varphi$ derived in part (b). Real $\varphi$ 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.