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For $V(x)=\alpha {\sum }_j\delta (x-ja)$ with $\alpha >0$, put $P=m\alpha a/{\hslash }^2$ and $x=ka$. The Floquet discriminant of the delta-comb Kronig-Penney model is $f(x)=\cos x+P\sin x/x$. Extended Bloch states require $|f(x)|\le 1$. Since $f(0)=1+P>1$, there is a positive lower band edge; a repulsive potential does not support negative-energy states in this energy convention.

The $j$th positive-energy band ends at $x=j\pi$. Its lower edge lies in $((j-1)\pi ,j\pi )$ and solves $f(x)=(-1{)}^{j-1}$. A finite periodic ring selects only the discrete allowed values of crystal momentum within these bands. Nonzero band gaps do not by themselves imply an insulator: the band filling must also place the Fermi energy in a gap.

For a potential of period $a$, define the Floquet matrix for a one-dimensional periodic potential byThe 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 whenThose energies form continuous allowed bands of bounded Bloch theorem solutions. Values with $|\mathrm{tr}M|>2$ have a growing and a decaying multiplier and form forbidden gaps. The band edges have $\mathrm{tr}M=\pm 2$, 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 byThusConsequently the Floquet discriminant of the delta-comb Kronig-Penney model isAll band edges are therefore determined byIn factorized form, the periodic edges $\sigma =1$ obeywhile the antiperiodic edges $\sigma =-1$ obey

To express the same edges through the single-barrier scattering data, the One-dimensional transfer matrix from scattering amplitudes givesAt an edge put $z=e^{-ika}$ and $\mathrm{tr}M=2\sigma$. Thenso the scattering-amplitude equations for one-dimensional band edges areExplicitly,For the delta barrier, $t-r=1$ andwhich reproduces the four factorized edge equations above.

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.