 Solution

Use the state vector $(\psi ,{\psi }^{\prime }{)}^T$ immediately to the right of a delta well. Free propagation to the next well and the derivative jump there areThus the Floquet matrix in this convention isChoosing the state on the other side of each well conjugates the matrix and leaves its eigenvalues unchanged.

The Bloch theorem requires a real quasimomentum $q$ with a Floquet eigenvalue $e^{iqa}$. Since $\det M=1$, this occurs exactly whenDividing the two inequalities by the positive $\sinh \kappa a$ givesThe upper boundary in $y$ is the unique intersection of the increasing line with the strictly decreasing $\coth y$, which runs from infinity to $1$. Call it $y_{-}$. The concave function $\tanh y$ starts with slope $1$ and approaches $1$. If $\lambda a\le 2$, the line has slope at least $1$ and stays above it for every $y>0$, so allowed $y$ run from $0$ to $y_{-}$. If $\lambda a>2$, there is exactly one positive crossing $y_{+}$ with $\tanh y$; below it the line is inadmissible, and above it the line stays above $\tanh y$. Because $\coth y>\tanh y$, $y_{+}<y_{-}$. This establishes the negative-energy band of an attractive delta comb: there is exactly one negative-energy band:where $E_{\pm }=-{\hslash }^2(2y_{\pm }/a{)}^2/(2m)$, without needing to evaluate the roots.

As $a\to \infty$ with $\lambda$ fixed, both positive boundary values of $\kappa$ tend to $\lambda$. The band collapses to the isolated delta-well bound state $\boxed{E=-{\hslash }^2{\lambda }^2/(2m)}$, as tunnelling between widely separated wells disappears.