graph takes the form of the Laplacian $-{{\mathrm d}}^2/{{\mathrm d}}x^2$ acting on the space of functions defined on the bonds of the graph that are twice-differentiable and satisfy the following matching conditions at the vertices: $$\begin{aligned}
\psi_j(0)=\psi_i(0)&=:&\Psi,\qquad j,i=1,\ldots,v\\
\sum_{j=1}^v \psi_j^{\prime}(0)&=&\frac{1}{\lambda}\Psi\\
\psi_j^{\prime}(L_j)&=&0,\qquad j=1,\ldots,v.\end{aligned}$$ Here $\psi_j$ is the component of the function defined on the $j^{\rm th}$ bond of the graph, and $\psi_j:[0,L_j]\to{\mathbb R}$ with the convention that $\psi_j(0)$ is the value of the function at the central vertex of the star graph. $\lambda$ is a parameter that allows us to vary the boundary conditions at the central vertex.

The Schrödinger operator so-defined is self-adjoint, so there exists a discrete unbounded set of values $0\leq k_0 < k_1 \leq k_2 \leq \cdots \to
\infty$ such that $k_n^2$ is an eigenvalue. It can be shown that $k=\pm k_n$ corresponds to an eigenvalue if and only if it is a solution of $Z(k,{{\mathbf L}})=0$, where $$Z(k,{{\mathbf L}}):=\sum_{j=1}^v \tan kL_j-\frac{1}{k\lambda}.
\label{spec:det}$$ We refer to $Z(k,{{\mathbf L}})$ as the [*spectral determinant*]{}. Note that $Z(k,{{\mathbf L}})$ has poles at $k=(2n+1)\pi/2L_j$ for each $n\in{\mathbb Z}$ and $j=1,\ldots,v$. The zeros and poles of $Z(k,{{\mathbf L}})$ interlace. The usual definition of the spectral determinant would require these poles to be 