# Quantum Graphs/Schrodinger equation on a metric graph

Schrödinger equation on a metric graph
Consider a graph $\Gamma =(V,E)$ , where $V$ is the set of vertices and $E$ is the set of edges. Each edge connects a pair of vertices; we allow more than one edge running between any two vertices. We also allow edges connecting vertices to themselves (loops). This freedom creates some notational difficulties, so we ask the reader to be flexible and forgiving.
Each edge $e$ is assigned a positive length $L_{e}$ and is thus identified with an interval $[0,L_{e}]$ (the direction is chosen arbitrarily and is irrelevant to the resulting theory). This makes $\Gamma$ a metric graph. Now a function on a graph is just a collection of functions defined on individual edges.
The eigenvalue equation for the Schrödinger operator is
${\begin{aligned}-{\frac {d^{2}f}{dx^{2}}}+V(x)f(x)=\lambda f(x),\end{aligned}}$
which is to be satisfied on every edge, in addition to the vertex matching conditions as follows
${\begin{aligned}f(x)\ {\text{is continuous}},\\\sum _{e\sim v}{\frac {df}{dx}}(v)=0.\end{aligned}}$
The continuity means that the values at the vertex agree among all functions living on the edges attached (or incident) to the vertex. In the second condition (often called current conservation condition), the sum is over all edges attached to the vertex and the derivative are all taken in the same direction: from the vertex into the edge. A looping edge contributes two terms to the sum, one for each end of the edge.
The function $V(x)$ is called the electric potential but we will set it identically to zero in all of the examples below. Vertex conditions \eqref{eq:cont}-\eqref{eq:current_cons} are called \emph{Neumann conditions}\footnote{Other names present in the
we give any more theory, let us consider some examples.

### Example: a trivial graph --- an interval

An interval $[0,L]$ is the simplest example of a graph; it has two vertices (the endpoints of the interval) and one edge. The continuity condition is empty at every vertex since there is only one edge. The current conservation condition at the vertex $0$ becomes
${\begin{aligned}f'(0)=0,\end{aligned}}$
and at the vertex $L$ becomes
${\begin{aligned}-f'(L)=0.\end{aligned}}$
The minus sign appeared because we agreed to direct the derivatives into the edge; of course it is redundant in this particular case.
Let $V(x)\equiv 0$ and consider first the positive eigenvalues, $\lambda >0$. The eigenvalue equation becomes
${\begin{aligned}-f''=k^{2}f,\end{aligned}}$
where for convenience we substituted $\lambda =k^{2}$. This is a second order linear equation with constant coefficients which for $k>0$ is readily solved by
${\begin{aligned}f(x)=C_{1}\cos(kx)+C_{2}\sin(kx).\end{aligned}}$
Applying the first vertex condition $f'(0)=0$ we get $C_{2}=0$ and $f(x)=C_{1}\cos(kx)$. The second vertex condition becomes
${\begin{aligned}C_{1}k\sin(kL)=0,\end{aligned}}$
which imposes a condition on $k$ but does nothing to determine $C_{1}$ (naturally we are not interested in the trivial solution $f(x)\equiv 0$). We thus get the \emph{eigenvalues} $\lambda =k^{2}=\left(\pi n/L\right)^{2}$, $n=1,2,\ldots$ with the corresponding eigenfunctions $f(x)=\cos \left(\pi nx/L\right)$ defined up to an overall constant multiplier (as befits eigenvectors and eigenfunctions).
There is one other eigenvalue in the spectrum that we missed: $\lambda =0$ with the eigenfunction $f(x)\equiv 1$. While this agrees with the above formulas with $n=0$, the premise of equation (\ref{eq:f_solution}) is no longer correct when $\lambda =0$.

#### Exercise 2

Integrate by parts the scalar product
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikibooks.org/v1/":): {\displaystyle \begin{aligned} \left\langle f, -f'' \right\rangle = \int_0^L \cc{f(x)} \left(-f''(x)\right) dx, \end{aligned} }
to obtain an expression that is obviously non-negative, and thus show that it is not necessary to solve \eqref{eq:eig_eq_V0} to conclude that there are no negative eigenvalues.
We did not try to look for complex eigenvalues. This is because the Schrödinger operator we defined is self-adjoint (see Thm 1.4.4 of \cite{BerKuc_graphs}) and therefore has real spectrum. The spectrum in the above example is discrete: all eigenvalues are isolated and of finite multiplicity. This is true for any graph which is compact (has finitely many edges, all of which have finite length), see Thm 3.1.1 of \cite{BerKuc_graphs}. The proof outlined in Exercise~\ref{hw:nonneg} works for general graphs with Neumann conditions at all vertices. The multiplicity of the eigenvalue 0 in the spectrum can be shown to equal the number of the connected components of the graph.

### Example: a trivializable graph with a vertex of degree two

\label{sec:fake_vertex}
Consider now a graph consisting of two consecutive intervals, $[0,L_{1}]$ and $[L_{1},L_{1}+L_{2}]$. We do not really have to parametrize the edges starting from 0, so in this example we will employ the natural parametrization.
Denote the components of eigenfunction living on the two intervals by $f_{1}$ and $f_{2}$ correspondingly. Solving the equation on the first edge and enforcing the Neumann condition at $0$ results in $f_{1}(x)=C\cos(kx)$. The conditions at the point $L_{1}$ are
${\begin{aligned}&f_{1}(L_{1})=f_{2}(L_{1}),\\&-f_{1}'(L_{1})+f_{2}'(L_{1})=0.\end{aligned}}$
Now, by uniqueness theorem for second order differential equations, the solution on the second edge is fully determined by its value at $L_{1}$ and the value of its derivative. Thus the solution is still $f_{2}(x)=C\cos(kx)$ and there is no change in the solution happening at $L_{1}$. We could have considered the interval $[0,L_{1}+L_{2}]$ without introducing the additional vertex at $L_{1}$. This obviously generalizes to the following rule: having a Neumann vertex of degree 2 is equivalent to having an uninterrupted edge.
This rule is useful, for example, for when one wants to program a looping edge but is troubled by the notational difficulties of loops or multiple edges. In this case a looping edge can be implemented as a triangle with two dummy vertices of degree two.

### Example: star graph with Neumann endpoints

\begin{figure}
\end{figure}
Consider now a first non-trivial example: a star graph with 3 edges meeting at a central vertex, see Fig.~\ref{fig:stargraph}. Parametrizing the edges from the endpoints towards the central vertex, we get
${\begin{aligned}&-f_{1}''=k^{2}f_{1},\quad -f_{2}''=k^{2}f_{2},\quad -f_{3}''=k^{2}f_{3},\\&f_{1}'(0)=0,\quad f_{2}'(0)=0,\quad f_{3}'(0)=0,\\&f_{1}(L_{1})=f_{2}(L_{2})=f_{3}(L_{3}),\\&-f_{1}'(L_{1})-f_{2}'(L_{2})-f_{3}'(L_{3})=0,\end{aligned}}$
where in addition to the already familiar equations \eqref{eq:star_equation} and \eqref{eq:neumann_per} (in three copies), we have continuity condition at the central vertex in equation \eqref{eq:cont_central} and current conservation at the central vertex in equation \eqref{eq:cur_cons_central}. Note that in equation \eqref{eq:star_equation}, the eigenvalue $k^{2}$ is the same on all three edges.
Equations \eqref{eq:star_equation}-\eqref{eq:neumann_per} are solved by
${\begin{aligned}f_{1}(x)=A_{1}\cos(kx),\quad f_{2}(x)=A_{2}\cos(kx),\quad f_{3}(x)=A_{3}\cos(kx),\end{aligned}}$
for some constants $A_{1}$, $A_{2}$ and $A_{3}$. Now the remaining two equations become, after a minor simplification,
${\begin{aligned}&A_{1}\cos(kL_{1})=A_{2}\cos(kL_{2})=A_{3}\cos(kL_{3}),\\&A_{1}\sin(kL_{1})+A_{2}\sin(kL_{2})+A_{3}\sin(kL_{3})=0.\end{aligned}}$
Dividing equation \eqref{eq:cur_cons_central_sub} by \eqref{eq:cont_central_sub} cancels the unknown constants, resulting in
${\begin{aligned}\tan(kL_{1})+\tan(kL_{2})+\tan(kL_{3})=0.\end{aligned}}$
Squares of the roots $k$ of this equation (which cannot be solved explicitly except when all $L$s are equal) are the eigenvalues of the star graph.

#### Exercise 3

${\begin{aligned}\sin(kL_{1})\cos(kL_{2})\cos(kL_{3})+\cos(kL_{1})\sin(kL_{2})\cos(kL_{3})+\cos(kL_{1})\cos(kL_{2})\sin(kL_{3})=0.\end{aligned}}$
Moreover, the order of the root $k$ of \eqref{eq:3star_robust} is equal to the dimension of the corresponding eigenspace.
For example, if $L_{1}=L_{2}=L_{3}=\pi /2$, the left-hand side of \eqref{eq:3star_robust} vanishes at $k=1$ to the second order. This corresponds to two linearly independent solutions,
${\begin{aligned}(f_{1},f_{2},f_{3})=(\cos(x),-\cos(x),0)\quad {\mbox{and}}\quad (\cos(x),0,-\cos(x)).\end{aligned}}$
There is actually a lot more that can be (and will be said) about this simple graph, but we first need to extend the set of possible vertex conditions that we consider.