Reducible Fermi Surface for Multi-layer Quantum Graphs Including Stacked Graphene

Abstract

We construct two types of multi-layer quantum graphs (Schrödinger operators on metric graphs) for which the dispersion function of wave vector and energy is proved to be a polynomial in the dispersion function of the single layer. This leads to the reducibility of the algebraic Fermi surface, at any energy, into several components. Each component contributes a set of bands to the spectrum of the graph operator. When the layers are graphene, AA-, AB-, and ABC-stacking are allowed within the same multi-layer structure. One of the tools we introduce is a surgery-type calculus for obtaining the dispersion function for a periodic quantum graph by joining two graphs together. Reducibility of the Fermi surface allows for the construction of local defects that engender bound states at energies embedded in the radiation continuum.

Appendix: Moving the Poles of the Dispersion Function

This appendix proves Proposition 1 in Sect. 2.1. The proof is based on the dotted-graph technique [41]. First consider the simple case of a quantum graph where the underlying graph E consists of two vertices and the edge $e\{v_1,v_2\}$ between them, identified with the x-interval [0, L]. With the operator $-d^2/dx^2-q(x)$ and Robin parameters $\alpha _1$ at $v_1 (x=0)$ and $\alpha _2$ at $v_2 (x=L)$, we obtain a quantum graph (E, Q). Let $\dot{E}$ be the graph obtained by placing an additional vertex v at the point of e corresponding to $x=\ell \in \;]0,L[$; thus $\dot{E}$ consists of three vertices and two edges $e_1\{v_1,v\}$ and $e_2\{v,v_2\}$, with $e_1$ identified with $[0,\ell ]$ and $e_2$ identified with $[\ell ,L]$.

Restricting the potential q to $e_1$ and $e_2$ and imposing the Neumann condition at v yields a quantum graph $(\dot{E},\dot{Q})$. The Neumann condition guarantees continuity of value and derivative across v, and therefore (E, Q) and $(\dot{E},\dot{Q})$ are essentially identical quantum graphs.

Denote the transfer matrices for $-d^2/dx^2 + q(x)$ on [0, L], on $[0,\ell ]$, and on $[\ell ,L]$ by

(8.1)

Considering $(\dot{E},\dot{Q})$ as one period of a d-periodic disconnected graph, its dispersion function is a meromorphic function of $\lambda$ alone, as its discrete reduction $\hat{\dot{Q}}(z,\lambda )$ is independent of z. When Robin conditions are imposed at both endpoints, denote this function by $\dot{h}^\text {RR}(\lambda ) = \det \hat{\dot{Q}}(z,\lambda )$,

$$\begin{aligned} \dot{h}^\text {RR}(\lambda ) \;=\; \det \left[ \begin{array}{ccc} -\frac{c_1(\lambda )}{s_1(\lambda )}-\alpha _1 &{} \frac{1}{s_1(\lambda )} &{} 0\\ \\ \frac{1}{s_1(\lambda )} &{} -\frac{s_1'(\lambda )}{s_1(\lambda )}-\frac{c_2(\lambda )}{s_2(\lambda )} &{} \frac{1}{s_2(\lambda )}\\ \\ 0 &{} \frac{1}{s_2(\lambda )} &{} -\frac{s_2'(\lambda )}{s_2(\lambda )}-\alpha _2 \end{array} \right] . \end{aligned}$$

(8.2)

When the homogeneous Dirichlet condition is imposed at either end (not both ends) of [0, L], with a Robin condition at the other end, one obtains

$$\begin{aligned} \dot{h}^\text {DR}(\lambda )= & {} \det \left[ \begin{array}{cc} -\frac{s_1'(\lambda )}{s_1(\lambda )}-\frac{c_2(\lambda )}{s_2(\lambda )} &{} \frac{1}{s_2(\lambda )}\\ \\ \frac{1}{s_2(\lambda )} &{} -\frac{s_2'(\lambda )}{s_2(\lambda )}-\alpha _2 \end{array} \right] , \nonumber \\ \dot{h}^\text {RD}(\lambda )= & {} \det \left[ \begin{array}{cc} -\frac{c_1(\lambda )}{s_1(\lambda )}-\alpha _1 &{} \frac{1}{s_1(\lambda )} \\ \\ \frac{1}{s_1(\lambda )} &{} -\frac{s_1'(\lambda )}{s_1(\lambda )}-\frac{c_2(\lambda )}{s_2(\lambda )}\end{array} \right] , \end{aligned}$$

(8.3)

and  $\dot{h}^\text {DD}(\lambda ) = -\frac{s_1'(\lambda )}{s_1(\lambda )}-\frac{c_2(\lambda )}{s_2(\lambda )}$ when the Dirichlet condition is imposed at both ends. Using the relation $T = T_2 T_1$, one computes that

$$\begin{aligned}&\dot{h}^\text {RR}= -\frac{c'+\alpha _1 s' + \alpha _2 c + \alpha _1\alpha _2 s}{s_1s_2}, \quad \dot{h}^\text {DR}= \frac{s'+\alpha _2 s}{s_1s_2},\nonumber \\&\dot{h}^\text {RD}= \frac{c + \alpha _1 s}{s_1s_2}, \quad \dot{h}^\text {DD}= -\frac{s}{s_1s_2}. \end{aligned}$$

(8.4)

For the un-dotted quantum graph (E, Q), one obtains these same expressions except with the denominator $s_1(\lambda )s_2(\lambda )$ replaced by $s(\lambda )$,

$$\begin{aligned}&h^\text {RR}= -\frac{c'+\alpha _1 s' + \alpha _2 c + \alpha _1\alpha _2 s}{s}, \quad h^\text {DR}= \frac{s'+\alpha _2 s}{s},\nonumber \\&h^\text {RD}= \frac{c + \alpha _1 s}{s}, \quad h^\text {DD}= -\frac{s}{s}. \end{aligned}$$

(8.5)

Observe that, given $\lambda _0$, one can guarantee that $s_1(\lambda _0)s_2(\lambda _0)\not =0$ by choosing the point $\ell$ not to be a root of any Dirichlet eigenfunction of $-d^2/dx^2+q(x)$ for $\lambda _0$ on [0, L].

These calculations show that the numerators in the expressions above contain the essential spectral information. In fact this is true of periodic quantum graphs in general. To go from the dispersion function for a quantum graph $(\Gamma ,A)$ to the dispersion function for a dotted version $({\dot{\Gamma }},\dot{A})$, one simply multiplies by a factor of the form $s(\lambda )/(s_1(\lambda )s_2(\lambda ))$ for each dotted edge.

Proof of Proposition 1

If $v_1$ and $v_2$ are not in the same ${\mathbb {Z}}^d$ orbit, we can assume that they both are in the vertex set ${\mathcal {V}}_0$ of the fundamental domain chosen for constructing ${\hat{A}}(z,\lambda )$, since $D(z,\lambda )$ is independent of that choice. Denote by ${\hat{A}}(z,\lambda )$ and $\hat{\dot{A}}(z,\lambda )$ the discrete reductions at energy $\lambda$ of the quantum graphs $(\Gamma ,A)$ and $({\dot{\Gamma }},\dot{A})$. Index the rows and columns of ${\hat{A}}(z,\lambda )$ so that the first two correspond to $v_1$ and $v_2$; then augment it with a $0^\text {th}$ column and a $0^\text {th}$ row consisting of a 1 in the leading entry and zeroes elsewhere. Call this matrix $\hat{{\tilde{A}}}(z,\lambda )$.

The matrix $\hat{{\tilde{A}}}(z,\lambda )$ has the block form

(8.6)

in which

$$\begin{aligned} \Sigma = \left[ \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} -cs^{-1} &{} s^{-1} \\ 0 &{} s^{-1} &{} -s's^{-1} \end{array} \right] , \end{aligned}$$

(8.7)

A and B have all zeroes in the first row, and A and C have all zeroes in the first column. The variable z does not appear in $\Sigma$ because v and w are both in the chosen fundamental domain. The matrix $\hat{\dot{A}}(z,\lambda )$ is obtained by replacing $\Sigma$ by a matrix ${\dot{\Sigma }}$, where ${\dot{\Sigma }}$ is obtained from $h^\text {RR}(\lambda )$ (Eq. 8.2) with $\alpha _1 = \alpha _2 = 0$ by switching the first two rows and the first two columns (that is, switching the order of the vertices from $(v_1,v,v_2)$ to $(v,v_1,v_2)$), to obtain

$$\begin{aligned} {\dot{\Sigma }} = \left[ \begin{array}{ccc} -s\,s_1^{-1}s_2^{-1} &{} s_1^{-1} &{} s_2^{-1} \\ s_1^{-1} &{} -c_1s_1^{-1} &{} 0 \\ s_2^{-1} &{} 0 &{} -s_2's_2^{-1} \end{array} \right] , \end{aligned}$$

(8.8)

where the relation $s=s_1c_2+s_1's_2$ is used in the upper left entry.

The $3 \times 3$ matrix $K = A-BD^{-1}C$ has all zeroes in its first row and first column. A computation using the relation $T=T_2T_1$ yields the key relation

$$\begin{aligned} s_1(\lambda )s_2(\lambda )\,\det ({\dot{\Sigma }}+K) \;=\; s(\lambda )\, \det (\Sigma +K), \end{aligned}$$

(8.9)

which holds for any matrix K whose first column and and first row vanish. Using this together with

$$\begin{aligned} \det \hat{\dot{A}} = \det D\,\det ({\dot{\Sigma }} + K), \qquad \det {\hat{A}} = \det \hat{{\tilde{A}}} = \det D \det (\Sigma +K) \end{aligned}$$

(8.10)

yields the statement of the theorem.

If $v_2=gv_1$ for some $g\in {\mathbb {Z}}^d$, the process above remains the same, except that

$$\begin{aligned} \Sigma \,=\, \frac{1}{s} \left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} -c-s'+z^g+z^{-g} \end{array} \right] , \qquad {\dot{\Sigma }} \,=\, \frac{1}{s_1s_2} \left[ \begin{array}{cc} -s &{} s_2+z^gs_1 \\ s_2+z^{-g}s_1 &{} -c_1s_2-s_2's_1 \end{array} \right] ,\nonumber \\ \end{aligned}$$

(8.11)

and K is a $2 \times 2$ matrix with its only nonzero entry being the lower right. In this case, one obtains (8.9) with an extra minus sign on one side. $\square$