Magnetic Fluxes

Abstract

We have already seen that the spectra of quantum graphs are independent of the particular form of the magnetic potential (5.8)—only integrals of the magnetic potential along the edges play a role. One may say that adding magnetic potential is equivalent to a special change of vertex conditions. The goal of this chapter is to look at this connection in more detail.

We have already seen that the spectra of quantum graphs are independent of the particular form of the magnetic potential (5.8)—only integrals of the magnetic potential along the edges play a role. One may say that adding magnetic potential is equivalent to a special change of vertex conditions. The goal of this chapter is to look at this connection in more detail.

The integrals along the cycles can be interpreted as the fluxes of the magnetic field. In particular, spectra of trees is independent of the magnetic potential. This phenomenon is well-known for physicists, who used to say that there is no magnetic field in one dimension. On the other hand Y. Aharonov and D. Bohm predicted that the movement of charged quantum particles in a ring is affected by the magnetic field though the ring, despite that it could be equal to zero on the ring itself. This phenomenon is known as Aharonov-Bohm effect. What we are going to do is an extension of these studies to the case of several cycles.

It will also be shown that even though the number of parameters the spectrum depends on is equal to the number $\beta _1$ of independent cycles, it might happen that dependence upon one of these parameters is suppressed, provided the other parameters are chosen in a special way. The reasons are pure topological, therefore we call this phenomenon by topological damping of Aharonov-Bohm effect.

16.1 Unitary Transformations via Multiplications and Magnetic Schrödinger Operators

Consider the magnetic Schrödinger operator $L_{q,a}^{{\mathbf {S}}} (\Gamma )$ defined following Sect. 3.8 on a connected finite compact metric graph $\Gamma$ by the differential expression

$$\displaystyle \begin{aligned} {} \tau_{q,a} u = \left( i \frac{d}{dx} + a(x) \right)^2 + q(x), \end{aligned} $$

(16.1)

on the functions satisfying vertex conditions (3.53)

$$\displaystyle \begin{aligned} {} i (S_m - I ) \vec{u}_m = (S_m + I ) \partial \vec{u}_m, \quad m = 1,2, \dots, M. \end{aligned} $$

(16.2)

In quantum mechanics unitarily equivalent operators define the same physical models and therefore are usually identified. On the other hand the probability density is given by the squared absolute value of the wave function $\rho (x) = \vert \psi (x) \vert ^2.$ Therefore if the configuration space is fixed, then it is natural to identify operators $\tilde {H}$ and $H$ connected via the unitary transformation given by multiplication by any unimodular function ${\mathbf {U}}(x) = \exp i \Theta (x) , \; \Theta (x) \in \mathbb R$:

$$\displaystyle \begin{aligned} {} \tilde{L} = {\mathbf{U}}^{-1} L {\mathbf{U}} = e^{-i \Theta(x)} L e^{i \Theta(x)}. \end{aligned} $$

(16.3)

We have already seen in Sect. 4.1 that magnetic potential on each edge can be eliminated, but the corresponding transformation affects the vertex conditions. In what follows we study this dependence systematically.

Let us first consider elementary examples when the function $\Theta$ is constant on the edges.

Special Case 1

If the function $\Theta (x)$ is chosen equal to a fixed real constant $\Theta _0 (x) \equiv \theta , \; \theta \in \mathbb R$ on the whole graph $\Gamma$, then the operator of multiplication ${\mathbf {U}}$ commutes with any $L_{q,a}^{\mathbf {S}}$ and therefore the transformed operator $L_{\tilde {q},\tilde {a}}^{\tilde {\mathbf {S}}}$ is equal to the original one:

$$\displaystyle \begin{aligned} L_{\tilde{q},\tilde{a}}^{\tilde{\mathbf{S}}} = L_{q,a}^{\mathbf{S}} .\end{aligned}$$

This case is trivial and may be ignored.

Special Case 2

Let the function $\Theta$ be chosen equal to a separate constant on each edge of $\Gamma$

$$\displaystyle \begin{aligned} {} \Theta_e (x) = \theta_n, \; \, x \in E_n, \end{aligned} $$

(16.4)

where $\theta _n$ are certain real parameters. With this choice of $\Theta$ the differential expressions $\tau _{\tilde {q}, \tilde {a}}$ and $\tau _{q,a}$ coincide on every edge $E_n$, but the corresponding operators may be different, since the vertex conditions at a vertex $V^m$ are affected if the phases $\theta (x_j)$ are different for $x_j \in V^m.$

More precisely, assume without loss of generality that the edges joined together at the vertex $V^m$ are enumerated as $E_1, E_2, \dots , E_{d_m}$ and that the functions from the domain of $L_{q,a}^{{\mathbf {S}}}$ satisfy vertex conditions (16.2). Consider the diagonal unitary matrix $U_m$ given by

$$\displaystyle \begin{aligned} U_m = \mathrm{diag}\, \left\{ e^{i\theta_1}, e^{i\theta_2}, \dots, e^{i\theta_{d_m}} \right\} ,\end{aligned}$$

where $\theta _n = \theta (x_j),$ provided $x_j \in E_n.$ Then the unitary matrices $\tilde {S}_m$ and $S_m$ associated with the two operators are connected via

$$\displaystyle \begin{aligned} {} \tilde{S}_m = \left( U_m\right)^{-1} S_m U_m. \end{aligned} $$

(16.5)

To see this assume that $u \in \mathrm {Dom}\,(L_{q,a}^{\tilde {\mathbf {S}}})$. Every such function after the transformation ${\mathbf {U}}$ is mapped to a function from the domain of $L_{q.a}^{{\mathbf {S}}}$ and therefore satisfies condition (16.2)

$$\displaystyle \begin{aligned} i (S_m - I) U_m \vec{u}_m = (S_m + I) U_m \partial \vec{u}. \end{aligned}$$

Multiplying both sides by $\left ( U_m \right )^{-1}$ we arrive at

$$\displaystyle \begin{aligned} i (\tilde{S}_m - I) \vec{u}_m = (\tilde{S}_m + I) \partial \vec{u}_m, \end{aligned}$$

using (16.5).

Summing up, multiplication by the function $e^{i \theta _e (x)}$ leads to magnetic Schrödinger operators given by the same differential expression, but the vertex conditions determined by matrices $\tilde {S}_m$ and $S_m$ are connected via (16.5).

General Case

Assume that the operators $L_{\tilde {q},\tilde {a}}^{\tilde {\mathbf {S}}}$ and $L_{q,a}^{{\mathbf {S}}}$ are related by the unitary transformation (16.3) above, where we assume that the function $\Theta (x)$ is continuously differentiable inside the edges. Applying formula

$$\displaystyle \begin{aligned} \left( i \frac{d}{dx} + a(x) \right) e^{i \Theta (x)} u (x) = e^{i \Theta (x)} \left( i \frac{d}{dx} + a(x) - \Theta'(x) \right) u(x)\end{aligned}$$

twice we obtain the following expression for the transformed differential operator

(16.6)

It follows that the electric potential is not affected but the magnetic potential is changed

$$\displaystyle \begin{aligned} \tilde{q} (x) = q(x) , \; \, \mathrm{and} \; \, \tilde{a} (x) = a(x) - \Theta'(x). \end{aligned} $$

(16.7)

The transformation of the vertex conditions is the same as the one discussed in the second special case. The only difference is that the phases $\theta _ n$ should be substituted with the limiting values $\Theta (x_j)$ where $x_j \in V^m.$ Note that the extended normal derivatives given by (2.26) are changed accordingly, since their definition depends on the value of the magnetic potential.

Problem 73

Prove that transformation (16.3) maps extended normal derivatives to extended normal derivatives and that new vertex conditions at a vertex $V^m$ are given by the matrix $\tilde {S}_m$ connected to $S_m$ via (16.5).

In particular, the magnetic potential on every edge is eliminated if one chooses:

$$\displaystyle \begin{aligned} \Theta (x) = \int_{x_0}^x a(y) dy.\end{aligned}$$

Observe that by eliminating the magnetic potential one introduces new phases in vertex conditions as given by (16.5). Hence in order to study spectral properties of magnetic Schrödinger operators on metric graphs it is enough to consider Schrödinger operators with zero magnetic potentials, but with different extra phases in the vertex conditions. We are going to call these phases simply vertex phases (see (16.10) below) and will in particular study the dependence upon these phases.

Special Case 3

If the graph $\Gamma$ is a tree, then the function $\Theta$ eliminating the magnetic potential can be chosen continuous on the whole metric graph

$$\displaystyle \begin{aligned} \Theta (x) = \int_{x_0}^x a(y) dy, \end{aligned} $$

(16.8)

where the point $x_0 \in \Gamma$ is arbitrary and integration is along the shortest path on $\Gamma$ connecting $x_0$ and $x .$ Note that integration along the edges forming the path should be carried out respecting their orientation: if the path goes along an edge in the positive direction, then the corresponding contribution should be taken with $+$ sign, otherwise—with $-$ sign. This is necessary since changing edge orientation the magnetic potential is multiplied by $-1.$

It follows that magnetic potential on a tree can be eliminated without changing the vertex conditions, i.e. the spectrum of a magnetic Schrödinger operator on a tree coincides with the spectrum of the non-magnetic Schrödinger operator with the same electric potential $q$ and the same vertex conditions.

Assume now that $\Gamma$ is not a tree, but contains cycles. Consider then any spanning tree $T$ on $\Gamma$ obtained by chopping precisely $\beta _1 = N-M+1$ vertices—one on each independent cycle in $\Gamma .$ Then the magnetic potential on $T$ can be eliminated as described above. Using the same function $\Theta$ to eliminate magnetic potential on $\Gamma$ leads to introducing at most $\beta _1$ vertex phases, since the values of $\Theta$ at different parts of the chopped vertices may be different. Assume that a vertex $V^m$ was divided into two vertices $V^{\prime }_m$ and $V^{\prime \prime }_m$ with one of the vertex degrees equal to 1. If $\Theta (V^{\prime }_m) \neq \Theta (V^{\prime \prime }_m)$ modulus $2 \pi$ then the matrix $S_m$ describing vertex conditions on $\Gamma$ is transformed by (16.5). The matrix $U_m$ contains factors $e^{i \Theta (V^{\prime }_m)}$ and $e^{i \Theta (V^{\prime \prime }_m)}$, but one common factor in the similarity transformation can be cancelled. Hence the matrix $\tilde {S}_m$ depends on the difference $\Theta (V^{\prime }_m) - \Theta (V^{\prime \prime }_m)$ which turns to be equal to the integral of the magnetic potential along the independent cycle to which $V^m$ belongs. We conclude that elimination of the magnetic potential on a graph with $\beta _1$ independent cycles $C_j$ leads to an operator with zero magnetic potential, the same electric potential and new vertex conditions containing at most $\beta _1$ phases equal to

$$\displaystyle \begin{aligned} {} \Phi_j = \int_{C_j} a(y) dy, \; \; j=1,2, \dots, g. \end{aligned} $$

(16.9)

These phases should be interpreted as magnetic fluxes—the fluxes of the magnetic field through the independent cycles.

We have in particular proven the following theorem which probably appeared first in [311]

Theorem 16.1

Consider the magnetic Schrödinger operator$L_{q,a}^{{\mathbf {S}}}$on a finite compact metric graph$\Gamma$with a fixed electric potential$q$and fixed vertex conditions (16.2). Then as the magnetic potential$a$varies, the spectrum of$L_{q,a}^{{\mathbf {S}}}$depends on at most$\beta _1 = N - M + 1$parameters which can be identified as the fluxes$\Phi _j$given by (16.9).

This theorem states that the spectrum depends on at most $\beta _1$ parameters. We have strong reasons to believe that if all vertex conditions are chosen properly connecting, then the spectrum does depend on all $\beta _1$ magnetic fluxes. Observe that in very special cases it might happen that the spectrum is independent of one of the fluxes, provided the other fluxes are chosen in a special way (see Sect. 16.3 below).

16.2 Vertex Phases and Transition Probabilities

We have seen that eliminating magnetic potential on the edges one obtains Schrödinger operator with the same electric potential, but with vertex conditions given by $\tilde {S}_m = (U_m)^{-1} S_m U_m$ instead of $S_m.$ The transformation above contains certain phases, which we are going to call vertex phases:

$$\displaystyle \begin{aligned} {} \theta_j = \Theta (x_j), \end{aligned} $$

(16.10)

where $\Theta (x)$ is the function used to define the unitary transformation ${\mathbf {U}} = \exp (i \Theta (x))$.

The matrices $\hat {S}_m$ and $S_m$ have the same absolute values of all entries:

$$\displaystyle \begin{aligned} \vert \left(\tilde{S}_m \right)_{ij} \vert^2 = \vert \left(S_m \right)_{ij} \vert^2. \end{aligned} $$

(16.11)

It follows that the waves penetrate such vertices with precisely equal transition probabilities given by $\rho _{ij} = \vert \left (S_m \right )_{ij} \vert ^2.$ Transition probabilities are physically relevant and in principle are much easier to be measured in experiments than the scattering coefficients. Therefore one may discuss the possibility to define the quantum graph model using just transition probabilities instead of vertex scattering coefficients.

It turns out that this point of view is not optimal, since a unitary matrix $S_m$ in general is not defined by transition probabilities $\vert (S_m)_{ij} \vert ^2.$ First of all it is trivial to see that the transformation

$$\displaystyle \begin{aligned} {} S_m \mapsto \tilde{S}_m = (D_m)^{-1} S_m D_m, \end{aligned} $$

(16.12)

where $D_m$ is a diagonal unitary matrix

$$\displaystyle \begin{aligned} D_m = \mathrm{diag} \; \left\{ e^{i \varphi_1}, e^{i \varphi_2}, \dots, e^{i \varphi_{d_m}} \right\}, \; \, \varphi_i \in \mathbb R, \end{aligned} $$

(16.13)

preserves the transition probabilities. Note that this transformation coincides with (16.5) if all phases $\varphi _i$ are chosen equal to the values of $\theta _j$ at the corresponding endpoints. Then we have $D_m = U_m$ and considering two quantum graphs with the vertex conditions defined by $\tilde {S}_m$ and $S_m$ is equivalent to reintroducing magnetic potentials.

Transformation (16.12) in general does not characterize all unitary matrices with the same transition probabilities. As an example, consider the following one parameter family of reflectionless equitransmitting matrices of order $6$ [358]

$$\displaystyle \begin{aligned} C_{1,1,1} = \frac{1}{\sqrt{5}} \left( \begin{array}{cccccc} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & -1 & -1 \\ 1 & 1 & 0 & -1 & e^{i \varphi} & - e^{i \varphi} \\ 1 & 1 & -1 & 0 & - e^{i \varphi} & e^{i \varphi} \\ 1 & 1 & e^{-i \varphi} & - e^{-i \varphi} & 0 & 1 \\ 1 & -1 & - e^{-i \varphi} & e^{-i \varphi} & 1 & 0 \end{array} \right),\end{aligned}$$

where $\varphi \in \mathbb R.$ All matrices from this family have the same reflection probability $\vert S_{jj} \vert ^2 = 0$ (reflectionless) and the same transition probabilities (equitransmitting) $\vert S_{ij} \vert ^2 = \frac {1}{5}, \, i \neq j.$ Of course, considered example is very special, since many of the entries in $C_{1,1,1}$ have the same absolute value. Adding the similarity transformation (16.12) one obtains a 6-parameter family of matrices with the same transition probabilities.

One may study quantum graph models fixing all $S_m$ up to the vertex phases described above, in other words considering vertex conditions given by $\tilde {S}_m = (D_m)^{-1} S_m D_m$ with $S_m$ fixed, $m= 1,2, \dots , M.$ Then with every vertex $V^m$ we have precisely $d_m -1$ arbitrary phases associated. The spectrum of the quantum graph depends on the vertex phases, but how many parameters are independent? Altogether we have $\sum _{m=1}^M (d_m -1) = 2N - M$ vertex phases. It turns out that only $\beta _1 = N-M+1$ parameters are independent. The proof is completely analogous to the proof of Theorem 16.1.

Let us study the following clarifying example.

Example 16.2

Consider the figure eight metric graph $\Gamma _{(2.4)}$ obtained by joining together two loops at a single vertex. The endpoints joined in the vertex $V^1$ are $x_1, x_2, x_3, x_4.$

Define the Laplace operator $L^{\tilde {S}}$ on $\Gamma _{(2.4)}$ on the functions from the Sobolev space $W_2^2 ([x_1, x_2] \cup [x_3, x_4])$ satisfying vertex conditions

$$\displaystyle \begin{aligned} {} i (\tilde{S} - I) \vec{\psi} (V^1) = (\tilde{S} + I) \partial \vec{\psi} (V^1), \end{aligned} $$

(16.14)

where $\tilde {S} = D^* S D$ with

$$\displaystyle \begin{aligned} {} S =\frac{1}{\sqrt{3+\cos^2\beta}}\left(\begin{array}{cccc} \cos\beta&1&1&1\\ 1&\cos\beta&-e^{i\beta}&-e^{-i\beta}\\ 1&-e^{-i\beta}&-\cos\beta&e^{-i\beta}\\ 1&-e^{i\beta}&e^{i\beta}&-\cos\beta \end{array}\right),\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \end{aligned} $$

(16.15)

and $D =\mathrm {diag}\, \left (1,e^{i\varphi _1}, e^{i\varphi _2}, e^{i\varphi _3}\right )$, $\varphi _j\in \left [-\pi ,\pi \right )$. For $a=-\frac {e^{i\beta }}{\sqrt {3+\cos ^2\beta }}$, $t=\frac {1}{\sqrt {3+\cos ^2\beta }}$ and $r=\frac {\cos \beta }{\sqrt {3+\cos ^2\beta }}$ we write the above matrix as follows

$$\displaystyle \begin{aligned} {} S =\left(\begin{array}{cccc} r&t&t&t\\ t&r&a&\overline{a}\\ t&\overline{a}&-r&-\overline{a}\\ t&a&-a&-r \end{array}\right). \end{aligned}$$

This is a zero-trace equitransmitting unitary Hermitian matrix constructed in [348].

The defined operator $L$ is determined by 4 real parameters: one in the matrix $S$ and three in $D$. The parameter included in $S$ determine different transition probabilities through the central vertex, while not all phases included in $D$ are important leading to unitarily equivalent operators. Let us calculate the spectrum explicitly in order to see this phenomena.

The solution to the Laplace equation $-\psi '' =k^2 \psi$ is given by

$$\displaystyle \begin{aligned} \psi \left(x\right)=\begin{cases} a_1e^{ik\left|x-x_1\right|}+ a_2e^{ik\left|x-x_2\right|},& x\in\left[x_1,x_2\right],\\[3mm] a_3e^{ik\left|x-x_3\right|}+a_4e^{ik\left|x-x_4\right|},& x\in\left[x_3,x_4\right]. \end{cases}\end{aligned}$$

Remember that $l_1=x_2-x_1$ and $l_2=x_4-x_3$. Since the unitary matrix $\tilde {S}$ is also Hermitian, it plays the role of the vertex scattering matrix $S$ connecting the amplitudes of the incoming and outgoing waves at the vertex. In other words, vertex conditions (16.2) imposed on $\psi$ imply that

$$\displaystyle \begin{aligned} \tilde{S} \left(\begin{array}{c} e^{ikl_1}a_2\\e^{ikl_1}a_1\\e^{ikl_2}a_4\\e^{ikl_2}a_3 \end{array}\right)= \left(\begin{array}{c} a_1\\a_2\\a_3\\a_4 \end{array}\right). \end{aligned} $$

(16.16)

This equation can be written as

$$\displaystyle \begin{aligned} \left( \tilde{S} S_{\mathbf{e}} - I \right) \left( \begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \end{array} \right) = 0, \end{aligned} $$

(16.17)

where

$$\displaystyle \begin{aligned} S_{\mathbf{e}} =\left(\begin{array}{cccc} 0&e^{ikl_1}&0&0\\[2mm] e^{ikl_1}&0&0&0\\[2mm] 0&0&0&e^{ikl_2}\\[2mm] 0&0&e^{ikl_2}&0 \end{array}\right). \end{aligned}$$

Hence the spectrum of $L$ is determined by the secular equation $\det \left (\tilde {S} S_{\mathbf {e}}- I\right )=0$.

We are going to show that the spectrum depends just on three parameters: $\beta$, the phase $\varphi _1$ and the difference $\varphi _3-\varphi _2$, i.e. one of the phase parameters can be eliminated if we are interested in the spectrum.

One may look at the secular equation directly:

$$\displaystyle \begin{aligned} \begin{aligned} &1+\left\{7t^4+2r^2t^2+rt^3\left(4e^{-i\beta}+e^{i\beta}\right)+\cos\left(2\beta\right)+rt^3e^{i\left(\beta-\varphi_1\right)}\right\}e^{2ik\left(l_1+l_2\right)}\\[3mm] &+2t\left\{\left(r^2-3t^2\right)e^{ik\left(2l_1+l_2\right)}+2te^{ik\left(l_1+l_2\right)}-e^{ikl_2}\right\}\cos\left(\beta-\varphi_3+\varphi_2\right)\\[3mm] &-2te^{ikl_1}\left(t^2e^{2ikl_2}+1\right)\cos\varphi_1-2t\left(2t^2-r^2+rt\cos\beta\right)e^{ik\left(l_1+2l_2\right)}\cos\varphi_1\\[3mm] &-2rt^2e^{ik\left(2l_1+l_2\right)}\left(\cos\varphi_1+\cos\left(2\beta-\varphi_3+\varphi_2\right)\right)=0. \end{aligned}\end{aligned}$$

Introducing $\phi _1 = \varphi _1, \phi _2=\varphi _3-\varphi _2$ we rewrite the equation as

$$\displaystyle \begin{aligned} \begin{aligned} &1+\left\{7t^4+2r^2t^2+rt^3\left(4e^{-i\beta}+e^{i\beta}\right)+\cos\left(2\beta\right)+rt^3e^{i\left(\beta-\phi_1\right)}\right\}e^{2ik\left(l_1+l_2\right)}\\[3mm] &+2t\left\{\left(r^2-3t^2\right)e^{ik\left(2l_1+l_2\right)}+2te^{ik\left(l_1+l_2\right)}-e^{ikl_2}\right\}\cos\left(\beta-\phi_2\right)\\[3mm] &-2te^{ikl_1}\left(t^2e^{2ikl_2}+1\right)\cos\phi_1-2t\left(2t^2-r^2+rt\cos\beta\right)e^{ik\left(l_1+2l_2\right)}\cos\phi_1\\[3mm] &-2rt^2e^{ik\left(2l_1+l_2\right)}\left(\cos\phi_1+\cos\left(2\beta-\phi_2\right)\right)=0. \end{aligned}\end{aligned}$$

It is clear that the spectrum is completely described by the three mentioned parameters.

Is it possible to see this using the unitary multiplication transformation described in the previous section? Consider the following function ${\mathbf {U}} (x)$ constant on each of the edges:

$$\displaystyle \begin{aligned} {\mathbf{U}} \psi\left(x\right)= \begin{cases} \psi\left(x\right),&x\in E_1 = \left[x_1,x_2\right], \\[2mm] e^{-i\varphi_2}\psi\left(x\right),&x\in E_2 = \left[x_3,x_4\right]. \end{cases}\end{aligned}$$

This unitary transformation does not change the differential operator but amends the vertex scattering matrix as follows

$$\displaystyle \begin{aligned} {} \hat{S}= U^{-1} \tilde{S} U= \mathrm{diag}\, \left(1,e^{-i\phi_1},1,e^{-i\phi_2}\right) \tilde{S} \; \mathrm{diag}\, \left(1,e^{i\phi_1},1,e^{i\phi_2}\right). \end{aligned} $$

(16.18)

The transformation ${\mathbf {U}}$ obviously does not change the edge scattering matrix $S_{\mathbf {e}} ,$

$$\displaystyle \begin{aligned} U^{-1} S_{\mathbf{e}} U = S_{\mathbf{e}},\end{aligned}$$

hence

$$\displaystyle \begin{aligned} \det\left(\hat{S} S_{\mathbf{e}}- I\right)=\det\left(U^{-1} \tilde{S} U S_{\mathbf{e}}- U^{-1} U\right)=\cdots=\det\left(\tilde{S} S_{\mathbf{e}} - I\right).\end{aligned}$$

Thus the secular equation remains unchanged despite one of the parameters having been eliminated.

The parameters $\phi _{1,2}$ are associated with the two loops forming $\Gamma _{(2.4)}$ and can be interpreted as fluxes of the magnetic field through the loops, since these phases disappear if the magnetic potential on the edges is chosen appropriately as described in the previous section.

16.3 Topological Damping of Aharonov-Bohm Effect

This section is devoted to just one concrete example of a magnetic Schrödinger operator showing that dependence of the spectrum upon some of the magnetic fluxes may be damped by choosing appropriate values of the other fluxes. We call this phenomenon topological damping as explained at the end of the section. This example is taken from our recent paper [352].

16.3.1 Getting Started

Consider again the figure eight metric graph $\Gamma _{(2.4)}$ given in Fig. 16.1 and magnetic Laplacian $L_{0,a}^S$ given by the differential expression

$$\displaystyle \begin{aligned} \tau_{0,a} = \left( i\frac{d}{dx} + a(x)\right)^2 \end{aligned} $$

(16.19)

assuming vertex conditions

$$\displaystyle \begin{aligned} {} i (S - {\mathbf{I}}) \vec{u} = (S + {\mathbf{I}}) \partial \vec{u}, \end{aligned} $$

(16.20)

where $\vec {u}$ and $\partial \vec {u}$ are the vectors of all limit values of the function and its normal extended derivatives at the vertex $V$

$$\displaystyle \begin{aligned} \vec{u} = \left( \begin{array}{c} u(x_1) \\ u(x_2) \\ u(x_3) \\ u(x_4) \end{array} \right), \; \; \; \partial \vec{u} = \left( \begin{array}{c} u'(x_1) - i a(x_1) u(x_1) \\ - \left(u'(x_2)- i a(x_2) u(x_2) \right) \\ u'(x_3) - i a(x_3) u(x_3)\\ - \left(u' (x_4) - i a(x_4) u(x_4) \right) \end{array} \right).\end{aligned}$$

The matrix $S$ is unitary and is used to parametrise all possible matching conditions making the operator $L$ self-adjoint in $L_2 (\Gamma _{(2.4)})$ when defined on all functions from the Sobolev space $W_2^2 (\Gamma \setminus V )$ satisfying (16.20).

Fig. 16.1

The figure eight graph $\Gamma _{(2.4)}$

Our main interest is the dependence of the spectrum upon magnetic fluxes through the two loops

$$\displaystyle \begin{aligned} \phi_j = \int_{x_{2j-1}}^{x_{2j}} a(x) dx , \; \; j=1,2. \end{aligned} $$

(16.21)

Using transformation

$$\displaystyle \begin{aligned} U_a : u(x) \mapsto \mathrm{exp}\; \left( i \int_{x_{2j-1}}^x a(y) dy \right) u(x) \end{aligned} $$

(16.22)

the magnetic Laplacian is mapped to the Laplacian $L^{S^{\phi _1, \phi _2}} = L_{0,0}^{S^{\phi _1, \phi _2}}$ defined on the set of functions satisfying vertex conditions

$$\displaystyle \begin{aligned} {} i (S^{\phi_1, \phi_2} - {\mathbf{I}} ) \vec{u} = (S^{\phi_1, \phi_2} + {\mathbf{I}}) \partial_n \vec{u}, \end{aligned} $$

(16.23)

which are obtained from (16.20) by substituting the matrix $S$ with

$$\displaystyle \begin{aligned} S^{\phi_1, \phi_2} = D^{-1} S D, \; \; D = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & e^{i \phi_1} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i \phi_2} \end{array} \right), \end{aligned} $$

(16.24)

and the vector of extended derivatives $\partial \vec {u}$—with the vector of normal derivatives

$$\displaystyle \begin{aligned} \partial_n \vec{u} = \left( \begin{array}{c} u'(x_1) \\ - u'(x_2) \\ u'(x_3) \\ - u' (x_4) \end{array} \right).\end{aligned}$$

In general the spectrum of the operator $L^{S^{\phi _1, \phi _2}}$ depends on the fluxes, but it might happen that only the sum of the fluxes counts. For example if

$$\displaystyle \begin{aligned} S = \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right)\end{aligned}$$

then the graph is equivalent to the loop of length ${\mathcal {L}} = x_2 - x_1 + x_4-x_3$ and the spectrum obviously depends on the sum of the fluxes $\phi _1 + \phi _2.$ This case is not interesting, since the anomalous behavior of the spectrum is due to the choice of vertex conditions that do not respect the geometry of the graph: the vertex $V$ can be divided into two vertices $V^1 = \{ x_1, x_4 \}$ and $V^2 = \{ x_2, x_3 \}$ and the vertex conditions connect separately the boundary values corresponding to the two new vertices. Such boundary conditions do not correspond to the figure eight graph but rather to the loop graph.

Another degenerate example is when

$$\displaystyle \begin{aligned} S = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) .\end{aligned}$$

In this case the eigenvalues can be divided into two series: each one depending on one of the fluxes $\phi _1$ or $\phi _2$ only. The vertex conditions connect together the pairs of endpoints $(x_1, \, x_2)$ and $(x_3, \, x_4)$ separately. The corresponding metric graph is not $\Gamma _{(2.4)}$ but rather two separate loops formed by the two edges. This case is not interesting for us either.

In what follows we study the magnetic Schrödinger operator corresponding to the vertex conditions given by the following vertex scattering matrix:

$$\displaystyle \begin{aligned} {} S = \left( \begin{array}{cccc} 0 & 0 & \alpha & \beta \\ 0 & 0 &- \beta & \alpha \\ \alpha & -\beta & 0 & 0 \\ \beta & \alpha& 0 & 0 \\ \end{array} \right), \; \; \, \alpha, \beta \in \mathbb R, \; \alpha^2 + \beta^2 = 1. \end{aligned} $$

(16.25)

This unitary matrix connects together boundary values at all four endpoints and therefore is properly connecting. One may visualize this by the following picture, where all possible scattering processes are indicated by curves (Fig. 16.2).

Fig. 16.2

Visual representation of the connections determined by the vertex conditions with $S$ given by (16.25): the curves indicate possible passages

It will be shown that interesting effects can be observed if the probabilities of these passages are equal, which corresponds to the choice $\alpha = \beta = 1/\sqrt {2}$:

$$\displaystyle \begin{aligned} {} S = \left( \begin{array}{cccc} 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 &- \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \end{array} \right). \end{aligned} $$

(16.26)

16.3.2 Explicit Calculation of the Spectrum

Our immediate goal is to derive the equation describing the spectrum of the operator $L^{S^{\phi _1, \phi _2}}$ depending on the fluxes $\phi _1$ and $\phi _2$ and parameters $\alpha$ and $\beta .$ The matrix $S$ and therefore the matrix $S^{\phi _1, \phi _2} = D^{-1} S D$ appearing in the vertex conditions is not only unitary, but also Hermitian. It follows that the corresponding vertex scattering matrix $S_v$ does not depend on the energy and coincides with

$$\displaystyle \begin{aligned} {} S^{\phi_1, \phi_2} = \left( \begin{array}{cccc} 0 & 0 & \alpha & e^{i \phi_2} \beta \\ 0 & 0 & - e^{- i \phi_1} \beta & e^{-i(\phi_1-\phi_2)} \alpha \\ \alpha & - e^{i\phi_1} \beta & 0 & 0 \\ e^{-i \phi_2} \beta & e^{i(\phi_1 - \phi_2)} \alpha & 0 & 0 \end{array} \right). \end{aligned} $$

(16.27)

The differential operator on the edges does not contain any electric or magnetic potential, hence the corresponding edge scattering matrix is

$$\displaystyle \begin{aligned} S_{\mathbf e} = \left( \begin{array}{cccc} 0 & e^{ik \ell_1} & 0 & 0 \\ e^{ik \ell_1} & 0 & 0 & 0 \\ 0 & 0 & 0 & e^{ik \ell_2} \\ 0 & 0 & e^{ik\ell_2} & 0 \end{array} \right), \end{aligned} $$

(16.28)

where $\ell _{j} = x_{2j}- x_{2j-1}, \; j= 1,2$ are the lengths of the edges. Then all nonzero eigenvalues are given by the solutions of the equation

$$\displaystyle \begin{aligned} \det \left( S_{\mathbf e} (k) S^{\phi_1, \phi_2} - \mathbb I \right) = 0, \end{aligned} $$

(16.29)

which is equivalent to

$$\displaystyle \begin{aligned} \det \left( \begin{array}{cccc} - 1 & 0 & - e^{ik\ell_1} e^{-i\phi_1} \beta & e^{ik\ell_1} e^{-i(\phi_1-\phi_2)} \alpha \\ 0 & -1 & e^{ik\ell_1} \alpha & e^{ik\ell_1} e^{i\phi_2} \beta \\ e^{ik\ell_2} e^{-i\phi_2} \beta & e^{ik\ell_2} e^{i (\phi_1 - \phi_2)} \alpha & -1 & 0 \\ e^{ik\ell_2} \alpha & - e^{ik\ell_2} e^{i\phi_1} \beta & 0 & -1 \end{array} \right) = 0 \end{aligned} $$

(16.30)

$$\displaystyle \begin{aligned} \Leftrightarrow 1 \!+\! \left(\alpha^2\!+\! \beta^2 \right)^2 e^{2 ik (\ell_1+ \ell_2) } \!+\! 2 \left( \cos (\phi_1 \!+\! \phi_2) \beta^2 \!-\! \cos (\phi_1 \!-\! \phi_2) \alpha^2 \right) e^{ ik (\ell_1+ \ell_2) } \!=\! 0. \end{aligned} $$

(16.31)

Taking into account that $\alpha ^2 + \beta ^2 = 1$ and $e^{ik (\ell _1 + \ell _2)} \neq 0$ we arrive at the following secular equation

$$\displaystyle \begin{aligned} {} \cos k (\ell_1+ \ell_2) = \alpha^2 \cos (\phi_1 - \phi_2) - \beta^2 \cos (\phi_1 + \phi_2). \end{aligned} $$

(16.32)

The right hand side of this equation is a real constant between $-1$ and $1$ and hence solutions to the equation form a periodic in $k$ sequence. The corresponding eigenvalues $\lambda = k^2$ in general depend on both magnetic fluxes $\phi _1$ and $\phi _2.$

Interesting phenomenon occurs if one choses $\alpha = \beta = 1/\sqrt {2}$, i.e. the matrix $S$ given by (16.26). The secular equation (16.32) takes the form

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \cos k (\ell_1+ \ell_2) & = & \displaystyle \frac{\displaystyle \cos (\phi_1 - \phi_2) - \cos (\phi_1 + \phi_2)}{2}\\[3mm] & = & \displaystyle \sin \phi_1 \sin \phi_2. \end{array} \end{aligned} $$

(16.33)

It follows, that if one of the magnetic fluxes is an integer multiple of $\pi$, then the spectrum is independent of the other flux. This is a trivial consequence of the secular equation (16.33), but we are interested in having an intuitive explanation of this phenomena. Aharonov-Bohm effect tells us that the spectrum of a system like magnetic Schrödinger operator on $\Gamma _{(2.4)}$ should depend on the magnetic fluxes. This dependence is damped only in very special cases. What is so special when one of the fluxes is zero? An explicit answer to this question is given in the following section. We use the trace formula connecting the spectrum of a quantum graph to the set of periodic orbits on the underlying metric graph.

In what follows we are interested in this special case, therefore without loss of generality let us assume that $\phi _1 = 0.$

Before proceeding, let us determine whether $\lambda = 0$ is an eigenvalue of the operator $L^{S^{\phi _1, \phi _2}}$ or not. $k= 0$ is a solution to the secular equation only if $\sin \phi _1 \sin \phi _2 = 1.$ If one of the fluxes is zero, then $k = 0$ is not a solution to the secular equation. It follows that the algebraic multiplicity¹$m_a (0)$ [332, 346] is zero.

Let us turn to calculation of the spectral multiplicity $m_s (0)$—the number of linearly independent solutions to the equation $L^{S^{\phi _1, \phi _2}} \psi = 0.$ In order to underline that only the lengths of the edges are important, let us parameterize the edges as follows

$$\displaystyle \begin{aligned} [x_1, x_2 ] = [0, \ell_1], \; \; [x_3, x_4 ] = [0, \ell_2] .\end{aligned}$$

All solutions to the differential equation are then given by:

$$\displaystyle \begin{aligned} \psi (x) = \begin{cases} a_1 x + b_1 &\text{If }x \in [x_1, x_2],\\ a_2 x + b_2 &\text{If }x \in [x_3, x_4]. \end{cases} {} \end{aligned} $$

(16.34)

Then at the endpoints we have

$$\displaystyle \begin{aligned} \vec{\psi} = \begin{pmatrix} b_1 \\ a_1 l_1 + b_1 \\ b_2 \\ a_2 l_2 + b_2 \end{pmatrix}, \quad \partial_n \vec{\psi} = \begin{pmatrix} a_1 \\ -a_1 \\ a_2 \\ -a_2 \end{pmatrix}. \end{aligned} $$

(16.35)

The matrix $S^{\phi _1, \phi _2}$ is unitary and Hermitian, hence its eigenvalues are just $\pm 1.$ Therefore the vertex conditions (16.23) are satisfied if and only if both the left and right hand sides are equal to zero:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left( \begin{array}{cccc} -\frac{1}{2} & 0 & \frac{1}{2 \sqrt{2}} & \frac{e^{-i \phi_2}}{2 \sqrt{2}} \\ 0 & -\frac{1}{2} & -\frac{e^{i \phi_1}}{2 \sqrt{2}} & \frac{e^{i \phi_1- i \phi_2}}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{e^{-i \phi_1}}{2 \sqrt{2}} & -\frac{1}{2} & 0 \\ \frac{e^{i \phi_2}}{2 \sqrt{2}} & \frac{e^{-i \phi_1+i \phi_2}}{2 \sqrt{2}} & 0 & -\frac{1}{2} \end{array} \right)& \begin{pmatrix} a_1 \\ -a_1 \\ a_2 \\ -a_2 \end{pmatrix} = 0, \end{array} \end{aligned} $$

(16.36)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left( \begin{array}{cccc} \frac{1}{2} & 0 & \frac{1}{2 \sqrt{2}} & \frac{e^{- i \phi_2}}{2 \sqrt{2}} \\ 0 & \frac{1}{2} & -\frac{e^{i \phi_1}}{2 \sqrt{2}} & \frac{e^{i \phi_1-i \phi_2}}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{e^{- i \phi_1}}{2 \sqrt{2}} & \frac{1}{2} & 0 \\ \frac{e^{i \phi_2}}{2 \sqrt{2}} & \frac{e^{- i \phi_1 + i \phi_2}}{2 \sqrt{2}} & 0 & \frac{1}{2} \end{array} \right)& \begin{pmatrix} b_1 \\ a_1 l_1 + b_1 \\ b_2 \\ a_2 l_2 + b_2 \end{pmatrix} = 0. \end{array} \end{aligned} $$

(16.37)

We omit tedious computations and give just a sketch. From the first equation we obtain $a_1 = a_2 = 0$, then plugging this result into the second equation we obtain $b_1 = b_2 = 0$ for any values of $\phi _2$. This proves that $\lambda = 0$ is not an eigenvalue and hence the spectral multiplicity (as well as the algebraic multiplicity) is zero in this case.

Summing up the spectrum of the magnetic Schrödinger operator on $\Gamma _{(2.4)}$ is given by the solutions of the secular equation

$$\displaystyle \begin{aligned} {} \cos k {\mathcal{L}} = 0, \; \, {\mathcal{L}} = \ell_1 + \ell_2, \end{aligned} $$

(16.38)

provided one of the magnetic fluxes is zero.

16.3.3 Topological Reasons for Damping

We have seen that in the case $\phi _1 = 0$ the spectrum does not depend on the flux $\phi _2$—the Aharonov-Bohm effect is damped which contradicts our intuition. The main goal of this section is to explain that this effect has a topological explanation. We are going to use trace formula (see Theorem 8.7) connecting the spectrum of a quantum graph to the set of periodic orbits on the metric graph. It will be shown that orbits that feel the magnetic flux $\phi _2$ give zero total contribution into the trace formula.

Under a periodic orbit we understand any oriented closed path on the graph $\Gamma _{(2.4)}$, which is allowed to turn back at the unique vertex only. Paths having opposite directions are considered to be different. We repeat the trace formula (8.20)

$$\displaystyle \begin{aligned} {} u(k) := 2 m_s (0) \delta(k) + \sum_{k_n\neq 0} \left( \delta(k-k_n) + \delta (k+k_n) \right) \end{aligned} $$

(16.39)

$$\displaystyle \begin{aligned} = \left(2 m_s (0) - m_a (0) \right) \delta(k) + \frac{\mathcal{L}}{\pi}+\frac{1}{\pi}\sum_{\gamma\in \mathcal{P}} l(\mathrm{prim}\,(\gamma)) S_v(\gamma) \cos kl(\gamma). \end{aligned}$$

The fluxes $\phi _1$ and $\phi _2$ are contained in the products $S_v(\gamma )$, since the entries of $S_v \equiv S^{\phi _1, \phi _2}$ depend on the fluxes (see formula (16.27)). Therefore it is natural to expect that the left hand side also depends on the fluxes as well. On the other hand the left hand side in (16.39) is determined by the spectrum of $L^{S^{\phi _1, \phi _2}}$ which in the case $\phi _1 = 0$ is independent of $\phi _2.$ More precisely, the spectrum is determined by $\cos k {\mathcal {L}} = 0$ (we have already shown that $\lambda = 0$ is not an eigenvalue in this case, $m_s (0) = 0$)

$$\displaystyle \begin{aligned} k_n = \frac{\pi}{2 {\mathcal{L}}} + \frac{\pi}{{\mathcal{L}}} n, \; \, n= 0, 1,2,3, \dots \end{aligned} $$

(16.40)

Then the left hand side of trace formula can be written as

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle u(k) & = & \displaystyle \sum_{k_n\neq 0} \left( \delta(k-k_n) + \delta (k+k_n) \right) \\ & = & \displaystyle \sum_{n\in \mathbb Z} \delta \left(k- \left(\frac{\pi}{2 {\mathcal{L}}} + \frac{\pi}{{\mathcal{L}}} n\right) \right) \\ & = & \displaystyle \sum_{m \in \mathbb Z} \delta \left(k- \frac{\pi}{2 {\mathcal{L}}} m \right) - \sum_{m \in \mathbb Z} \delta \left(k- \frac{\pi}{{\mathcal{L}}} m \right) . \end{array}\end{aligned}$$

We use now Poisson summation formula

$$\displaystyle \begin{aligned} \sum_{n\in \mathbb Z} \delta(x- T n ) = \frac{1}{T} \sum_{m \in \mathbb Z} e^{\displaystyle - i 2 \pi \frac{m}{T} x}\end{aligned}$$

and rewrite the last expression as follows:

$$\displaystyle \begin{aligned} {} u(k) = \sum_{n\in \mathbb Z} \delta \left(k- \left(\frac{\pi}{2 {\mathcal{L}}} + \frac{\pi}{{\mathcal{L}}} n\right) \right) = \frac{2 {\mathcal{L}}}{\pi} \sum_{m \in \mathbb Z} e^{-i 4 {\mathcal{L}} m k} - \frac{{\mathcal{L}}}{\pi} \sum_{m \in \mathbb Z} e^{-i 2 {\mathcal{L}} m k} . \end{aligned} $$

(16.41)

This formula represents the distribution $u(k)$ as a formal exponential series. This series is independent of $\phi _2$, while the series on the right hand side of (16.39) formally contain $\phi _2$, since $S_v(\gamma )$ depends on the second flux. Let us examine the series over all periodic orbits in more detail in order to understand the reason why all terms containing $\phi _2$ cancel.

As we have shown, both the algebraic and spectral multiplicities of $k = 0$ are equal to zero, hence the right hand side of trace formula can be written as

$$\displaystyle \begin{aligned} u(k) = \frac{\mathcal{L}}{\pi}+\frac{1}{\pi}\sum_{\gamma \in \mathcal{P}} l(\mathrm{prim}\,(\gamma)) S_v(\gamma) \cos kl(\gamma).\end{aligned}$$

Let us note first that the sum in the trace formula contains contributions from the paths that go around the left and right loops equally many times. This is due to the fact that the coefficients $12, 21, 34,$ and $43$ in the vertex scattering matrix are zero

$$\displaystyle \begin{aligned} (S)_{12} = (S)_{21} = (S)_{34} = (S)_{43} = 0 .\end{aligned}$$

Therefore the length of each path with nontrivial $S_v(\gamma )$ is an integer multiple of the total length ${\mathcal {L}} := \ell _1 + \ell _2.$

The sum over all paths is taken over all closed paths and $l (\mathrm {prim}\,(\gamma ))$ is the length of the corresponding primitive path. It will be convenient for us to distinguish paths with different starting edges—the first edges the path comes across. Then the sum $\sum _{\gamma \in \mathcal {P}} l(\mathrm {prim}\,(\gamma )) S_v(\gamma ) \cos kl(\gamma )$ can be written as two sums—over the paths that go around the left loop first and over the paths that go around the right loop first:

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle u(k) & = & \displaystyle \frac{\mathcal{L}}{\pi}+\frac{1}{\pi} \sum_{\gamma \in \mathcal{P}} l(\mathrm{prim}\,(\gamma)) S_v(\gamma) \cos kl(\gamma) \\ & = & \displaystyle \frac{\mathcal{L}}{\pi}+\frac{ \ell_1}{\pi} \sum_{\gamma \in \mathbb{P}_l} S_v(\gamma) \cos kl(\gamma) +\frac{ \ell_2}{\pi} \sum_{\gamma \in \mathbb{P}_r} S_v(\gamma) \cos kl(\gamma) , \end{array} \end{aligned} $$

(16.42)

where $\mathbb P_{l,r}$ denote the sets of paths where paths with different starting edges are considered different. The lower indices $l$ and $r$ indicate whether the path goes around the left or the right loop first. Each of the two sums can be treated in a similar way.

Let us consider first the series $\sum _{\gamma \in \mathbb {P}_l} S_v(\gamma ) \cos kl(\gamma )$ over all paths starting by going into the left edge. After going around the left loop the path should go around the right loop and then again around the left one: the left and right loops appear one after another. Every such path can be uniquely parametrised by a series of indices $\nu _j = \pm$ indicating whether the path goes around the left or right path in the positive ($+$) (clockwise following the orientation of the edges) or negative ($-$) (anti clockwise) direction. All odd indices correspond to the left loop, all even—to the right loop. The number of signs is even, which reflects the fact that every such path goes around the left and right loops equally many times. For example the path indicted on Fig. 16.3 is parametrised as $(+,-,+,+).$

Fig. 16.3

A path of length $2(l_1+l_2)$

Let us turn to the calculation of the coefficients $S_v(\gamma ).$ The path indicated on Fig. 16.3 has coefficient $S_v(\gamma )$ equal to

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle S_v(\gamma) & = & \displaystyle ( S_{v})_{14} (S_{v})_{32} (S_{v})_{13} (S_{v})_{42} \\[3mm] & = & \displaystyle e^{i \phi_2} \beta \cdot (- e^{-i \phi_1} \beta) \cdot \alpha \cdot e^{i(\phi_1 - \phi_2)} \alpha \\ & = & \displaystyle - e^{2i \phi_1} \alpha^2 \beta^2 = \frac{-1}{4} e^{i 2\phi_1}. \end{array} \end{aligned}$$

One may calculate the same product using the original vertex scattering matrix (16.26), but taking into account that each time when the path goes along the left or right loop the product gains the phase coefficient $e^{\pm i \phi _1}$ or $e^{\pm i \phi _2}$, respectively. The sign corresponds to positive or negative direction. Each time when the path crosses the middle vertex, $S_v(\gamma )$ gets an extra term $\pm \frac {1}{\sqrt {2}}.$ Note that only coefficients corresponding to the transitions $2 \rightarrow 3$ and $3 \rightarrow 2$ have minus sign, all other coefficients are positive

It will be convenient to see the product $S_v(\gamma )$ corresponding to the path $(\nu _1, \nu _2, \dots , \nu _{2n})$ divided into three factors

• the product of all phase factors

  $$\displaystyle \begin{aligned} e^{i \sum_{j=1}^n \nu_{2j-1} \phi_1} \cdot e^{i \sum_{j=1}^n \nu_{2j} \phi_2};\end{aligned}$$

• the product of absolute values of scattering coefficients

  $$\displaystyle \begin{aligned} \left( \frac{1}{\sqrt{2}} \right)^{2n} = \frac{1}{2^n};\end{aligned}$$

• the product of sign factors $\pm 1 .$

Our next claim is that only paths that every second time go around the right loop in a different direction give a contribution into the trace formula. Consider the path $p'$ that contains the sequence Then the contribution from the path $p''$ obtained from $p'$ by reversing the edge with the number $2m+1$, i.e. given by , cancels the contribution from $p'.$ Really, the phase contributions from $p'$ and $p''$ are the same, the absolute values are also the same, while the product of signs for $p'$ contains $(-1) \times 1$ corresponding to transitions $3 \rightarrow 2$ and $1 \rightarrow 4$, in contrast to the product $1 \times 1$ appearing in the product for $p''$ (corresponds to the transitions $3 \rightarrow 1$ and $2 \rightarrow 4$, all other coefficients are the same). Similarly contributions from the paths given by and cancel each other.

Assume now that the path $p'$ contains the sequence then the contribution from $p''$ corresponding to is just the same. It follows that only the paths of the form $(\nu _1, +, \nu _3, - ,\nu _5,+, \nu _7, -, \dots )$ and $(\nu _1, -, \nu _3, + ,\nu _5,-, \nu _7, +, \dots )$ survive in the series. Every such path has discrete length (the number of edges it comes across) being multiple of $4.$ The phase contribution from such paths is zero, since we assumed $\phi _1 = 0.$ It follows that the sum over the periodic paths starting with the left loop does not depend on $\phi _2.$ Similar result holds for the other sum explaining the reason why the spectrum of the magnetic Schrödinger operator on $\Gamma _{(2.4)}$ does not depend on $\phi _2,$ provided $\phi _1 = 0.$

Let us continue to calculate the sum over the periodic orbits. We have seen that only orbits of lengths $2 n {\mathcal {L}}$ (discrete length $4 n$) make a contribution to the series. Consider for example the orbits of length $2 {\mathcal {L}} .$ Only the orbits of the form $(\nu _1, +, \nu _3, -)$ and $(\nu _1, -, \nu _3, + )$ give nonzero contributions. The phase contribution is $e^{ i 0 } = 1.$ The absolute value contribution is $\left ( \frac {1}{\sqrt {2}} \right )^4 = \frac {1}{4}.$ The sign contribution is $-1.$ Altogether there are $4 \times 2$ such orbits, since the signs $\nu _1, \nu _3$ can be chosen freely. So the total contribution to the series is: $- \ell _1 2 \cos k 2 {\mathcal {L}}.$

Similarly contribution from the orbits of length $2 n {\mathcal {L}}$ is $\ell _1 2 (-1)^n \cos k 2 {\mathcal {L}} n .$ Taking into account contribution from the paths from $\mathbb P_r$ (starting by first going around the right loop) we get the following expression

which coincides with the expression (16.41) obtained using Poisson summation formula.

The calculations carried out above show the reason, why Aharonov-Bohm effect is not present if one of the fluxes is zero: contributions from the periodic orbits going with non-zero flux cancel each other. On the other hand, if one of the fluxes is not zero, then the spectrum depends on the other flux as can be seen from Eq. (16.32). Similar result holds even if one of the fluxes is an integer multiple of $\pi .$ It is clear that figure eight graph is not unique and other graphs exhibiting the same effect can be found. Trace formula together with our explicit calculations provides a recipe to construct such graphs.

Problem 74

Investigate whether topological damping of Aharonov-Bohm effect can be observed for an equilateral flower graph.