Spectrum and gauge periodicity of a periodic flux operator

Set $A_0=-\kappa {\mathrm{\partial }}_{\varphi }^2$ on $H_{\mathrm{p}\mathrm{e}\mathrm{r}}^2(S^1)$. In the boundary coordinates ${\mathrm{\Gamma }}_{0}f=(f(0),f(2\pi ))$ and ${\mathrm{\Gamma }}_{1}f=(f^{\mathrm{\prime }}(0),-f^{\mathrm{\prime }}(2\pi ))$, the unitary swap matrix in [F1] makes the boundary equation equivalent to $f(0)=f(2\pi )$ and $f^{\mathrm{\prime }}(0)=f^{\mathrm{\prime }}(2\pi )$: adding and subtracting its two scalar equations gives these two equalities. Hence $-{\mathrm{\partial }}_{\varphi }^2$ is self-adjoint on this domain, and multiplication by the positive real $\kappa$ shows that $A_0$ is self-adjoint. Let $P=-i{\mathrm{\partial }}_{\varphi }$. Integration by parts [F5] and equality of the endpoint traces show that $P$ is symmetric on $H_{\mathrm{p}\mathrm{e}\mathrm{r}}^2$. For $f={\sum }_n{c}_n{e}_n$ in this domain, integration by parts gives $⟨Pf,e_n⟩=n{c}_n$ and $⟨A_{0}f,e_n⟩=\kappa n^2{c}_n$. Parseval for [F2] therefore implies $\mathrm{\parallel }Pf{\mathrm{\parallel }}_2^2={\sum }_n{n}^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$ and $\mathrm{\parallel }{A}_{0}f{\mathrm{\parallel }}_2^2={\kappa }^2{\sum }_n{n}^4\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$.

On $D(A_0)$ write $T_{\alpha }=A_0+B_{\alpha }$ with $B_{\alpha }=-2\kappa \alpha P+\kappa {\alpha }^{2}I$. This operator is symmetric because $P$ is symmetric and $\alpha ,\kappa$ are real. For every $\epsilon >0$ and integer $n$, $\mathrm{\mid }n\mathrm{\mid }\le \epsilon n^2+(4\epsilon {)}^{-1}$; the coefficient identities in step 1.1 and the triangle inequality give $\mathrm{\parallel }{B}_{\alpha }f{\mathrm{\parallel }}_2\le a\mathrm{\parallel }{A}_{0}f{\mathrm{\parallel }}_2+b\mathrm{\parallel }f{\mathrm{\parallel }}_2$ with any prescribed $a>0$ after choosing $\epsilon$ sufficiently small and then a finite $b$. Applying [F3] with $a<1$ proves that $T_{\alpha }$ is self-adjoint on the stated periodic domain.

Each $e_n$ lies in the periodic domain and direct differentiation gives $T_{\alpha }{e}_n=\kappa (n-\alpha {)}^2{e}_n$. Completeness [F2] makes this the full orthonormal eigenbasis. The associated spectral projections are therefore the sums of the character projections whose eigenvalues lie in the chosen Borel set, so every scalar spectral measure is supported on $\{\kappa (n-\alpha {)}^2:n\in \mathbb{Z}\}$. This set has no finite accumulation point because $\kappa >0$ and $\mathrm{\mid }n-\alpha \mathrm{\mid }\to \mathrm{\infty }$. If $z$ is outside it, ${\inf }_n\mathrm{\mid }{\lambda }_n-z\mathrm{\mid }>0$. For $f={\sum }_n{c}_n{e}_n$, where $c_n=⟨f,e_n⟩$, the diagonal map $R_{z}f={\sum }_n{c}_n({\lambda }_n-z{)}^{-1}{e}_n$ is bounded on $H$. Also ${\sup }_n\mathrm{\mid }{\lambda }_n\mathrm{/}({\lambda }_n-z)\mathrm{\mid }<\mathrm{\infty }$, so the domain characterization in [F4] shows that $R_z$ maps into $D(T_{\alpha })$. Coefficientwise, $(T_{\alpha }-z)R_z=I$ and $R_z(T_{\alpha }-z)=I$ on $D(T_{\alpha })$, so $z$ is in the resolvent. If $z={\lambda }_n$ then $e_n$ is a nonzero kernel vector. Hence the spectrum is exactly the displayed set and there is no continuous spectral measure.

Let $j\in \mathbb{Z}$ and $U_{j}f=e^{ij\varphi }f$. It is unitary, preserves the periodic $H^2$ domain, and $(P-\alpha -j)U_{j}f=U_j(P-\alpha )f$ by differentiation; squaring gives $T_{\alpha +j}{U}_j=U_j{T}_{\alpha }$. If a smooth periodic unit-modulus function $u$ satisfies the displayed conjugacy, then $(P-\alpha )u=0$, or $u^{\mathrm{\prime }}=i\alpha u$. The solutions are $u(\varphi )=c{e}^{i\alpha \varphi }$ with $\mathrm{\mid }c\mathrm{\mid }=1$, which are periodic exactly when $e^{2\pi i\alpha }=1$, equivalently $\alpha \in \mathbb{Z}$. Conversely, for integral $\alpha$ this multiplier is single-valued and periodic.