Twisted circle momentum and its interval boundary matrix

The first supplier [F1] gives $P_{\theta }^0$ self-adjoint with the stated eigenfamily and eigenvalues. Since $\mathrm{\hslash }$ is a positive real scalar, $P_{\theta }=\mathrm{\hslash }{P}_{\theta }^0$ has the same domain and is self-adjoint: the adjoint identity [F4] gives $(\mathrm{\hslash }{P}_{\theta }^0{)}^{\ast }=\mathrm{\hslash }(P_{\theta }^0{)}^{\ast }$. Moreover $P_{\theta }{\varphi }_n=\mathrm{\hslash }(2\pi n+\theta ){\varphi }_n\mathrm{/}L$, so the same complete orthonormal family has the scaled eigenvalues in the Statement.

By [F5], $f\in D(P_{\theta }^2)$ exactly when $f\in D(P_{\theta })$ and $P_{\theta }f\in D(P_{\theta })$. The latter condition requires $Df\in H^1$, hence $f\in H^2$ by [F3], and its twisted trace is $-i\mathrm{\hslash }{f}^{\mathrm{\prime }}(L)=e^{i\theta }(-i\mathrm{\hslash }{f}^{\mathrm{\prime }}(0))$, equivalently $f^{\mathrm{\prime }}(L)=e^{i\theta }{f}^{\mathrm{\prime }}(0)$. Thus $D(P_{\theta }^2)$ is exactly the stated second-order domain, and $P_{\theta }^2\mathrm{/}{\mathrm{\hslash }}^2=-D^2$ there.

For a function in this domain write $c=f(0)$ and $d=f^{\mathrm{\prime }}(0)$. Its boundary vectors are $a=cp$ and $b=\mathrm{\ell }dq$, where $p=(1,e^{i\theta })$ and $q=(1,-e^{i\theta })$. These vectors are orthogonal and have equal norm, since $⟨p,q⟩=1-e^{i\theta }{e}^{-i\theta }=0$ and $\mathrm{\parallel }p{\mathrm{\parallel }}^2=\mathrm{\parallel }q{\mathrm{\parallel }}^2=2$. The columns of the displayed matrix $U_{\theta }$ are $(0,e^{i\theta })$ and $(e^{-i\theta },0)$, which are orthonormal, so $U_{\theta }\in U(2)$. It sends $p$ to $p$ and $q$ to $-q$, so for $x=a+ib=cp+i\mathrm{\ell }dq$ it gives $U_{\theta }x=cp-i\mathrm{\ell }dq=a-ib$. Therefore $(U_{\theta }-I_2)a+i(U_{\theta }+I_2)b=0$. Conversely, for any $f\in H^2(0,L)$ satisfying this equation, expand its boundary vectors as $a=\alpha p+\beta q$ and $b=\gamma p+\delta q$; the orthogonal nonzero vectors $p,q$ form a basis of ${\mathbb{C}}^2$. The equation becomes $-2\beta q+2i\gamma p=0$, so $\beta =\gamma =0$. Hence $f(L)=e^{i\theta }f(0)$ and, since $\mathrm{\ell }>0$, $f^{\mathrm{\prime }}(L)=e^{i\theta }{f}^{\mathrm{\prime }}(0)$. By [F2], this is exactly the self-adjoint boundary realization $T_{U_{\theta }}$, which has the same domain and action as $P_{\theta }^2\mathrm{/}{\mathrm{\hslash }}^2$.