We see that the adjoint of \( \hat { X } ^ { ( \alpha ) } \) is defined as \( i ( d / d k ) \) in the momentum basis provided that \|φ) also obeys the boundary condition \( \tilde { \phi } ( \Omega ) \, = \, e ^ { - i \alpha } \tilde { \phi } ( - \Omega ) \). Therefore we see \( D ( \hat { X } ^ { ( \alpha ) } ) = D ( \hat { X } ^ { ( \alpha ) \dagger } ) \), and so \( \hat { X } ^ { ( \alpha ) } \) is self-adjoint.

Now, since the operators \( \hat { X } ^ { ( \alpha ) } \) are self-adjoint, they have spectral decompositions. The spectrum of the oper-ator \( \bar { X } ^ { ( \alpha ) } \)(for fixed \( \alpha \)) is discrete,\( \mathrm { s p e c } ( \hat { X } ^ { ( \alpha ) } ) = \{ x _ { n } ^ { ( \alpha ) } : = \)\( \textstyle { \frac { 2 \pi n + \alpha } { 2 \Omega } } \} _ { n \in \mathbb { Z } } \), and describes a one-dimensional lattice. The corresponding eigenvectors, \( \{ | x _ { n } ^ { ( \alpha ) } \rangle \} _ { n \in \mathbb { Z } } \), are represented in the momentum eigenbasis as

\[ ( k | x _ { n } ^ { ( \alpha ) } ) = \frac { 1 } { \sqrt { 2 \Omega } } e ^ { - i k x _ { n } ^ { ( \alpha ) } } . \eqno ( \mathrm { A 5 } ) \]

These eigenvectors are orthogonal and admit a resolu-tion of the identity:

\[ \sum _ { n \in \mathbb { Z } } | x _ { n } ^ { ( \alpha ) } ) ( x _ { n } ^ { ( \alpha ) } | = \mathbb { 1 } \qquad \qquad \qquad ( { \mathrm { A 6 } } ) \]

. Crucially, the position eigenvectors from different self-adjoint extensions are not orthogonal:

\[ ( x _ { n } ^ { ( \alpha ) } | x _ { n ^ { \prime } } ^ { ( \alpha ^ { \prime } ) } ) = \mathrm { s i n c } \left[ ( x _ { n } ^ { ( \alpha ) } - x _ { n ^ { \prime } } ^ { ( \alpha ^ { \prime } ) } ) \Omega \right] . \qquad ( \mathrm { A 7 } ) \]

Note that the union of the spectra of the entire family of self-adjoint extensions provides a covering of R. There-fore, it is possible to construct an overcomplete contin-uum basis by taking the union of eigenbases of the family of self-adjoint extensions, i.e., \|x) := \|x(α)n \( ) \iff x \)\( \therefore \)=\( \begin{array} { r } { x _ { n } ^ { ( \alpha ) } : = \frac { 2 \pi n + \alpha } { 2 \Omega } } \end{array} \).

It is then simple to write down the Shannon sampling theorem for a bandlimited function \( \psi \):

\[ \begin{array} { l l } { { \displaystyle \psi ( x ) = ( x | \psi ) } } & { { \displaystyle ( \mathrm { A } 8 ) } } \\ { { \displaystyle ~ ~ ~ ~ ~ = \sum _ { n \in \mathbb { Z } } ( x | x _ { n } ^ { ( \alpha ) } ) ( x _ { n } ^ { ( \alpha ) } | \psi ) } } & { { \displaystyle ( \mathrm { A } 9 ) } } \\ { { \displaystyle ~ ~ ~ ~ ~ = \sum _ { n \in \mathbb { Z } } \mathrm { s i n c } \left[ ( x - x _ { n } ^ { ( \alpha ) } ) \Omega \right] \psi ( x _ { n } ^ { ( \alpha ) } ) . } } & { { \displaystyle ( \mathrm { A 1 0 } ) } } \end{array} \]

Therefore, we see that the function \( \psi \) is determined at any point \( x \in \mathbb { R } \) from its values on one of the lattices \( \{ x _ { n } ^ { ( \alpha ) } \} _ { n } . \).

Also, we obtain an overcomplete resolution of identity,

\[ \frac { \Omega } { \pi } \int _ { \mathbb { R } } d x \ | x ) ( x | = 1 \qquad \qquad \quad ( \mathrm { A 1 1 } ) \]

where \( \pi / \Omega \) is the density of degrees of freedom. We can use the resolution of identity for the continuum basis to show that the space of bandlimited functions has a reproducing kernel,

\[ ( x | \psi ) = \frac { \Omega } { \pi } \int d x ^ { \prime } \; ( x | x ^ { \prime } ) ( x ^ { \prime } | \psi ) \; \; \; \; \; \; \; \; \; \; \; ( \mathrm { A 1 2 } ) \]

\[ \psi ( x ) = \int d x \; K ( x , x ^ { \prime } ) \psi ( x ^ { \prime } ) , \; \; \; \; \; \; \; \; \; \; \; ( \mathrm { A 1 3 } ) \]

where \( K ( x , x ^ { \prime } ) : = ( \Omega / \pi ) ( x | x ^ { \prime } ) = ( \Omega / \pi ) \operatorname { s i n c } [ ( x - x ^ { \prime } ) \Omega ] \).

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