Lemma: Deficiency count

The expression ${\ell }_{\nu }$ is limit circle at zero exactly when $0\le \nu <1$, and is limit point at infinity. The transverse minimal operator has deficiency indices $(d_{\beta },d_{\beta })$, where

In particular:

1. $d_0=1$, with deficient mode $n=0$.

2. For $0<\beta <1$, only $n=0,-1$ can be deficient. They are deficient respectively when $\beta <\alpha$ and $1-\beta <\alpha$.

3. Two deficient modes occur precisely when $1-\alpha <\beta <\alpha$, which requires $\alpha >1/2$.

The full minimal operator on $N$ has infinite deficiency indices if $d_{\beta }>0$, and is essentially self-adjoint if $d_{\beta }=0$.

Proof. At zero the solutions at spectral parameter zero are $r^{\nu },r^{-\nu }$ for $\nu >0$, and $1,\log r$ for $\nu =0$. Both are square integrable with weight $r$ exactly for $\nu <1$. At infinity the expression is limit point, for example after conjugating by $f\mapsto r^{1/2}f$ to the half-line Schrödinger expression

Weyl's alternative gives the radial assertion ReedSimonII,DerezinskiRichard. Counting the angular modes gives the transverse indices.

Fourier transformation in $z$ gives the radial expressions shifted by ${k_z}^2$. When a deficient transverse mode exists, its boundary data may vary over an infinite-dimensional space of longitudinal functions. If no such mode exists, the direct integral of the unique self-adjoint fibre closures is the closure of the full operator.