# When can we get discrete spectrum?

Suppose that $T$ is a densely defined closed operator on a separable Hilbert space $H$. Form $N = T^*T$. Assume further that $T$ has a finite dimensional kernel and satisfies the commutation relation $$TT^* - T^*T = 1$$ Is it always true that $N$ has discrete countable spectrum (namely, $\sigma(N) = \{0,1,2,3,...\}$) and $N+1$ is the inverse of a compact bounded operator? (It seems this is true for most interesting cases in QM)

• There's a problem just defining the difference because the domain $\mathcal{D}(TT^{\star})$ is not generally the same as $\mathcal{D}(T^{\star}T)$. Is there anything special you had in mind there? – DisintegratingByParts May 28 '14 at 3:49
• The setting I have in mind is sections of complex line bundle on a torus. So $H$ should be $L^2$ sections and $T$ is the complexified covariant derivative. In this case the domain of $TT^*$, $T^*T$, and $T$ all contain smooth sections. But the case of harmonic oscillator in the usual quantum mechanics setting works as well. – user44442 May 28 '14 at 17:32