Weyl Operators: Spectrum

Question

Given a CCR-algebra $\mathcal{A}_{CCR}(\mathcal{H})$ over a Hilbert space $\mathcal{H}$.

Then the Weyl operators are unitary: $$W(f)^*=W(-f)=W(f)^{-1}$$ Thus, their spectrum lies on the unit circle: $$\sigma(W(f))\subseteq\mathbb{S}$$ But why even all of it: $$\sigma(W(f\neq0))=\mathbb{S}$$

Answer 1

The spectrum is invariant under unitary transformations, thus it is rotational invariant: $$\sigma(W(f\neq0))=\sigma(W(g)W(f\neq0)W(g)^*)=e^{-i\sigma(g,f\neq0)}\sigma(W(f\neq0))$$ Note that this excludes the special case: $$\sigma(W(0))=\sigma(1)=\{1\}$$