Spectrum of the modular operator

Question

In the above proof, how to check that $\lambda \notin sp(\Delta_{\varphi})$.

If $\lambda \in sp(\Delta_{\varphi})$, we have $\lambda ^{it} \in sp(\Delta_{\varphi}^{it})=\{c\}$, where $c\in \Bbb C$.

How to derive the contradiction?

Answer 1

You always have $1\in\sigma(\Delta_\varphi)$. So $1\in\sigma(\Delta^{it}_\varphi)$. As $\Delta^{it}_\varphi$ is a scalar, you get that $\lambda^{it}=1$ for all $t$. This forces $\lambda=1$, which is the contradiction.