Is $\sigma(-\Delta)=\sigma_{\mathrm{ess}}(-\Delta)$? Or under which conditions do we have this?

Question

Let $\Delta: H^2(\mathbb{R}^n)\subseteq L^2(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)$ be the Laplace operator in the weak sense.

A Lemma in the book of Borthwick (Spectral Theory) says:

It is proved, by showing that $\sigma_{ess}(-\Delta)=\sigma_{ess}(-\Delta+V)$. Then it is said that the claim follows.

But why? Is $\sigma_{ess}(-\Delta)=\sigma(-\Delta)$?

So far I know $\sigma(\Delta)=(-\infty,0]$.