# What is "trivial" about the trivial topological superconducting phase?

Once more I am stuck on my favorite word: "trivial". I am reading a bunch of stuff about topological superconductors at the moment and people keep talking about having to distinguish between the topologically "trivial" and "nontrivial" phases.

To have the easiest possible concrete example, consider the one dimensional p-wave superconductor:

$H=H_{metal} + H_{pairing}\\ H_{metal}= \sum_p c_p^\dagger\left(\frac{p^2}{2m}-\mu\right)c_p\\ H_{pairing}= p\frac{1}{2}\left(\Delta c_p^\dagger c_{-p}^\dagger + \Delta^* c_{-p}c_{p} \right)$

which has the following energy spectrum

$E_{\pm}=\pm \sqrt{\left(\frac{p^2}{2m}-\mu\right)^2+|\Delta|^2p^2}$

The book I am reading (Topological Insulators and Topological Superconductors, by B. Andrei Bernevig and Taylor L. Hughes), refers to the $\mu<0, \; \Delta=0$ case as the trivial insulating limit (page 198, bottom).

I think I do understand why the $\mu<0$ and $\mu>0$ are topologically distinct. However I can't deduce why the $\mu<0, \; \Delta=0$ case

• is said to be trivial. What is the trivial referring to?
• is the insulating limit. I thought $H_{metal}$ describes a 1D metal of spinless fermions.

I am most grateful for any clarifying answers.

"Trivial" is used because the equation reduces to something a bit simpler. Take the equation $E_{\pm}=\pm \sqrt{\left(\frac{p^2}{2m}-\mu\right)^2+|\Delta|^2p^2}$
Setting $\Delta$ to $0$ gets rid of the last term, and making $\mu<0$ means that the first term will reduce to $$E_1=\pm \left(\frac{p^2}{2m}-\mu\right)$$ which is pretty simple. That's what triviality boils down to. From Wikipedia,