Hardwiring the output in quantum black box separation


In this paper, while using a diagonalization argument in Section $ 5$ , the authors write:

Fix some enumeration over all $ poly(n)$ -size quantum verifiers $ M_{1}, M_{2},…$ which we can do because the number of such machines is countably infinite (by the Solovay-Kitaev theorem [15]). Some of these verifiers may try to decide a language by trivially “hardwiring” its outputs; for example, by returning 1 independent of the input. We start by fixing a unary language $ L$ such that no machine $ M_{i}$ hardwires the language. We can always do this because there are more languages than $ poly(n)$ -sized machines.

Maybe I am missing something very basic, but:

  1. Why do we care if some of the verifiers have the output hardwired?
  2. How can we find an $ L$ , like in the description? Why will this $ L$ even be decidable? I know there are more languages than there are Turing machines; but those extra languages aren’t decidable.