# 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.