Shannon formula, why square the channel gain?
The Shannon channel capacity equation, as I remember is

$$\begin{align}
R &amp;= W \log_2 \left( 1 + \frac{S}{N} \right) \\
\\
  &amp;= W \log_2 \left( \frac{S+N}{N} \right) \\
\\
  &amp;= W \big( \log_2(S+N) \, - \, \log_2(N) \big) \\
\\
  &amp;= \frac{W}{6.02} \big( (S+N)_\text{dB} \, - \, (N)_\text{dB} \big) \\
\end{align}$$

where $S$ is the signal power and $N$ is the noise power (both over the same bandwidth $W$).

Noise power for a given bandwidth (thin enough that noise power density $N_0$ is considered constant is

$$ N = N_0 W $$

and the received signal power is the transmitted power, $P_t$, times the power gain of the channel (assuming both are constant over the bandwidth $W$).  The power gain of the channel is the square of the gain in amplitude because instantaneous power is the square of instantaneous amplitude.  So the received signal power is

$$ S = |H|^2 P_t $$

assuming all of the transmitted power is in the channel bandwidth $W$ and the channel gain, $H$, is constant over the bandwidth $W$.