Rayleigh distribution
This is for normalization purposes. When you have noise following a circularly-symmetric complex normal distribution, (i.e the real and imaginary part are independent from each other and have the same variance) with a variance of 1 in both parts, the absolute value will be Rayleigh-distributed with a variance of 2. Therefore, when you introduce the factor, the real and imaginary part will have a variance of 1/2 each, resulting in a variance of 1 for the Rayleigh-distributed absolute value.

Edit to adress the question in the comment: Lets call the unit-variance Rayleigh-diastributed noise \$ \tilde{n} \$. Then you can directly set the variance by scaling with a factor \$ \sigma \$:
\$ n = \sigma \cdot \tilde{n} \Rightarrow  \mathrm{var}(n) = \sigma^2 \cdot \mathrm{var}(\tilde{n}) = \sigma^2 \cdot 1  = \sigma^2\$.
Thus, by scaling to unit variance first, you do not have to worry about the original variance anymore and can just scale to the variance you want to have.