What is the variance of this unbiased estimator for a normal distribution?

I have been spending some time trying to make sense of:

$$ \widehat{\sigma_2^2} = \frac{1}{2}(Y_1-Y_2)^2 $$

But I don't understand it, specifically the squared part, I don't think I can apply the variance properties if it's squared.


\begin{align} \frac14E[(Y_1-Y_2)^4] &=\frac14E[((Y_1-\mu)-(Y_2-\mu))^4] \\ &=\frac14\sum_{i=0}^4\binom{4}{i}E[ (Y_1-\mu)^i]E[(Y_2-\mu)^{4-i}] \\ &=\frac14 \left(3\sigma^4 +6(\sigma^2)(\sigma^2) +3\sigma^4 \right)\\ &=3\sigma^4 \end{align}

Hence

$$Var\left( \frac12 (Y_1-Y_2)^2\right)=3\sigma^4 - \sigma^4 = 2\sigma^4$$