The square which agrees with $ x^2-x-1 $ except for the constant term is $ \left(x-\frac 12\right)^2 $, which is equal to $ x^2-x+\frac 14 $, and thus to $ (x^2-x-1) + \frac 54 $. Therefore, by adding $ \frac 54 $ to each side, Sviatoslav rewrote the equation $ x^2-x-1 = 0 $ as $$\left(x-\frac 12\right)^2 = \frac 54.$$We have $ a=-\frac 12 $ and $ b=\boxed{\frac 54} $.