Solution:
Let the roots be $r_1$ and $r_2$ with $r_1 - r_2 = 1$ (without loss of generality, $r_1 > r_2$).

From Vieta's formulas:
$$\begin{align*} r_1 + r_2 &= a \\ r_1 r_2 &= 1 \end{align*}$$

Since $r_1 - r_2 = 1$, we can write:
$$r_1 = r_2 + 1$$

Substitute into the sum:
$$(r_2 + 1) + r_2 = a \implies 2r_2 + 1 = a$$

From the product:
$$(r_2 + 1) r_2 = 1 \implies r_2^2 + r_2 - 1 = 0$$

Solve for $r_2$:
$$r_2 = \frac{ -1 \pm \sqrt{1 + 4} }{2 } = \frac{ -1 \pm \sqrt{5} }{2 }$$

Since $r_1$ and $r_2$ are real, both values are possible.

Now, $a = 2r_2 + 1$:

For $r_2 = \frac{ -1 + \sqrt{5} }{2 }$:
$$a_1 = 2 \left( \frac{ -1 + \sqrt{5} }{2 } \right ) + 1 = ( -1 + \sqrt{5} ) + 1 = \sqrt{5}$$

For $r_2 = \frac{ -1 - \sqrt{5} }{2 }$:
$$a_2 = 2 \left( \frac{ -1 - \sqrt{5} }{2 } \right ) + 1 = ( -1 - \sqrt{5} ) + 1 = -\sqrt{5}$$

But the problem asks for positive values of $a$, so the only answer is:
$$a = \sqrt{5}$$