# Can I compute the fidelity between two states without having to diagonalise them?

So I have been given the following quantum states:

$$\rho = \frac{I}{2} + \frac{\bar{s}.\bar{\sigma}}{2}$$ $$\pi = \frac{I}{2} + \frac{\bar{r}.\bar{\sigma}}{2}$$

How do I calculate the fidelity between the two?

I know that the general formula for fidelity is: $$tr\sqrt{\rho^{1/2}\pi\rho^{1/2}}$$

I want to know if there's any way to calculate the fidelity other than to go through the mess of diagonalizing the respective matrices of $$\rho \text{ and } \pi$$.

If these are qubit states, the formula in your question simplifies dramatically to $$F'(\rho,\pi)=\sqrt{\text{tr}(\rho \pi) + 2 \sqrt{\text{det}(\rho) \text{det}(\pi)}}.$$
If you consider the components of the vectors, $$\vec s = (s_1, s_2, s_3)$$ and $$\vec r = (r_1, r_2, r_3)$$, this can be expressed simply as $$F'(\rho, \pi) = \frac{1}{\sqrt{2}} \left[1+ \sum \limits_{i=1}^3 s_i r_i + \sqrt{(1-\vert \vec s \vert^2)(1-\vert \vec r \vert^2)} \right]^\frac{1}{2}.$$
Note that the formula you provided is normally referred to as quantity fidelity ($$F'$$), defined in terms of fidelity ($$F$$) by $$F'\equiv\sqrt{F}$$.