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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 $\rho \text{ and } \pi$.

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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 $F'\equiv\sqrt{F}$.

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  • $\begingroup$ Why unit vectors? $\endgroup$ – Norbert Schuch Feb 4 at 3:16
  • $\begingroup$ @NorbertSchuch I misread the question thinking that was implicit. I edited to correct. Thanks for catching that. $\endgroup$ – Jonathan Trousdale Feb 4 at 5:43

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