Square root of density matrix in terms of Pauli matrices Let $\rho$ a density matrix such that:
$$\rho=\frac12(I + \vec{r}\cdot \vec{\sigma})$$
Where $\vec{r}$ is a vector with the property $|\vec{r}|$ is less than unit.
And $\vec{\sigma} = (\sigma_x,\sigma_y,\sigma_z)$ are Pauli matrices.
The problem is express $\rho^{\frac12}$ in terms of Pauli matrices. Using spectral decomposition for this problem is very boring. Is there any elegant way?
 A: Here is your serotonin: can you solve
$$ \frac12 I + \frac{r}{2} \hat{r}\cdot \vec{\sigma}= (x I+ y ~ \hat{r}\cdot \vec{\sigma} )^2 =(x^2+y^2)I+ 2xy~ \hat{r}\cdot \vec{\sigma}, $$
so,
$$  x^2+y^2=1/2, \qquad 2xy=r/2 $$
for x and y?
A: I would compute the matrix $\sqrt{\rho}$ in terms of $\vec{r}=(r_x,r_y,r_z)$ following, for example, https://en.wikipedia.org/wiki/Square_root_of_a_2_by_2_matrix. Then, you can define the operator space given by the Pauli matrices and the identity with basis elements $\{B_i\}=\{I,\sigma_x,\sigma_y,\sigma_z\}$ (that is the definition you use for $\rho$) where the elements fulfill $\text{Tr}(B^{\dagger}_iB_j)=\delta_{ij}$ (they are orthogonal respect to the Hilbert-Smidt norm). Then, you can express $\sqrt{\rho}$ in this basis as $\sqrt{\rho}=\sum_i a_i B_i$, where $a_i=\text{Tr}(\sqrt{\rho} B_i)$. So you just have to compute $\text{Tr}(\sqrt{\rho} I)$, $\text{Tr}(\sqrt{\rho} \sigma_x)$, $\text{Tr}(\sqrt{\rho} \sigma_y)$ and $\text{Tr}(\sqrt{\rho} \sigma_z)$ to write it in terms of Pauli matrices.
I hope this can help you!
