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In Demkowicz-Dobrzanski et al. (arXiv:2001.11742), the authors mention in Eq. (74), page 22, that the symmetric logarithmic derivatives (SLDs) for pure states parametrised in the usual way via the Bloch sphere are $$L_\theta = \boldsymbol\sigma\cdot\partial_\theta\mathbf r\equiv \sum_{i=1}^3 \sigma_i \partial_\theta r_i, \qquad L_\varphi = \boldsymbol\sigma\cdot\partial_\varphi\mathbf r \equiv \sum_{i=1}^3 \sigma_i \partial_\varphi r_i,$$ where $\sigma_i$ are the Pauli matrices, and $\mathbf r$ is (I assume) the unit vector pointing in the direction $(\theta,\varphi)$, that is: $$\mathbf r=\begin{pmatrix}\sin\theta \cos\varphi \\ \sin\theta\sin\varphi \\ \cos\theta\end{pmatrix}.$$ More explicitly, we're parametrising the pure qubit states as $$\rho_{(\theta,\varphi)} = |\psi_{(\theta,\varphi)}\rangle\!\langle\psi_{(\theta,\varphi)}| = \frac{ I+ \sin(\theta)\cos(\varphi)X+\sin(\theta)\sin(\varphi)Y+\cos(\theta)Z}{2}, \\ \lvert\psi_{(\theta,\varphi)}\rangle \equiv \cos(\theta/2)|0\rangle+\sin(\theta/2)e^{i\varphi}|1\rangle.$$ The SLDs are defined by the relations: $$\frac12\{L_\theta,\rho_{(\theta,\varphi)}\}\equiv \frac{L_\theta \rho_{(\theta,\varphi)}+\rho_{(\theta,\varphi)} L_\theta}{2} = \partial_\theta \rho_{(\theta,\varphi)}, \\ \frac12\{L_\varphi,\rho_{(\theta,\varphi)}\} = \partial_\varphi \rho_{(\theta,\varphi)}. $$ I can directly verify by computation that the given forms for $L_\theta,L_\varphi$ do indeed satisfy these relations: for example, from $$\frac12\left\{\frac{I+\mathbf r\cdot\boldsymbol\sigma}{2},\boldsymbol\sigma\cdot\partial\mathbf r\right\} = \frac12\boldsymbol\sigma\cdot\partial\mathbf r+\frac14\{\boldsymbol\sigma\cdot \mathbf r,\boldsymbol\sigma\cdot\partial\mathbf r\} \\= \frac12\boldsymbol\sigma\cdot\partial\mathbf r+\frac12 \mathbf r \cdot\partial\mathbf r =\frac12\boldsymbol\sigma\cdot\partial\mathbf r = \partial\left(\frac{I+\mathbf r\cdot\boldsymbol\sigma}{2}\right).$$ But what would be direct ways to get to these handy formulas for the LSDs? I'm aware of the general recipe to compute LSDs via the relation $\frac12(\lambda_i+\lambda_j) L_{ij}=(\partial \rho)_{ij}$ with $\lambda_i$ eigenvalues of $\rho$ and the operators expressed in the eigenstates of $\rho$, but this gets rather unwieldy in this case. Is there a more direct approach (or an easy way to follow this approach)?

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These results are specific to qubits and your direct verification is the best way to do it. You can see this because they required the property of Pauli matrices that $\{\pmb{v}\cdot\pmb{\sigma},\pmb{w}\cdot\pmb{\sigma}\}=2\pmb{v}\cdot \pmb{w}$. If you tried writing your state in some basis of generalized Pauli matrices like $\rho=(I+\pmb{v}\cdot\pmb{\Sigma})/d$ with $\partial \rho=\partial\pmb{v}\cdot\pmb{\Sigma}/d=\partial\pmb{v}\cdot\pmb{\Sigma}/d+\alpha \partial \pmb{v}\cdot \pmb{v}$ and tried to match it with some $\{(I+\pmb{v}\cdot\pmb{\Sigma})/d,\pmb{w}\cdot\pmb{\Sigma}\}/2=\pmb{w}\cdot\pmb{\Sigma}/2+\{\pmb{v}\cdot\pmb{\Sigma}/d,\pmb{w}\cdot\pmb{\Sigma}\}/2$, we wouldn't have a direct way to simplify because of a lack of general relationship $\{\Sigma_i,\Sigma_j\}$.


To generalize this, we go in a different direction. The trick, when dealing with pure states, is to write them as a parametrized unitary acting on a fixed initial state. Then we can express the SLDs in terms of the generators of the unitary and be done.

For a pure state pointing in direction $(\theta,\phi)$ on the surface of the Bloch sphere, we can consider this to have started at (say) the north pole, rotated along the $x$-axis by polar angle $\theta$, then rotated around the $z$-axis by azimuthal angle $\phi$. Alternative constructions are of course possible that achieve this in a single rotation by angle $\theta$ around a $\phi$-dependent axis (something like $x\cos\phi+ y\sin\phi$, but it will pay to be lazy here and not compute it). We thus take $$|\psi_{(\theta,\phi)}\rangle=e^{-i\phi \sigma_z/2}e^{-i\theta \sigma_y/2}|\psi_{(0,0)}\rangle\equiv U(\theta,\phi)|\psi_0\rangle.$$

The SLDs for pure states undergoing unitary transformations are $$L_j=2i[|\psi_{(\theta,\phi)}\rangle\langle \psi_{(\theta,\phi)}|,G_j];\qquad G_j=i\frac{\partial U}{\partial \theta_j}U^\dagger$$ (I don't normally compute these, because quantum Fisher information just needs the $G_j$ in the end). Now the only math we need is to compute the generators: $$G_\theta=U \sigma_y U^\dagger /2\qquad \mathrm{and}\qquad G_\phi=(\sigma_z/2) \, U U^\dagger.$$ Note that these were easy because I broke $U$ into the two terms and some things commute like $\sigma_y$ and $e^{i\sigma_y}$. The factors of two come from the relative normalization of Pauli matrices versus generators of SU(2) (all of this works for any spin, not just qubits). To be clear, I'll explicitly write $$G_\theta=e^{-i\phi\sigma_z/2} \sigma_y e^{i\phi\sigma_z/2} /2=(-\sigma_x\sin\phi+\sigma_y \cos\phi)/2\qquad \mathrm{and}\qquad G_\phi=\sigma_z/2.$$ This is the direct way to get the SLDs. It can actually be used for mixed qubit states too because they can be written as some unitary acting on a state whose Bloch vector points toward the north pole but is not unit length (for higher spins, it is not enough to describe the state by a vector length and orientation).

Now, their expressions have derivatives - can we get those? All of the terms in the SLDs here are commutators between angular momentum operators (Pauli matrices making up the $G_j$) and the state. For qubits, that's easy because we know how to do commutators of Pauli matrices with Pauli matrices, but for higher spins the state is not just going to be a linear combination of angular momentum generators, so we won't be able to simplify in the same way. We can do an example: $$L_\phi=i[\rho,\sigma_z]=i[\frac{I+\pmb{r}\cdot\pmb{\sigma}}{2},\sigma_z]=i^2(r_2,-r_1,0)\cdot\pmb{\sigma}=\partial_\phi \pmb{r}\cdot \pmb{\sigma}.$$ This comes about from the cross product $[(0,0,1)\cdot\pmb{\sigma},\pmb{r}\cdot\pmb{\sigma}]=2i[(0,0,1)\times\pmb{r}]\cdot\pmb{\sigma}$ and then identifying $[(0,0,1)\times\pmb{r}]=\partial_\phi \pmb{r}$.


Other avenues: We could do some math to rewrite $$U(\theta,\phi)=\exp[-i(\boldsymbol{\sigma}\cdot \boldsymbol{v})/2]$$ for a vector $\boldsymbol{v}$ that depends on $(\theta,\phi)$. If I recall correctly it is something like $\boldsymbol{v}=\theta(-\sin\phi,\cos\phi,0)$ (I think it should be perpendicular to your $r$). But then $$G_j\neq \boldsymbol{\sigma}\cdot\frac{\partial \boldsymbol{v}}{\partial \theta_j}$$ because in general $[\boldsymbol{\sigma}\cdot\frac{\partial \boldsymbol{v}}{\partial \theta_j},U]\neq 0.$ So this does not directly give a method for identifying where the partial derivatives of $\pmb{r}$ come from. Instead, the trick is just to compute all of this in some representation of SU(2) (ie use Pauli matrices) and then show it will have the same format for any spin.

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