Directly get tangent vector of Bloch sphere from quantum state (qubit)?

Question

We know that Bloch sphere is a good way to represent a qubit(two energy quantum systems). Now I want to know the tangent vector in Bloch sphere, e.g. for states $\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1\\ e^{i\varphi}\\ \end{array} \right) $, or equivalently with $x,y,z$ coordinate:$\left( \begin{array}{c} \cos\varphi\\ \sin\varphi\\ 0\\ \end{array} \right) $. We can calculate the tangent vector by $\partial _{\varphi}\left( \begin{array}{c} \cos\varphi\\ \sin\varphi\\ 0\\ \end{array} \right) =\left( \begin{array}{c} -\sin\varphi\\ \cos\varphi\\ 0\\ \end{array} \right) $.

My question is, is there a way to calculate a quantity similar to $\left( \begin{array}{c} -\sin\varphi\\ \cos\varphi\\ 0\\ \end{array} \right) $ without refer to $x,y,z$ coordinates? Because I want to see what the tangent vector correspond to $n$-qubits instead of single qubit case, in that case, we can't seek help from $x,y,z$ coordinates.

Answer 1

I think the crucial point is that why we could represent a qubit by a sphere is that they could be represented as coherent state: any state is just $e^{i\hat{n}\cdot S}\left|\uparrow\right\rangle_z$, when as a qubit you choose $S_i=\frac{1}{2}\sigma_i$ to be Pauli matrix and $\left|\uparrow\right\rangle_z=\left(\begin{array}{c}1\\ 0\end{array}\right)$and as a vector you choose $\left|\uparrow\right\rangle_z=\left(\begin{array}{c}1\\ 0\\0\end{array}\right)$ and $S$ the generator of rotating vectors -- the Lie algebra of $SU(2)$ is the same as $SO(3)$, which means that the commutation relation of $S_i$ in these two cases are the same. Thus when you calculate $\partial_\varphi |\psi\rangle$ in the two case, treat $S$ as some abstract operators with commutation relation and finally put its true form back. Thus the result should be the same.