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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.

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  • $\begingroup$ What do you mean by "tangent vector"? Tangent to what? $\endgroup$ Commented May 1, 2022 at 15:55
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    $\begingroup$ are you asking how to compute a basis of tangent vectors to a (point on a) sphere in general? That's doable, but note that in higher dimensions the "Bloch representation" of quantum states won't look like a (hyper)sphere. See e.g. quantumcomputing.stackexchange.com/a/24422/55, quantumcomputing.stackexchange.com/q/8416/55, and links therein $\endgroup$
    – glS
    Commented May 2, 2022 at 8:07
  • $\begingroup$ @glS Thanks for the refs. I just don't quit understand how $\partial _{\varphi}\left( \begin{array}{c} \cos \varphi\\ \sin \varphi\\ 0\\ \end{array} \right) $ can be connected with $\partial _{\varphi}|\psi _{\varphi}\rangle $, where $\left( \begin{array}{c} \cos \varphi\\ \sin \varphi\\ 0\\ \end{array} \right) $ is the bloch vector of $|\psi _{\varphi}\rangle $. $\endgroup$
    – Sherlock
    Commented May 4, 2022 at 9:58
  • $\begingroup$ Related : My answer here Understanding the Bloch sphere. I know nothing about Quantum Computation but as you could see in my answer a state on the Bloch sphere may be represented by $$ \vert\psi\rangle =\cos\left(\dfrac{\theta_3}{2}\right)\vert u_3\rangle + e^{i\phi_3}\sin\left(\dfrac{\theta_3}{2}\right)\vert d_3\rangle $$ $\endgroup$
    – Frobenius
    Commented May 16, 2022 at 19:50
  • $\begingroup$ A unit vector tangent to the sphere (in the sense that it will be orthogonal to $\vert\psi\rangle$) could be produced by differentiation with respect to $\theta_3$ $$ \vert\chi_\theta\rangle=2\dfrac{\partial \vert\psi\rangle}{\partial \theta_3} =-\sin\left(\dfrac{\theta_3}{2}\right)\vert u_3\rangle + e^{i\phi_3}\cos\left(\dfrac{\theta_3}{2}\right)\vert d_3\rangle $$ represented by a point diametrically opposite to that of $\vert\psi\rangle$. $\endgroup$
    – Frobenius
    Commented May 16, 2022 at 20:00

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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.

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  • $\begingroup$ Can you give some concrete example? I can't quite understand... $\endgroup$
    – Sherlock
    Commented May 15, 2022 at 3:01

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