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

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.

• What do you mean by "tangent vector"? Tangent to what? May 1, 2022 at 15:55
• 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
– glS
May 2, 2022 at 8:07
• @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 $|\psi _{\varphi}\rangle$. May 4, 2022 at 9:58
• 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 $$\vert\psi\rangle =\cos\left(\dfrac{\theta_3}{2}\right)\vert u_3\rangle + e^{i\phi_3}\sin\left(\dfrac{\theta_3}{2}\right)\vert d_3\rangle$$ May 16, 2022 at 19:50
• 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 $\vert\psi\rangle$. May 16, 2022 at 20:00

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.