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I know that for a given spin-j quantum state, say $\vert\psi\rangle = (\psi_0 , \psi_1 , \cdots , \psi_{2j})$, we can construct a polynomial as follows

$ w(z) = \sum_{k = 0}^{2j} (-1)^k \psi_k \sqrt{\binom{2j}{k} } z^{2j-k} $

and by means of the inverse stereographic projection of the roots of $w(z)$, the Majorana's representation of $\vert\psi\rangle$ is obtained on sphere. I also know that the Majorana's representation for the eigen-state of $\mathbf{J.n}$ operator with eigen-value $m$ is the configuration of points on sphere in which there are $j+m$ points in the $\mathbf{n}$ direction and other $j-m$ points in the antipodal point (equivalently $-\mathbf{n}$ direction).

My question is why the Majorana's representation of $e^{i\mathbf{n.J} \theta} \vert \psi \rangle$ is just that of $\vert\psi\rangle$ rotated around $\mathbf{n}$ by an angle of $\theta$? (This is transparent when $\vert \psi \rangle$ is an eigen-vector of angular momentum in some direction but I do not see why this is true in general.)

Thank you

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  • $\begingroup$ This type of question might be more appropriate over in the physics stack exchange. To my knowledge the Majorana representation has nothing to do with the bloch sphere. $\endgroup$
    – Condo
    Aug 4, 2020 at 13:53
  • $\begingroup$ Have you seen Borel-Weil-Bott? $\endgroup$
    – AHusain
    Aug 4, 2020 at 19:51

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