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$\newcommand{\ket}[1]{\left|#1\right>}$ Normally the matrix representation of $J_x$ and $J_y$ are derived using the ladder operators. But is a better intuition possible by starting with $J_z$ alone ? One just needs to find the eigenvectors of these operators in the basis of eigenvectors of $J_z$, but how do you do that ?

For $j$ = 1, the basis states are $\ket{j=1, m_z = +1}$, $\ket{j=1, m_z = 0}$ and $\ket{j=1, m_z = -1}$

if $\ket{j, m_y}$ are the eigenstates of $J_y$ then the basis just as above would be $\ket{j=1, m_y = +1}$, $\ket{j=1, m_y = 0}$ and $\ket{j=1, m_y = -1}$

To represent say $\ket{1, m_y = +1}$ in $J_z$ basis : $$\ket{1, m_y = +1} = \langle{1, m_z = +1}\ket{1, m_y = +1} \ket{1, m_z = +1} + ...$$ and so on for all the three $J_z$ eigenvectors

But how do you intuitively figure out those inner products. For $j = 1$, the momentum vector $\vec{J}$ in real space looks like this:

Momentum

Then what is the intuition behind $\langle{1, m_z = +1}\ket{1, m_y = +1}$ ?

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But is a better intuition possible by starting with $J_z$ alone ?

I doubt that. There would be an infinity of solutions, as any rotation between 𝐽𝑥 and 𝐽𝑦 would provide an equally acceptable set of such. In any case, I'll just provide the conventional ladder operator stick-in-the mud answers for $j$ = 1.

In the same z-basis you started with, $$\ket{j=1, m_z = +1} = (1,0,0)^T,\\ \ket{j=1, m_z = 0}= (0,1,0)^T,\\ \ket{j=1, m_z = -1}=(0,0,1)^T, $$ you have the obvious $$ J_z=\operatorname{diag}(1,0,-1), \qquad J_y= \frac{i}{\sqrt 2} \begin{bmatrix}0&-1&0\\ 1&0& -1\\ 0&1&0 \end{bmatrix}, $$ $=(J_+-J_-)/2i$, (which you might also obtain by blood-sweat and tears through a π/2 rotation around x if you had the rotation matrix for that, and converted all in the spherical basis... don't go there.)

It is then evident by inspection of the eigenvectors of $J_y$ that $$\ket{j=1, m_y = +1} = (1,i\sqrt{2},-1)^T/2,\\ \ket{j=1, m_y = 0}= (1,0,1)^T/\sqrt{2},\\ \ket{j=1, m_y = -1}=(1,-i\sqrt{2},-1)^T/2. $$ Do not neglect the requisite conjugation in the dot products of the spherical basis.

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  • $\begingroup$ "which you might also obtain by blood-sweat and tears through a π/2 rotation around x if you had the rotation matrix for that, and converted all in the spherical basis... don't go there" are you talking about obtaining $J_y$ ? How is the rotation matrix related to the angular momentum operators ? could you point to a resource where that is elaborated ? $\endgroup$ Commented Nov 25, 2022 at 4:30
  • $\begingroup$ $R_x(\pi/2) J_z R_x(-\pi/2)=\exp(i\pi J_x/2) ~J_z \exp(-i\pi J_x/2)$=J_y$, easily provable by Campbell/Hadamard's Lemma, easy to illustrate in the doublet representation, and quite intuitive. Should be in Rose, Edmonds, Mamermesh, or Biedenharn & Louck... $\endgroup$ Commented Nov 25, 2022 at 12:16
  • $\begingroup$ It is assumed you are comfortable with Wigner’s d matrix. $\endgroup$ Commented Nov 25, 2022 at 13:29
  • $\begingroup$ I have seen this answer : physics.stackexchange.com/a/20230/275718 where they derive the pauli spin matrices. Is something similar possible for the case when j = 1 ? $\endgroup$ Commented Nov 25, 2022 at 13:39
  • $\begingroup$ Whether you see it or not, for any spin. the reps are even by the standard ladder expression provided there. All derivations are logically traceable to this, whether apparent or not. $\endgroup$ Commented Nov 25, 2022 at 14:37

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