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In addition of angular momenta, does both the relations depict the same thing?

$$ \vec{J} = (\vec{J_1}\otimes 1 +1\otimes \vec{J_2}) $$

$$ \vec{J} = (\vec{J_2}\otimes 1 +1\otimes \vec{J_1}) $$

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In essence yes, the only difference is in the labels you give to each subspace - 1 or 2.

This could mean simply giving different labels to particle 1 and particle 2, each of which has certain angular momentum. Or if the system is one particle with orbital and spin contributions to its angular momentum its just a different assignation of labels one and two to each part. The tensor product and its decomposition goes through the same since it is symmetric in subspaces 1 and 2.

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Physically yes but technically the order matters in the sense that the phase of the state of "good" $J$ actually depend on the ordering. Writing \begin{align} \vert J M_J\rangle =\sum_{m_1m_2} C_{j_1m_1;j_2m_2}^{JM_J}\vert j_1m_1\rangle \vert j_2m_2\rangle \end{align} where $C_{j_1m_1;j_2m_2}^{JM_J}$ is a Clebsch-Gordan coefficient, and using the symmetry property \begin{align} C_{j_1m_1;j_2m_2}^{JM_J}=(-1)^{j_1+j_2-J}C_{j_2m_2;j_1m_1}^{JM_J} \end{align} shows that inverting the role of $j_1$ and $j_2$ may introduce an overall phase $(-1)^{j_1+j_2-J}$ in the construction of $\vert JM_J\rangle$, i.e. \begin{align} \vert J M_J\rangle =\sum_{m_1m_2} C_{j_1m_1;j_2m_2}^{JM_J}\vert j_1m_1\rangle \vert j_2m_2\rangle = (-1)^{j_1+j_2-J}\sum_{m_1m_2} C_{j_1m_1;j_2m_2}^{JM_J}\vert j_2m_2\rangle\vert j_1m_1\rangle \end{align}

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