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This is probably a standard question but I couldn't found it anywhere online, so I thought it might be a good ideal to add it in Physics exchange.

Modern Quantum Mechanics Second Edition J.J. Sakurai Jim Napolitano Equation 3.8.36

$$(J_z-J_{1z}-J_{2z})|j_1,j_2; jm\rangle =0$$

However, the textbook doesn't exactly explained that where did this expression come from, i.e. although that $J\equiv J_1\otimes 1+1\otimes J_2$, it's not necessarily such that $J_z= J_{1z}\otimes 1+1\otimes J_{2z}$. Especially, no relationship was given in the textbook such that $J_z-J_{1z}-J_{2z}$ was understood.

Could you show that why $(J_z-J_{1z}-J_{2z})|j_1,j_2; jm\rangle =0$ ?

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    $\begingroup$ What do you think the definition of $J_z$ is? $\endgroup$
    – knzhou
    Commented Nov 30, 2019 at 5:23
  • $\begingroup$ @knzhou technically it wasn't proven, so you might need to prove and show that as well, which doesn't seem to be a straightforward job, unless you want to say:"by conversion"/"it's angular momentum". I think with SO(3) group representation it might be easier and pretty casual, but that's sort of cheap. $\endgroup$
    – ShoutOutAndCalculate
    Commented Nov 30, 2019 at 5:27
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    $\begingroup$ No, I’m not asking about any proofs, I’m asking how you think the symbol “$J_z$” is defined in the first place. If it’s defined the usual way, the question is trivial. So you must be imagining we start from a different definition, but then to answer your question we have to know what that definition is. $\endgroup$
    – knzhou
    Commented Nov 30, 2019 at 5:31
  • $\begingroup$ @knzhou Like I said, "technically it wasn't proven". Basically the textbook just showed up the equation(the one similar to the one in the post) and never explained how he got it, but "in a sense" "of course" "and so on"... $\endgroup$
    – ShoutOutAndCalculate
    Commented Nov 30, 2019 at 5:45
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    $\begingroup$ Related : Total spin of two spin-1/2 particles. $\endgroup$
    – Voulkos
    Commented Nov 30, 2019 at 8:54

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It may not be obvious, but it is definitional. For both J1 and J2, we define the z axis to point in the same direction (is it's an external axis) and from your definition of J it follows that $J_z=J_{z1}+J_{z2}$. Does that help?

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  • $\begingroup$ Welcome to PSE. I think your answer does not help. There exists a whole theory about product states, product (Hilbert) spaces, product transformations (operators) etc under which the addition of two angular momenta in QM is well-defined without contradictions.... $\endgroup$
    – Voulkos
    Commented Nov 30, 2019 at 19:12
  • $\begingroup$ .... Think for a moment this : if $j_{1}$ and $j_{2}$ are (nonnegative) integers or half-integers representing angular momenta living in the $\;\left(2j_{1}+1\right)-$ dimensional and $\;\left(2j_{2}+1\right)-$ dimensional spaces $\;\mathcal{H}_{\boldsymbol{1}}\;$ and $\;\mathcal{H}_{\boldsymbol{2}}\;$ respectively, expressions like this \begin{equation} J_{z}=J_{1z}+J_{2z} \tag{01} \end{equation} have no sense since $J_{1z}$ and $J_{2z}$ are operators acting on different spaces and if $j_{1}\ne j_{2}$ of different dimensions too. $\endgroup$
    – Voulkos
    Commented Nov 30, 2019 at 19:12
  • $\begingroup$ I didn't downvote your answer. $\endgroup$
    – Voulkos
    Commented Nov 30, 2019 at 21:58
  • $\begingroup$ Even in product spaces, you're still talking about vector quantities that add. The allowed values of $J_z$ depend on the value of j, but the form of the operator doesn't. $J_{1z}$ only acts on the angular momentum space of particle 1, so I disagree that it makes no sense. Sorry, but I'd be happy to discuss further and revise or delete my answer $\endgroup$
    – James Johns
    Commented Nov 30, 2019 at 21:59
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    $\begingroup$ I suggest you to see my answers here : Total spin of two spin-1/2 particles. These are my efforts to understand all this stuff in the past. $\endgroup$
    – Voulkos
    Commented Nov 30, 2019 at 22:03

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