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

ShoutOutAndCalculate
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    What do you think the definition of $J_z$ is? – knzhou Nov 30 '19 at 05:23
  • @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. – ShoutOutAndCalculate Nov 30 '19 at 05:27
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    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. – knzhou Nov 30 '19 at 05:31
  • @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"... – ShoutOutAndCalculate Nov 30 '19 at 05:45
  • Physicists tend to treat some things as “obvious”. – G. Smith Nov 30 '19 at 05:52
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    Related : Total spin of two spin-1/2 particles. – Frobenius Nov 30 '19 at 08:54

1 Answers1

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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?

James Johns
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  • 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.... – Frobenius Nov 30 '19 at 19:12
  • .... 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. – Frobenius Nov 30 '19 at 19:12
  • I didn't downvote your answer. – Frobenius Nov 30 '19 at 21:58
  • 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 – James Johns Nov 30 '19 at 21:59
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    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. – Frobenius Nov 30 '19 at 22:03
  • Hi Frobenius, Apologies for taking so long to get back to you, I was traveling. I agree with everything in that answer, it's standard theory for product spaces, but there you also use that $S_z^{tot} = S_{1z}+S_{2z}$, which comes from the simple vectorial addition of the spin angular momentum along the z axis. So I don't see the conflict. I think the original question was from someone who was confused about why $J_z=J_{1z}+J_{2z}$ which doesn't require sophisticated treatment to get at the right answer – James Johns Dec 02 '19 at 03:17