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I'm reading through J. J. Sakurai's Modern Quantum Mechanics book and currently looking at the "Angular-momentum addition" part.

Here, it says you have two options and that one option is to construct simultaneous eigenket $\vert j_1j_2;m_1m_2\rangle$ of $J_1^2, J_2^2, J_{1z}, J_{2z}$ since the four operators commute with each other. I understand that $J_1^2$ and $J_{1z}$ commute, but I'm not sure how $J_1^2$ and $J_{2z}$ can commute intuitively.

"commute" means that one can measure both at once right? But total angular momentum of spin 1 and angular momentum of spin 2 are independent.

Where am I wrong here?

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  • $\begingroup$ One way of seeing it is, $ \mathbf{J} := \mathbf{r} \times \mathbf{p}$, since we postulate $[x_i,x_j]=[x_i,p_j]=[p_i,p_j]=0$, for $i \neq j$. $\endgroup$
    – user26143
    Jun 12, 2014 at 4:53

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From your statement "measure both at once", it seems you've misunderstood what's meant by "simultaneous measurement". It does not mean that you can run a single experiment to measure both "independent" values together. Rather, it means that in principle, you can measure one quantity without "ruining" the results of measurement the other, so both quantities can be obtained.

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Technically when two operators $A$ and $B$ commute it means that $AB = BA$, but from a physical standpoint yes it means that both observables can be measured simultaneously, and in that respect you kinda answered your own question. They're independent so they commute. For an analogy think of the spin of the electron in a hydrogen atom and its angular momentum about the CM, or think of the spins of two separate electrons. You can simultaneously measure these two quantities, why should they have anything to do with each other? Using the definition of commutativity it's a simple exercise to verify this: just operate both $J_1^2$ and $J_{2z}$ on the kets you've written and see what you get.

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  • $\begingroup$ Oh! I think I have had the "wrong" concept of what "commute" really is. By heisenberg's uncertainty relation if it does not commute it means it's dependent to each other, not being able to measure both exactly at once. Okays until here. Then in the case of J^2=(J_1)^2+(J_2)^2+2(J_1z)(J_2z)+(J+J-)+(J-J+), even if the equation J^2 contains (J_1)^2 in the equation and therfore related to eachother(or dependent), [J^2,(J_1)^2]=0? Is this because of other terms "off the diagnal" of the matrice? $\endgroup$
    – user50352
    Jun 12, 2014 at 6:08
  • $\begingroup$ Can you please use $\LaTeX$ formatting in your comments and posts? The last two terms in your equation are missing $1$ and $2$ subscripts. It seems like you're asking if $J^2$ and $J_1^2$ commute, and the answer is no because of the raising and lowering operators in the last two terms. I'm not sure what you mean regarding off-diagonal matrix elements. $\endgroup$
    – Jordan
    Jun 12, 2014 at 6:18
  • $\begingroup$ Sorry, I just joined here and not really used to those formatting..I should learn that right away.. Yes you've understood correctly, but in fact they DO commute. So I was asking how to understand this with regards to what I have just understood about the meaning "commute". I first thought that those two would not commute because the equation shows that they are related(dependent), but it commutes! I think I'm in some trap now.. $\endgroup$
    – user50352
    Jun 12, 2014 at 6:26
  • $\begingroup$ Right, that's my bad. I was confused there. To see if something commutes the most straightforward way is to apply the operators to a state and see what happens. The criterion of expressing one operator in terms of another is not a good way to see if the operators commute. The physical reason that $J_2^2$ and $J_{1z}$ commute is that these two quantities have nothing to do with each other, but you can still write an equation relating the two of them. $\endgroup$
    – Jordan
    Jun 12, 2014 at 6:56

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