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While introducing Clebsch-Gordan coefficients, they state that the operators: $$ \vec{J_1}^2,\vec{J_2}^2,J_{1z},J_{2z}$$ form a complete set of compatible observables. Which means that there is no degeneracy in their common eigenspaces.

What I wonder is, how it follows that (if the above operators form indeed a complete set), that also $$\vec{J}^2,\vec{J_1}^2,\vec{J_2}^2,J_z$$ with $$\vec{J} = \vec{J_1}+\vec{J_2}$$ forms a complete set. Which is what they claim in the textbook. I see that the operators in both sets are compatible (i.e. they commute) and that $J_{1z}$ and $J_{2z}$ are not compatible with the second set. Does it follow from this information that the second set of observables is complete, if the first is?

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It does, yes. The reason for this is central to the mathematical theory which undergirds quantum mechanics.

We are used to thinking about vectors as little arrows with lengths and directions, but that is not complete nor is it always useful. A particle's wave function (and indeed its spin) are also vectors but in a more general way. The defining characteristic of vectors that we should hold on to is that they "behave the same" in every reference frame (i.e. if you rotate or translate a collection of vectors, the resulting vectors will be expressed differently, but their relative lengths and directions will be the same).

The same basic principle applies in quantum mechanics. The state of a system can be described in "momentum space" or "position space" or "energy space" but the state itself does not depend on how it is expressed. You may be familiar with the "identity operator" $\Bbb I=\sum_n |\psi_n\rangle\langle\psi_n|$, where $|\psi_n\rangle$ can be any orthonormal basis you like. This is called the identity operator because by applying it to some state all you accomplish is expanding that state in the $\psi$-basis.

The same idea works for spin. You have some state of the system described by the individual total and $z$-component angular momenta of two particles ($|l_1,l_2,m_1,m_2\rangle$), and you are going to express it in another basis ($|l_{tot},l_1,l_2,m_{tot}\rangle$). Your question about whether the basis is complete could be rephrased this way "Is all of the information I had about the system in my first basis still available in my second basis?"

To show that this is the case, you only need to show that every operator of the new basis commutes with every other operator in the new basis and that the total number of operators in the new basis is the same number as before (i.e. that the state is specified by the same number of quantum numbers). In this case, both of these conditions are trivial to check. (This is the same basic procedure as when you change coordinate bases in a regular vector space: you have to make sure that the dimension of the new basis is the same as the dimension of the old basis.) Another way to check that the new basis works is to show that there does not exist any other operator that commutes with every operator of the basis. While that's possible, it is also tedious.

By way of pointing out that this analogy to traditional vector spaces is actually exact in this case (it isn't actually an analogy), realize two things:

First, the spin-state space is a finite-dimensional vector space since spin is quantized (i.e. spin states can be exactly represented as column vectors).

Second, the Clebsch-Gordan Coefficients are exactly those numbers which characterize the following transformation: $$ |l_{tot},m_{tot},l_1,l_2\rangle=\sum_{m_1, m_2}C \ |l_1,m_1,l_2,m_2\rangle $$ And by recalling the identity operator that I mentioned above, we can deduce that the Clebsch-Gordan Coefficients must be given by $$ C=\langle l_1,m_1,l_2,m_2|l_{tot},m_{tot},l_1,l_2\rangle $$

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Thank you, that was really helpful! –  Rayman Dec 22 '13 at 20:04
"you only need to show that every operator of the new basis commutes with every other operator in the new basis and that the total number of operators in the new basis is the same number as before" - I feel like something is missing. At the very least, you have to make sure your operators aren't secretly the same. But also what about the two spin-3/2 particles case: "$4 \otimes 4 = 7 \oplus 5 \oplus 3 \oplus 1$"? It feels like I went from 4 operators to 7 or 8. Or maybe I'm misunderstanding? –  Chris White Dec 23 '13 at 7:41
I guess that's a good point, I implicitly assumed that all of the operators would be different. To be fair though, if two of them are secretly the same, they will have the same eigenvalues (which should be a pretty big give-away since that will screw up the matrix representation). In your example you are confusing an important distinction. In either representation, a pair of spin-3/2 particles is described by 4 operators and 16 basis vectors; however, in a "direct product space" the particles are mathematically uncoupled (and thus described by individual operators). (cont.) –  Geoffrey Dec 23 '13 at 16:28
On the other hand, in a "direct sum space" the particles are intrinsically coupled and can no longer be treated as individual particles (each with their own pair of spin operators). The quasi-particles that are described by the odd-dimensional spaces in the coupled (direct sum) representation are just that: quasi-particles. They are merely blocks (7x7, 5x5, etc.) from the block-diagonalized angular momentum operator matrix, but they do not have their own, separate set of spin operators because their individual spin makes no sense outside of the context of the spin of the other quasi-particles. –  Geoffrey Dec 23 '13 at 16:30
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