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When hermitian operators $L_1, L_2, L_3$ follow the commutation relations:

$$ [L_1,L_2]=i\;L_3 \\ [L_2,L_3]=i\;L_1 \\ [L_3,L_1]=i\;L_2 $$

one can show that, assuming they are in finite number, their eigenvalues are integer or half-integer. It turns out that the basis of the Lie algebra of $SU(2)$ follows these relations, and after identifying the angular momentum to these operators, it constitutes a proof of the quantization of angular momentum.

Now, my mathematical knowledge does not go much further, but the basis of the Lie algebra of $SO(3)$ follows the same commutation relations, with the exception of the $i$ coefficient. In that case, I cannot find any way to obtain quantized eigenvalues. How come I read here and there that $SO(3)$ and $SU(2)$ have isomorphic Lie algebras, and that they basically lead to the same quantization? If so, how do I derive that quantization using only $SO(3)$ operators? And if the result is the same, why would we bother using $SU(2)$ when we can use real 3x3 rotation operators?

Qmechanic
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fffred
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    SU(2) isn't isomorphic to SO(3), it's a double covering; however their Lie algebras are. – Mozibur Ullah Aug 05 '17 at 15:18
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    possible duplicate: Lie Algebra Conventions: Hermitian vs. anti-Hermitian. – AccidentalFourierTransform Aug 05 '17 at 15:18
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    The Lie algebras $so(3)\cong su(2)$ are isomorphic; the Lie groups $SO(3)$ and $SU(2)$ are not isomorphic. – Qmechanic Aug 05 '17 at 15:21
  • I corrected the question according to your comments. – fffred Aug 05 '17 at 15:23
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    Possible duplicates: http://physics.stackexchange.com/q/96045/2451 , http://physics.stackexchange.com/q/96569/2451 , http://physics.stackexchange.com/q/96542/2451 , http://physics.stackexchange.com/q/47740/2451 , http://physics.stackexchange.com/q/78536/2451 , http://physics.stackexchange.com/q/129340/2451 , and links therein. – Qmechanic Aug 05 '17 at 15:25
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    The "i" is irrelevant and purely conventional to guarantee that the generators are hermitian. – ZeroTheHero Aug 05 '17 at 15:26
  • As much important as the algebra itself is the algebra representation. Once you pick an algebra and a particular representation, you fix the associated Lie group. For example, if you choose the doublet of the algebra $\mathfrak{su}(2)$ you end up with the group $SU(2)$ whereas by choosing the triplet you get $SO(3)$. The former symmetry group can give spin transformations and the later can give orbital angular momentum transformations. – Diracology Aug 05 '17 at 15:33
  • I have read in several comments and links therein, that the "i" factor is irrelevant. This is precisely my question. How is it possible to obtain quantization without this factor? The "ladder" proof does not seem to hold. In fact, I am still able to do the ladder proof, but eigenvalues become imaginary. – fffred Aug 05 '17 at 16:07

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su(2) and so(3) Lie algebra are homomorpic, so if you redefine L by a factor "-$i$" then you get the so(3). But the group SU(2) is cover group to SO(3). its odd dimension representation matrices correspondences to all SO(3) representation matrices, which means SO(3) has no even dimension representations. So for paricles with integer angular momentun of course you can use SO(3), but for half integer spins, Like spin-1/2 fermions you have to use SU(2) representation. Altogether, using SU(2) is valid for all cases while SO(3) not. So this may provide you some idea?

Kangle
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