The algebraic method to find the irreducible representation of the $SU(2)$ group makes use of the operators: $$J_z\\J_+=\frac{1}{\sqrt{2}}(J_x+iJ_y)\\J_-=\frac{1}{\sqrt{2}}(J_x-iJ_y)$$

In the book Georgi (Lie Algebra in particle physics, 1999) this set of operators is said to be a set of generators of a $SU(2)$ subalgebra (chapter 6.5, pag 93).

So, if this were a set of generators, I would say that those operators satisfy a this algebra:

$$[J_z,J_\pm]=\pm J_{\pm}\\ [J_+,J_-]= J_z$$

which is clearly different from the $SU(2)$ algebra: $$[J_a,J_b]=i \varepsilon_{abc}J_c $$

So is it correct to say that operators $J_z,J_+,J_-$ are NOT a set of generators of the $SU(2)$ group?

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    $\begingroup$ Notice that the ladder operators aren't hermitian, like the generators. However, sometime it is still said that they are elements of the lie algebra, because they are in the same (complexified) tangent space. $\endgroup$ – lionelbrits Feb 26 '15 at 16:50
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    $\begingroup$ Yes I noticed that. But Georgi says "$J_{\pm}, J_3$ are generators of SU(2) subalgebra" at page 93. I do not know anything about the tangent space of an operator space... $\endgroup$ – Caos Feb 26 '15 at 16:53
  • $\begingroup$ It is the tangent space to the identity of the Lie group (which is also a manifold), i.e. the Lie algebra. $\endgroup$ – Phoenix87 Feb 26 '15 at 17:01
  • $\begingroup$ You can obviously generate any element of SU(2) by exponentiating a linear combination of $J_z,J\pm$, since this is also a linear combination of $J_z,J_x,J_y$. So in this sense these operators do generate the group, although they are not a conventional choice. $\endgroup$ – Mark Mitchison Feb 26 '15 at 17:06
  • $\begingroup$ What is the difference between real and complex Lie Algebra? $\endgroup$ – Saurabh U. Shringarpure Aug 14 at 10:51

I think I have found a rigorous answer (http://en.wikipedia.org/wiki/Representation_theory_of_SU%282%29).

This is the complexified algebra of SU(2) $$[J_z,J_\pm]=\pm J_{\pm}\\ [J_+,J_-]= J_z$$

The complex Lie algebra (i.e. the complexification of the Lie algebra) doesn't affect the representation theory. So both the real $[J_a,J_b]=i\varepsilon_{abc}J_c$ and the complexified algebra (above) originates the same representation of the SU(2) group.

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    $\begingroup$ That is indeed the case. The complexification of $su(2)$ is $sl(2,\mathbb{C})$. A very nice and understandable account of the finite-dimensional irreps of semisimple Lie-Algebras and -groups can be found in "Brian C. Hall - Lie Groups, Lie Algebras and Representations" $\endgroup$ – Nephente Feb 26 '15 at 18:31

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