6
$\begingroup$

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 H. Georgi's book Georgi Lie Algebras in Particle Physics, 1999, this set of operators is said to be a set of generators of a $SU(2)$ subalgebra (Chap. 6.5, p. 93).

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

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ 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
    Commented Feb 26, 2015 at 16:50
  • 1
    $\begingroup$ 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... $\endgroup$
    – Chaos
    Commented Feb 26, 2015 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
    Commented Feb 26, 2015 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
    Commented Feb 26, 2015 at 17:06
  • $\begingroup$ What is the difference between real and complex Lie Algebra? $\endgroup$
    – Saurabh Shringarpure
    Commented Aug 14, 2020 at 10:51

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ 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
    Commented Feb 26, 2015 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.