2
$\begingroup$

This question is continuation to the previous post. The lie algebra of $ \mathfrak{so(3)} $ is real Lie-algebra and hence, $ L_{\pm} = L_1 \pm i L_2 $ don't belong to $ \mathfrak{so(3)} $.

However, when constructing a representation for $\mathfrak{so(3)} $, one uses these operators and take them to be endomorphisms (operators) defined on some vector space $V$. Let $\left|lm \right> \in V $,then

$$ L_3\left|lm \right> = m \left|lm \right> \;\;\;\;\; L_{\pm}\left|lm \right> = C_{\pm}\left|l(m\pm1) \right> $$

Now, how do we justify these two things ? If $L_{\pm} \notin \mathfrak{so(3)}$, then how is this kind of a construction of the representation possible ?

I belive similar is the case with $\mathfrak{su(n)}$ algebras, where the group is semi simple and algebra is defined over a real LVS.

Improve this question
$\endgroup$
10
  • $\begingroup$ I might be misunderstanding something here, so let me raise a point: Without judging if the operators do or do not lie in the algebra, why does your question arise anyway? In my ear, it sounds similar to "I want to study the properties of consecutive derivatives and people use abstract algebra to do it. How is that justified?" Why not? If you study how $a\mapsto\mathrm{e}^{i\phi}a$ affects elements of $\mathbb C$, is there a reason you would you restrict your study by demanding not to use complex conjugation on $\mathbb C$? $\endgroup$
    – Nikolaj-K
    Apr 4, 2014 at 8:38
  • $\begingroup$ Sorry, I do not understand why $L_\pm |l \:m \rangle = C_\pm |l\: (m\pm 1)\rangle$ should require that $L_\pm$ belongs to a representation of the (real) Lie algebra of $so(3)$ or $su(2)$. $\endgroup$
    – Valter Moretti
    Apr 4, 2014 at 8:39
  • $\begingroup$ @V.Moretti : Neither do I ! But I am not able to convince myself that if they don't belong this, then how can I use them in construction of the representation ?? $\endgroup$
    – user35952
    Apr 4, 2014 at 8:43
  • $\begingroup$ @NiftyKitty95 : Thanks for that point, although your analogy has not gotten on me yet. Will ponder again with this. $\endgroup$
    – user35952
    Apr 4, 2014 at 8:45
  • $\begingroup$ Ok, so does it mean that when I construct a representation of this algebra using its operation on a Linear Vector space(LVS), only few legitimate operators on this LVS belong to algebra and not all the operators defined over the LVS ? $\endgroup$
    – user35952
    Apr 4, 2014 at 8:50

1 Answer 1

Reset to default
3
$\begingroup$

They do not lie in $\mathfrak{so}(3)$ but they lie in its complexification, which would be $A_1$ in the usual mathematical classification. Much of Lie representation theory is set up this way: you work at the level of the complexification then go back to the real form. For compact groups it's not a big deal; for non-compact groups extra care is needed.

So while $L_{\pm}$ do not make sense as elements of $\mathfrak{so}(3)$, they make sense in the complexification. You can revert back to $\mathfrak{so}(3)$ by using $L_1=\frac{1}{2}(L_++L_-)$ and $L_2=\frac{1}{2i}(L_+-L_-)$. (Be careful: the basis where $L_0$ is diagonal is a complex combination of the real basis vectors.)

Improve this answer
$\endgroup$

Your Answer

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

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