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Where do $L_+$ and $L_-$ live, if not in $\mathfrak{so(3)}$?

+ 2 like - 0 dislike
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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.


This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952

asked Apr 4, 2014 in Mathematics by user35952 (155 points) [ revision history ]
recategorized Apr 13, 2014 by dimension10
Most voted comments show all comments
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 $\mathbb C$?

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user NiftyKitty95
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 $su(2)$.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user V. Moretti
@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 ??

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952
@NiftyKitty95 : Thanks for that point, although your analogy has not gotten on me yet. Will ponder again with this.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952
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 ?

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952
Most recent comments show all comments
@V.Moretti : Also, is the invariant subspace of $V$, the space over which the Casimir Operators of the Lie algebra is defined ?

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952
If $V$ is irreducible in addition to be invariant, it is an eigenspace of the Casimir operators, indeed.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user V. Moretti

1 Answer

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$L^\pm$ live in the complexification of the Lie algebra $so(3)$, which is $sl(2)$. The complexification comes with a natural involution $*$ which takes the conjugate of all coefficients of a representation in a basis of the (real) $so(3)$. Thus the $so(3)$ is recovered as the Lie subalgebra of all Hermitian opersators in the complexification. 

The same works in general. indeed, the classification of semisimple Lie algebras is generally done first for the complexified version (where one can introduce raising and lowering operators), and after having the complex Lie algebras one finds the real ones by looking for the possible involutions. 

answered Oct 24, 2016 by Arnold Neumaier (12,355 points) [ no revision ]

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