The irreducible representations of of su(2): Highest weight method

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Summary:

I need to understand this passage from "An Elementary Introduction to Groups and Representations" Brian C. Hall

Main Question or Discussion Point

Hi all I need to understand the following passage from Hall link page 78 :

Some notation first:

Basis for ##sl(2;C)##:

##H=\begin{pmatrix} 1&0\\0&−1\end{pmatrix} ;X=\begin{pmatrix} 0&1\\0&0\end{pmatrix} ;Y=\begin{pmatrix} 0&0\\1&0\end{pmatrix} ##

which have the commutation relations

##[H,X] = 2X ~ ~, [H,Y] =−2Y ~ , [X,Y] =H ##


##π(X)## acts as the raising operator such that:

##π(H)π(X)u= (α+ 2)π(X)u ##

##π(Y)## acts as the lowering operator such that:

##π(H)π(Y)u= (α−2)π(Y)u ##

There is some N≥0 such that ##π(X)^Nu \neq 0##

but ##π(X)^{N+1}u= 0 ##

We define ##u_0=π(X)^Nu ## then

##(H)u_0=λu_0##

##π(X)u_0= 0 ##



Now, by definition

##u_{k+1}=π(Y)u_k##

Using ##π(H)u_k= (λ−2k)u_k## and induction we have


##π(X)u_{k+1}=π(X)π(Y)u_{k}
\\= (π(Y)π(X) +π(H))u_k
\\=π(Y) [kλ−k(k−1)]u_{k−1}+ (λ−2k)u_k
\\= [kλ−k(k−1) + (λ−2k)]u_k##


I don't understand how to get ##kλ−k(k−1)]## at the third passage and why ## (λ−2k)u_k## should be zero to get

##π(X)u_{k+1}= [kλ−k(k−1)]u_k##.
 
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  • #2
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I have difficulties to understand how the representation is finally defined. There seem to be some mistakes in what you have written. E.g. the lines where you explain the action of ##\pi(X)## is actually the description of ##\pi(H)##, which leaves ##\pi(X).u## invariant. And I do not see where ##k^2## comes from, and ##(\lambda - 2k)u_k## isn't zero, why should it be?

I hope I gave a better description of the theorem here:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/
 
  • #3
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please read page 77 of the pdf at the link....the full proof is laid out..... the theorem is clear and I don't see mistakes yet that passage to prove the lemma is obscure. Wikipedia refers to that book...https://en.wikipedia.org/wiki/Representation_theory_of_SU(2) ...
"Weights and the structure of the representation"
 
  • #4
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So the key is Lemma 5.11. I will omit the ##\pi## and write shortly ##\pi(X)(v)=X.v##.

As always, let us list what we have:
##u_0:=X^N.u## such that ##H.u_0=\lambda u_0=(2N+\alpha)u_0## and ##X.u_0=0\,.##
##u_k:=Y^k.u_0## and thus ##H.u_k=(\lambda-2k)u_k\,.##

The only equation which is used repeatedly is ##A.B.v=[A,B].v+B.A.v## where ##[A,B]=C## is known, so ##A.B.v=C.v+B.A.v## This is the general procedure. In Lemma 5.11. we want to know, what ##X.u_k## is.

##X.u_0 = 0## per definition of ##u_0\,.##
## X.u_1=X.Y^1.u_0= [X,Y].u_0+Y.X.u_0= H.u_0+Y.0=\lambda u_0 \,.##
Now per induction we have
\begin{align*}
X.u_{k+1}&=X.Y^{k+1}.u_0 =X.Y.(Y^k.u_0)=[X,Y].(Y^k.u_0) +Y.X.(Y^k.u_0)\\
&=H.Y^k.u_0+Y.(X.Y^k.u_0)=H.u_k+Y.(X.u_k)\\&=(\lambda-2k)u_k+[k\lambda-k(k-1)]Y.u_{k-1}\\
&=\left(\lambda - 2k + k \cdot \lambda - k^2 +k\right)u_k\\&=\left[(k+1)\lambda - k^2 - k\right]u_k\\
&=\left[(k+1)\lambda - (k+1)k\right]u_k
\end{align*}
 
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  • #5
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It was so obvious......Thank you for your help and I will read your insight.
 
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