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A vector under rotation has the following property: [^Ji,^Vj]=iϵijk^Vk. Furthermore the projection lemma is defined as the following:

k;j,mj|V|k;j,mj=k;j,mj|(VJ)J|k;j,mj2j(j+1).

On page 24 of this note it is claimed that k;j,mj|Sz|k;j,mj=mjk;j,mj|(SJ)|k;j,mj2j(j+1).

I have been starring at this for hours, and it is not obvious where the m is coming from, given that it is Jz|j,mj=mj|j,mj.

Secondly, shouldn't (SJ)=SzJz, since we are only analyzing the Sz part and not the full S vector? I feel I may be misinterpreting notation, so any help is appreciated.

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    ˆS=(ˆSˆJ)ˆJˆJ2 so ^Sz=(ˆSˆJ)^JzˆJ2 Commented Jun 13 at 15:54

2 Answers 2

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Indeed, you are misreading the notation involved. (SJ)=SxJx+SyJy+SzJz, and not just the last term, as you are imagining—how could you? These operators are well-defined and non-vanishing.

Plugging into (2.5.15), then, collapses to k;j,mj|ˆSz|k;j,mj=k;j,mj|(ˆSˆJ)ˆJz|k;j,mj2j(j+1),

amounting to (2.5.17).

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k;j,mj|V|k;j,mj=k;j,mj|(VJ)J|k;j,mj2j(j+1).

On page 24 of this note it is claimed that k;j,mj|Sz|k;j,mj=mjk;j,mj|(SJ)|k;j,mj2j(j+1).

I have been starring at this for hours, and it is not obvious where the m is coming from, given that it is Jz|j,mj=mj|j,mj.

Your Eq. (2.5.15) holds for each cartesian component individually. So, k;j,mj|Vx|k;j,mj=k;j,mj|(VJ)Jx|k;j,mj2j(j+1)

and k;j,mj|Vy|k;j,mj=k;j,mj|(VJ)Jy|k;j,mj2j(j+1)
and k;j,mj|Vz|k;j,mj=k;j,mj|(VJ)Jz|k;j,mj2j(j+1)
=mjk;j,mj|(VJ)|k;j,mj2j(j+1),
where, yes, you are right, you use Jz|k;j,mj=mj|k;j,mj.

If you take Vz=Sz in Eq. (B) above you get k;j,mj|Sz|k;j,mj=mjk;j,mj|(SJ)|k;j,mj2j(j+1)

Secondly, shouldn't (SJ)=SzJz,

No. The (SJ) is the usual dot product, nothing more.

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