Stenger writes "Any vector operator J whose components above the angular momentum commutation rules [Jx,Jy] = iħJz will have the following eigenvalue equations: J^2|j,m> = j(j+1)ħ^2|j,m> where J^2 = Jx^2 + Jy^2 + Jz^2."
Landau and Lifshitz volume 3 had a demonstration of this that I could follow.
For the purposes of this demonstration, I drop the factor ħ to keep the paper clean. We can stick it back in at the end easily enough.
The solution begins with the introduction of two operators:
J+ = Jx + iJy
J- = Jx - iJy
Now consider the operator JzJ+
JzJ+
= Jz( Jx + iJy )
= JzJx + iJzJy
= (JxJz + iJy) + i(JyJz - iJx)
= JxJz + i JyJz + Jx + i Jy
= (Jx + iJy)( Jz + 1 )
= J+ (Jz + 1)
Now apply this operator to an eigenstate of Jz
JzJ+|m>
= J+ (Jz+1) |m>
= J+ ( Jz|m> + |m> )
= J+ ( m|m> + |m> )
= J+ ( m+1 |m> )
= (m+1) J+|m>
In short, J+|m> is an eigenstate of Jz, and is proportional to |m+1>. In other words, J+ is akin to a raising operator. A similar argument determines that J- is a lowering operator.
Now, the next bit is to consider J-J+. Careful! Jx and Jy don't commute.
J-J+
= (Jx - iJy) (Jx + iJy )
= Jx^2 + i JxJy - i JyJx + Jy^2
= Jx^2 + i (JxJy-JyJx) + Jy^2
= Jx^2 + i (iJz) + Jy^2
= Jx^2 - Jz + Jy^2
= Jx^2 + Jy^2 + Jz^2 - Jz^2 - Jz
= J^2 - Jz^2 - Jz
One last twist. Consider j = the largest m such that Jz|m> = m|m>. I didn't find the explanation in L&L complete satisfactory, but let it go for now. This implies that J+|j> = 0, and it follows that J-J+|j> is also 0. Therefore
J^2|j>
= Jz^2|j> + Jz|j>
= j^2|j> + j|j>
= j(j+1)|j>
A final note - what about ħ? The J^2 and Jz^2 terms both produce ħ^2, where Jz only gives us ħ. So where is the missing factor? We hid it when commuting - [Jx,Jy] = iħJz, so that factor properly sits in front of Jz in our final equation, and the units for all of the terms agree.
March 14, 2004 6:30 PM
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