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Addition of Angular Momenta

 
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Jan14-13, 02:50 PM   #1
 

Addition of Angular Momenta


Hey!

While I was reading some book in Quantum Mechanics, I ran across the following, and couldn't
know how can this be true or actually how was it assumed.

How by adding equation (7.91)and (7.92), we get (7.110), see attachment.
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Jan14-13, 03:21 PM   #2
 
Well isn't
[tex]
\vec{J}=\vec{J}_1+\vec{J}_2
[/tex]?
Then you can work component by component and obtain the result.
Jan14-13, 03:53 PM   #3
 
Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..
Jan14-13, 04:21 PM   #4
 

Addition of Angular Momenta


Quote by M. next View Post
Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..
But [itex]\vec{J}[/itex] may be defined in this way on the space defined as the direct sum of the spaces where 1 and 2 act, or not?
Jan14-13, 04:27 PM   #5
 
Please read carefully what's written in the attachment.
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Jan14-13, 05:32 PM   #6
 
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But J may be defined in this way on the space defined as the direct sum of the spaces where 1 and 2 act, or not?
They do act on different subspaces. But actually it's not the direct sum, it's the direct product. To be technical about it, J1 is really J1 ⊗ I, and J2 is really I ⊗ J2, and J = J1 + J2 = J1 ⊗ J2.

Now if you focus on two of the components, say x and y components, and look at their commutator,

[Jx, Jy] = [J1x, J1y] ⊗ [J2x, J2y] = i J1z ⊗ J2z = i Jz
Jan15-13, 02:37 AM   #7
 
Quote by Bill_K View Post
They do act on different subspaces. But actually it's not the direct sum, it's the direct product. To be technical about it, J1 is really J1 ⊗ I, and J2 is really I ⊗ J2, and J = J1 + J2 = J1 ⊗ J2.

Now if you focus on two of the components, say x and y components, and look at their commutator,

[Jx, Jy] = [J1x, J1y] ⊗ [J2x, J2y] = i J1z ⊗ J2z = i Jz
Yes, the direct product, I messed up my operations.
Jan15-13, 09:48 AM   #8
 
Thanks! This was helpful.
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