Addition of Angular Momenta


by M. next
Tags: addition, angular, momenta
M. next
M. next is offline
#1
Jan14-13, 02:50 PM
P: 354
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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kevinferreira
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#2
Jan14-13, 03:21 PM
P: 123
Well isn't
[tex]
\vec{J}=\vec{J}_1+\vec{J}_2
[/tex]?
Then you can work component by component and obtain the result.
M. next
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#3
Jan14-13, 03:53 PM
P: 354
Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..

kevinferreira
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#4
Jan14-13, 04:21 PM
P: 123

Addition of Angular Momenta


Quote 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?
M. next
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#5
Jan14-13, 04:27 PM
P: 354
Please read carefully what's written in the attachment.
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Bill_K
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#6
Jan14-13, 05:32 PM
Sci Advisor
Thanks
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P: 3,861
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
kevinferreira
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#7
Jan15-13, 02:37 AM
P: 123
Quote 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.
M. next
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#8
Jan15-13, 09:48 AM
P: 354
Thanks! This was helpful.


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