How can we add Angular Momenta in Quantum Mechanics?

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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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Well isn't
<br /> \vec{J}=\vec{J}_1+\vec{J}_2<br />?
Then you can work component by component and obtain the result.
 
Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..
 
M. next said:
Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..

But \vec{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?
 
Please read carefully what's written in the attachment.
 

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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
 
Bill_K said:
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.
 
Thanks! This was helpful.
 

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