by M. next
 P: 378 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. Attached Thumbnails
 P: 123 Well isn't $$\vec{J}=\vec{J}_1+\vec{J}_2$$? Then you can work component by component and obtain the result.
 P: 378 Yes, but this is not the 'real' addition, each of the operators you've listed belong to different spaces..
P: 123

 Quote by M. next 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?
 P: 378 Please read carefully what's written in the attachment. Attached Thumbnails
Thanks
P: 4,160
 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
P: 123
 Quote by Bill_K 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.
 P: 378 Thanks! This was helpful.

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