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If we have the raising and lowering operators for separate particles $J_1^\pm$ and $J_2^\pm$, we may define the total angular momentum operator $J$ and the corresponding raising and lowering operators as $J^\pm = J_1^\pm + J_1^\pm$. The intuition is difficult for me here since the left side lowers the z component of spin by one unit while the right side seems to do so by two units.

Now, say this acts on the two spin half particles in the state $\vert{\uparrow\uparrow}\rangle$. To preserve the norm, I need to have a factor of $\frac{1}{\sqrt 2}$ somewhere since I should get the state $\frac{1}{\sqrt 2} (\vert{\uparrow\downarrow}\rangle + \vert{\downarrow\uparrow}\rangle)$.

Yet the definition of any addition of angular momentum operator simple adds the two corresponding operators without normalization e.g. $J_z = J^1_z + J^2_z$. What gives?

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    $\begingroup$ Raising and lowering operators don't preserve normalization. $\endgroup$ – Chris Jan 21 '18 at 9:34
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$J_{\rm tot}^-$ only lowers the z component by 1 unit, not 2. In $J_{tot} =J_1+J_2$ the $J_1$ acts only on the $j_1,m_1$ entries in $|J_1,m_1,j_2,m_2\rangle$ and the $J_2$ only on the $j_2,m_2$.
Thus

It's better to write (as a mathematician would) $$ J_{\rm tot}= J_1\otimes {\mathbb I}+ {\mathbb I}\otimes J_2 $$ and $$ |J_1,m_1,j_2,m_2\rangle= |j_1,m_1\rangle\otimes |j_2,m_2\rangle $$ that makes what's going on more obvious. Each operator only acts on its own Hilbert space, and the total Hilbert space is the tensor product of the two spaces for the two separate spins. See section 6.3.4 in https://courses.physics.illinois.edu/phys509/sp2018/bmaster.pdf

The "missing" $\sqrt 2$ is correct as $|\uparrow\uparrow\rangle$ is the $|j=1,m=1,\rangle$ state in a spin-one multiplet where $J^-|j=1,m=1\rangle = \sqrt 2|j=1,m=0\rangle$.

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