Spin operators on two spin-$\frac{1}{2}$ particle system In a text I am using (Introduction to Quantum Mechanics by Griffiths), this is an introductory text hence much of rigor is omitted. The following is stated:
Suppose that we have two spin-$\frac{1}{2}$ particles- for example the proton and electron in the ground state of hydrogen. Each can have spin up or spin down, so there are  four possibilities in all. Let $$\hat{S} := \hat{S}^{(1)} + \hat{S}^{(2)}.$$ Considering the eigenstates of $\hat{S}_z$ we have $$\hat{S}_z = (\hat{S_z}^{(1)} + \hat{S_z}^{(2)})\chi_1 \chi_2 = (\hat{S_z}^{(1)} \chi_1) \chi_2 + \chi_1 (\hat{S_{z}}^{(2)} \chi_2) = (\hbar m_1 \chi_1) \chi_2 + \chi_1 (\hbar m_2 \chi_2) = \hbar (m_1 + m_2) \chi_1 \chi_2$$
Question:
What is the mathematically rigorous (or more complete) justification  for why $\hat{S}^{(1)}$  acts only on $\chi_1$ and $\hat{S}^{(2)}$ acts only $\chi_2$? Also why do we define the operator as $\hat{S} := \hat{S}^{(1)} + \hat{S}^{(2)}?$ I can see it is natural to consider this operator but I am interested in if there is a more general motivation for a definition of this kind?
Thanks for any assistance.
 A: Given two vector spaces $V_1$ and $V_2$, we can define their tensor product $V_1 \otimes V_2$, which is a new vector space. Product states in this vector space can be written
$$ v_1 \otimes v_2 $$
where $v_1 \in V_1$ and $v_2 \in V_2$. Not every element of $V_1 \otimes V_2$ can be written like this as a separable product state, but you can form a basis of $V_1 \otimes V_2$ using the product states, and then the most general state is some linear combination of those product states. (This is the idea of entanglement: an entangled state is one that cannot be written as a product state.)
Furthermore, given two linear operators $T_1: V_1 \rightarrow W_1$ and $T_2: V_2 \rightarrow W_2$, we can define their tensor product as a new operator
$$T_1 \otimes T_2: V_1 \otimes V_2 \rightarrow W_1 \otimes W_2 $$
by
$$ T_1 \otimes T_2 (v_1 \otimes v_2) = T_1(v_1) \otimes T_2(v_2). $$
This definition of $T_1 \otimes T_2$ can be extended for non-product states by linearity (remember that $V_1 \otimes V_2$ has a basis of product states).
Beware that the symbol $\otimes$ is being used in three distinct ways! It can denote a tensor product of vector spaces, of vectors, or of operators. Physicists often omit the $\otimes$ notation entirely and just use juxtaposition, but hopefully including the tensor products explicitly will answer your questions.
Also note that mathematicians have more sophisticated definitions of tensor products, but for the our purposes this should be sufficient.
Now, if we have two electrons, we have two spin spaces. We consider their tensor product, whose basis states might look like
$$ \left| \chi_1 \right> \otimes \left| \chi_2 \right> $$
where $\left| \chi_1 \right>$ is the state of one electron and $\left| \chi_2 \right>$ the state of the other electron. Then we define the spin operator on this space by
$$ S = S_1 \otimes I + I \otimes S_2 $$
where $I$ is the identity operator. Using the definition of the tensor product of operators, it follows that
$$ S \left( \left| \chi_1 \right> \otimes \left| \chi_2 \right> \right) = S_1 \left| \chi_1 \right> \otimes \left| \chi_2 \right> + \left| \chi_1 \right> \otimes S_2 \left| \chi_2 \right>. $$
With this notation it should be clear why $S_1$ only acts on $\left| \chi_1 \right>$ and $S_2$ only acts on $\left| \chi_2 \right>$. For example, if $\left| \chi_1 \right> = \left| \uparrow \right>$ and $\left| \chi_2 \right> = \left| \uparrow \right>$, then
\begin{align}S_z \left( \left| \uparrow \right> \otimes \left| \uparrow \right> \right) &= S_{1z} \left| \uparrow \right> \otimes \left| \uparrow \right> + \left| \uparrow \right> \otimes S_{2z} \left| \uparrow \right> \\
&= \frac{\hbar}{2} \left| \uparrow \right> \otimes \left| \uparrow \right> + \left| \uparrow \right> \otimes \frac{\hbar}{2} \left| \uparrow \right> \\
&= \hbar \left( \left| \uparrow \right> \otimes \left| \uparrow \right> \right).
\end{align}
We see that our definition of $S$ seems to be sensible.
