Combining two 1D Hamiltonians: How to construct the new Hamiltonian properly? Lets say I have two 1D Hilbert spaces $\mathcal H_A, \mathcal H_B$, for example two 1D harmonic oscillators. Each space comes with an orthonormal basis $B_A=\{\phi^A_n \}, \ B_B=\{\phi^B_m \}, \ n,m\geq 0$ where each function is a eigenfunction of the respective 1D-Hamilton operator. Now I'd like to construct the combined system, without any coupling. My understanding is that we take the tensor product of both spaces to obtain the new space $\mathcal H_c$,
$$
\mathcal H_C = \mathcal H_A\otimes \mathcal H_B
$$
A basis for our new space is  $B_C = \{\phi_n^A\otimes\phi_m^B \}$ such that
$$\begin{aligned}
\hat H_C \phi^C_{nm} &= E_{nm}\phi^C_{nm} \\
(\hat H_A\otimes \hat 1 + \hat 1\otimes \hat H_B)(\phi_n^A\otimes\phi_m^B)  &= (E^A_n+E_m^B)\phi_n^A\otimes\phi_m^B \\
\end{aligned}$$
For everything to work out like this we need
$$
\hat H_C = \hat H_A\otimes \hat 1 + \hat 1\otimes \hat H_B
$$
But it is not a-priori clear to me that it should be so. Why is the operator not given by
$$
\hat H_C = \hat H_A\otimes \hat H_B  \quad ?
$$
When are operators in the new space of the form $$\hat O_A\otimes \hat 1+ \hat 1\otimes 
 \hat O_B $$
and when do operators take on the form
$$
\hat O_A\otimes \hat O_B.
$$
The parity operator for example is of this form. Is there a simple way to tell how operators are "transferred" to a product space ?
 A: A first answer in the case of the Hamiltonian is dimensional analysis : $H_A\otimes H_B$ has dimension of energy squared, so it is not a good candidate hamiltonian.
A more profound answer is that unitary operators extend using the tensor product, while hermitian operator extend using the sum rule (like the Hamiltonian).
For example, the time evolution operator $U(t) = e^{-i\hat Ht /\hbar}$ solves the Schrödinger equation. If you are given two solutions $|\psi_A(t)\rangle = U_A(t) |\psi_A(0)\rangle$ and $|\psi_B(t)\rangle= U_B(t) |\psi_B(0)\rangle$, you expect (since you are not introducing any coupling between the two subsystems, that $|\psi_A(t)\rangle \otimes |\psi_B(t)\rangle$ is a solution of the Schrödinger equation for the combined system.
That is :
$$U_{AB}(t) = U_A(t)\otimes U_B(t)$$
Since since $i\hbar\frac{d}{dt} U(t)|_{t=0} = H$, by taking a time derivative at $t=0$, you get  :
$$H_{AB} = \hat H_A \otimes \mathbb I_B + \mathbb I_A \otimes H_B$$
More generally, symmetry operators (eg translations, rotations, parity, etc.) will extend using the tensor product. For continuous symmetries, taking a derivative will mean that the generators (momentum, angular momentum, spin, etc.) will extend using the sum rule.
