Eigenstates/vectors of the sum of non-commuting Hamiltonians Suppose I have two Hamiltonians $H_1$ and $H_2$, and they're both two-level systems (they do not commute, such as pauli $X$ and $Z$). $H_1$ has eigenstates $|\psi_{11}\rangle$ and $|\psi_{12}\rangle$, with energy eigenvalues $E_{11}$ and $E_{12}$; $H_2$ has eigenstates $|\psi_{21}\rangle$ and $|\psi_{22}\rangle$, with energy eigenvalues $E_{21}$ and $E_{22}$. Now if I have a Hamiltonian:
$$
H=aH_1+bH_2
$$
How can I find the energy eigenvalues and the corresponding eigenstates? To me, I think one of the eigenvalues will be
$$
\frac{1}{\sqrt{a^2+b^2}}(E_{11}+E_{21})
$$
and the corresponding eigenstate is
$$
\frac{1}{\sqrt{a^2+b^2}}(|\psi_{11}\rangle+|\psi_{21}\rangle)
$$
Is that correct? What if the plus sign is replaced by cdot? Thanks for the help:)
 A: No, linear algebra doesn't work like this. Since you mentioned these non-commuting Pauli matrices, just consider
$$
H=3\sigma_1+4\sigma_3= \begin{pmatrix} 4&3\\3&-4\end{pmatrix},
$$
with unnormalized eigenvectors:  $(3,1)^T$ for eigenvalue 5; and $(1,-3)^T$ for eigenvalue -5.
Your conjecture is false.
A: Talking in general,
$$H^{(1)\otimes (2)}=a(H_1^{(1)}\otimes I^{(2)})+b(I^{(1)}\otimes H^{(2)}_2)$$
You can write eigenvalues equation
$$H^{(1)\otimes (2)}|E\rangle =E|E\rangle $$
$$(H^{(1)\otimes (2)}-E)|E\rangle =0$$
Writing
$$|E\rangle =\sum_{i,j}C_{ij}|E_{1i}\rangle \otimes |E_{2j}\rangle $$
and striking with $|E_{1m}\rangle\otimes |E_{2n}\rangle $ from the left
$$\left(\langle E_{1m}|\otimes  \langle  E_{2n}|\right)(H^{(1)\otimes (2)}-E)\sum_{i,j}C_{ij}|E_{1i}\rangle \otimes |E_{2j}\rangle=0$$
It looks quite complicated but it's not. The first term is just the matrix element of  Hamiltonian.

In the present case, the product space would be four-dimensional. So all you need to do is write the Hamiltonian in $4\times 4$ matrix and then solve the eigenvalue problem as usual.
Let's see one of the elements (I will suppress the notation):
$$\langle  E_{11},E_{21}|H|E_{11},E_{21}\rangle =\langle  E_{11},E_{21}|aH_1+bH_2|E_{11},E_{21}\rangle$$
$$=a\langle  E_{11},E_{21}|H_1|E_{11},E_{21}\rangle+b\langle  E_{11},E_{21}|H_2|E_{11},E_{21}\rangle=aE_1+bE_2$$
