System of two spin-1/2 particles I am trying to understand a system involving two particles both with spin-1/2. The particles are In an electric field and hence we can write a Hamiltonian in the following form 
$\hat{H_0}=\frac{g\mu B}{\hbar}(\hat{S_{1z}}+\hat{S_{2z}}) \tag{1}$
eigenvalues of $\hat{S_{1z}}$ and $\hat{S_{2z}}$ are $\hbar m_1$ and $\hbar m_2$ respectively I've seen similar systems to mine elsewhere however and this may be a small query, but none the less a query I don't have the answer to. I keep seeing mentions of the "basis of states" and it appears to be labeled $|m_1m_2\rangle$. Can someone explain why this is true? if this is true how can I show the $|m_1m_2\rangle$ states are eigenstates of the Hamiltonian $\hat{H_0}$? and if they are in fact eigenstates how do I then use this information to find the energy eigenvalues?
Am I correct in thinking these eigenvalues can be written as $E^{0}_{m_1m_2}$?
 A: The notation $\vert m_1 m_2\rangle$ is a shorthand for $\vert m_1\rangle\vert m_2\rangle$, where $\vert m_1\rangle$ is a ket for the first spin alone, and $\vert m_2\rangle$ is a ket for the second spin alone.
Now the states of your first system are $\vert m_1\rangle$, and the states of your second systems are $\vert m_2\rangle$.  For every fixed state $\vert \bar{m}_1\rangle$ your second system can be in any state $\vert m_2\rangle$.  Thus, the possible states of your combined system are of the form $\vert m_1\rangle\vert m_2\rangle$.
Note that, in your Hamiltonian, $S_{1z}$ acts only on $\vert m_1\rangle$ and $S_{2z}$ only on $\vert m_2\rangle$, and $\vert m_1\rangle$ and $\vert m_2\rangle$ are assumed to be eigenkets of $S_{1z}$ and $S_{2z}$ respectively.
Thus:
$$
S_{1z}\vert m_1m_2\rangle=\Bigl(S_{1z}\vert m_1\rangle\Bigr)\vert m_2\rangle = \hbar m_1 \vert m_1\rangle\vert m_2\rangle=\hbar m_1
\vert m_1m_2\rangle\, ,
$$
and simlarly $S_{2z}\vert m_1 m_2\rangle=\hbar m_2\vert m_1m_2\rangle$ so
that
$$
\left(S_{1z}+S_{2z}\right)\vert m_1m_2\rangle= 
S_{1z}\vert m_1m_2\rangle + S_{2z}\vert m_1m_2\rangle=
\hbar(m_1+m_2)\vert m_1m_2\rangle
$$
You can appreciate how the product form of basis states arises by analogy with the usual method of separation of variables: if you have a partial differential equation of the form - say - 
$$
\left(\frac{\partial}{\partial x}+2\frac{\partial}{\partial y}\right)F(x,y)=F(x,y),
$$
you would naturally seek solutions in the product form $F(x,y)=X(x)Y(y)$, so that $\partial/\partial x$ acts only on $X(x)$ and $\partial/\partial y$ acts only on $Y(y)$.  The solution you can find using $F(x,y)=e^{\lambda_1 x}e^{\lambda_2 y}$.  Then
\begin{align}
\frac{\partial}{\partial x}e^{\lambda_1 x}=\lambda_1e^{\lambda_1 x} ,\\
\frac{\partial}{\partial y}e^{\lambda_2 y}=\lambda_2e^{\lambda_2 y}
\end{align}
so that 
$$
\left(\frac{\partial}{\partial x}+2\frac{\partial}{\partial y}\right)
X(x)Y(y)=(\lambda_1+2\lambda_2)X(x)Y(y)= X(x)Y(y)
$$
from which you see that the sum $\lambda_1+2\lambda_2$ should be $1$. As you can see, although the solution is a product of functions, the eigenvalue is the sum of individual eigenvalues.
In your system, the Hamiltonian is a sum of Hamitonians $S_{1z}$ and $S_{2z}$, each acting on its part of the state, and the total eigenvalue is the sum of the individual $\hbar m_1$ and $\hbar m_2$ eigenvalues.
A: You find the average energy of a state via:
$\langle m_1m_2|H|m_1m_2\rangle$, 
which for an eigenstate simplifies to:
$H|m_1m_2\rangle=E|m_1 m_2\rangle$.
Per @ZeroTheHero, you know how $H$ acts on $|m_1m_2\rangle$; there are only 4 states for $||m_i|| = 1/2$. Write each state down and apply the Hamiltonian operator to them.
