Coupled and decoupled matrix representations of spin-spin interactions I'm very confused about how to conceptually think about generating matrices and changing between coupled and uncoupled basis.
The question I'm looking at in particular involves two electrons in a molecular environment ($H_{1}$) and with a magnetic field applied ($H_{2})$ and a Hamiltonian given by $$H=\frac{D}{\hbar^{2}}\overrightarrow{s_{1}}\cdot\overrightarrow{s_{2}}+\frac{\mu_{B}B}{\hbar}(g_{1}s_{1z}+g_{2}s_{2z})=H_{1}+H_{2}$$ where the uncoupled basis is $|m_{s1}m_{s2}\rangle$ and the coupled is $|SM_{S}\rangle$; $D$ is a constant.
I've applied $H_1$ to the coupled vector and got
 $$\begin{bmatrix}
 \frac{-3D}{4}& 0 & 0 &0 \\ 
 0& \frac{D}{4} & 0 &0 \\ 
 0& 0 & \frac{D}{4}  &0 \\ 
0 &0  &0  &  \frac{D}{4}
\end{bmatrix}.$$
I then applied $H_2$ to the uncoupled vector and got
$$\begin{bmatrix}
 \frac{g_{1}+g_{2}}{2} & 0 & 0 &0 \\ 
 0& \frac{g_{1}-g_{2}}{2} & 0 &0 \\ 
 0& 0 & \frac{-g_{1}+g_{2}}{2}  &0 \\ 
0 &0  &0  &  \frac{-g_{1}-g_{2}}{2}
\end{bmatrix}.$$
So, I guess, my result is that $H_{1}$ only operates on the coupled basis and $H_{2}$ only on the uncoupled?  I'm still very confused on how this makes sense physically, and how to conceptually think about this.  I'd love any feedback on how to think about this, or what I'm doing wrong.
 A: Both hamiltonian pieces operate on both mutually equivalent bases--it's just that their action on each is different, as in "non diagonal". 
Given both diagonal pieces in some basis, however, you may easily reconstruct the non-diagonal pieces you skipped in the respective bases. No further work is required. (I suppose the time hiatus ensures I am not 
doing your homework for you.)
Your coupled basis vectors (call them v), are the eigenstates of $S$ and $S_z$, whence $H_1$ as well. While your uncoupled-basis vectors, w, consist of the eigenstates of $s_{1z}\otimes s_{2z}$, whence $H_2$ as well,
$$
v=\begin{bmatrix}
 \frac{\uparrow \downarrow -\downarrow \uparrow}{\sqrt 2}  \\ 
  \uparrow \uparrow    \\ 
   \frac{\uparrow \downarrow +\downarrow \uparrow}{\sqrt 2}  \\ 
  \downarrow \downarrow \\
\end{bmatrix}   ,\qquad 
w=\begin{bmatrix}
  \uparrow \uparrow    \\ 
     \uparrow \downarrow  \\ 
   \downarrow \uparrow \\ 
   \downarrow \downarrow
\end{bmatrix}.
$$
The two bases are, of course, interconvertible through the orthogonal Clebsch-Gordan matrix, which I trust your instructor has introduced you to,
$$
Uw=v, \qquad  \begin{bmatrix}
0& \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 2}  &0 \\ 
 1&0& 0 &0 \\ 
 0& \frac{1}{\sqrt 2}  &  \frac{1}{\sqrt 2}   &0 \\ 
0 &0  &0  &1
\end{bmatrix}, \qquad U^T U=\mathbb{1} ~.
$$
So, then, setting $\hbar=1$ and $\mu_B B=1$ as you all but did, you may write your Hermitean hamiltonian in either basis, always with only a piece of it diagonal, but never both, as you implicitly observe:
$$  H_v= \frac{D}{4} \begin{bmatrix}
 -3& 0 & 0 &0 \\ 
 0&  1 & 0 &0 \\ 
 0& 0 &  1  &0 \\ 
0 &0  &0  &   1
\end{bmatrix} 
+\frac{1}{2}\begin{bmatrix}
 0&0& g_{1}-g_{2}  & 0   \\ 
 0&  g_{1}+g_{2}  & 0 &0 \\ 
    g_{1}-g_{2}   &0&0&0 \\ 
0 &0  &0  &   -g_{1}-g_{2} 
\end{bmatrix},$$
$$H_w=U^T H_v U=\frac{D}{4} \begin{bmatrix}
  1 & 0 & 0 &0 \\ 
 0&  -1 & 2 &0 \\ 
 0& 2 &  -1  &0 \\ 
0 &0  &0  &  1
\end{bmatrix} +\frac{1}{2}\begin{bmatrix}
  g_{1}+g_{2}  & 0 & 0 &0 \\ 
 0&  g_{1}-g_{2}  & 0 &0 \\ 
 0& 0 &  -g_{1}+g_{2}   &0 \\ 
0 &0  &0  &   -g_{1}-g_{2} 
\end{bmatrix}.$$
Both Hermitean, of course, and only diagonal in their leading and trailing pieces, respectively, as you evidently encountered. $H_{2v}$, naturally, was found from the $UH_{2w}U^T$ you computed.
In the coupled basis, you see the non-diagonal $H_2$ leaves the $S_z=\pm 1$ components alone, up to multiplication by these eigenvalues. But permutes the $S=1$ with the $S=0$ components that share a common $S_z=0$.  
Analogously, in the uncoupled basis $H_1$, again the $|\uparrow \uparrow\rangle$ and $|\downarrow \downarrow\rangle$ components are left alone, but the singlet and triplet $S_z=0$ components are mixed thoroughly.
A: 
So I guess my result is that H1 only operates on the coupled basis and H2 only on the uncoupled? I'm still very confused on how this makes sense physically, and how to conceptually think about this. I'd love any feedback on how to think about this....

If two or more physical systems have conserved angular momenta, it can be useful to combine these momenta to a total angular momentum of the combined system— a conserved property of the total system.
The   angular momentum coupling  is the building of eigen states of the total conserved angular momentum from the angular momentum eigen states of the individual sub-systems 
Application of angular momentum coupling is useful when there is an interaction between subsystems that, without interaction, would have conserved angular momentum.
By the very interaction the spherical symmetry of the subsystems is broken, but the angular momentum of the total system remains a constant of motion. 
one can expand the quantum states of composed systems basis state  which are made of tensor products of quantum  states (which  describe the sub  systems individually)  .
One can assume that the states of the sub - system  can be chosen as eigen states of their angular momentum operators (and of their component along any arbitrary z axis).
The subsystems are therefore correctly described by a set of  possible eigen states.
When there is interaction among the subsystems, the total Hamiltonian contains terms that do not commute with the angular operators acting on the subsystems only.
 However, these terms do commute with the total angular momentum operator. One  refers to the non-commuting interaction terms in the Hamiltonian as angular momentum  coupling term , because they do involve  the angular momentum coupling.
