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.
Best Answer
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.