Suppose that we have two identical monomers placed side by side, sharing one electron. Naively, we would expect that a sensible basis here would be two states, $\ket \phi$ and $\ket \psi$ , each representing the electron localised to either monomer. By symmetry, we expect that the Hamiltonian to like like $ H = E_0 \ket \phi \bra \phi + E_0 \ket \psi \bra \psi + J \ket \phi \bra \psi + J \ket \psi \bra \phi $ where $E_0$ is the energy of the electron being localised at either state. The interpretation of $J$ is a little tricker, but it's the probability amplitude of jumping from one monomer to the other, which is referred to as an exciton coupling. In this framework, the Hamiltonian can be represented in this basis as follows. $ \hat{H} = \begin{pmatrix} E_0 & J \\ J & E_0 \end{pmatrix} $ If we were to compute the dipoles using this basis of states localised to each monomer, we would get, since the two dipoles that identical $\text{TDM} = [ \mathbf \mu \quad \mathbf \mu ] \in \mathbb R ^{ 3 \times 2 } $ But in practice, ab initio calculations only spit out a diagonalised hamiltonian. In this case, running an SCF on the scenario above is likely to give a Hamiltonian that looks like $ H ' = \begin{pmatrix} E_ 0 + J & 0 \\ 0 & E_0 - J \end{pmatrix} $ In reality we would only get **numerical** values on the diagonal, and not know how it splits! In addition, similar to the case with H2, we also expect non bonding and bonding orbitals, which means that we either get a dipole that's constructive, or cancels out. This means that we expect the TDM to look like this: $\text{TDM}' = [ 2\mathbf \mu \quad 0 ] \in \mathbb R ^{ 3 \times 2 } $ and this is what would be given by an ab initio calculation. So if we were to do a SCF on a dimer system naively, we would get $H'$ and $TDM'$ , with no knowledge about the parameters we really want: $E_0$ and $J$. So, we use a trick, and change basis. We find a change of basis $M$ that takes $TDM'$ to $TDM$. In practice this is done numerically. But in our case the change of basis is given by $ \boldsymbol{\mu}_{0 \to \text{adi}} = \begin{bmatrix} \boldsymbol{\mu} & \boldsymbol{\mu} \end{bmatrix} M= \begin{bmatrix} \boldsymbol{\mu} & \boldsymbol{\mu} \end{bmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\[2mm] 1 & 1 \end{pmatrix} = \begin{bmatrix} \sqrt{2}\, \boldsymbol{\mu} & 0 \end{bmatrix}. $ And then, we find that $H =M H ' M $ And so, we've recovered the couplings we desire in the diabatic basis.