Title: ETH_Sem5 / AdvQC / 08_Interaction_Two_Electrons

Q: Write down the ansatz for the Dirac Hamiltonian of two electrons and resting nuclei!
A: \[\begin{aligned} \left\{\left[-\mathrm{i} \hbar \frac{\partial}{\partial t_1}+c \bm{\alpha}_1 \cdot\left(\bm{p}_1-\frac{q_1}{c} \bm{A}^{(2)}\right) \right.\right. & \\ +\beta_1 m_e c^2 + \left.q_1 \phi^{(2)}+V_\mathrm{nuc}\left(r_1\right)\right] \otimes \mathbf{1}_4 & \\ +\mathbf{1}_4 \otimes\left[-\mathrm{i} \hbar \frac{\partial}{\partial t_2}+c \bm{\alpha}_2 \cdot\left(\bm{p}_2-\frac{q_2}{c} \bm{A}^{(1)}\right)\right. &\\ \left.\left.+\beta_2 m_e c^2+q_2 \phi^{(1)}+V_\mathrm{nuc}\left(r_2\right)\right]\right\} \\ \Psi\left(c t_1, \bm{r}_1, c t_2, \bm{r}_2\right)=0 \end{aligned} \]
Here, two (\(16\times16\))-matrices result after evaluation of the tensor products and are then added. Hence, the exact total wave function consists then of \(4\otimes4=16\) components arranged in vector form according to
\[\begin{aligned} \Psi\left(c t_1, \bm{r}_1, c t_2, \bm{r}_2\right)&=\psi_1\left(c t_1, \bm{r}_1\right) \otimes \psi_2\left(c t_2, \bm{r}_2\right)\\&=\left(\begin{array}{c}\psi_1^{L_1}\left(c t_1, \bm{r}_1\right) \psi_2\left(c t_2, \bm{r}_2\right) \\ \psi_1^{L_2}\left(c t_1, \bm{r}_1\right) \psi_2\left(c t_2, \bm{r}_2\right) \\ \psi_1^{S_1}\left(c t_1, \bm{r}_1\right) \psi_2\left(c t_2, \bm{r}_2\right) \\ \psi_1^{S_2}\left(c t_1, \bm{r}_1\right) \psi_2\left(c t_2, \bm{r}_2\right)\end{array}\right) \end{aligned}\]
Note that the Pauli principle has not yet been taken into account.

Q: What to do with the two different time derivatives?
A: We adopt the single absolute time frame, \(t_1, t_2 \rightarrow t\) of non-relativistic theory,
\[\frac{\partial}{\partial t_1} \otimes \mathbf{1}_4+\mathbf{1}_4 \otimes \frac{\partial}{\partial t_2} \equiv \frac{\partial}{\partial t_1} \mathbf{1}_4 \otimes \mathbf{1}_4+\mathbf{1}_4 \otimes \frac{\partial}{\partial t_2} \mathbf{1}_4 \rightarrow \frac{\partial}{\partial t}\left(\mathbf{1}_4 \otimes \mathbf{1}_4\right)\]
so that we may write
\[\begin{aligned} \left[\left(-\mathrm{i} \hbar \frac{\partial}{\partial t_1}\right) \otimes \mathbf{1}_4+\mathbf{1}_4 \otimes\left(-\mathrm{i} \hbar \frac{\partial}{\partial t_2}\right)\right] \psi_1\left(c t_1, \bm{r}_1\right) \otimes \psi_2\left(c t_2, \bm{r}_2\right) \longrightarrow \cdots \\ \cdots \longrightarrow-\mathrm{i} \hbar\left[\frac{\partial \psi_1\left(t, \bm{r}_1\right)}{\partial t} \psi_2\left(t, \bm{r}_2\right)+\psi_1\left(t, \bm{r}_1\right) \frac{\partial \psi_2\left(t, \bm{r}_2\right)}{\partial t}\right] \\ =-\mathrm{i} \hbar \frac{\partial}{\partial t}\left[\psi_1\left(t, \bm{r}_1\right) \psi_2\left(t, \bm{r}_2\right)\right]\end{aligned}\]

Q: How are the scalar and vector potentials of the electrons to be chosen? How can we translate them to operators?
A: With this expression for the scalar potential, we obtain the unretarded expression for the corresponding vector potential of the moving electron 2:
\[\phi_{2,\text{unret.}}\left(\bm{r}_1, \bm{r}_2\right)=\frac{q_2}{r_{12}} \thus \bm{A}_{2,\text {unret.}}\left(\bm{r}_1, \bm{r}_2, \dot{\bm{r}}_2\right)=\frac{\dot{\bm{r}}_2}{c} \phi_{2,\text{unret.}}=\frac{\dot{\bm{r}}_2}{c} \frac{q_2}{r_{12}}\]
According to the correspondence principle, the velocities \(\dot{\pos}_1\) and \(\dot{\pos}_2\) are then to be substituted by the Dirac velocity operators \(c\bm{\alpha}_1\) and \(c\bm{\alpha}_2\). Note that one might also think about introducing momentum operators instead of velocity operators. Obviously, that would yield quite different expressions since no \(\bm{\alpha}\) matrices are then introduced at all.
However, if the Dirac Hamiltonian defined previously should obey the symmetry condition
\[c\bm{\alpha}_i\frac{q_i}{c}\bm{A}^{(j)} = c\bm{\alpha}_j\frac{q_j}{c}\bm{A}^{(i)}\]
only the velocity operator is valid to be introduced.
We arrive at the Dirac equation in a more compact form:
\[\begin{aligned} &{\left[c \bm{\alpha}_1 \cdot \bm{p}_1+c \bm{\alpha}_2 \cdot \bm{p}_2+\beta_1 m_e c^2+\beta_2 m_e c^2-q_1 q_2 \frac{\bm{\alpha}_1 \bm{\alpha}_2}{r_{12}}\right.} \\ &\left.\quad+\frac{q_1 q_2}{r_{12}}+V_\mathrm{n u c}\left(r_1\right)+V_\mathrm{n u c}\left(r_2\right)\right] \Psi\left(t, \bm{r}_1, \bm{r}_2\right)=\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi\left(t, \bm{r}_1, \bm{r}_2\right) \end{aligned}\]

Q: What is the Gaunt operator?
A: The unretarded classical interaction energy is given by:
\[V_{12}=\frac{q_1 q_2}{r_{12}}\left(1-\frac{\dot{\bm{r}}_1 \cdot \dot{\bm{r}}_2}{c^2}\right)\]
where the first term is the electrostatic Coulomb energy, while the second is due to the magnetic interaction of the moving electrons.
If we now substitute the velocities by their quantum analogs as before,
\[\hat{V}_{12}=\frac{q_1 q_2}{r_{12}}\left(1-\bm{\alpha}_1 \bm{\alpha}_2\right)\]
we identify the Coulomb operator as the first term and then the so-called Gaunt operator,
\[{G_0(1,2)=-q_1 q_2 \frac{\bm{\alpha}_1 \bm{\alpha}_2}{r_{12}}=-\frac{q_1 q_2}{r_{12}} \frac{4}{\hbar^2}\left(\begin{array}{cc}0 & \bm{s}_1 \bm{s}_2 \\ \bm{s}_1 \bm{s}_2 & 0\end{array}\right)}\]
Hence, we understand that the instantaneous magnetic interaction of moving electrons shows up in four-component first-quantized Dirac theory as the interaction of the two electron spin momenta. Since the Gaunt term originates from the unretarded magnetic interaction in classical electrodynamics, the Gaunt operator is also considered to describe (such) magnetic interactions.

Q: What is the Breit operator?
A: The interaction energy of two moving charges is, affected not only by the unretarded but also by the retarded electromagnetic fields as in the classical case derived by Darwin. The operator for the classical potential energy of the retarded interaction of the two electrons can again be obtained from the correspondence principle applied to Darwin's approximation of the classical relativistic interaction energy assigned to two moving charges.
Again, for consistency reasons we choose the velocity operator expression \(\dot{\pos} \rightarrow c\bm{\alpha}\) (instead of the momentum operator) and obtain the interaction energy operator
\[{\hat{V}_{12} \approx \frac{q_1 q_2}{r_{12}} \underbrace{-\frac{q_1 q_2}{2}\left[\frac{\bm{\alpha}_1 \bm{\alpha}_2}{r_{12}}+\frac{\left(\bm{r}_{12} \cdot \bm{\alpha}_1\right)\left(\bm{r}_{12} \cdot \bm{\alpha}_2\right)}{r_{12}^3}\right]}_{\equiv B_0(1,2)}}\]
The term marked as \(B_0\) is the Breit operator. The approximation sign indicates that \(B_0(1,2)\) contains only retardation effects to leading order as in the Darwin interaction energy, which was the starting point.

Q: The Breit operator is not damped by \(c^{-2}\) compared to the Coulomb operator. Does this mean, that the magnetic retardation effects are of the same magnitude as the electrostatic interactions in the quantum world in contrast to the classical regime?
A: No, because the \(\bm{\alpha}_i\) operators have zero block-diagonal entries and, hence, couple only large and small components of the wave function.
\[\begin{aligned} \biggl \langle {\Psi_1\Psi_2} \bigg| \frac{\bm{\alpha_1}\bm{\alpha_2}}{r_{12}} \bigg| {\Psi_1\Psi_2} \biggr \rangle &= \sum_{i=1}^3 \biggl \langle {\Psi_1\Psi_2} \bigg| \frac{\bm{\alpha_{1,i}}\bm{\alpha_{2,i}}}{r_{12}} \bigg| {\Psi_1\Psi_2} \biggr \rangle \\ &= \sum_{i=1}^3 \biggl \langle {\Psi_1\Psi_2} \bigg| {\frac{1}{r_{12}}} \bigg| {\begin{pmatrix}\sigma_{1,i}\Psi_1^S \\ \sigma_{1,i}\Psi_1^L\end{pmatrix}\begin{pmatrix}\sigma_{2,i}\Psi_2^S \\ \sigma_{2,i}\Psi_2^L\end{pmatrix}} \biggr \rangle \\ &= \langle {\Psi_1^L\Psi_2^L} | {\frac{1}{r_{12}}} | {\sigma_{1,i}\Psi_1^S\sigma_{2,i}\Psi_2^S} \rangle + \langle {\Psi_1^S\Psi_2^S} | {\frac{1}{r_{12}}} | {\sigma_{1,i}\Psi_1^L\sigma_{2,i}\Psi_2^L} \rangle \end{aligned}\]
Matrix elements of the Breit operator \(B_0\) are therefore damped by \(c^{-2}\) because of the kinetic balance relation
\[\Psi^S\approx\frac{\bmsigma\cdot\bm{\pi}}{2m_ec}\Psi^L\]
Note that the two-electron Breit operator couples, of course, large and small components from the spinors of two electrons so that the kinetic balance condition hits twice and produces \(c^{-2}\). For a more rigorous explanation see exercise 8.

Q: Does one need to employ additional vector potentials for magnetic interactions that arise from the moving electrons? When are Breit-type operators used in quantum chemistry?
A: No, as these magnetic interactions are already covered by the Darwin and therefore by the Breit interaction, although no vector potential explicitly arises in these interaction terms!
Breit-type operators are usually not employed in quantum chemistry.
This is because of the \(c^{-2}\) damping which reduces the effect of the Breit correction compared to the Coulomb interaction and which makes this term even vanish in the limit of instantaneous interactions, \(c\rightarrow\infty\).
Non-negligible effects may be encountered in accurate calculations of excitation energies and in spin- or magnetic-field-dependent properties.