Title: ETH_Sem5 / AdvQC / 09_Relativistic_Many_Electron_Theory

Q: Write down the first-quantized electronic Hamiltonian!
A: The quasi-relativistic many-particle Hamiltonian reads for freezed nuclei (BO-approx)
\[{H_\mathrm{el}^{\prime}=\sum_i^N h^D(i)+\sum_{i<j}^N g(i, j)+V_0} \thus H_\mathrm{el}^{\prime}\Psi_{\mathrm{el},A} = E_{\mathrm{el},A}^{\prime}\Psi_{\mathrm{el},A}\]
with the single one-electron external-field-free Hamiltonian \(h^D(i)\)
\[h^D(i)=c \bm{\alpha}_i \cdot \bm{p}_i+\left(\beta_i-1\right) m_e c^2+V_\mathrm{nuc}\left(\bm{r}_i\right)\]
The \(-m_ec^2\) term is introduced to make the energies comparable to the non-relativistic ones. As it corresponds to the rest energy of one single electron, it has to be substracted from each electron!
The external electrostatic scalar potential energy for the interaction of electron \(i\) with all atomic nuclei reads
\[V_\mathrm{nuc}\left(\bm{r}_i\right)=-\sum_I^M \frac{Z_I e^2}{\left|\bm{r}_i-\bm{R}_I\right| } \thus V_0\equiv\sum_{J<I}^M \frac{Z_IZ_Je^2}{\left|\bm{R}_I-\bm{R}_J\right| }\]
Note that this implies a neglect of magnetic effects on the interaction of electrons and nuclei and on the interaction of nuclei among themselves!

Q: Of which form are the electron-electron repulsion operators in \(h^D(i)\)?
A: In the case of the Dirac-Coulomb many-electron Hamiltonian, \(g(i,j)\) is to be replaced by
\[g(i, j)\rightarrow V_C(i, j)=\frac{e^2}{\left|\bm{r}_i-\bm{r}_j\right| }\]
and in the case of the Dirac-Coulomb-Breit many-electron Hamiltonian
\[g(i, j)\rightarrow V_C(i, j)+B_0(i, j)=\frac{e^2}{\left|\bm{r}_i-\bm{r}_j\right| }+B_0(i, j)\]
where also the unretarded Gaunt interaction \(G_0(i,j)\) may be employed instead of \(B_0(i,j)\).
The many-electron Dirac-Coulomb or Dirac-Coulomb-Breit Hamiltonian is not Lorentz covariant.

Q: What is the problem with the one-electron Dirac Hamiltonian, if used along with the variational principle? How can one resolve this issue?
A: The unboundedness of the one-electron Dirac Hamiltonian, i.e., the fact that states with infinitely large negative energy are possible, prohibits the use of the operator in any variational approach. The Hamiltonian does not possess a bound ground state!
In principle, we are only interested in the upper part of the spectrum of the Dirac Hamiltonian, which describes electronic bound states as well as positive-energy continuum states. Hamiltonians for the upper part of the spectrum (pair-creation processes can no longer occur) are called no-pair Hamiltonians.
In order to produce such a Hamiltonian, projection operators, \(\Lambda_i^{(+)}\), which may be expressed in terms of the sought-for electronic states \(\psi_i(\pos)\) as
\[\Lambda_i^{(+)} = \ket{\psi_i(\pos)}\bra{\psi_i(\pos)}\]
to yield vanishing contributions whenever they hit a ``positronic'' state, are introduced for the one- and two-electron terms
\[\begin{aligned} \sum_i^N h^D(i) \longrightarrow \sum_i^N \Lambda_i^{(+)} h^D(i) \Lambda_i^{(+)} \\ \sum_{i<j}^N g(i, j) \longrightarrow \sum_{i<j}^N \Lambda_i^{(+)} \Lambda_j^{(+)} g(i, j) \Lambda_j^{(+)} \Lambda_i^{(+)} \end{aligned}\]
All subsequent 4D Hamiltonians are to be understood as no-pair Hamiltonians.

Q: What is the continuum dissolution (Brown-Ravenhall disease)?
A: Because of the negative-energy continuum in the spectrum of the one-\(\mathrm{e}^{-}\) Dirac Hamiltonian \(h^D(i)\), a two-\(\mathrm{e}^{-}\) equation becomes meaningless as there is no ground state.
Imagine two non-interacting \(\mathrm{e}^{-}\). The continuous spectrum of the Hamiltonian for two non-interacting \(\mathrm{e}^{-}\) can be easily obtained from the combination of all energies of the one-\(\mathrm{e}^{-}\) Dirac spectrum! For any energy of an \(\mathrm{e}^{-}\) in the positive energy continuum, it is thus possible to find an energy in the negative continuum, so that the sum of the two energies exactly matches the energy of the bound state.
The atom could thus decay, one \(\mathrm{e}^{-}\) being ejected in the positive energy continuum, and the other in the negative energy continuum.
Any interaction between the two \(\mathrm{e}^{-}\), which couple the two continua, e.g. the Coulomb interaction, will then lead to the decay of the atom (continuum dissolution).
The projection-operator ``trick'' also solves this problem.

Q: How can the variational principle still be applied formally and in practice to determine parameters of out trial wavefunction (one-electron spinors and CI coefficients)?
A: Formally: The projection operators are applied in variational procedures to ensure variational stability.
Approach in practice:
There is no explicit representation of the projection operators. They may be thought of as being hidden in the algorithmic structure of the actual variational procedure employed. In the basis-set approach the projectors may be thought of as being applied when the Fock equation is diagonalized to produce electronic bound and continuum states that are orthogonal to the ``positronic'' states. The ``positronic'' states are then not considered for the construction of the density matrix needed to construct the Fock operator in every SCF step.
In contrast to numerical four-component calculations, basis set expansions need to obey the kinetic balance condition (i.e. the basis set of the small component is the derivative of the large component's basis set) to prevent variational collapse.
Formal Approach:
A formal solution is the so-called minimax principle by Talman (1986), which states that the problem of variational collapse is avoided by determining the minimum of the electronic energy with respect to the large component of the spinor, while guaranteeing a maximum of the energy with respect to the small component \(\rightarrow\) SCF energy has a saddle point w.r.t the small-component (max) and large-component parameter (min).

Q: Write down the energy density functional of the 4-current!
A: When discussing the two-electron Breit interaction we neglected exchange effects. Hence, according to non-relativistic DFT we propose an energy expression for the Current-Density Functional Theory:
\[{E_\mathrm{el}\left[j^\mu\right]=T\left[j^\mu\right]+V_{\text {nuc}}\left[j^\mu\right]+J\left[j^\mu\right]+E_{\mathrm{xc}}\left[j^\mu\right]}\]
where \(T\left[j^\mu\right]\) is the kinetic energy functional, \(V_{\text {nuc}}\left[j^\mu\right]\) is the external potential energy of the interaction of the electronic 4-current \(j^\mu\) with all nuclei, and \(J\left[j^\mu\right]\) is the classical Coulomb repulsion energy of the electrons as already deduced for two electrons as
\[\begin{aligned} V_{ee} &= \frac{1}{2}\frac{1}{c}\int\mathrm{d}^3r_1j_c^\mu(1)A_\mu(1) \\ &= \frac{1}{2}\frac{1}{c^2}\int\mathrm{d}^3r_1\int\mathrm{d}^3r_2 \frac{j_c^\mu(1)j_{c,\mu}(2)}{r_{12}}\end{aligned}\]
\(E_{\mathrm{xc}}\left[j^\mu\right]\) denotes an exchange-correlation current-density functional, which contains all quantum, i.e., exchange and correlation effects.

Q: What is the \(N\)-electron current density?
A: According to the Ehrenfest theorem, we get (note that all other terms of the Hamiltonian commute with \(\hat{\rho}_{\pos}\))
\[\begin{aligned} \pdv{}{t} \braket{\hat{\rho}_{\pos}} &= \frac{\imag}{\hbar}\braket{\comm{\sum_{i=1}^N (c\bm{\alpha}_i\cdot\mom_i)}{\hat{\rho}_{\pos}}} = \frac{\imag}{\hbar}\braket{\comm{\sum_{i=1}^N (c\bm{\alpha}_i\cdot\mom_i)}{\sum_{j=1}^N\delta^{(3)}(\pos_j-\pos)}} \\ &= \frac{\imag}{\hbar}\braket{\sum_{i=1}^N\comm{c\bm{\alpha}_i\cdot\mom_i}{\delta^{(3)}(\pos_j-\pos)}} = \frac{\imag}{\hbar}N\braket{\comm{c\bm{\alpha}_1\cdot\mom_1}{\delta^{(3)}(\pos_1-\pos)}} \\ &= cN\braket{\bm{\alpha}_1\cdot\nabla_1\delta^{(3)}(\pos_1-\pos) - \delta^{(3)}(\pos_1-\pos)\bm{\alpha}_1\cdot\nabla_1}{\Psi} \\ &= -cN\left(\langle{\nabla_1\Psi}|{\bm{\alpha}_1\delta^{(3)}(\pos_1-\pos)}|{\Psi} \bigr \rangle + \langle{\Psi}|{\bm{\alpha}_1\delta^{(3)}(\pos_1-\pos)}|{\nabla_1\Psi}\rangle\right)\\ &= -\nabla_1\cdot cN \braket{\bm{\alpha}_1\delta^{(3)}(\pos_1-\pos)}{\Psi} \end{aligned}\]
Where we exploited the indistinguishability of the electrons and the antihermiticity of \(\nabla\). Taking the sum back into the expectation value yields the most general form of a continuity
equation for an \(N\)-electron system for which we may define the \(N\)-electron current-density as
\[{\bm{j} \equiv \braket{c\bm{\alpha}\hat{\rho}_{\bm{r}}}{\Psi}}\]

Q: What are the four-component single-determinant density and current density? Why are they useful in the context of the Kohn-Sham theory?
A: In the case of a single Slater determinant (SD) containing a set of \(N\) (molecular) spinors \(\{\psi_i\}\) both read
\[{\begin{aligned} \rho^{\mathrm{SD}}(\bm{r}) &=\sum_i \psi_i^{\dagger}(\bm{r}) \cdot \psi_i(\bm{r}) \\ \bm{j}^{\mathrm{SD}}(\bm{r}) &=c \sum_i \psi_i^{\dagger}(\bm{r}) \cdot \bm{\alpha} \cdot \psi_i(\bm{r}) \end{aligned}}\]
Rajagopal and Callaway (1973) gave a relativistic generalization of the Hohenberg-Kohn theorem, in which the 4-current plays the role of the electron density.
As in non-relativistic Kohn-Sham theory, one may define a non-interacting
reference system for which a single Slater determinant is the correct state
function so that density and current density are given by the above expressions.
Then, \(E_{\mathrm{xc}}\left[j^\mu\right]\) remains as the only unknown functional which needs to be
approximated.

Q: Write down the four-component Kohn-Sham equations!
A: The 4-current can be calculated after solution of the relativistic Kohn-Sham equations
\[{\left[c \bm{\alpha} \cdot\left(\bm{p}-\frac{q_e}{c} \bm{A}_{\mathrm{eff}}(\bm{r})\right)+\beta m_e c^2+q_e \phi_{\mathrm{eff}}(\bm{r})\right] \psi_i(\bm{r})=\epsilon_i \psi_i(\bm{r})}\]
where \(\psi_i(\bm{r})\) are the Kohn-Sham orbitals. Note the exact accordance with the stationary Dirac equation for a single electron!
The effective potential terms depend on the electronic 4-current \(j^\mu\)
\[\begin{aligned} \phi_{\mathrm{eff}}(\bm{r}) &= \phi_{\mathrm{ext}}(\bm{r}) + \underbrace{\frac{q_e}{c} \int \mathrm{d}^3 r^{\prime} \frac{j^0\left(\bm{r}^{\prime}\right)}{\left|\bm{r}-\bm{r}^{\prime}\right|}}_{\phi_{\mathrm{Coulomb}}(\bm{r})} + \underbrace{c\frac{\delta E_{\mathrm{xc}}\left[j^\mu\right]}{\delta j^0(\bm{r})}}_{\phi_{\mathrm{xc}}(\bm{r})} \\ \bm{A}_{\mathrm{eff}}(\bm{r}) &=\bm{A}_{\mathrm{ext}}(\bm{r})+\frac{q_e}{c} \int \mathrm{d}^3 r^{\prime} \frac{\bm{j}\left(\bm{r}^{\prime}\right)}{\left|\bm{r}-\bm{r}^{\prime}\right|}+c \frac{\delta E_{\mathrm{xc}}\left[j^\mu\right]}{\delta \bm{j}(\bm{r})} \end{aligned}\]
The \(\phi_{\mathrm{eff}}(\bm{r})\) term is completely analogous to the non-relativistic case, whereas the \(\bm{A}_{\mathrm{eff}}(\bm{r})\) is new.