Title: ETH_Sem5 / AdvQC / 01_Non-Relativistic_QC

Q: Why is a relativistic description of quantum chemistry even needed?
A: Basic assumption of the reductionist’s approach to the theory of chemistry:
All chemistry can be deduced to the dynamics of electrons and nuclei!
But what is the correct mechanical theory for this description? Clearly, it is not Newton’s mechanics, but also not Schrödinger’s theory . . .
Only the correct theory can describe quantitatively all experimentally observed facts such as:
- bond lengths in molecules with heavy atoms
- spin-orbit splitting observed in spectroscopy
- the yellow color of gold
- the melting point of mercury

Q: Write down the (non-relativistic) Hamiltonian operator and the corresponding Schrödinger equation for a general molecule!
A: For a molecule of N electrons and M atomic nuclei (Hartree atomic units):
\[\hat H = \hat T + \hat V = \hat T_e + \hat T_n + \hat V_{ee} + \hat V_{en} + \hat V_{nn}\]\[\hat T_e = \sum_{i=1}^N \left(-\frac{1}{2}\nabla_i^2 \right)\]\[\hat V_{en} = - \sum_{i=1}^N \sum_{I=1}^M \frac{Z_I}{|r_i - R_I|}\]\[\hat V_{nn} = \sum_{I=1}^M\sum_{J>1}^M \frac{Z_I Z_J}{|R_I - R_J|}\]Schrödinger's equation depends on both nuclear and electronic coordinates:
\[\hat H (\{r_i\}, \{R_I\}) \Psi_k(\{r_i\}, \{R_I\}) = E_k \Psi_k(\{r_i\}, \{R_I\})\]\(\rightarrow\) separation into nuclear and electronic Schrödinger equation via the introduction of a basis set.

Q: How is the electronic Schrödinger equation defined? What is the exact ansatz for the total wavefunction?
A: Definition of the electronic S.E.:
\[\hat H_{el} = \hat H - \hat T_n = \hat T_e + \hat V_{ee} + \hat V_{en} + \hat V_{nn}\]\[\hat H_{el}\Psi^{\{R_I\}}_{el,n}(\{r_i\}) = E^{\{R_I\}}_{el,n}\Psi^{\{R_I\}}_{el,n}(\{r_i\})\]Parametrical dependence on nuclear coordinates:
The electronic Schrödinger equation is solved for a given set of nuclear coordinates \(\{R_I\}\), indicated by a superscript, which reduces then the number of dynamical coordinates over which one has to integrate.
Exact ansatz:
The electronic wave functions \(\Psi_{el, n}\) form a complete basis set in which the total wave function \(\Psi_k\) may be expanded.
\[\Psi_k(\{r_i\}, \{R_I\}) = \sum_n \chi_{k,n}(\{R_I\}) \cdot \Psi_{el,n}(\{r_i\}, \{R_I\})\]for an exact representation of the total wave function for the state \(k\), a linear combination of all electronic states \(\{\Psi_{el,n} \}\) is required.

Q: What is the Born-Oppenheimer approximation?
A: Within the BO approximation, the total wave function is expressed as a product of a single electronic and a nuclear wave function.\[\Psi_k(\{r_i\}, \{R_I\}) \approx \chi_{k,n}(\{R_I\}) \cdot \Psi_{el,n}^{\{R_I\}} (\{r_i\})\]Insertion of ansatz for the BO approximation into Schrödinger equation yields
\[\begin{aligned} & \hat H \chi_{k,n} \Psi_{el,n} = E_{k} \chi_{k,n} = E_{k} \chi_{k,n} \Psi_{el,n} \\ & \thus \hat T_n \chi_{k,n} \Psi_{el,n} + \hat H_{el} \chi_{el,n}\Psi_{el,n} = E_k \chi_{k,n} \Psi_{el,n} \end{aligned}\]Assuming that \(\hat H_{el}\) acts only on \(\Psi_{el,n}\) and \(\hat T_n\) only on \(\chi_{k,n}\) (approximation) gives
\[\hat T_n \chi_{k,n} \Psi_{el,n} + \hat H_{el} \Psi_{el,n} \chi_{k,n} = E_k \chi_{k,n} \Psi_{el,n}\]Multiplication with \(\Psi_{el,n}^{\ast}\) and integration over electronic coordinates:
\[\hat T_n \chi_{k,n} \braket{\Psi_{el,n}|\Psi_{el,n}} + \bra{\Psi_{el,n}}\hat H_{el}(\{r_i\}, \{R_I\}) \ket{\Psi_{el,n}} \chi_{k,n} = E_k \chi_{k,n} \braket{\Psi_{el,n}|\Psi_{el,n}}\]This yields the nuclear S.E. within the BO approximation:\[\hat H_n \chi_{k,n} = [\hat T_n + E_{el,n}(\{R_I\}) ] \chi_{k,n} = E_k \chi_{k,n}\]

Q: Which ansatz/model can be used to solve the electronic S.E.?
A: The independent particle model is the standard approximation:
\[\Psi_{el} \approx \chi = \prod_{i=1}^N \phi_i(r_i)\]The orbitals \(\phi_i(r_i)\) are one-electron functions!
Can one efficiently use two-electron functions instead?
Problem: Hartree product does not fulfill the Pauli principle.
Pauli Principle: For fermionic systems, the wave function must be antisymmetric with respect to the exchange of two identical particles (and symmetric for bosonic systems)
\(\implies\)Any pair exchange of two electrons results in a change of the sign of the wave function. A determinantal generalization of the Hartree product fulfills the "sign change upon pair exchange" requirement by construction!
The general ansatz for a normalized \(N\)-electron wavefunction the reads\[\Phi_{el}(r_1, r_2, \dots, r_N) = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi_1(r_1) & \phi_2(r_1) & \cdots & \phi_N(r_1) \\ \phi_1(r_2) & \phi_2(r_2) & \cdots & \phi_N(r_2)\\ \vdots & \vdots & \ddots & \vdots \\ \phi_1(r_N) & \phi_2(r_N) & \cdots & \phi_N(r_N)\\ \end{vmatrix}\]

Q: How can the orbitals be obtained?
A: - Find an optimal \(\Phi_{el}\), i.e., the optimal orbitals \(\{\phi_i\}\) that minimize the electronic energy expectation value
- Calculate \(E_{el}^{HF} \equiv \bra{\Phi_{el}} \hat H_{el} \ket{\Phi_{el}}\) (variational principle)
- This approach with a single Slater determinant is called Hartree-Fock (HF) theory
Next step:
Rewrite the high-dimensional integral \(E_{el}^{HF} \equiv \bra{\Phi_{el}} \hat H_{el} \ket{\Phi_{el}}\) as an expression containing only three- and six-dimensional integrals (Slater-Condon rules; or more advanced: second quantization - the latter bookkeeping scheme originates from the field quantization in QED).

Q: Derive the non-relativistic Hartree-Fock energy!
A: The electronic Hamiltonian in terms of one- and two-electron operators reads\[\hat H_{el} = \sum_{i=1}^N (\hat t_{e,i} + \hat V_{en,i}) + \sum_{j<i} \frac{1}{r_{ij}} + \hat V_{nn} = \sum_{i=1}^N \hat h(i) + \sum_{j<i}^N \hat g(i,j) + \hat V_{nn}\]\[\hat h(i) = \hat t_{e,i} + \hat V_{en,i} = - \frac{\Delta_i}{2} + \hat V_{en,i} \qquad \hat g(i,j) = \frac{1}{r_{ij}}\]The Hartree-Fock energy is therefore given by
\[\begin{aligned} E_{el}^{HF} & \equiv \bra{\Phi_{el}} \hat H_{el} \ket{\Phi_{el}} = \bra{\hat A_{\chi}} \hat H_{el} \ket{\hat A_{\chi}} \\ &= \bra{\chi} \hat A^T \hat H_{el} \hat A \ket \chi = \bra \chi \hat H_{el} \hat A \hat A^T \ket \chi \\ &= \bra \chi \hat H_{el} \hat A^2 \ket \chi = \sqrt{N!} \bra \chi \hat H_{el} \hat A \ket \chi \\ &= \bra \chi \hat H_{el} \sum_{\nu=1}^{N!} \hat P_{\nu} \ket \chi \\ &=\sum_{i=1}^N \bra{\phi_i(r)} \hat h(r) \ket{\phi_i(r)} \\ &+\sum_{j<i}^N \left[ \bra{\phi_i(r_1)\phi_j(r_2)} \hat g(1,2) \ket{\phi_i(r_1)\phi_j(r_2)} \\ - \bra{\phi_i(r_1)\phi_j(r_2)} \hat g(1,2) \ket{\phi_i(r_2) \phi_j(r_1)} \right] \end{aligned}\]

Q: Write down the closed-shell Hartree-Fock energy!
A: For closed-shell systems, \(E_{el}^{HF}\) simplifies to:
\[E_{el}^{HF} = \sum_{i=1}^N h_i' + \frac{1}{2}\sum_{i=1}^N \sum_{j=1, j \neq i}^N [J'_{ij} - K\_{ij}]\]\[E_{el}^{HF, closed-shell}=2\sum_{i=1}^{N/2}\sum_{j=1}^{N/2}[2J_{ij}-K_{ij}]\]\(h'_i, J'_{ij}, K'_{ij}\) are integrals over spin orbitals \(\{\phi_i'(r, s)\},\) the others are integrals over spatial orbitals.
Possible only because Schrödinger Hamiltonian does not depend on spin!
Number of exchange integrals is divided by two, due to spin integration:
\(\rightarrow\) for exchange orbitals with different spin components: \(\braket{\alpha | \beta} = 0\)
\(\rightarrow\) exchange interaction only between orbitals with equal spin components

Q: From what equation are the orbitals calculated?
A: \(E_{\mathrm{el}}^{\mathrm{HF}}\) cannot be varied freely, because the orthonormality condition of the orbitals must be fulfilled as a constraint as this was a condition in the derivation of \(E_{\mathrm{el}}^{\mathrm{HF}}\left(\left\{\phi_i\right\}\right)\).
\(\Rightarrow\) Lagrange method of undetermined multipliers
\[\mathcal{L}^{\mathrm{HF}}=E_{\mathrm{el}}^{\mathrm{HF}}-\sum_{i j}^N \varepsilon_{i j}\left(\left\langle\phi_i(\bm{r}) \mid \phi_j(\bm{r})\right\rangle-\delta_{i j}\right)\]
Making the Lagrangian stationary, \(\delta \mathcal{L}^{\mathrm{HF}}=0\), yields the \(\mathrm{HF}\) eqs.:
\[\hat{h} \phi_i+\sum_{j=1}^N\left(\left\langle\phi_j\left|\frac{1}{r_{12}}\right| \phi_j\right\rangle_2 \phi_i-\left\langle\phi_j\left|\frac{1}{r_{12}}\right| \phi_i\right\rangle_2 \phi_j\right)-\sum_{j=1}^N \varepsilon_{i j} \phi_j \stackrel{!}{=} 0\]
Definition of Coulomb, Exchange and Fock operators yields conditional equations for orbitals in compact form:
\[\begin{aligned} & \hat{J} \equiv \sum_{j=1}^N \hat{j}_j \ , \\ & \hat{j}_j \phi_i=\left\langle\phi_j\left|\frac{1}{r_{12}}\right|\phi_j\right\rangle_2 \phi_i \quad\text{ and } \\ & \hat{K} \equiv \sum_{j=1}^N \hat{k}_j \ , \\ & \hat{k}_j \phi_i=\left\langle\phi_j\left|\frac{1}{r_{12}}\right| \phi_i\right\rangle_2 \phi_j \end{aligned}\]
\[\Rightarrow \quad \hat{f}\phi_i = [\hat{h}+\hat{J}-\hat{K}]\phi_i = \sum_{j=1}^N \varepsilon_{i j} \phi_j\]

Q: How can the HF equations be solved?
A: Fock operator \(\hat{f}\) depends on all orbitals, i.e., calculation of a new orbital requires a pre-existing full set of orbitals.
\(\rightarrow\) Equations are solved iteratively — self-consistent field (SCF):
- Guess orbitals \(\{\phi_i^{(0)}\}\) (e.g., Hückel or extended Hückel guess or core-Hamiltonian \(\hat{f} \rightarrow \hat{h}\) )
- Calculate Fock operator from orbitals: \(\hat{f}^{(n)}=\hat{f}^{(n)}[\{\phi_i^{(n)}\}]\)
- Solve Hartree-Fock equations:
\[\hat{f}^{(n)} \phi_i^{(n+1)}=\sum_{j=1}^N \varepsilon_{i j}^{(n+1)} \phi_j^{(n+1)}\] \(\rightarrow\) new set of orbitals \(\{\phi_i^{(n+1)}\}\)
If new set of orbitals \(\{\phi_i^{(n+1)}\}\) is equal to old one \(\{\phi_i^{(n)}\}\) (self consistency), stop; otherwise continue with step 2.

Q: What is the meaning of the Lagrangian multipliers?
A: Hartree-Fock equation for the orbital \(\phi_i\) :
\[\hat{f} \phi_i=\sum_{j=1}^N \varepsilon_{i j} \phi_j\]
Find a new set of orbitals \(\tilde{\phi}_i=\sum_{j=1}^N U_{i j} \phi_j\), which diagonalizes the matrix of the Lagrangian multipliers.
\(\Rightarrow\) yields a Schrödinger-like pseudo-eigenvalue equation.
\[\hat{f} \tilde{\phi}_i=\tilde{\varepsilon}_{i i} \tilde{\phi}_i\]
The Fock operator is invariant with respect to unitary transformations.
The new equation is called the canonical Hartree-Fock equation and directly yields the orbital energy as being the transformed Lagrange multipliers \(\tilde{\varepsilon}_{ii}\).

Q: What are the Roothaan-Hall equations?
A: The orbitals can be represented either
- fully numerically on a grid of points (mostly atoms, DFT).
- or by a basis-set expansion into a set of well known functions (standard):
\[ \phi_i(\bm{r})=\sum_{j=1}^{\infty} c_j^{(i)} \chi_j(\bm{r}) \approx \sum_{j=1}^m c_j^{(i)} \chi_j(\bm{r})\]
Required: the basis set for which \(m\) is as small as possible (typically \(m=1000\)).
\(\longrightarrow\) yields the matrix form of the Hartree-Fock eqs. \(\equiv\) Roothaan-Hall equations:
\[ \bm{f} \cdot \bm{c} = \bm{S} \cdot \bm{c} \cdot \bm{\varepsilon}\]

Q: Can the HF energy reach the exact non-relativistic energy for a hypothetical complete basis set?
A: No, as the ansatz of independent particles and the use of a one-electron product basis is wrong \(\rightarrow\) correlation energy is neglected!

Q: Which methods can be used to consider the electron correlation effects?
A: - Møller-Plesset perturbation theory
For ground states! What about excited states?
- Wave function based correlation methods:
  - configuration interaction (CI): full CI / truncated CI
  - multi-configuration self-consistent-field methods
  - coupled-cluster methods
All these methods define a many-electron determinantal basis; are there other options?
- Density functional theory (DFT) — introduction of correlation potentials into the Fock operator of a HF-like formulation
Requires the density, what about the spin density?

Q: Explain the Møller-Plesset perturbation theory!
A: Define \(\hat{H}^{(0)}\) as the Hamiltonian for which the single HF determinant is the exact solution.
\[\begin{aligned} \hat{H}^{(0)} &=\sum_{i=1}^N \hat{f}(i) \Rightarrow \Psi_0^{(0)}=\Phi_{\mathrm{el}}^{\mathrm{HF}} \equiv \Phi \ \text { (Slater) } \\ & \hat{f}(i) =\hat{h}(i)+\sum_j\left[\hat{j}_j(i)-\hat{k}_j(i)\right] \end{aligned}\]
\(\hat{H}^{(0)}\) is not the exact Hamiltonian operator of the system. \(\hat{f}_i\) is a one-electron operator.
Zeroth-order energy:
\[\begin{aligned}E_{\mathrm{el}}^{\mathrm{MP} 0} & =E^{(0)}=\langle\Phi|\hat{H}^{(0)}| \Phi\rangle \\ &=\sum_{i\in\,\text{Konfig.}\,0} \varepsilon_{i i}\end{aligned}\]
\(E^{(0)}\)is not the HF energy.
First order:
\[\hat{H}^{(1)} =\sum_{j<i}^N \frac{1}{r_{i j}}-\sum_{i j}^N\left[\hat{j}_j(i)-\hat{k}_j(i)\right] \quad E^{(1)} = \langle\Phi|\hat{H}^{(1)}| \Phi \rangle\]
\[ E_{\mathrm{el}}^{\mathrm{MP} 1}=E^{(0)}+\lambda E^{(1)}=\langle\Phi|\underbrace{\hat{H}^{(0)}+\hat{H}^{(1)}}_{\hat{H}_{\mathrm{el}}}| \Phi\rangle=E_{\mathrm{el}}^{\mathrm{HF}}\]
Correlation energy is approximated by second order energy correction:
\[ E^{(2)}=\sum_{b=a+1}^m \sum_{a=N+1}^m \sum_{i=j+1}^N \sum_{j=1}^N \frac{|\langle\phi_a\langle\phi_b|\frac{1}{r_{12}}| \phi_i\rangle \phi_j\rangle-\langle\phi_a\langle\phi_b|\frac{1}{r_{12}}| \phi_j\rangle \phi_i\rangle|^2}{\varepsilon_i+\varepsilon_j-\varepsilon_a-\varepsilon_b}\]
\[\begin{aligned} E_{\mathrm{el}}^{\mathrm{MP} 2} &=E^{(0)}+E^{(1)}+E^{(2)} \\ &= E_{\mathrm{el}}^{\mathrm{HF}}+E^{(2)} \quad \\ & \Rightarrow\quad E_{\mathrm{korr}} \approx E^{(2)} \end{aligned}\]

Q: What is the basic idea of configuration interaction? How the number of Slater determinants be calculated?
A: The complete many-electron basis approach just encountered in MP can be extended to approach the exact solution:
expand electronic wave function in a complete set of determinants
\[\Psi_A^\mathrm{CI} = \sum_I C_{IA}\Phi_I\]
\(A\) denotes the \(A\)-th electronic state. The HF Slater determinant \(\Phi_\mathrm{el}\) is such an \(N\)-electron basis function.
How can one construct these basis functions? \(\rightarrow\) make use of the basis set introduced in MP perturbation theory (there: eigenfunction of the non-interacting system; here: use as basis functions).
Including all possible excited Slater determinants for a finite or infinite one-electron basis set leads to the so-called full CI (FCI) approach. Number of Slater determinants \(n_\mathrm{SD}\) for \(N\) spin orbitals, chosen from a set of \(m\) spin orbitals (slang: \(N\) electrons in \(m\) orbitals):
\[n_\mathrm{SD} = \binom{m}{N} = \frac{m!}{N!(m-N)!}\]
FCI calculations employing the complete set of Slater determinants are only feasible for small basis sets. For practical \(\rightarrow\) truncated CI.

Q: What is truncated CI?
A: For practical reasons an infinite number of Slater determinants cannot be dealt with, only a restricted set of determinants is feasible \(\rightarrow\) truncated CI.
The accuracy of truncated CI is determined by two factors. It is the better, the larger the
- Size of the one-electron basis set \(\{\chi_\nu\} \rightarrow \{\phi_i\}\)
- Size of the \(N\)-electron basis set \(\{\Phi_I\}\)
The determinants are classified by the number of ``excitations'' (= substitutions in the HF reference determinant).
If all singly excited determinants are included, the wave function is classified as CIS (where S stands for singles)
\[\Psi_A^{\mathrm{CIS}}=C_{0 A} \Phi_0+\sum_{(a i)} C_{(a i) A} \Phi_i^a\]
If all doubly excited determinants are included, the wave function is classified as CISD (where D stands for doubles)
\[\Psi_A^{\mathrm{CISD}}=C_{0 A} \Phi_0+\sum_{(a i)} C_{(a i) A} \Phi_i^a+\sum_{(a i)(b j)} C_{(a i, b j) A} \Phi_{i j}^{a b}\]
\[C_{0 A}, C_{(a i) A}, C_{(a i, b j) A} \in\left\{C_{I A}\right\}\]

Q: How are the CI expansion coefficients determined?
A: The CI expansion coefficients can be determined by the variational
principle:
- write down the expectation value for the energy
- introduce the determinantal basis set
- vary the energy in order to minimize it
Expectation value for CI energy:
\[ E_{\mathrm{el}, A}^{\mathrm{CI}}=\frac{\left\langle\Psi_A^{\mathrm{CI}}\left|\hat{H}_{\mathrm{el}}\right| \Psi_A^{\mathrm{CI}}\right\rangle}{\left\langle\Psi_A^{\mathrm{CI}} \mid \Psi_A^{\mathrm{CI}}\right\rangle}\]
Insert expansion of Slater determinants:
\[ E_{\mathrm{el}, A}^{\mathrm{CI}}=\frac{\sum_{K, L} C_{K A}^* C_{L A}\left\langle\Phi_K\left|\hat{H}_{\mathrm{el}}\right| \Phi_L\right\rangle}{\sum_{K, L} C_{K A}^* C_{L A}\left\langle\Phi_K \mid \Phi_L\right\rangle}\]
Calculate all derivatives \(\partial E_{\mathrm{el}, A}^{\mathrm{CI}} / \partial C_{K A}^*\) and set it equal to zero, which yields the CI eigenvalue problem:
\[\bm{H} \cdot \bm{C}_A=E_{\mathrm{el}, A} \bm{C}_A\]
By solving the \(\mathrm{CI}\) eigenvalue problem, ground and excited electronic states of the system are obtained:
\[\bm{H} \cdot \bm{C}=\bm{C} \cdot \bm{E}_{\mathrm{el}}\]
\(\bm{E}_{\mathrm{el}}\) is a diagonal matrix with the total energies of all electronic states that can be expressed in the basis given (\(K\) determinants yield \(K\) electronic states).

Q: What is the second quantization in the context of quantum chemistry? Write down the many-electron Hamiltonian, the CI wave function and the CI energy all in second quantization! Why is this useful?
A: Operators and wave functions are expressed in terms of creation and annihilation operators to implement the Slater-Condon rules directly into the formalism.
\[\begin{aligned} \hat{H}_{\mathrm{el}} &= \sum_{i j}\left\langle\phi_i|\hat{h}(i)| \phi_j\right\rangle \hat{a}_i^{\dagger} \hat{a}_j + \frac{1}{2} \sum_{i j k l}\left\langle\phi_i(1)\left\langle\phi_k(2)|\hat{g}(1,2)| \phi_l(2)\right\rangle \phi_j(1)\right\rangle \hat{a}_i^{\dagger} \hat{a}_k^{\dagger} \hat{a}_l \hat{a}_j \Psi_A^{\mathrm{FCI}} \\ &= C_{0 A} \Phi_0+\sum_{(a i)} C_{(a i) A} \hat{a}_a^{\dagger} \hat{a}_i \Phi_0+\sum_{(a i)(b j)} C_{(a i, b j) A} \hat{a}_b^{\dagger} \hat{a}_j \hat{a}_a^{\dagger} \hat{a}_i \Phi_0 \cdots E_{\mathrm{el}, A}^{\mathrm{CI}} \\ &= \left\langle\Psi_A^{\mathrm{CI}}\left|\hat{H}_{\mathrm{el}}\right| \Psi_A^{\mathrm{CI}}\right\rangle \\ &= \sum_{i j}^n \underbrace{\sum_{K L}^n C_{K A}^* C_{L A} t_{i j}^{K L}}_{\gamma_{i j}^A} \underbrace{\left\langle\phi_i(1)|\hat{h}(1)| \phi_j(1)\right\rangle}_{\equiv h_{i j}} \\ &+ \frac{1}{2} \sum_{i j k l}^n \underbrace{\sum_{K L}^n C_{K A}^* C_{L A} T_{i j k l}^{K L L}}_{\Gamma_{i j k l}^A} \hphantom{=}\ \times \underbrace{\left\langle\phi_i(1)\left\langle\phi_k(2)|\hat{g}(1,2)| \phi_l(2)\right\rangle \phi_j(1)\right\rangle}_{g_{i j k l}} \\ &= \sum_{i j}^n \gamma_{i j}^A h_{i j}+\frac{1}{2} \sum_{i j k l}^n \Gamma_{i j k l}^A g_{i j k l}\end{aligned}\]
If \(t_{i j}^{K L} = \bra{\hat{a}_i\Phi_K}\ket{\hat{a}_j\Phi_L}\) or \(T_{i j k l}^{K L} = \bra{\hat{a}_k\hat{a}_i\Phi_K}\ket{\hat{a}_l\hat{a}_j\Phi_L}\) is equal to zero, one does not need to compute the corresponding integrals. Density matrix element \(\gamma_{i j}^A\) is a generalized occupation number.

Q: What is coupled cluster?
A: In coupled cluster (CC) an advanced CI-type wavefunction is used.
Ansatz:
\[\Psi_{\mathrm{el}}^{\mathrm{CC}}=\exp{\hat{T}} \Phi_{\mathrm{el}}^{\mathrm{HF}}\]
Excitation operator:
\[\hat{T}=\hat{T}_1+\hat{T}_2+\hat{T}_3+\cdots\]
where
\[\begin{aligned} \hat{T}_\alpha &= \sum_{\underbrace{ab\cdots}_{\alpha\text{ times}}\underbrace{ij\cdots}_{\alpha\text{ times}}} \overbrace{t_{ij\cdots}^{ab\cdots}}^{\text{cluster-amplitudes}} \underbrace{\cdots\hat{a}_b^\dagger\hat{a}_j\hat{a}_a^\dagger\hat{a}_i}_{\alpha\text{ pairs }\hat{a}^\dagger\hat{a}} \\ & \thus \hat{T}_1 = \sum_{ai}t_i^a\hat{a}_a^\dagger\hat{a}_i \end{aligned}\]
Notation:
\[\begin{aligned}&\operatorname{CCS}\left(\hat{T}=\hat{T}_1\right),\\&\operatorname{CCSD}\left(\hat{T}=\hat{T}_1+\hat{T}_2\right),\\ &\operatorname{CCSDT}\left(\hat{T}=\hat{T}_1+\hat{T}_2+\hat{T}_3\right),\\ &\cdots\end{aligned}\]
Coupled cluster improves on truncated \(\mathrm{CI}\), because certain (disconnected) higher excited configurations (e.g., \(t_i^a \hat{a}_a^{\dagger} \hat{a}_i t_{j k}^{b c} \hat{a}_c^{\dagger} \hat{a}_k \hat{a}_b^{\dagger} \hat{a}_j\) ) are included. The CC-conditional equations for the cluster-amplitudes are solved by projection against the HF determinant and not in a variational manner! CC is a pure ground state method.

Q: How can truncated CI be improved?
A: In truncated CI, the CI expansion coefficients are optimized, whereas the orbitals remain unchanged. Is there an optimal set of orbitals, which results in the shortest possible expansion of Slater determinants?
\(\Rightarrow\) vary \(E_\mathrm{el}\) with respect to \(\{C_{IA}\}\) and \(\{\phi_i\}\)
Multi-configuration self-consistent-field (MCSCF) methods rely on an alternating/simultaneous optimization with respect to the expansion coefficients and the spin orbitals.

Q: What is the problem with first-generation MCSCF and how can it be solved?
A: First-generation MCSCF suffers from bad or no convergence.
In addition to the gradient information, also the second derivatives w.r.t. the variational parameters are required \(\Rightarrow\) Second-generation MCSCF.
As a first step, the optimization of the orbitals is rewritten as an application of a unitary transformation \(\bm{U}=e^{-\bm{\kappa}}\)
\[\left(\begin{array}{c} \tilde{\phi_1} \vdots \tilde{\phi_m} \end{array}\right)=\bm{U} \cdot\left(\begin{array}{c} \phi_1 \vdots \phi_m \end{array}\right)\]
\(\bm{U}\) is unitary for \(\left(\bm{\kappa}^T\right)^*=-\bm{\kappa}\)
With orbital rotation parameters \(\kappa_{p q} \ \left(\bm{\kappa}=\left\{\kappa_{p q}\right\}\right)\).

Q: Which method is used to calculate the CI-expansion coefficients and the orbital rotation parameters in second-generation MCSCF?
A: The Newton-Raphson Method is used. If the electronic energy is expressed in terms of a set of parameters \(\bm{p}\), it can be Taylor expanded using the gradient \(\bm{g}\) and the Hessian \(\bm{H}\)
\[\begin{aligned} E_{\mathrm{el}}(\bm{p}) &=E_{\mathrm{el}}(0)+\bm{g}^{\dagger} \bm{p}+\frac{1}{2} \bm{p}^{\dagger} \bm{H} \bm{p}+\cdots \\ \text{with} \qquad & g_i=\left(\frac{\partial E_{\mathrm{el}}}{\partial p_i}\right)_0 \quad H_{i j}=\left(\frac{\partial^2 E_{\mathrm{el}}}{\partial p_i \partial p_j}\right)_0 \end{aligned}\]
The set of parameters is then optimized in Newton-Raphson steps
\[\bm{p}=-\bm{H}^{-1} \cdot \bm{g}\]
\[\Rightarrow \quad \left(\begin{array}{c} \bm{C}_A \bm{\kappa} \end{array}\right)=-\left(\begin{array}{ll} \left\{E_{I J}^{(2)}\right\} & \left\{E_{I, p q}^{(2)}\right\} \left\{E_{p q, I}^{(2)}\right\} & \left\{E_{p q, r s}^{(2)}\right\} \end{array}\right)^{-1} \cdot\left(\begin{array}{c} \left\{E_I^{(1)}\right\} \left\{E_{p q}^{(1)}\right\} \end{array}\right)\]
\[\def\arraystretch{3} \begin{array}{>{\displaystyle}r >{\displaystyle}l} E_I^{(1)} \equiv \frac{\partial E_{\mathrm{el}}^{\mathrm{MCSCF}}}{\partial C_I^*} , \\ E_{p q}^{(1)} \equiv \frac{\partial E_{\mathrm{el}}^{\mathrm{MCSCF}}}{\partial \kappa_{p q}^*} \\ E_{I J}^{(2)} \equiv \frac{\partial^2 E_{\mathrm{el}}^{\mathrm{MCSCF}}}{\partial C_I^* \partial C_J^*}, \\ E_{I, p q}^{(2)} \equiv \frac{\partial^2 E_{\mathrm{el}}^{\mathrm{MCSCF}}}{\partial C_I^* \partial \kappa_{p q}^*}, \\ E_{p q, r s}^{(2)} \equiv \frac{\partial^2 E_{\mathrm{el}}^{\mathrm{MCSCF}}}{\partial \kappa_{p q}^* \partial \kappa_{r s}^*} \end{array}\]
Note: computationally very demanding, but extremely efficient.

Q: What is CASSCF?
A: Defining a restricted active space and including all determinants that are obtained from substitutions within this restricted space yields another variant of MCSCF, called complete active space SCF (CASSCF).
In the CASSCF approach the orbitals and the expansion coefficients are optimized in each iteration.
Some definitions:
An active space containing 4 electrons in 6 spatial orbitals is denoted as CAS(4,6).
The computational feasible limit is given by (18,18). Everything larger cannot be calculated in reasonable time.

Q: What is the basic idea of DFT?
A: Up to now: The electronic energy has been calculated from the wave function, which depends on \(3 N\) spatial and \(N\) spin coordinates,
\[\Psi=\Psi\left(\bm{r}_1, s_1, \bm{r}_2, s_2, \cdots, \bm{r}_N, s_N\right)\]
Density functional theory: (DFT)
Electron density is the basic quantity instead of the wave function. The electron density \(\rho\left(\bm{r}_1\right)\) only depends on 3 spatial coordinates.
\[\rho\left(\bm{r}_1\right)=N \iint \cdots \int\left|\Psi\left(\left\{\bm{r}_i\right\},\left\{s_i\right\}\right)\right|^2 \mathrm{~d} s_1 \mathrm{~d}^3 r_2 \mathrm{~d} s_2 \cdots \mathrm{d}^3 r_N \mathrm{~d} s_N\]
Normalization condition:
\[\int \rho\left(\bm{r}_1\right) \mathrm{d}^3 r_1=N\]
The electron density \(\rho\left(\bm{r}_1\right)\) is an observable.
But what about the current density?

Q: Does the electron density contain sufficient information to uniquely determine a quantum mechanical system?
A: Hohenberg-Kohn theorem:
1. The ground state electron density \(\rho_0(\pos)\) determines the external potential \(\nu_\mathrm{ext}\) up to a constant. Therefore, the ground state wavefunction \(\Psi_0\) is also determined uniquely by \(\rho_0(\pos)\).
2. Every normalized test density \(\tilde{\rho}(\pos)\), which is positive everywhere in space, satisfies
E. Bright Wilson argument. \(E_\mathrm{el}\) has to follow only from \(\rho(\pos)\) because:
1. The number of electrons \(N\) follows from \(\rho_0\) (normalization condition of \(\rho_0\))
2. The cusp positions equal the positions of the atomic nuclei \(\{R_A\}\)
3. At every cusp the atomic number \(Z_A\) is obtained by the derivative of the density (Kato cusp condition)
\(\Rightarrow\) Therefore \(H_\mathrm{el}\) is uniquely defined and in principle one also knows its eigenvalues and therefore \(E_\mathrm{el,0}\)

Q: What is the problem of DFT? How is it resolved?
A: The functional which expresses the total energy in terms of the electron density is NOT known.
How to get the density (or at least a guess density)?
\(\Rightarrow\) Consider a system of non-interacting electrons
\[\hat{H}_{\mathrm{s}}=\hat{T}_{\mathrm{s}}+\hat{V}_{\mathrm{s}}=\underbrace{\sum_i\left[-\frac{1}{2} \Delta_i+v_{\mathrm{s}}\left(\bm{r}_i\right)\right]}_{\text {sum of one-electron operators }} \]
\(\Rightarrow \) no \(\ce{e^-}-\ce{e^-}\) interaction
The exact solution for such a system is known: a single Slater determinant. Calculate KS orbitals from a one-electron Schrödinger equation
\[\begin{aligned} \left[-\frac{1}{2} \Delta+v_{\mathbf{s}}(\bm{r})\right] \phi_i(\bm{r})=\varepsilon_i \phi_i(\bm{r}) \\ \text{(Kohn-Sham equation)} \end{aligned}\]
Electron density: \(\rho_{\mathrm{s}}(\bm{r})=\sum_i^{\mathrm{occ.}}\left|\phi_i(\bm{r})\right|^2\)
Exact kinetic energy is known for the non-interacting reference system:
\[T_{\mathrm{s}}=\sum_i^{\text {occ. }}\left\langle\phi_i\left|-\frac{1}{2} \Delta\right| \phi_i\right\rangle\]

Q: How are the Kohn-Sham equations obtained from the system of non-interacting electrons?
A: Assumption of Kohn and Sham:
There exists a potential \(v_s(\bm{r})\) for which the ground-state electron density \(\rho_{\mathrm{s}}(\bm{r})\) is equal to the ground-state electron density \(\rho(\bm{r})\) of the interacting system.
\(\Rightarrow\) use kinetic energy of the non-interacting reference system \(T_{\mathrm{s}}[\rho]\) as an approximation for the kinetic energy of the interacting system \(T[\rho]\).
All unknowns are then collected in the exchange-correlation functional:
\[\begin{aligned} E_{\mathrm{xc}}[\rho] &= \left(T[\rho]-T_{\mathrm{s}}[\rho]\right)+V_{\mathrm{ee}, \mathrm{nk}}[\rho] \\ \thus \ E_{\mathrm{el}, 0}[\rho] &=T_s[\rho]+J[\rho]+V_{\mathrm{eK}}[\rho]+E_{\mathrm{xc}}[\rho] \end{aligned}\]
\[\begin{aligned} v_{\mathrm{eff}}(\bm{r}) &= v_{\mathrm{coul}}(\bm{r})+v_{\mathrm{eK}}(\bm{r})+v_{\mathrm{xc}}(\bm{r}) v_{\mathrm{coul}}(\bm{r}) \\ &= \int \frac{\rho\left(\bm{r}^{\prime}\right)}{\left|\bm{r}-\bm{r}^{\prime}\right|} \mathrm{d}^3 r^{\prime} = \frac{\delta J_{\mathrm{xc}}[\rho]}{\delta \rho(\bm{r})} v_{\mathrm{eK}}(\bm{r}) \\ &= -\sum_A \frac{Z_A}{\left|\bm{r}-\bm{R}_A\right|} = \frac{\delta V_{\mathrm{ek}}[\rho]}{\delta \rho(\bm{r})}v_{\mathrm{xc}}(\bm{r}) \\ &= \frac{\delta E_{\mathrm{xc}}[\rho]}{\delta \rho(\bm{r})} \end{aligned}\]
Therefore \(v_s(\bm{r})\) has to be equal to \(v_\mathrm{eff}(\bm{r})\).

Q: Compare the Kohn-sham equations to the Hartree-Fock equations!
A: Note the very similar structure:
Kohn-Sham equations:
\[\Big[-\frac{1}{2} \Delta+v_{\mathrm{eK}}(\bm{r})+v_{\mathrm{coul}}(\bm{r})+v_{\mathrm{xc}}(\bm{r})\Big] \phi_i^{\mathrm{KS}}(\bm{r})=\varepsilon_i^{\mathrm{KS}} \phi_i^{\mathrm{KS}}(\bm{r})\]
Hartree-Fock equations:
\[\Big[-\frac{1}{2} \Delta+v_{\mathrm{eK}}(\bm{r})+\underbrace{\hat{J}(\bm{r})}_{=v_{\mathrm{coul}}(\bm{r})}-\hat{K}(\bm{r})\Big] \phi_i^{\mathrm{HF}}(\bm{r})=\varepsilon_i^{\mathrm{HF}} \phi_i^{\mathrm{HF}}(\bm{r})\]
Both equations differ in the description of the electron-electron interaction. The exchange operator is replaced by the \(x\) potential (exchange operator: non-local; xc potential: local).
Note that the xc potential also includes electron correlation effects.