Rotations of states, operators and expectation values

Any electronic state \(\require{physics} \ket{\Psi}\) is an element of the Fock space, which means that it can be written as a linear combination of the occupation number vectors \(\ket{\vb{k}}\): \[\begin{equation} \ket{\Psi} = \sum_{\vb{k}} c_{\vb{k}} \ket{\vb{k}} \thinspace . \end{equation}\] Realizing that we can write the occupation number vectors as strings of elementary creation operators on top of the vacuum (cfr. equation \(\eqref{eq:ON_creators_vacuum}\)), we can show that the rotated state is \[\begin{equation} \label{eq:rotated_state} \ket*{\Psi'} = \exp(-\hat{\kappa}) \ket{\Psi} \thinspace , \end{equation}\] which in particular means that the vacuum state is invariant upon rotations: \[\begin{equation} \ket{\text{vac}'} = \exp(-\hat{\kappa}) \ket{\text{vac}} = \ket{\text{vac}} \thinspace . \end{equation}\]

A general operator \(\hat{A}\) can also be rotated. Since it can be written as a linear combination of products of elementary second quantization operators (with the restriction that no coefficient in the linear combination depends on the one- or two-electron integrals in the MO basis), we can then show that the rotated operator is given by \[\begin{equation} \hat{A}' % = \exp(-\hat{\kappa}) % \thinspace \hat{A} \thinspace % \exp(\hat{\kappa}) % \thinspace . \end{equation}\] This means commutation and anticommutation relations that hold in the old spinor basis also hold in the new spinor basis: \[\begin{equation} \comm{ \hat{A}' }{ \hat{\bar{B}}' }_{\pm} % = \comm{ \hat{A} }{ \hat{B} }_{\pm} % \thinspace . \end{equation}\]

What about expectation values? Using the previously discussed rotations, we can calculate the `energy after rotation’, which is the expectation value of the Hamiltonian in the rotated state: \[\begin{equation} \label{eq:rotated_energy} E(\boldsymbol{\kappa}) = % \ev{ % \exp(\hat{\kappa}) % \hat{\mathcal{H}} % \exp(-\hat{\kappa}) % }{\Psi} % \thinspace . \end{equation}\] We can also show that \[\begin{equation} \exp(\hat{\kappa}) % \hat{\mathcal{H}} % \exp(-\hat{\kappa}) % = \sum_{PQ}^M h'_{PQ} \hat{E}_{PQ} % + \frac{1}{2} \sum_{PQRS}^M % g'_{PQRS} \thinspace \hat{e}_{PQRS} + h_{\text{nuc}} % \thinspace , \end{equation}\] which is just the original Hamiltonian in which we replace the one- and two-electron integrals by their corresponding values in the new MO basis.

Let us now also clear up any misconceptions regarding rotated DMs. The form of the 1-DM of the rotated state in the original MO basis is \[\begin{align} \bar{D}_{PQ}(\boldsymbol{\kappa}) % &= \ev{ % \exp(\hat{\kappa}) % \thinspace \hat{E}_{PQ} \thinspace % \exp(-\hat{\kappa})% }{\Psi} \\ &= \sum_{RS}^M U_{PR} \bar{D}_{RS} U^\dagger_{SP} % \thinspace , \end{align}\] whereas the 1-DM of the old state in the rotated MO basis is \[\begin{align} \bar{D}'_{PQ}(\boldsymbol{\kappa}) % &= \ev{ % \exp(-\hat{\kappa}) % \thinspace \hat{E}_{PQ} \thinspace % \exp(\hat{\kappa})}{\Psi} \\ &= \sum_{RS}^M U^\dagger_{PR} \bar{D}_{RS} U_{SP} \thinspace . \end{align}\] This means that diagonalizing the 1-DM, i.e. \[\begin{equation} \bar{\boldsymbol{\Delta}} = \vb{V}^\dagger \bar{\vb{D}} \vb{V} \end{equation}\] in which \(\vb{V}\) then contains the so-called natural orbitals (i.e. the eigenvectors of the 1-DM) as columns, is equivalent to finding the unitary transformation for which the density matrix of the old state in the rotated MO basis is diagonal.