Linear response derivatives for GHF
In order to find the parameter response for the GHF wave function model when applying an external magnetic field, we have to solve the linear response equations. The wave function model parameters are the non-redudant occupied-virtual orbital rotation generators \(\{ {}^{\text{R}} \boldsymbol{\kappa}, {}^{\text{I}} \boldsymbol{\kappa} \}\), with the two sets of real parameters \(\{ {}^{\text{R}} \boldsymbol{\kappa} \}\) and \(\{ {}^{\text{I}} \boldsymbol{\kappa} \}\), such that \[\begin{equation} \require{physics} \kappa_{AI} = {}^{\text{R}} \kappa_{AI} + i {}^{\text{I}} \kappa_{AI} \thinspace . \end{equation}\]
The linear response equations (in this case also called the CPHF, coupled perturbed Hartree-Fock, equations) then admit a blocked-out form: \[\begin{equation} \begin{pmatrix} {}^{\text{R}} \vb{F}_{\boldsymbol{\kappa}} \\ {}^{\text{I}} \vb{F}_{\boldsymbol{\kappa}} \end{pmatrix} + \begin{pmatrix} {}^{\text{RR}} \vb{k}_{\boldsymbol{\kappa}} & {}^{\text{RI}} \vb{k}_{\boldsymbol{\kappa}} \\ {}^{\text{IR}} \vb{k}_{\boldsymbol{\kappa}} & {}^{\text{II}} \vb{k}_{\boldsymbol{\kappa}} \\ \end{pmatrix} \begin{pmatrix} {}^{\text{R}} \vb{x} \\ {}^{\text{I}} \vb{x} \end{pmatrix} = \begin{pmatrix} \vb{0} \\ \vb{0} \end{pmatrix} \thinspace . \end{equation}\] In full, they can be written as: \[\begin{equation} \begin{split} 0 = & \eval{ \pdv{ E(\boldsymbol{\eta}, \boldsymbol{\kappa}) }{ {}^{\text{R}} \kappa_{AI} }{\eta_m} }_{ \boldsymbol{\eta}_0, \boldsymbol{\kappa}_0(\boldsymbol{\eta}_0) } \\ & + \sum_{BJ} \qty( \eval{ \pdv{ E(\boldsymbol{\eta}_0, \boldsymbol{\kappa}) }{ {}^{\text{R}} \kappa_{AI} }{ {}^{\text{R}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\boldsymbol{\eta}_0)} ) \qty( \eval{ \pdv{ {}^{\text{R}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \\ & + \sum_{BJ} \qty( \eval{ \pdv{ E(\boldsymbol{\eta}_0, \boldsymbol{\kappa}) }{ {}^{\text{R}} \kappa_{AI} }{ {}^{\text{I}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\boldsymbol{\eta}_0)} ) \qty( \eval{ \pdv{ {}^{\text{I}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \thinspace , \end{split} \end{equation}\] and \[\begin{equation} \require{physics} \begin{split} 0 = & \eval{ \pdv{ E(\boldsymbol{\eta}, \boldsymbol{\kappa}) }{ {}^{\text{I}} \kappa_{AI} }{\eta_m} }_{ \boldsymbol{\eta}_0, \boldsymbol{\kappa}_0(\boldsymbol{\eta}_0) } \\ & + \sum_{BJ} \qty( \eval{ \pdv{ E(\boldsymbol{\eta}_0, \boldsymbol{\kappa}) }{ {}^{\text{I}} \kappa_{AI} }{ {}^{\text{R}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\boldsymbol{\eta}_0)} ) \qty( \eval{ \pdv{ {}^{\text{R}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \\ & + \sum_{BJ} \qty( \eval{ \pdv{ E(\boldsymbol{\eta}_0, \boldsymbol{\kappa}) }{ {}^{\text{I}} \kappa_{AI} }{ {}^{\text{I}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\boldsymbol{\eta}_0)} ) \qty( \eval{ \pdv{ {}^{\text{I}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) \thinspace . \end{split} \end{equation}\] The elements of the response force are given by: \[\begin{align} {}^{\text{R}} F_{AI, m} = \eval{ \pdv{ E(\boldsymbol{\eta}, \boldsymbol{\kappa}) }{ {}^{\text{R}} \kappa_{AI} }{\eta_m} }_{ \boldsymbol{\eta}_0, \boldsymbol{\kappa}_0(\boldsymbol{\eta}_0) } &= \ev{ \comm{ \hat{E}^-_{AI} }{\qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} )} }{\text{core}} \\ &= - \qty[ \mathscr{F}_{IA} \qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) + \mathscr{F}^*_{IA} \qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) ] \thinspace , \end{align}\] and \[\begin{align} {}^{\text{I}} F_{AI, m} = \eval{ \pdv{ E(\boldsymbol{\eta}, \boldsymbol{\kappa}) }{ {}^{\text{I}} \kappa_{AI} }{\eta_m} }_{ \boldsymbol{\eta}_0, \boldsymbol{\kappa}_0(\boldsymbol{\eta}_0) } &= i \ev{ \comm{ \hat{E}^+_{AI} }{\qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} )} }{\text{core}} \\ &= i \qty[ \mathscr{F}_{IA} \qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) - \mathscr{F}^*_{IA} \qty( \eval{ \pdv{ \hat{\mathcal{H}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) ] \thinspace . \end{align}\]
where we have done some simplifications due to the GHF Fockian and have assumed that the first-order perturbational partial derivative of the Hamiltonian is again Hermitian. The elements of the response force constant \(\vb{k}_{\boldsymbol{\kappa}}\) are exactly those of the GHF Hessian.
External magnetic field without London orbitals
Let us discuss the case where the external perturbation is a uniform external magnetic field \(\vb{B}_{\text{ext}}\). If we are not employing London orbitals, we may treat the spinor basis metric as perturbation-independent and can therefore proceed with the simplest form of the perturbation-dependent Hamiltonian.
In the Pauli Hamiltonian there are only one-electron terms affected by the uniform external magnetic field and since it only appears inside a commutator, we may use the following form of the first-order partial perturbational derivative of the Hamiltonian: \[\begin{equation} \eval{ \pdv{ \hat{\mathcal{H}}(\vb{B}_{\text{ext}}, \vb{G}) }{B_{\text{ext}, m}} }_{\vb{B}_{\text{ext}, 0}} = \sum_{PQ}^M \eval{ \pdv{ h_{PQ}(\vb{B}_{\text{ext}}, \vb{G}) }{B_{\text{ext}, m}} }_{\vb{B}_{\text{ext}, 0}} \hat{E}_{PQ} \thinspace , \end{equation}\] where the two-electron terms are not affected by the perturbation and the nuclear term can be ignored due to the aforementioned commutator-related discussion. The physical interactions that are included, therefore enter the calculations through the (derivative of the) one-electron integrals only. The derivative operator that should be quantized is can be calculated as: \[\begin{equation} \eval{ \pdv{ h^c(\vb{B}_{\text{ext}}, \vb{G}) }{B_{\text{ext}, m}} }_{\vb{B}_{\text{ext}, 0}} = -\frac{i}{2} \qty[ (\vb{r} - \vb{G}) \cross \grad ]_m \vb{I}_2 + \frac{1}{2} \sigma_m \thinspace , \end{equation}\] where only the first part is considered in (Keith and Bader 1993) and (Ruud et al. 1993).
Following (Ruud et al. 1993), if we may use real spinors for the field-free optimization, and use an imaginary wave function response for the first-order perturbation. In the previous equations, this means that we set the real first-order response to zero: \[\begin{equation} \qty[ {}^{\text{R}} \vb{x} ]_{BJ, m} = \qty( \eval{ \pdv{ {}^{\text{R}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) = 0 \thinspace , \end{equation}\] which indeed leads to analogous equations as in (Ruud et al. 1993): \[\begin{equation} \eval{ \pdv{ E(\vb{B}_{\text{ext}}, \vb{G}, \boldsymbol{\kappa}) }{B_{\text{ext}, m}}{ \kappa_{AI} } }_{ \vb{B}_{\text{ext}, 0}, \boldsymbol{\kappa}_0(\vb{B}_{\text{ext}, 0}) } + \sum_{BJ} \qty( \eval{ \pdv{ E(\vb{B}_{\text{ext}, 0}, \boldsymbol{\kappa}) }{ \kappa_{AI} }{ {}^{\text{I}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\vb{B}_{\text{ext}, 0})} ) \qty( \eval{ \pdv{ {}^{\text{I}} \kappa_{BJ}^\star(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} ) = 0 \thinspace , \end{equation}\] with the response force: \[\begin{equation} \eval{ \pdv{ E(\vb{B}_{\text{ext}}, \vb{G}, \boldsymbol{\kappa}) }{B_{\text{ext}, m}}{ \kappa_{AI} } }_{ \vb{B}_{\text{ext}, 0}, \boldsymbol{\kappa}_0(\vb{B}_{\text{ext}, 0}) } = 2 \ev{ \comm{ \hat{E}_{AI} }{\qty( \eval{ \pdv{ \hat{\mathcal{H}}(\vb{B}_{\text{ext}}, \vb{G}) }{B_{\text{ext}, m}} }_{\vb{B}_{\text{ext}, 0}} )} }{\text{core}} \end{equation}\] and the response force constant being a combination of two parts of the GHF orbital hessian: \[\begin{equation} \eval{ \pdv{ E(\vb{B}_{\text{ext}, 0}, \boldsymbol{\kappa}) }{ \kappa_{AI} }{ {}^{\text{I}} \kappa_{BJ} } }_{\boldsymbol{\kappa}_0(\vb{B}_{\text{ext}, 0})} = 2i \ev*{ \comm*{ \hat{E}_{AI} }{ \comm*{ \hat{E}^+_{BJ} }{ \hat{\mathcal{H}}(\vb{B}_{\text{ext}, 0}) } } }{ \text{core} } \thinspace . \end{equation}\]