Seniority-preserving operators
Using the seniority operator \[\begin{equation} \require{physics} \hat{\Omega} = \sum_p^K \hat{\Omega}_p = \sum_p^K ( \hat{N}_{p \alpha} - \hat{N}_{p \beta} )^2 \thinspace , \end{equation}\] we can write (Henderson, Bulik, and Scuseria 2015) the electronic Hamiltonian (in a spin-orbital basis) as subdivided in three parts: \[\begin{equation} \hat{\mathcal{H}}_{\text{elec}} = \hat{\mathcal{H}}^{\delta\Omega=0}_\text{elec} + \hat{\mathcal{H}}^{\delta\Omega=2}_\text{elec} + \hat{\mathcal{H}}^{\delta\Omega=4}_\text{elec} \thinspace , \end{equation}\] in which \(\hat{\mathcal{H}}^{\delta\Omega=0}_\text{elec}\) can be used to calculate elements within a given seniority sector, and \(\hat{\mathcal{H}}^{\delta\Omega=2}_\text{elec}\) and \(\hat{\mathcal{H}}^{\delta\Omega=4}_\text{elec}\) couple ONVs with a seniority number that differs by 2 or 4, respectively. A term of the form \(\hat{\mathcal{H}}^{\delta\Omega=6}_\text{elec}\) does not exist, since the Hamiltonian contains at most two-body interactions, which can maximally lead to a difference in seniority of 4. In the article, we find an expression for the seniority-preserving part of the Hamiltonian \(\hat{\mathcal{H}}^{\delta\Omega=0}_\text{elec}\) (Henderson, Bulik, and Scuseria 2015): \[\begin{equation} \begin{split} \hat{\mathcal{H}}^{\delta\Omega=0}_\text{elec} = & \sum_p^K h_{pp} \hat{N}_p + \sum_{pq}^K g_{pqpq} \hat{P}^+_p \hat{P}^-_q + \frac{1}{2} \sum_{p \neq q}^K g_{ppqq} \hat{N}_p \hat{N}_q \\ &- \frac{1}{2} \sum_{p \neq q}^K g_{pqqp} \qty( \hat{N}_{p \alpha} \hat{N}_{q \alpha} + \hat{N}_{p \beta} \hat{N}_{q \beta} ) - \frac{1}{2} \sum_{p \neq q}^K g_{pqqp} \qty( \hat{S}^+_p \hat{S}^-_q + \hat{S}^-_p \hat{S}^+_q ) \thinspace . \end{split} \end{equation}\]
An alternative derivation of the seniority-preserving part of the Hamiltonian
We can already feel intuitively that the one- and two-electron singlet excitation operators won’t commute with the seniority operator, as most generally the excitation operators break pairs. Let’s check the commutator with the singlet one-electron excitation operator: \[\begin{align} \comm{\hat{E}_{pq}}{\hat{\Omega}} & = 2 \sum_\sigma \hat{E}^{\sigma}_{pq} ( \hat{N}_{p \bar{\sigma}} - \hat{N}_{q \bar{\sigma}} ) \\ &= 2 \qty( \hat{E}^\alpha_{pq} ( \hat{N}_{q \beta} - \hat{N}_{p \beta} ) + \hat{E}^\beta_{pq} ( \hat{N}_{q \alpha} - \hat{N}_{p \alpha} ) ) \thinspace , \end{align}\] in which \(\bar{\sigma}\) denotes the opposite spin of \(\sigma\). For the separate spin parts, we thus have: \[\begin{equation} \comm{ \hat{\Omega} }{ \hat{E}^\sigma_{pq} } = 2 \hat{E}^\sigma_{pq} ( \hat{N}_{q \bar{\sigma}} - \hat{N}_{p \bar{\sigma}} ) \thinspace . \end{equation}\]
Using some of the operators that are introduced elsewhere, we can calculate \[\begin{equation} \hat{E}^2_{pq} = 2 \hat{P}^+_p \hat{P}^-_q \end{equation}\] and \[\begin{equation} \hat{E}_{pq} \hat{E}_{qp} = \hat{N}_p - \qty( \hat{N}_{p \alpha} \hat{N}_{q \alpha} + \hat{N}_{p \beta} \hat{N}_{q \beta} ) - (\hat{S}^+_p \hat{S}^-_q + \hat{S}^-_p \hat{S}^+_q) \thinspace , \end{equation}\] which are some relations that we will need in the followin derivations.
The commutator of the seniority operator and the he singlet two-electron excitation operator can then be written in terms of the commutator with the singlet one-electron excitation operator: \[\begin{align} \comm{ \hat{e}_{pqrs} % }{ \hat{\Omega} } &= \comm{ \hat{E}_{pq} \hat{E}_{rs} }{ \hat{\Omega} } - \delta_{qr} \comm{ \hat{E}_{ps} }{ \hat{\Omega} } \\ &= \hat{E}_{pq} \comm{ \hat{E}_{rs} }{ \hat{\Omega} } + \comm{ \hat{E}_{pq} }{ \hat{\Omega} } \hat{E}_{rs} - \delta_{qr} \comm{ \hat{E}_{ps} }{ \hat{\Omega} } \thinspace , \end{align}\] All things considered, this means that seniority is not a good quantum number (Bytautas et al. 2011), as the seniority operator does not commute with the Hamiltonian.
It therefore is a natural question to ask ourselves ‘which parts’ of these singlet excitation operators preserve seniority, i.e. those parts don’t break electron pairs. For reference, we can look at the derivation in the article (Henderson, Bulik, and Scuseria 2015), but let us here present a different approach. Preserving seniority in essence means that the commutator with the seniority operator should vanish. For the singlet one-electron excitation operator, it is easy to see from equation \(\eqref{eq:commutator_E_pq_seniority}\) that it vanishes if \(p=q\), such that we find \[\begin{equation} \hat{E}_{pq} \xrightarrow{\delta\Omega=0} \delta_{pq} \hat{N}_p \thinspace , \end{equation}\] in which the \(\delta\Omega=0\) over the arrow means ‘the seniority preserving part of’. We could then do the same thing for the singlet two-electron operator, leading to \[\begin{equation} \label{eq:seniority_preserving_e_pqrs_initial} \begin{split} \hat{e}_{pqrs} \xrightarrow{\delta\Omega=0} & \delta_{pq} \delta_{pr} \delta_{ps} 2 \hat{N}_{p \alpha} \hat{N}_{p \beta} + \delta_{pq} \delta_{rs} (1 - \delta_{qr}) \hat{N}_p \hat{N}_r \\ &+ \delta_{pr} \delta_{qs} (1 - \delta_{qr}) 2 \hat{P}^+_p \hat{P}^-_q \\ &- \delta_{ps} \delta_{qr} (1 - \delta_{pq}) \qty( \hat{S}^+_p \hat{S}^-_q + \hat{S}^-_p \hat{S}^+_q + \hat{N}_{p \alpha} \hat{N}_{q \alpha} + \hat{N}_{p \beta} \hat{N}_{q \beta} ) \thinspace , \end{split} \end{equation}\] in which we will now present the previous terms chronologically.
If we set every index equal to one another, i.e. \(p=q=r=s\), we end up with commutators of orbital number operators and the seniority operator (as the diagonal of the singlet excitation operator is the orbital number operator, see equation \(\eqref{eq:diagonal_E_pp}\)). Since these commutators are equal to zero (cfr. equation \(\eqref{eq:commutator_seniority_on}\)), the total commutator is also equal to zero.
If we were to set \(r=s\) and \(p=q\) and \(q \neq r\), then the first two terms vanish because they are reduced to orbital number operators (cfr. equation \(\eqref{eq:diagonal_E_pp}\)) as in the previous case, and the Kronecker delta makes sure the last term also vanishes.
If we were to set \(p=r\) and \(q=s\) and \(q \neq r\), then the Kronecker delta assures that we end up with the form \[\begin{equation} \comm{ \hat{E}^2_{pq} }{ \hat{\Omega} } \thinspace . \end{equation}\] As \(\hat{E}^2_{pq}\) can be written (cfr. equation \(\eqref{eq:E^2_pq}\)) in terms of operators that are all seniority-preserving (cfr. equation \(\eqref{eq:commutator_seniority_P}\)), the commutator in question vanishes.
If we let \(p=s\) and \(q=r\), we end up with \[\begin{equation} \comm{ \hat{E}_{pq} \hat{E}_{qp} }{ \hat{\Omega} } \thinspace . \end{equation}\] Since \(\hat{E}_{pq} \hat{E}_{qp}\) can be written (cfr. equation \(\eqref{eq:E_pq_E_qp}\)) in terms of seniority-preserving operators (cfr. equations \(\eqref{eq:commutator_seniority_S}\) and \(\eqref{eq:commutator_seniority_on}\)), the commutator in question vanishes. However, we must also impose \(p\neq q\), since if not this would be no different from case number 1. \end{enumerate}
The first and the third term can together also be simplified, finally leading to \[\begin{equation} \label{eq:seniority_preserving_e_pqrs} \begin{split} \hat{e}_{pqrs} \xrightarrow{\delta\Omega=0} & \delta_{pr} \delta_{qs} 2 \hat{P}^+_p \hat{P}^-_q + \delta_{pq} \delta_{rs} (1 - \delta_{qr}) \hat{N}_p \hat{N}_r \\ &- \delta_{ps} \delta_{qr} (1 - \delta_{pq}) \qty( \hat{S}^+_p \hat{S}^-_q + \hat{S}^-_p \hat{S}^+_q + \hat{N}_{p \alpha} \hat{N}_{q \alpha} + \hat{N}_{p \beta} \hat{N}_{q \beta} ) \thinspace . \end{split} \end{equation}\] Using the previous formulas, we can subsequently find the seniority-preserving part of the electronic Hamiltonian: \[\begin{equation} \label{eq:seniority-preserving_hamiltonian} \begin{split} \hat{\mathcal{H}}^{\delta\Omega=0}_\text{elec} = &\sum_p^K h_{pp} \hat{N}_p + \sum_{pq}^K g_{pqpq} \hat{P}^+_p \hat{P}^-_q + \frac{1}{2} \sum_{p \neq q}^K g_{ppqq} \hat{N}_p \hat{N}_q \\ &- \frac{1}{2} \sum_{p \neq q}^K g_{pqqp} \qty( \hat{N}_{p \alpha} \hat{N}_{q \alpha} + \hat{N}_{p \beta} \hat{N}_{q \beta} ) - \frac{1}{2} \sum_{p \neq q}^K g_{pqqp} \qty( \hat{S}^+_p \hat{S}^-_q + \hat{S}^-_p \hat{S}^+_q ) \thinspace , \end{split} \end{equation}\] which is the result also found in (Bytautas et al. 2011).
Spin-separated parts of seniority-preserving operators
We would now like to repeat the process in the previous section, but for the separate spin contributions of the singlet one- and two-electron excitation operators. It is of course unnecessary to repeat this whole procedure if we were only interested in the seniority-preserving part of the molecular electronic Hamiltonian, but since we are interested in the separate spin contributions for the 1- and 2-DM, we are required to do the derivation in more general terms.
We start by calculating the generalized versions of the commutators \(\eqref{eq:commutator_E_pq_seniority}\) and \(\eqref{eq:commutator_e_pqrs_seniority}\), for which we find \[\begin{equation} \label{eq:commutator_E_pq_seniority_generalized} \comm{ % \hat{a}^\dagger_{p \sigma} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } % = 2 \delta_{\sigma \alpha} % \hat{a}^\dagger_{p \alpha} \hat{a}_{q \alpha} ( % \hat{N}_{p \beta} % - \hat{N}_{q \beta} % ) % + 2 \delta_{\sigma \beta} ( % \hat{N}_{p \alpha} - \hat{N}_{q \alpha} % ) \hat{a}^\dagger_{p \beta} \hat{a}_{q \beta} \end{equation}\] and \[\begin{equation} \label{eq:commutator_e_pqrs_seniority_generalized} \comm{ % \hat{a}^\dagger_{p \sigma} \hat{a}^\dagger_{r \tau} % \hat{a}_{s \tau} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } % = \hat{a}^\dagger_{p \sigma} \hat{a}_{q \sigma} % \comm{ % \hat{a}^\dagger_{r \tau} \hat{a}_{s \tau} % }{ % \hat{\Omega} % } % + \comm{ % \hat{a}^\dagger_{p \sigma} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } \hat{a}^\dagger_{r \tau} \hat{a}_{s \tau} % - \delta_{qr} \delta_{\sigma \tau} % \comm{ % \hat{a}^\dagger_{p \sigma} \hat{a}_{s \sigma} % }{ % \hat{\Omega} % } % \thinspace . \end{equation}\]
The one-electron commutator (cfr. equation \(\eqref{eq:commutator_E_pq_seniority_generalized}\)), vanishes for \(p=q\), so we immediately get \[\begin{equation} \label{eq:seniority_preserving_E_pq_generalized} \hat{a}^\dagger_{p \sigma} \hat{a}_{q \sigma} % \xrightarrow{\delta\Omega=0} % \delta_{pq} \hat{N}_{p \sigma} % \thinspace . \end{equation}\] The two-electron commutator (cfr. equation \(\eqref{eq:commutator_e_pqrs_seniority_generalized}\)) vanishes in the following cases:
If we let \(p=q=r=s\), we again end up with orbital occupation number operators, which commute with the seniority operator.
If we let \(p=q\) and \(r=s\), but \(q \neq r\), the first two commutators vanish because they involve orbital occupation number operators and the seniority operator, and and third term vanishes because of the Kronecker delta.
If we let \(p=r\) and \(q=s\), but \(q \neq r\), we immediately have that the last term vanishes because of the Kronecker delta. We then end up with the commutator \[\begin{equation} \comm{ % \hat{a}^\dagger_{p \sigma} \hat{a}^\dagger_{p \tau} % \hat{a}_{q \tau} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } \thinspace . \end{equation}\] If \(\tau = \sigma\), then this commutator vanishes anyways due to the Pauli principle. If we have \(\tau \neq \sigma\), then we must have \(\tau = \bar{\sigma}\), so we can simplify the expression to \[\begin{equation} (1 - \delta_{\sigma \tau}) % \comm{ % \hat{a}^\dagger_{p \sigma} % \hat{a}^\dagger_{p \bar{\sigma}} % \hat{a}_{q \bar{\sigma}} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } % \thinspace . \end{equation}\] \(\sigma\) is either \(\alpha\) or \(\beta\), so this means that we identify pair operators, so we get: \[\begin{equation} (1 - \delta_{\sigma \tau}) % \comm{ % \hat{P}^+_p \hat{P}^-_q % }{ % \hat{\Omega} % } % \thinspace , \end{equation}\] which vanishes because pair operators commute with the seniority operator.
If we let \(p=s\) and \(q=r\) (and additionally \(p \neq q\) to differ from case \(\ref{en:case_pqrs}\)), the commutator becomes \[\begin{equation} \comm{ % \hat{a}^\dagger_{p \sigma} % \hat{a}^\dagger_{q \tau} % \hat{a}_{q \tau} \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } % =- \comm{ % \hat{a}^\dagger_{p \sigma} % \hat{a}_{q \tau} % \hat{a}^\dagger_{q \tau} % \hat{a}_{q \sigma} % }{ % \hat{\Omega} % } % \thinspace . \end{equation}\]
In the case that \(\sigma = \tau\), the commutator vanishes because we end up with occupation number operators, which commute with \(\hat{\Omega}\). In the case \(\sigma \neq \tau\), which means \(\tau = \bar{\sigma}\), we end up with either \(\hat{S}^+_p\) and \(\hat{S}^-_q\) (in the case \(\sigma = \alpha\)) or \(\hat{S}^-_p\) and \(\hat{S}^+_q\) (in the case \(\sigma = \beta\)). In any of both cases, the commutator is zero, by equation \(\eqref{eq:commutator_seniority_S}\)). \end{enumerate} We thus get \[\begin{equation} \label{eq:seniority_preserving_e_pqrs_generalized} \begin{split} \hat{a}^\dagger_{p \sigma} \hat{a}^\dagger_{r \tau} % \hat{a}_{s \tau} \hat{a}_{q \sigma}% \xrightarrow{\delta\Omega=0} % &\delta_{pq} \delta_{pr} \delta_{ps} % \hat{N}_{p \sigma} % ( \hat{N}_{p \tau} - \delta_{\sigma \tau} ) \\ &+\delta_{pq} \delta_{rs} (1 - \delta_{qr}) % \hat{N}_{p \sigma} \hat{N}_{r \tau} \\ &+\delta_{pr} \delta_{qs} % (1 - \delta_{qr}) % (1 - \delta_{\sigma \tau}) % \hat{P}^+_p \hat{P}^-_q \\ &-\delta_{ps} \delta_{qr} (1 - \delta_{pq}) % \hat{a}^\dagger_{p \sigma} % \hat{a}_{p \tau} % \hat{a}^\dagger_{q \tau} % \hat{a}_{q \sigma} % \thinspace . \end{split} \end{equation}\] We can check and verify that summing equation \(\eqref{eq:seniority_preserving_e_pqrs_generalized}\) over \(\sigma\) and \(\tau\) yields equation \(\eqref{eq:seniority_preserving_e_pqrs_initial}\).
Raising and lowering operators
Let’s say we have a wave function \(\ket{n}\), characterized by the quantum number \(n\). This means that \(n\) is an eigenfunction of an operator \(\hat{N}\): \[\begin{equation} \hat{N} \ket{n} = n \ket{n} \thinspace . \end{equation}\] If we furthermore have an operator \(\hat{X}^+\), whose action is to raise the quantum number \(n\) of \(\ket{n}\) by \(c\): \[\begin{equation} \hat{X}^+ \ket{n} = \ket{n+c} \thinspace , \end{equation}\] then we can show that \(\hat{N}\) and \(\hat{X}^+\) obey the following commutation relation: \[\begin{equation} \label{eq:commutator_counter_raiser} \comm{\hat{N}}{\hat{X}^+} = c \hat{X}^+ \thinspace . \end{equation}\]
If furthermore \(\hat{N}\) is Hermitian, then from this commutator follows that the adjoint of \(\hat{X}^+\), defined as \[\begin{equation} \hat{X}^- = \qty(\hat{X}^+)^\dagger \end{equation}\] obeys \[\begin{equation} \comm{\hat{N}}{\hat{X}^-} = -c \hat{X}^- \thinspace , \end{equation}\] and thus its action is to lower the quantum number \(n\) of \(\ket{n}\) by \(c\).
Seniority raising and lowering operators
In the partitioning of the Hamiltonian according to seniority, cfr. equation \(\eqref{eq:seniority_components_Hamiltonian}\), we did not make a distinction between seniority-raising parts and seniority-lowering parts of the Hamiltonian. If we do, we should be able to write the Hamiltonian as \[\begin{equation} \label{eq:Hamiltonian_seniority_partitioning} \hat{\mathcal{H}}_\text{elec} % = \sum_{\nu = 0, \pm 2, \pm 4} % \hat{ \mathcal{H} } % _{\text{elec}} % ^{\delta \Omega = \nu} % \thinspace . \end{equation}\]
Since the Hamiltonian is a singlet operator, we only have to focus on the singlet excitation operators \(\hat{E}_{pq}\) and possibly \(\hat{e}_{pqrs}\). Let us begin with the beginning, and explore the one-electrong singlet excitation operator \(\hat{E}_{pq}\). We already know the following commutator: \[\begin{equation} \comm{ % \hat{\Omega} % }{ % \hat{E}^\sigma_{pq} % } % = 2 \hat{E}^\sigma_{pq} ( % \hat{N}_{q \bar{\sigma}} - \hat{N}_{p \bar{\sigma}} % ) % \tag{\ref{eq:commutator_E_pq_seniority_spin}} \thinspace , \end{equation}\] but we see that doesn’t quite have the required form (cfr. equation \(\ref{eq:commutator_counter_raiser}\)). It is pretty close, though, so let’s explore a more elaborate \(\hat{E}^\sigma_{pq}\)-operator. If we would like to raise the seniority number by 2, we should start from a doubly occupied orbital, and excite to an empty orbital. We could then proceed by suggesting the operator \(\hat{\Omega}^{2+}_{pq, \sigma}\) as \[\begin{equation} \hat{\Omega}^{2+}_{pq, \sigma} % = \hat{E}^\sigma_{pq} % (1 - \hat{N}_{p \alpha}) % (1 - \hat{N}_{p \beta}) % \hat{N}_{q \alpha} % \hat{N}_{q \beta} % \thinspace . \end{equation}\] First of all, we can immediately rewrite this operator as \[\begin{equation} \hat{\Omega}^{2+}_{pq, \sigma} % = \hat{E}^\sigma_{pq} % (1 - \hat{N}_{p \sigma}) % (1 - \hat{N}_{p \bar{\sigma}}) % \hat{N}_{q \sigma} % \hat{N}_{q \bar{\sigma}} % \label{eq:seniority_raising_2_first} \thinspace , \end{equation}\] because there are only two possibilities (\(\alpha\) or \(\beta\)) for \(\sigma\) to take values from. Since we further more have: \[\begin{equation} \hat{E}^\sigma_{pq} \hat{N}_{p \sigma} = 0 % \tag{\ref{eq:E_N_zero}} \end{equation}\] and \[\begin{equation} \hat{E}^\sigma_{pq} \hat{N}_{q \sigma} = \hat{E}^\sigma_{pq} % \tag{\ref{eq:E_N_E}} \thinspace , \end{equation}\] we can simplify our suggested operator to \[\begin{equation} \hat{\Omega}^{2+}_{pq, \sigma} % = \hat{E}^\sigma_{pq} % (1 - \hat{N}_{p \bar{\sigma}}) % \hat{N}_{q \bar{\sigma}} % \label{eq:seniority_raising_2} \thinspace , \end{equation}\] for the case where \(p \neq q\). To be compatible with the first definition \(\eqref{eq:seniority_raising_2_first}\), we require our operator \(\hat{\Omega}^{2+}_{pq, \sigma}\) to take the form of \(\eqref{eq:seniority_raising_2}\) and to vanish when \(p = q\), which is what we would intuitively would expect: we can’t raise seniority by two if we only have one orbital to our disposal. Effectively, we can thus write: \[\begin{equation} \hat{\Omega}^{2+}_{pq, \sigma} % = (1 - \delta_{pq}) \hat{E}^\sigma_{pq} % (1 - \hat{N}_{p \bar{\sigma}}) % \hat{N}_{q \bar{\sigma}} % \label{eq:seniority_raising_2} \thinspace . \end{equation}\] We will now proceed to calculate the all-important commutator \[\begin{equation*} \comm{ % \hat{\Omega} % }{ % \hat{\Omega}^{2+}_{pq, \sigma} % } % \thinspace . \end{equation*}\] In order to do so, we immediately notice that the factors to the right of the excitation operator only consist of number operators, which commute with the seniority operator. Therefore, we immediately have: \[\begin{equation} \comm{ % \hat{\Omega} % }{ % \hat{\Omega}^{2+}_{pq, \sigma} % } % = (1 - \delta_{pq}) % \comm{ % \hat{\Omega} % }{ % \hat{E}_{pq}^\sigma % } % (1 - \hat{N}_{p \bar{\sigma}}) % \hat{N}_{q \bar{\sigma}} % \thinspace . \end{equation}\] Plugging in the known commutator \(\eqref{eq:commutator_E_pq_seniority_spin}\), and recognizing that the spin orbital number operators are idempotent, we indeed find that \(\hat{\Omega}^{2+}_{pq, \sigma}\) raises seniority by two because \[\begin{equation} \comm{ % \hat{\Omega} % }{ % \hat{\Omega}^{2+}_{pq, \sigma} % } % = 2 \thinspace \hat{\Omega}^{2+}_{pq, \sigma} % \thinspace . \end{equation}\] This means that if we define \[\begin{equation} \hat{\Omega}^{2+}_{pq} % = \sum_\sigma \hat{\Omega}^{2+}_{pq, \sigma} \thinspace , \end{equation}\] we have found an operator that raises seniority by two, irrespective of spin, in the sense that it obeys a seniority-raising commutator: \[\begin{equation} \comm{ % \hat{\Omega} % }{ % \hat{\Omega}^{2+}_{pq} % } = 2 \thinspace \hat{\Omega}^{2+}_{pq} % \thinspace . \end{equation}\] Furthermore, since the seniority operator is Hermitian, according to the previous section, an operator that lowers the seniority by two can then be written as: \[\begin{align} \hat{\Omega}^{2-}_{pq} % = \qty( % \hat{\Omega}^{2+}_{pq} % )^\dagger \\ &= \sum_\sigma % \hat{\Omega}^{2-}_{pq, \sigma} \thinspace , \end{align}\] in which we have defined the seniority-lowering operator \[\begin{align} \hat{\Omega}^{2-}_{pq, \sigma} % &= \qty( % \hat{\Omega}^{2+}_{pq, \sigma} % )^\dagger \\ &= (1 - \delta_{pq}) \hat{E}^\sigma_{qp} % (1 - \hat{N}_{p \bar{\sigma}}) % \hat{N}_{q \bar{\sigma}} % \thinspace , \end{align}\] which means that for the excitation of an \(\alpha\)-electron from orbital \(p\) to \(q\), orbital \(p\) should be empty for \(\beta\) (i.e. \(p\) is occupied only with an \(\alpha\)-electron) and \(q\) should be only occupied for \(\beta\), which is something we would intuitively expect if we would try to lower seniority by \(2\).
We can then show that the operator \(\hat{\Omega}^{4+}_{pqrs}\), defined as: \[\begin{equation} \hat{\Omega}^{4+}_{pqrs} = % \hat{\Omega}^{2+}_{pq} \hat{\Omega}^{2+}_{rs} \end{equation}\] raises seniority by four, since the required commutator holds: \[\begin{equation} \comm{ % \hat{\Omega} % }{ % \hat{\Omega}^{4+}_{pqrs} % } % = 4 \thinspace \hat{\Omega}^{4+}_{pqrs} % \thinspace . \end{equation}\]
Open questions and ideas
We were able to calculate the seniority-preserving Hamiltonian by requiring that the commutator \[\begin{equation} \comm{ % \hat{\mathcal{H}} % }{ % \hat{\Omega} % } = 0 \end{equation}\] vanishes. Should we, according to the discussion in section \(\ref{sec:raising_lowering}\) require similar commutators, such as \[\begin{equation} \comm{ % \hat{\mathcal{H}}^{\delta\Omega=2}_\text{elec} % }{ % \hat{\Omega} % } % = 2 \hat{\mathcal{H}}^{\delta\Omega=2}_\text{elec} \end{equation}\] and \[\begin{equation} \comm{ % \hat{\mathcal{H}}^{\delta\Omega=4}_\text{elec} % }{ % \hat{\Omega} % } % = 4 \hat{\mathcal{H}}^{\delta\Omega=4}_\text{elec} % \thinspace ? \end{equation}\]
If we have found these seniority-raising and seniority-lowering operators, how can we smuggle them inside the electronic Hamiltonian?
Does a generalization of the seniority number as \[\begin{equation} \hat{\Omega}_{pq} = (\hat{N}_{p \alpha} - \hat{N}_{q \beta})^2 \end{equation}\] help?
Does the requirement \[\begin{equation} \ev{\comm*{\hat{\mathcal{H}}}{\hat{\Omega}}}{\Psi} = 0 \end{equation}\] help?
After a discussion with Jeppe Olsen at ESQC 2019, he suggested that we could play around with the \[\begin{equation*} (1 - \hat{N}_{p \bar{\sigma}}) \end{equation*}\] part using the relation \[\begin{equation} \hat{N}_{p \sigma} + (1 - \hat{N}_{p \sigma}) = 1 % \thinspace . \end{equation}\]
Another route that Jeppe Olsen at ESQC2019 proposed would be to determine \(\hat{\Omega}^{4+}\) and \(\hat{\Omega}^{4-}\) and thus \(\hat{\mathcal{H}}^{\delta\Omega=4}_\text{elec}\) first, and then use the Hamiltonian expression \(\eqref{eq:seniority_components_Hamiltonian}\) to yield the unknown \(\hat{\mathcal{H}}^{\delta\Omega=2}_\text{elec}\).
We can calculate that \[\begin{equation} \sum_{pq}^K \qty( % \hat{\Omega}^{2+}_{pq, \alpha} % + \hat{\Omega}^{2-}_{pq, \alpha} % ) % = \sum_{pq}^K % \hat{E}^\alpha_{pq} % (\hat{N}_{p \beta} - \hat{N}_{q \beta})^2 \end{equation}\] and also more generally: \[\begin{equation} \sum_{pq}^K \qty( % \hat{\Omega}^{2+}_{pq} + \hat{\Omega}^{2-}_{pq} % ) % = \sum_{pq}^K \sum_\sigma \hat{E}^\sigma_{pq} % (\hat{N}_{p \bar{\sigma}} - \hat{N}_{q \bar{\sigma}})^2 % \thinspace . \end{equation}\]
We can now try smuggling in these seniority-raising operators inside the Hamiltonian, which is an open research problem. From what I have already calculated, we can write the \(\alpha\)-part of the one-electron Hamiltonian as follows: \[\begin{align} \hat{h}^\alpha % &= \sum_{pq}^K h_{pq} \hat{E}^\alpha_{pq} \\ &= \sum_p^K h_{pp} \hat{N}_{p\alpha} % + \sum_{pq}^K h_{pq} \qty( % \hat{\Omega}^{2+}_{pq, \alpha} % + \hat{\Omega}^{2-}_{qp, \alpha} % ) \notag \\ & \hspace{12pt} + \sum_{pq}^K h_{pq} (1 - \delta_{pq}) % \hat{E}^\alpha_{pq} % ( % 1 - \hat{N}_{q \beta} % + 2 \hat{N}_{p \beta} \hat{N}_{q \beta} % ) % \thinspace , \end{align}\] in which we have already smuggled in \(\hat{\Omega}^{2+}_{pq, \alpha}\) and \(\hat{\Omega}^{2-}_{qp, \alpha}\), but we are still left with some kind of `rest term’ that I don’t know what to do with.
Maybe we don’t have to partition the Hamiltonian as in equation \(\eqref{eq:Hamiltonian_seniority_partitioning}\) per se, but maybe the following is what we need: \[\begin{equation} \matrixel{\mu}{ % \hat{\mathcal{H}} % }{\nu} % = \matrixel{\mu}{ % \hat{\mathcal{H}}^{\delta \Omega = \mu - \nu} % }{\nu} % \thinspace , \end{equation}\] where we have labeled bras and kets with their seniority numbers \(\mu\) and \(\nu\), and we are looking for the part of the Hamiltonian \(\hat{\mathcal{H}}^{\delta \Omega = \mu - \nu}\) that actually couples these determinants.