Spin tensor operators

A spin tensor operator of rank \(S\) is a set of \(2S+1\) operators \(\hat{T}^{S,M}\), where \[\begin{equation} \require{physics} M= -S, -S+1, \ldots, S-1, S \end{equation}\] that fulfills the commutators \[\begin{equation} \comm{\hat{S}^\pm}{ \hat{T}^{S,M} } % = \sqrt{S(S+1) - M(M \pm 1)} \thinspace \hat{T}^{S, M \pm 1} % \thinspace , \end{equation}\] and \[\begin{equation} \comm{\hat{S}^z}{\hat{T}^{S,M}} = % M \thinspace \hat{T}^{S,M} % \thinspace , \end{equation}\] assuming that \(\hat{T}^{S, S+1} = 0\) and \(\hat{T}^{S, -S-1} = 0\). A spin tensor operator with \(S=0\) is called a singlet operator, with \(S=1/2\) a doublet operator and with \(S=1\) a triplet operator, and so on. Combining these commutators, we find that \[\begin{equation} \begin{split} \comm{\hat{S}^2}{ \hat{T}^{S,M} } % = &\sqrt{S(S+1) - M(M+1)} \thinspace \hat{S}^- \hat{T}^{S, M+1} \\ &+ \sqrt{S(S+1) - M(M-1)} \thinspace \hat{T}^{S, M-1} \hat{S}^+ \\ &+ \hat{T}^{S,M} \qty( M(M+1) + 2M \hat{S}^z ) % \thinspace . \end{split} \end{equation}\] We can thus classify the operators according to their commutation with \(\hat{S}^z\) and \(\hat{S}^2\). Spin tensor operators with \(M=0\) necessarily commute with \(\hat{S}^z\): \[\begin{equation} \comm{\hat{S}^z}{ \hat{T}^{S,0} } = 0 % \thinspace . \end{equation}\] Moreover, any singlet operator \(\hat{T}^{0,0}\) (with \(S=0\) and thus \(M=0\)) commutes with both \(\hat{S}^z\) and \(\hat{S}^2\): \[\begin{align} & \comm{\hat{S}^2}{ \hat{T}^{0,0} } = 0 \\ % & \comm{\hat{S}^z}{ \hat{T}^{0,0} } = 0 % \thinspace . \end{align}\]

Spin tensor operators are, in general, not Hermitian. Given a spin tensor operator \(\hat{T}^{S,M}\), let us therefore introduce the operator \[\begin{equation} \label{eq:hermitian_spin_tensor_operator} \hat{U}^{S,M} = (-1)^{S+M} (\hat{T}^{S,-M})^\dagger \thinspace , \end{equation}\] which turns out to be a spin tensor operator as well. Taking the Hermitian adjoint of a spin tensor operator thus leads to a new tensor operator, but with the order of projections reversed.

We can use these spin tensor operators to generate a set of spin eigenfunctions with total spin \(S\) and projected spin \(M\), by letting them act on the vacuum state: \[\begin{equation} \hat{T}^{S,M} \ket{\text{vac}} \thinspace . \end{equation}\] The following equations then define a spin tensor state of rank \(S\): \[\begin{align} & \hat{S}^{\pm} \qty( % \hat{T}^{S,M} \ket{\text{vac}} % ) % = \sqrt{S(S+1) - M(M \pm 1)} \qty( % \hat{T}^{S,M\pm1} \ket{\text{vac}} % ) \\ % & \hat{S}_{z} \qty( % \hat{T}^{S,M} \ket{\text{vac}} % ) % = M \qty( % \hat{T}^{S,M} \ket{\text{vac}} % ) \\ % & \hat{S}^2 \qty( % \hat{T}^{S,M} \ket{\text{vac}} % ) % = S(S+1) \qty( % \hat{T}^{S,M} \ket{\text{vac}} % ) % \thinspace . \end{align}\]

Given the following commutators \[\begin{align} & \comm{ % \hat{S}^{\pm} % }{ % \hat{a}^\dagger_{p, m_s} % } % = \sqrt{\frac{3}{4} - m_s\qty(m_s \pm 1)} \thinspace % \hat{a}^\dagger_{p, m_s \pm 1} \\ % & \comm{ % \hat{S}^z % }{ % \hat{a}^\dagger_{p, m_s} % } % = m_s \hat{a}^\dagger_{p, m_s} % \thinspace , \end{align}\] we can see that the elementary creation operators \(\hat{a}^\dagger_{p \alpha}\) and \(\hat{a}^\dagger_{p \beta}\) form a set of doublet operators, as does the pair \(\qty{-\hat{a}_{p \beta}, \hat{a}_{p \alpha}}\), according to equation \(\eqref{eq:hermitian_spin_tensor_operator}\).

For two-body creation operators, it can be shown that \[\begin{equation} \hat{Q}^{0,0}_{pq} % = \frac{1}{\sqrt{2}} \qty( % \hat{a}^\dagger_{p \alpha} \hat{a}^\dagger_{q \beta} % - \hat{a}^\dagger_{p \beta} \hat{a}^\dagger_{q \alpha} % ) \end{equation}\] creates a two-electron open-shell singlet state by acting on the vacuum state, and that \[\begin{equation} \hat{Q}^{0,0}_{pp} % = \sqrt{2} \hat{a}^\dagger_{p \alpha} \hat{a}^\dagger_{p \beta} \end{equation}\] creates a two-electron closed-shell singlet by acting on the vacuum state. The two-body triplet creation operators are \[\begin{align} & \hat{Q}^{1,1}_{pq} % = \hat{a}^\dagger_{p \alpha} \hat{a}^\dagger_{q \alpha} \\ % & \hat{Q}^{1,0}_{pq} % = \frac{1}{\sqrt{2}} \qty( % \hat{a}^\dagger_{p \alpha} \hat{a}^\dagger_{q \beta} % + \hat{a}^\dagger_{p \beta} \hat{a}^\dagger_{q \alpha} % ) \\ % & \hat{Q}^{1,-1}_{pq} % = \hat{a}^\dagger_{p \beta} \hat{a}^\dagger_{q \beta} % \thinspace . \end{align}\] We can then write the following mixed-spin products of elementary creation operators as: \[\begin{align} & \hat{a}^\dagger_{p \alpha} \hat{a}^\dagger_{q \beta} % = \frac{1}{\sqrt{2}} \qty( % \hat{Q}^{1,0}_{pq} % + \hat{Q}^{0,0}_{pq} ) \\ & \hat{a}^\dagger_{p \beta} \hat{a}^\dagger_{q \alpha} % = \frac{1}{\sqrt{2}} \qty( % \hat{Q}^{1,0}_{pq} % - \hat{Q}^{0,0}_{pq} ) % \thinspace . \end{align}\]

For excitation operators, the true singlet operator is \[\begin{equation} \hat{S}^{0,0}_{pq} % = \frac{1}{\sqrt{2}} \qty( % \hat{a}^\dagger_{p \alpha} \hat{a}_{q \alpha} % + \hat{a}^\dagger_{p \beta} \hat{a}_{q \beta}% ) % \thinspace , \end{equation}\] and thus the one that we normally use in writing down the Hamiltonian is related to it by: \[\begin{equation} \hat{E}_{pq} = \sqrt{2} \hat{S}^{0,0}_{pq} % \thinspace . \end{equation}\] The triplet components of the excitation operators are given by: \[\begin{align} & \hat{T}^{1,1}_{pq} = - \hat{a}^\dagger_{p \alpha} \hat{a}_{q \beta} \\ % & \hat{T}^{1,0}_{pq} = \frac{1}{\sqrt{2}} \qty( % \hat{a}^\dagger_{p \alpha} \hat{a}_{q \alpha} % - \hat{a}^\dagger_{p \beta} \hat{a}_{q \beta}% ) \\ % & \hat{T}^{1,-1}_{pq} = \hat{a}^\dagger_{p \beta} \hat{a}_{q \alpha} % \thinspace . \end{align}\] We can therefore write that \[\begin{align} & \hat{a}^\dagger_{p \alpha} \hat{a}_{q \alpha} % = \frac{1}{2} \hat{E}_{pq} % + \frac{1}{\sqrt{2}} \hat{T}^{1,0}_{pq} \\ % & \hat{a}^\dagger_{p \beta} \hat{a}_{q \beta} % = \frac{1}{2} \hat{E}_{pq} % - \frac{1}{\sqrt{2}} \hat{T}^{1,0}_{pq} % \thinspace , \end{align}\] which are formulas that are suitable to change from elementary-operator-based expressions to singlet-and-triplet-based expressions.

Furthermore, the Hermitian adjoints of the triplet excitation operators are \[\begin{equation} \qty(\hat{T}^{1,M})^\dagger = (-1)^M \hat{T}^{1,-M} % \thinspace . \end{equation}\]

The diagonal elements of the triplet excitation operators can be linked to the spin operators: \[\begin{align} & \hat{T}^{1,1}_{pp} = - \hat{S}^+_p \\ % & \hat{T}^{1,0}_{pp} = \frac{1}{\sqrt{2}} \hat{S}^z_p \\ % & \hat{T}^{0,0}_{pp} = \hat{S}^-_p % \thinspace . \end{align}\]

We also have a singlet two-electron excitation operator: \[\begin{equation} \hat{e}_{pqrs} = % \sum_{\sigma \tau} % \hat{a}^\dagger_{p \sigma} % \hat{a}^\dagger_{r \tau} % \hat{a}_{s \tau} % \hat{a}_{q \sigma} % \thinspace , \end{equation}\]