Restricted Open-Shell Hartree-Fock References

A single-determinant restricted open-shell Hartree-Fock (ROHF) wavefunction describing a high-spin open-shell system will be an eigenfunction of ${\hat S}^2$ (i.e., a CSF). This is easy to verify by direct application of the ${\hat S}^2$ operator, which is

$${\hat S}^2 {\hat S} \cdot {\hat S} = \sum_i^N \sum_j^N {\hat s}(i) \cdot {\hat s}(j)$$ =

$${\hat S}_{-} {\hat S}_{+} + {\hat S}_{z} + {\hat S}_z^2,$$ = (46)

where

$${\hat S}_z \sum_i^N {\hat s}_z(i)$$ = (47)

$${\hat S}_z^2 \sum_i^N {\hat s}_z^2(i)$$ = (48)

$${\hat S}_{\pm} \sum_i^N {\hat s}_{\pm}(i)$$ = (49)

and

$$\begin{array}{ccc} {\hat s}_z(i) a_{i}^{\dagger} \vert \rangle= \... ...verline{i}}^{\dagger} \vert \rangle= a_{i}^{\dagger} \vert \rangle. \end{array}$$ (50)

The high-spin ROHF wavefunction can be written as

$$\vert \Phi_{ROHF} \rangle = \prod_{i}^{\rm docc} a_{i}^{\da... ...{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle,$$ (51)

where t and u will denote open-shells. Applying ${\hat S}^2$, this becomes

$${\hat S}^2 \vert \Phi_{ROHF} \rangle = \left[ {\hat S}_{-}... ...{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle.$$ (52)

This is easy to evaluate:

$${\hat S}_z \prod_{i}^{\rm docc} a_{i}^{\dagger} a_{\overline{i}}^{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle \left[ \frac{1}{2}(N_{\alpha} - N_{\beta}) \right] \prod_{i}^{\rm... ...} a_{\overline{i}}^{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle$$ = (53)

$${\hat S}_z^2 \prod_{i}^{\rm docc} a_{i}^{\dagger} a_{\overline{i}}^{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle \left[ \frac{1}{2}(N_{\alpha} - N_{\beta}) \right]^2 \prod_{i}^{\... ... a_{\overline{i}}^{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle.$$ = (54)

The raising operator ${\hat S}_{+}$ yields zero when acting on $\vert \Phi_{ROHF} \rangle$: raising operators applied to $\alpha$ electrons always yield zero, and raising operators applied to the $\beta$electrons yield $\alpha$ spin orbitals which are already occupied (so the determinant vanishes by the Pauli principle). Hence the final result is

$${\hat S}^2 \vert \Phi_{ROHF} \rangle \left[ \frac{1}{4} (N_{\alpha} - N_{\beta})^2 + \frac{1}{2} (N_{\alpha} - N_{\beta}) \right] \vert \Phi_{ROHF} \rangle$$ =

$$\left[ \frac{1}{2} N_s \left( \frac{1}{2} N_s + 1 \right) \right] \vert \Phi_{ROHF} \rangle,$$ = (55)

where $N_s = N_{\alpha} - N_{\beta}$.

Now it is worthwhile to consider how to form CSFs for the single excitations. Determinants which promote electrons from the singly-occupied space to the virtual space, as well as determinants which promote $\beta$ electrons in the doubly-occupied orbitals to the singly-occupied orbitals, are already spin-adapted (the proof is completely analogous to that above for the ROHF reference determinant). The only other relevant single excitations are those which move an electron from a doubly-occupied orbital to a virtual orbital. These determinants are not spin-adapted, as we will proceed to demonstrate. Consider the action of ${\hat S}^2$ on the determinant $\vert \Phi_i^a \rangle$:

$$\left[ \hat{S}_{-} \hat{S}_{+} + {\hat S}_z + {\hat S}_z^2 \r... ...}^{\dagger} \prod_t^{\rm socc} a_{t}^{\dagger} \vert \rangle.$$ (56)

The result of ${\hat S}_z + {\hat S}_z^2$ is easily determined to be

$$\left[ {\hat S}_z + {\hat S}_z^2 \right] \vert \Phi_i^a \rang... ...ft( \frac{N_s}{2} + 1 \right) \right] \vert \Phi_i^a \rangle.$$ (57)

Now all that remains is the raising and lowering operators. These are somewhat more involved and require that attention be paid to the sign.

$${\hat S}_{+} \vert \Phi_i^a \rangle = {\hat S}_{+} a_{a}^{\da... ...{\dagger} \prod_{t}^{\rm socc} a_{t}^{\dagger} \vert \rangle.$$ (58)

By arguments similar to those presented above, all raising operators yield zero except for ${\hat s}_{+}(\overline{i})$. Using the anticommutation relations for creation and annihilation operators,

$${\hat s}_{+}(\overline{i}) a_{a}^{\dagger} a_{i} \vert \Phi_0 \rangle {\hat s}_{+}(\overline{i}) \left( a_{\overline{i}}^{\dagger} a_{\... ...} a_{\overline{i}}^{\dagger} \right) a_{a}^{\dagger} a_{i} \vert \Phi_0 \rangle$$ =

$${\hat s}_{+}(\overline{i}) a_{\overline{i}}^{\dagger} a_{\overline{i}} a_{a}^{\dagger} a_{i} \vert \Phi_0 \rangle$$ =

$$a_{i}^{\dagger} a_{\overline{i}} a_{a}^{\dagger} a_{i} \vert \Phi_0 \rangle$$ =

$$- a_{a}^{\dagger} a_{\overline{i}} \vert \Phi_0 \rangle.$$ = (59)

The ${\hat S}_{-}$ operator can now affect electrons in any of the following orbitals: a, i, and any of the open-shell orbitals.

 

$${\hat S}_{-} \left[ - a_{a}^{\dagger} a_{\overline{i}} \vert \Phi_0 \rangle \right] - a_{\overline{a}}^{\dagger} a_{\overline{i}} \vert \Phi_0 \rangl... ...\dagger} a_{t} a_{\overline{t}}^{\dagger} a_{\overline{i}} \vert \Phi_0 \rangle$$ =

$$\vert \Phi_i^a \rangle - \vert \Phi_{\overline{i}}^{\overline{a}} \rangle + \sum_t^{\rm socc} \vert \Phi_{t \overline{i}}^{a \overline{t}} \rangle.$$ = (60)

Thus overall,

$${\hat S}^2 \vert \Phi_i^a \rangle = \left[ \frac{N_s}{2} \l... ...rm socc} \vert \Phi_{t \overline{i}}^{a \overline{t}} \rangle.$$ (61)

The analogous equation for $\vert \Phi_{\overline{i}}^{\overline{a}} \rangle$is

$${\hat S}^2 \vert \Phi_{\overline{i}}^{\overline{a}} \rangle =... ...rm socc} \vert \Phi_{t \overline{i}}^{a \overline{t}} \rangle.$$ (62)

Clearly a spin eigenfunction can be constructed as

$$\vert ^{(N_s+1)}\Phi_i^a \rangle = \frac{1}{\sqrt{2}} \left... ...e + \vert \Phi_{\overline{i}}^{\overline{a}} \rangle \right).$$ (63)

Then the operation of ${\hat S}^2$ is

$${\hat S}^2 \vert ^{(N_s+1)}\Phi_i^a \rangle = \left[ \frac{N... ...N_s}{2} + 1 \right) \right] \vert ^{(N_s+1)}\Phi_i^a \rangle.$$ (64)

Now that the relevant CSFs have been obtained, they can be used to define the ROHF convergence criteria: the final ROHF wavefunction will not mix with any of the singly substituted CSFs. Thus

$$\begin{array}{cc} \langle \Phi_{ROHF} \vert {\hat H} \vert \Phi_t... ...N_s+1)}\Phi_i^a \rangle = 0 & \mbox{docc $\rightarrow$\space virt}, \end{array}$$ (65)

which implies the following conditions on the Fock matrix elements

F_ta = 0 (66)

$$F_{\overline{i} \overline{t}}$$ = 0 (67)

$$-F_{\overline{i} \overline{a}}.$$ F_ia = (68)

Using these results, we can write down expressions for the $\sigma$vectors. Since determinants $\vert \Phi_i^a \rangle$ must enter with the same coefficients as $\vert \Phi_{\overline{i}}^{\overline{a}} \rangle$, $\sigma_0=0$ once again. Furthermore, since the ROHF reference cannot mix with any other singly excited configurations, the c₀ contribution to $\sigma_{ia}$ and $\sigma_{\overline{i}\overline{a}}$ may be safely ignored. We will therefore consider four cases: $\sigma_i^a$, $\sigma_{\overline{i}}^{\overline{a}}$, $\sigma_t^a$, and $\sigma_{\overline{i}}^{\overline{t}}$, where once again t,u represent singly occupied orbitals.

The equation for $\sigma_i^a$ is readily seen to be

$$\sigma_i^a \sum_{jb} c_j^b \left[ F_{ab} \delta_{ij} - F_{ij} \delta_{ab} + ... ...rline{j}}^{\overline{b}} \langle a\overline{j} \vert\vert i\overline{b} \rangle$$ =

$$\sum_{tb} c_t^b \left[ -F_{it} \delta_{ab} + \langle at \vert\ver... ...line{j}}^{\overline{t}} \langle a\overline{j} \vert\vert i\overline{t} \rangle.$$ + (69)

Separating the Fock operator terms from the two-electron integrals, and integrating out spin, this yields

$$\sigma_i^a \sum_{jb} c_j^b \left[ F_{ab} \delta_{ij} - F_{ij} \delta_{ab} \r... ...t} \delta_{ab} + \sum_{jb} c_j^b \left[ 2 (ai \vert jb) - (ab \vert ji) \right]$$ =

$$\sum_{tb} c_t^b \left[ 2 (ai \vert tb) - (ab \vert ti) \right] + \sum_{\overline{j}\overline{t}} c_{\overline{j}}^{\overline{t}} (ai \vert jt),$$ + (70)

where we have used the relation $c_j^b = c_{\overline{j}}^{\overline{b}}$. The analogous equation for $\beta$ spins is

$$\sigma_{\overline{i}}^{\overline{a}} \sum_{\overline{j}\overline{b}} c_{\overline{j}}^{\overline{b}} \... ... c_{\overline{j}}^{\overline{b}} \left[ 2 (ai \vert jb) - (ab \vert ji) \right]$$ =

$$\sum_{tb} c_t^b (ai \vert tb) + \sum_{\overline{j}\overline{t}} c_{\overline{j}}^{\overline{t}} \left[ 2 (ai \vert jt) - (at \vert ji) \right],$$ + (71)

If the equality $c_j^b = c_{\overline{j}}^{\overline{b}}$ is to be maintained, we must have $\sigma_i^a = \sigma_{\overline{i}}^{\overline{a}}$. It is then computationally convenient to form these quantities as

$$\sigma_i^a = \sigma_{\overline{i}}^{\overline{a}} = \frac{1}{2} ( \sigma_i^a + \sigma_{\overline{i}}^{\overline{a}}).$$ (72)

Thus

$$2 \sigma_i^a = 2 \sigma_{\overline{i}}^{\overline{a}} \sum_{b} c_i^b \left[ F_{ab} + F_{\overline{a}\overline{b}} \righ... ...sum_{\overline{t}} c_{\overline{i}}^{\overline{t}} F_{\overline{a}\overline{t}}$$ =

$$\sum_{jb} c_j^b \left[ 2 (ai \vert jb) - (ab \vert ji) \right] + ... ... c_{\overline{j}}^{\overline{b}} \left[ 2 (ai \vert jb) - (ab \vert ji) \right]$$ +

$$\sum_{tb} c_t^b \left[ 2 (ai \vert tb) - (ab \vert ti) \right] + ... ...c_{\overline{j}}^{\overline{t}} \left[ 2 (ai \vert jt) - (at \vert ji) \right].$$ + (73)

Now it is clear that the two-electron integrals can be treated all together. To begin condensing the notation once again, let us define the following quantities:

$${\tilde P}_{\lambda \sigma}^{DV} \sum_{jb} C_{\lambda j}^* c_j^b C_{\sigma b}$$ = (74)

$${\tilde P}_{\lambda \sigma}^{\overline{DV}} \sum_{\overline{j}\overline{b}} C_{\lambda \overline{j}}^* c_{\overline{j}}^{\overline{b}} C_{\sigma \overline{b}}$$ = (75)

$${\tilde P}_{\lambda \sigma}^{SV} \sum_{tb} C_{\lambda t}^* c_t^b C_{\sigma b}$$ = (76)

$${\tilde P}_{\lambda \sigma}^{\overline{DS}} \sum_{\overline{j}\overline{t}} C_{\lambda \overline{j}}^* c_{\overline{j}}^{\overline{t}} C_{\sigma \overline{t}}$$ = (77)

$${\tilde P}_{\lambda \sigma}^{+} \frac{1}{2} \left[ {\tilde P}_{\lambda \sigma}^{DV} + {\tilde P}_... ... P}_{\lambda \sigma}^{SV} + {\tilde P}_{\lambda \sigma}^{\overline{DS}} \right]$$ = (78)

where ${\tilde P}_{\lambda \sigma}^{+}$ the same as that defined in eq. (17) of Maurice and Head-Gordon [2]. Then $\sigma_i^a$ can be evaluated as

$$\sigma_i^a = \sigma_{\overline{i}}^{\overline{a}} = \frac{1}... ..._{\overline{a}\overline{t}} \right\} + {\tilde F}_{ia}^{+},$$ (79)

where

$${\tilde F}_{ia}^{+} \sum_{\mu \nu} C_{\mu i}^{*} {\tilde F}_{\mu \nu} C_{\nu a}$$ = (80)

$${\tilde F}_{\mu \nu}^{+} \sum_{\lambda \sigma} \left[ 2 (\mu \nu \vert \lambda \sigma) - (\mu \sigma \vert \lambda \nu) \right] {\tilde P}_{\lambda \sigma}^{+}$$ = (81)

Next consider the term $\sigma_t^a$:

$$\sigma_t^a \sum_{jb} \left[ \langle aj \vert\vert tb \rangle - F_{tj} \delta... ... a \overline{j} \vert\vert t \overline{u} \rangle c_{\overline j}^{\overline u}$$ =

$$\sum_{jb} c_j^b \left[ -F_{tj} \delta_{ab} + (at \vert jb) - (ab ... ... \sum_{\overline{j} \overline{b}} c_{\overline{j}}^{\overline{b}} (at \vert jb)$$ =

$$\sum_{ub} c_u^b \left[ (at \vert ub) - (ab \vert ut) + F_{ab} \de... ... \sum_{\overline{j} \overline{u}} c_{\overline{j}}^{\overline{u}} (at \vert ju)$$ +

$${\tilde F}_{ta} - \sum_{j} c_j^a F_{tj} + \sum_b c_t^b F_{ab} - \sum_u c_u^a F_{tu}$$ = (82)

It is computationally more efficient to evaluate $\sigma_t^a$ at the same time as $\sigma_i^a$. The value of $\sigma_t^a$ computed in this manner must of course be corrected for the difference in the formulas for $\sigma_t^a$ and $\sigma_i^a$, but this correction scales as only ${\cal O}(N^2)$ (see Maurice and Head-Gordon [2]). The two-electron part is thus computed by

$${\tilde F}_{ta} = {\tilde F}_{ta}^{+} - \frac{1}{2} \sum_{u... ...} \overline{u}} c_{\overline{j}}^{\overline{u}} (at \vert ju)$$ (83)

Similar considerations apply to $\sigma_{\overline{i}}^{\overline{t}}$, which is

$$\sigma_{\overline{i}}^{\overline{t}} \sum_{jb} c_j^b \langle \overline{t} j \vert\vert \overline{i} b ... ... \overline{t} \overline{j} \vert\vert \overline{i} \overline{u} \rangle \right]$$ =

$$\sum_{\overline{b}} c_{\overline{i}}^{\overline{b}} F_{\overline{... ...}} c_{\overline{j}}^{\overline{b}} \left[ (ti \vert jb) - (tb \vert ji) \right]$$ =

$$\sum_{ub} c_u^b (ti \vert ub) + \sum_{\overline{j} \overline{u}} c_{\overline{j}}^{\overline{u}} \left[ (ti \vert ju) - (tu \vert ji) \right]$$ +

$$\sum_{\overline{b}} c_{\overline{i}}^{\overline{b}} F_{\overline{... ...line{t}} F_{\overline{i} \overline{j}} + {\tilde F}_{\overline{i} \overline{t}}$$ = (84)

where ${\tilde F}_{\overline{i} \overline{t}}$ is actually evaluated according to

$${\tilde F}_{\overline{i} \overline{t}} = {\tilde F}_{\overl... ... \overline{u}} c_{\overline{j}}^{\overline{u}} (tu \vert ji)$$ (85)