Olsen's Full CI $\sigma$ Equations

We now turn our attention to the general formulation of the full CI problem in terms of alpha and beta strings. Later on, we will consider how to modify our results for RAS CI's. We begin by describing Olsen's expressions for $\sigma$, which is the action of the Hamiltonian $\hat{H}$ on the CI vector (or matrix) $C(I_{\alpha}, I_{\beta})$. First we express $\hat{H}$ in second-quantized  form (see section 5).

$$\hat{H} = \sum_{kl}^{n} h_{kl} \hat{E}_{kl} + \frac{1}{2} ... ... \hat{E}_{ij} \hat{E}_{kl} - \delta_{jk} \hat{E}_{il} \right)$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred16 H = _kl^n h_kl Ê_kl + 12 _ijkl^n (ij|kl) ( Ê_ij Ê_kl - _jk Ê_il ) )

where $\hat{E}_{kl}$ is the shift operator

$$\hat{E}_{kl} = \hat{E}^{\alpha}_{kl} + \hat{E}^{\beta}_{kl}$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred17 Ê_kl = Ê^_kl + Ê^_kl )

Again, $\sigma$ is given by

$$\sigma(I_{\alpha}, I_{\beta}) = \sum_{J_{\alpha}, J_{\beta}}... ...(I_{\alpha}) \beta(I_{\beta}) \rangle C(J_{\alpha}, J_{\beta})$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred18 (I_, I_) = _J_, J_ (J_) (J_) | H | (I_) (I_) C(J_, J_) )

As we will proceed to show, $\sigma$ can be split up into three terms: one involving only beta components of the linear group generators ($\sigma _1$), one involving only alpha components of the generators ($\sigma_2$), and one involving mixtures of the two ($\sigma _3$). Inserting equation (6.16) into equation (6.18) yields

 

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}19}\sigma(I_{\alpha}, I_{\beta}) \sum_{J_{\alpha}, J_{\beta}} \langle \beta(J_{\beta}) \alpha(J_{\alpha}) \vert \sum_{kl}^{n} h_{kl} \hat{E}_{kl}$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred19(I_, I_) &=& _J_, J_ (J_) (J_) | _kl^n h_kl Ê_kl )

$$\frac{1}{2} \sum_{ijkl}^{n} (ij\vert kl) \left( \hat{E}_{ij} \hat... ...ght) \vert \alpha(I_{\alpha}) \beta(I_{\beta}) \rangle C(J_{\alpha}, J_{\beta})$$ +

Now expanding the shift operators according to equation (6.17), we write $\sigma$ as a sum of three terms

$$\sigma(I_{\alpha}, I_{\beta}) = \sigma_1(I_{\alpha}, I_{\be... ...ma_2(I_{\alpha}, I_{\beta}) + \sigma_3(I_{\alpha}, I_{\beta})$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred20(I_, I_) = _1(I_, I_) + _2(I_, I_) + _3(I_, I_) )

where

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}21}\sigma_1(I_{\alpha}, I_{\beta}) \sum_{J_{\alpha}, J_{\beta}} \langle \beta(J_{\beta}) \alpha(J_{\alpha}) \vert \sum_{kl}^{n} h_{kl} \hat{E}^{\beta}_{kl}$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred21_1(I_, I_) &=& _J_, J_ (J_) (J_) | _kl^n h_kl Ê^_kl )

$$\frac{1}{2} \sum_{ijkl}^{n} (ij\vert kl) \left( \hat{E}^{\beta}_{... ...ght) \vert \alpha(I_{\alpha}) \beta(I_{\beta}) \rangle C(J_{\alpha}, J_{\beta})$$ +

and

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}22}\sigma_2(I_{\alpha}, I_{\beta}) \sum_{J_{\alpha}, J_{\beta}} \langle \beta(J_{\beta}) \alpha(J_{\alpha}) \vert \sum_{kl}^{n} h_{kl} \hat{E}^{\alpha}_{kl}$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred22_2(I_, I_) &=& _J_, J_ (J_) (J_) | _kl^n h_kl Ê^_kl )

$$\frac{1}{2} \sum_{ijkl}^{n} (ij\vert kl) \left( \hat{E}^{\alpha}_... ...ght) \vert \alpha(I_{\alpha}) \beta(I_{\beta}) \rangle C(J_{\alpha}, J_{\beta})$$ +

and

$$\sigma_3(I_{\alpha}, I_{\beta}) = \sum_{J_{\alpha}, J_{\beta}... ...(I_{\alpha}) \beta(I_{\beta}) \rangle C(J_{\alpha}, J_{\beta})$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred23_3(I_, I_) = _J_, J_ (J_) (J_) | 12 _ijkl^n (ij|kl) ( Ê^_ij Ê^_kl + Ê^_ij Ê^_kl ) | (I_) (I_) C(J_, J_) )

Obviously, these expressions can be simplified further. First, we observe that the expression for $\sigma _1$ contains no $\alpha$operators. Therefore, $\sigma_1(I_{\alpha}, I_{\beta}) = 0$ unless $J_{\alpha} = I_{\alpha}$. Using this fact, and integrating out the $\alpha$ part, we obtain

$$\sigma_1(I_{\alpha}, I_{\beta}) = \sum_{J_{\beta}} \langle \... ...ight) \vert \beta(I_{\beta}) \rangle C(I_{\alpha}, J_{\beta})$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred24_1(I_, I_) = _J_ (J_) | _kl^n h_kl Ê^_kl + 12 _ijkl^n (ij|kl) ( Ê^_ij Ê^_kl - _jk Ê^_il ) | (I_) C(I_, J_) )

Taking out the Kronecker delta term and rearranging, we have

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}25}\sigma_1(I_{\alpha}, I_{\beta}) \sum_{J_{\beta}} \sum_{kl}^{n} h_{kl} \langle \beta(J_{\beta}) \vert \hat{E}^{\beta}_{kl} \vert \beta(I_{\beta}) \rangle C(I_{\alpha}, J_{\beta})$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred25_1(I_, I_) &=& _J_ _kl^n h_kl (J_) | Ê^_kl | (I_) C(I_, J_) )

$$\frac{1}{2} \sum_{J_{\beta}} \sum_{ijl}^{n} (ij\vert jl) \langle ... ...rt \hat{E}^{\beta}_{il} \vert \beta(I_{\beta}) \rangle C(I_{\alpha}, J_{\beta})$$ -

$$\frac{1}{2} \sum_{J_{\beta}} \sum_{ijkl}^{n} (ij\vert kl) \langle... ...j} \hat{E}^{\beta}_{kl} \vert \beta(I_{\beta}) \rangle C(I_{\alpha}, J_{\beta})$$ +

Now the second term can be combined with the first to give equation (9a) of reference [16].

 

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}26}\sigma_1(I_{\alpha}, I_{\beta}) \sum_{J_{\beta}} \sum_{kl}^{n} \langle \beta(J_{\beta}) \vert \ha... ...[ h_{kl} - \frac{1}{2} \sum_j^{n} (kj\vert jl) \right] C(I_{\alpha}, J_{\beta})$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred26_1(I_, I_) &=& _J_ _kl^n (J_) | Ê^_kl | (I_) [ h_kl - 12 _j^n (kj|jl) ] C(I_, J_) )

$$\frac{1}{2} \sum_{J_{\beta}} \sum_{ijkl}^{n} \langle \beta(J_{\be... ...beta}_{kl} \vert \beta(I_{\beta}) \rangle (ij\vert kl) C(I_{\alpha}, J_{\beta})$$ +

Similarly, $\sigma_2$ can be simplified to

 

$${tex2html_deferred}{{tex2html_deferred}6.{tex2html_deferred}27}\sigma_2(I_{\alpha}, I_{\beta}) \sum_{J_{\alpha}} \sum_{kl}^{n} \langle \alpha(J_{\alpha}) \vert ... ...[ h_{kl} - \frac{1}{2} \sum_j^{n} (kj\vert jl) \right] C(J_{\alpha}, I_{\beta})$$ = (tex2html_deferredtex2html_deferred6.tex2html_deferred27_2(I_, I_) &=& _J_ _kl^n (J_) | Ê^_kl | (I_) [ h_kl - 12 _j^n (kj|jl) ] C(J_, I_) )

$$\frac{1}{2} \sum_{J_{\alpha}} \sum_{ijkl}^{n} \langle \alpha(J_{\... ...ha}_{kl} \vert \alpha(I_{\alpha}) \rangle (ij\vert kl) C(J_{\alpha}, I_{\beta})$$ +

For efficient implementation, it is convenient to precompute the quantities

$$h'_{kl} = h_{kl} - \frac{1}{2} \sum_j^{n} (kj\vert jl)$$ (tex2html_deferredtex2html_deferred6.tex2html_deferred28h'_kl = h_kl - 12 _j^n (kj|jl) )

Finally, we simplify $\sigma _3$. It may be rewritten as

$$\sigma_3(I_{\alpha}, I_{\beta}) \frac{1}{2} \sum_{J_{\alpha},J_{\beta}} \sum^n_{ijkl} \langle \be... ...lpha(I_{\alpha}) \beta(I_{\beta}) \rangle (ij\vert kl) C(J_{\alpha}, J_{\beta})$$ =

$$\frac{1}{2} \sum_{J_{\alpha},J_{\beta}} \sum^n_{ijkl} \langle \be... ...lpha(I_{\alpha}) \beta(I_{\beta}) \rangle (ij\vert kl) C(J_{\alpha}, J_{\beta})$$ +

Since we sum over all ijkl, we can permute i and j with k and l. We can also swap $\hat{E}^{\alpha}_{ij}$ and $\hat{E}^{\beta}_{kl}$as can easily be verified. This yields equation (9c) from reference [16].

$$\sigma_3(I_{\alpha}, I_{\beta}) \sum_{J_{\alpha},J_{\beta}} \sum^n_{ijkl} \langle \beta(J_{\beta}... ...ha}_{kl} \vert \alpha(I_{\alpha}) \rangle (ij\vert kl) C(J_{\alpha}, J_{\beta})$$ =

Thus we have written the action of the Hamiltonian on the current CI vector in terms of alpha and beta strings and alpha and beta shift operators. The product $\sigma$ is written as a sum of three terms: the first ($\sigma _1$) involves only beta shift operators, the second ($\sigma_2$) involves only alpha shift operators, and the third ($\sigma _3$) involves both alpha and beta shift operators. Note that, except for the factor $C(I_{\alpha}, J_{\beta})$, $\sigma _1$ is independent of $I_{\alpha}$ so that the algorithm for computing $\sigma _1$ is vectorizable (see below). Analogous results hold for $\sigma_2$. This situation does not obtain in the computation of $\sigma _3$, however, which is the rate-limiting step.