6
$\begingroup$

I know how to show that the Coulomb operator of the Fock operator is invariant under a unitary transformation of the orbitals, as on page 121 of Szabo and Ostlund, but the indices in my proof for the exchange operator are just not working. What might I be missing? I have

\begin{align} \sum_i\hat{K}_i'(1) &=\sum_i\int dr_2\chi_i'^*(2)\chi_j'(2)/r_{12} \\ &=\sum_i\int dr_2\sum_kU_{ki}^*\chi_k^*(2)\sum_lU_{lj}\chi_l(2)/r_{12}\\ &=\sum_{ikl}\int dr_2U_{ki}^*U_{lj}\chi_k^*(2)\chi_l(2)/r_{12}\\ &=\sum_{i}\int dr_2\chi_i^*(2)\chi_i(2)/r_{12}\\ &\neq\sum_{i}\int dr_2\chi_i^*(2)\chi_j(2)/r_{12}. \end{align}

$\endgroup$
1
  • $\begingroup$ How are you contracting $U_{ki}^* U_{lj}$? $\endgroup$
    – user37142
    Feb 5, 2018 at 12:18

1 Answer 1

7
$\begingroup$

The exchange operator is not a operator by itself, it is only defined with the orbital it is working on: $% \newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle} \newcommand{\lb}{\left|}\newcommand{\rb}{\right|} \newcommand{\op}[1]{\mathbf{#1}}$ $$\begin{align} && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\ \text{with}&& \op{J}_j\lb \phi_i\rr &= \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\ \text{and}&& \op{K}_j\lb \phi_i\rr &= \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr. \end{align}$$

While in the Coulomb operator case the $\lb \phi_i\rr$ doesn't do anything, so you can take it through the proof as a constant:* \begin{align} \sum_j\op{J}_j\lb \phi_i\rr &= \sum_j \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\ &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj} \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\ &= \sum_k \sum_l \delta_{kl} \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\ &= \sum_k \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_k(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\ &= \sum_k \op{J}_k \lb \phi_i\rr \end{align}

That is not the case in the Exchange operator, as it switches the orbitals: \begin{align} \sum_j\op{K}_j\lb \phi_i\rr &= \sum_j \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr\\ &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj} \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\ &= \sum_k \sum_l \delta_{kl} \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\ &= \sum_k \ll \phi_k(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_k(\op{x}_2) \rr\\ &= \sum_k \op{K}_k \lb \phi_i\rr \end{align}


* Let $$ \mathbb{U}=\left(\begin{matrix} U_{11} & U_{12} & \dots \\ U_{21} & U_{22} & \dots \\ \vdots & \vdots & \ddots\\ \end{matrix}\right); \mathbb{U}^\dagger\mathbb{U}=\mathbb{E} \implies U_{ij}^* U_{kl}= \delta_{ij} = \begin{cases} 1; & i = j = k = l \\ 0; & i \neq j \dots\\ \end{cases} .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.