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Chemistry LibreTexts

22.6.6: v. Exercise Solutions

  • Page ID
    85563
  • Q1

    \[F \phi_i = \varepsilon_i\phi_j = h \phi_i + \sum_j \left[ J_j - K_j \right] \phi_i \]
    Let the closed shell Fock potential bewritten as:
    \begin{align}V_{ij} &= \sum_k \left( 2\langle ik | jk \rangle - \langle ik | kj \rangle \right) \text{, and the 1e}^- \text{component as:} \\ h_{ij} &= \langle \phi_i - \dfrac{1}{2} \nabla^2 - \sum\limits_A \dfrac{Z_A}{|r - R_A|}| \phi_j \rangle \text{, and the delta as:} \\ \delta_{ij} &= \langle i|j \rangle \text{, so that: } h_{ij} + V_{ij} = \delta_{ij}\varepsilon_{i} \\ \text{using: } \phi_i &= \sum\limits_\mu C_{\mu i}\chi_{\mu}, \phi_j = \sum\limits_\nu C_{\nu j}\chi_\nu \text{, and } \phi_k = \sum\limits_\gamma C_{\gamma k} \chi_\gamma \end{align}
    , and transforming from the mo to ao basis we obtain:
    \begin{align} V_{ij} &= \sum\limits_{k\mu\gamma\nu k} C_{\mu i} C_{\gamma k}C_{\nu j} C_{\kappa k} \left( 2\langle \mu\gamma |\nu\kappa \rangle - \langle \mu\gamma | \kappa \nu \rangle \right) \\ &= \sum\limits_{k \mu \gamma \nu\kappa} \left(C_{\gamma k} C_{\kappa k}\right)\left(C_{\mu i}C_{\nu j}\right) \left( 2\langle \mu\gamma | \nu\kappa \rangle - \langle \mu\gamma | \kappa\nu \rangle \right) \\ &= \sum\limits_{\mu\nu} \left( C_{\mu i} C_{\nu j} \right) V_{\mu \nu} \text{ where, } \\ V_{\mu\nu} &= \sum\limits_{\gamma\kappa} P_{\gamma\kappa} \left( 2\langle \mu\gamma | \nu\kappa \rangle - \langle \mu\gamma | \kappa\nu \rangle \right) \text{, and } P_{\gamma\kappa} = \sum\limits_k \left( C_{\gamma k}C_{\kappa k} \right) , \\ h_{ij} &= \sum\limits_{\mu\nu} \left( C_{\mu i}C_{\nu j} \right) h_{\mu\nu} \text{ , where } \\ h_{\mu\nu} &= \langle \chi_\mu | - \dfrac{1}{2} \nabla^2 - \sum\limits_A \dfrac{Z_A}{|r - R_A|} | \chi_\nu \rangle \text{ , and } \\ \delta_{ij} &= \langle i|j \rangle = \sum\limits_{\mu\nu} \left( C_{\mu i} S_{\mu\nu} C_{\nu j} \right). \end{align}
    SO, \( h_{ij} + V_{ij} = \delta_{ij}\varepsilon_j\) becomes:
    \begin{align} & \sum\limits_{\mu\nu} \left( C_{\mu i}C_{\nu j} \right) h_{\mu\nu} + \sum\limits_{\mu\nu} \left( C_{\mu i}C_{\nu j}\right) V_{\mu\nu} = \sum\limits_{\mu\nu} \left( C_{\mu i}S_{\mu\nu}C_{\nu j} \right) \varepsilon_j, \\ & \sum\limits_{\mu\nu} \left( C_{\mu i}S_{\mu\nu} C_{\nu j} \right) \varepsilon_ - \sum\limits_{\mu\nu} \left( C_{\mu i}C_{\nu j}\right) h_{\mu\nu} - \sum\limits_{\mu\nu} \left( C_{\mu i}C_{\nu j} \right) V_{\mu\nu} = 0 \text{ for all i,j } \\ & \sum\limits_{\mu\nu} C_{\mu i} \left[ \varepsilon_jS_{\mu\nu} - h_{\mu\nu} - V_{\mu\nu} \right] C_{\nu j} = 0 \text{ for all i,j} \\ \text{Therefore, } & \\ \sum\limits_\nu \left[ h_{\mu\nu} + V_{\mu\nu} - \varepsilon_jS_{\mu\nu} \right] C_{\nu j} = 0 \end{align}
    Tis is FC = SCE.

    Q2

    The Slater Condon rule for zero (spin orbital) difference with N electrons in N spin orbitals is:
    \begin{align} E &= \langle |H + G|\rangle = \sum\limits_i^N \langle \phi_i |h|\phi_i \rangle + \sum\limits_{i>j}^N \left( \langle \phi_i\phi_j |g| \phi_i\phi_j \rangle - \langle \phi_i\phi_j |g| \phi_j\phi_i \rangle \right) \\ &= \sum\limits_i h_{ii} + \sum\limits_{i>j} \left( g_{ijij} - g_{ijji} \right) \\ &= \sum\limits_i h_{ii} + \dfrac{1}{2}\sum\limits_{ij} \left( g_{ijij} - g_{ijji} \right) \end{align}
    If all orbitals are doubly occupied and we carry out the spin integration we obtain:
    \[ E = 2\sum\limits_i^{occ} h_{ii} + \sum\limits_{ij}^{occ} \left( 2g_{ijij} - g_{ijji} \right), \]
    where i and j now refer to orbitals (not spin-orbitals).

    Q3

    If the occupied orbitals obey \( F\phi_k = \varepsilon_k\phi_k\), then the expression for E in problem 2 above can be rewritten as.
    \[ E = \sum\limits_{i}^{occ} \left( h_{ii} + \sum\limits_j^{occ} \left( 2g_{ijij} - g_{ijji} \right) \right) + \sum\limits_{i}^{occ} h_{ii} \]
    We recognize the closed shell Fock operator expression and rewrite this as
    \[ E = \sum\limits_i^{occ} F_{ii} + \sum\limits_i^{occ} h_{ii} = \sum\limits_i^{occ} \left( \varepsilon_i + h_{ii} \right) \]