Slater-Condon rule for a one-electron operator

The Slater-Condon rule for a one-electron operator is an expression of the expectation value of the one-electron operator involving Slater determinants.

Let’s consider the expectation value of the one-electron Hamiltonian operator: $\langle\Psi_a\vert\hat h\vert\Psi_a\rangle$ , where $\hat h=\sum_{i=1}^n\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)$ and $\Psi_a$ is given by eq66.

Question

Show that the product of $\hat h$ and $\hat A$ is Hermitian.

Answer

$(\hat h\hat A)_{ij}^{\dagger}=[(\hat h\hat A)_{ij}^T]^*=(\hat h\hat A)_{ji}^*=\left(\sum_kh_{jk}a_{ki}\right)^*=\left[\sum_k(h^T)_{kj}(a^T)_{ik}\right]^*$ $=\left[\sum_k(a^T)_{ik}(h^T)_{kj}\right]^*=(\hat A^T\hat h^T)_{ij}^*=(\hat A^{\dagger}\hat h^{\dagger})_{ij}$

or $(\hat h\hat A)_{ij}^{\dagger}=(\hat A^{\dagger}\hat h^{\dagger})_{ij}$ .

With reference to eq64 and the fact that $\hat h=\hat h^{\dagger}$ , we have $(\hat h\hat A)^{\dagger}=\hat A^{\dagger}\hat h^{\dagger}=\hat A\hat h$ . Using eq65, we have $(\hat h\hat A)^{\dagger}=\hat h\hat A$ , i.e. the product of $\hat h$ and $\hat A$ is Hermitian. This implies that $\hat h\hat O$ , where $\hat O=\hat A^2$ is also Hermitian.

Substitute eq59 in $\langle\Psi_a\vert\hat h\vert\Psi_a\rangle$ and using

i) eq65
ii) the Hermitian property of $\hat h\hat A$
iii) the Hermitian property of $\hat h\hat A^2$ twice

we have

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\langle\hat A\Psi_s\vert\hat h\hat A\vert\Psi_s\rangle=\langle\hat h\hat A^2\Psi_s\vert\Psi_s\rangle=\langle\Psi_s\vert\hat h\hat A^2\vert\Psi_s\rangle$

Substitute eq62 and eq63 in the above equation,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\langle\Psi_s\vert\hat h\sum_{t=1}^{n!}(-1)^t\hat P_t\Psi_s\rangle\;\;\;\;\;\;\;\;(85)$

Substitute $\Psi_s=\phi_1(x_1)\phi_2(x_2)\cdots\phi_x(x_n)$ and $\hat h=\sum_{i=1}^n\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)$ in eq85 and simplifying,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\sum_{i=1}^n\int\phi_1^*(x_1)\phi_1(x_1)dx_1\cdots\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$ $\cdots\int\phi_n^*(x_n)\phi_n(x_n)dx_n$

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\sum_{i=1}^{n}H_i\;\;\;\;\;\;\;\;(86)$

where $H_i=\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$ .

Given that any expectation value of a physical property of an electron is invariant to a unitary transformation of the electron’s wavefunction that is expressed as a Slater determinant, we substitute eq74 and eq75 in eq86 to give

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\sum_{q=1}^n\sum_{i=1}^nu_{iq}^*u_{ir}\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_q'(x)dx\;\;\;\;\;\;\;\;(87)$

where we have changed the dummy variable from $x_i$ to $x$ .

Substitute eq82 in eq87,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\sum_{q=1}^n\delta_{qr}\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_q'(x)dx$

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_r'(x)dx\;\;\;\;\;\;\;\;(88)$

Eq88 is used in the derivation of the Hartree-Fock equations.

Slater-Condon rule for a one-electron operator

Question

Answer

Next article: Slater-Condon rule for a two-electron operator

Previous article: Slater determinant of unitarily transformed spin orbitals

Content page of computational chemistry

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply