In Szabo and Ostlund's Modern Quantum Chemistry, the matrix elements for the hamiltonian of a two-electron system, with an operator $\hat O$, are written on pages 64-66 as

$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\int dx_1 dx_2 [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]^* \cdot \hat O [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]$$

for a spin orbital $\chi$ of electron x defined as the product of a spacial orbital $\psi(r)$ and spin $\alpha(\omega)$ or $\beta(\omega)$, e.g., $\chi(x)=\psi(r)\alpha(\omega)$.

This is multiplied out into

$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\frac{1}{2} \int dx_1 dx_2 \{\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)+\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)-\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)-\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)\}$$

This expanded equation is what I will base my question on.

For the core-hamiltonian (one-electron integrals),

$$\hat O=h(r_1)+h(r_2)$$

and for the electron repulsion (two-electron integrals),

$$\hat O=\frac{1}{r_{12}}$$

For the core hamiltonian, you can integrate out the electron which is not being acted upon by the operator to reduce the expanded equation to a one-electron integral.

For the electron repulsion, on page 66 the two-electron integral is simplified by stating the first two terms of the expanded equation are equal. This means:

$$\int \chi_1^*(x_1)\chi_2^*(x_2)\frac{1}{r_{12}} \chi_1(x_1)\chi_2(x_2)= \int \chi_2^*(x_1)\chi_1^*(x_2)\frac{1}{r_{12}} \chi_2(x_1)\chi_1(x_2)$$

This is what I do not follow. How is this derived? I can permute the electrons to achieve the same form, but that would introduce a negative sign.

Furthermore, for the equality to be true, does it not imply one of two scenarios: that either the spatial functions are degenerate, i.e.,

$$\chi_1^*(x_1)=\chi_2^*(x_1)$$ $$\chi_2^*(x_2)=\chi_1^*(x_2)$$ $$\chi_1(x_1)=\chi_2(x_1)$$ $$\chi_2(x_2)=\chi_1(x_2)$$

or otherwise that the electrons have the same spin and coordinate such that $x_1=x_2$

$$\chi_1^*(x_1)=\chi_1^*(x_2)$$ $$\chi_2^*(x_1)=\chi_2^*(x_2)$$ $$\chi_1(x_1)=\chi_1(x_2)$$ $$\chi_2(x_1)=\chi_2(x_2)$$

neither of which would make sense.


I believe your statement

I can permute the electrons to achieve the same form, but that would introduce a negative sign.

is mistaken. The exchange integrals arise from permuting one pair of integration variable instances, not two. The latter is the case necessary here. Szabo and Ostlund state on the same page:

Since $r_{12} = r_{21}$, we can interchange the dummy variables of integration in the second term [...].

where the second term appears to be you are troubled with.

  • $\begingroup$ I wasn't sure on what they meant by the dummy variables of integration, or why $r_{12}=r_{21}$ is a relevant property for this interchange. What precisely is being permuted here? I only have the one pair here, $x_1$ and $x_2$, to interchange, correct? $\endgroup$ – Blaise Jul 21 '19 at 19:19
  • $\begingroup$ Basically, one is interested in the distance between the electrons. (That is a directionless quantity, thus $r_{12} = r_{21}$.) For the integral, it is then without consequence how the integration variables are chosen or ordered. (That's why Szabo and Ostlund call them dummy variables.) $\endgroup$ – TAR86 Jul 22 '19 at 4:42

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