Mathematically, how does the exchange integral for a closed-shell system reduce to zero?

I am looking at the Hartree-Fock method of approximation for a closed-shell two-electron system.

I have the basis functions

$$\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(s_1) \\ \chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(s_2)$$

where $\vec{x}_i$ is the combined spatial and spin coordinates. The opposite spins are illustrated by the $\alpha$ and $\beta$ labels.

The above gives the total wave function to be the following Slater determinant

$$\Psi(\vec{r}_1,\vec{r}_2,s_1,s_2) = \begin{vmatrix} \psi_1(\vec{r}_1)\alpha(s_1) & \psi_1(\vec{r}_1)\beta(s_1) \\ \psi_1(\vec{r}_2)\alpha(s_2) & \psi_1(\vec{r}_2)\beta(s_2) \end{vmatrix}$$

Let's bypass a lot of math, and just look at the exchange integral, where the spin and spatial parts have been separated:

$$\int ds_1 \alpha^*(s_1)\beta(s_1) \cdot \int ds_2 \beta^*(s_2)\alpha(s_2) \cdot \int d\vec{r}_1 \int d\vec{r}_2 \psi_1^*(\vec{r}_1) \psi_1^*(\vec{r}_2) \dfrac{1}{r_{12}} \psi_1(\vec{r}_2) \psi_1(\vec{r}_1)$$

So the spin integrals do not survive in this case, due to the closed shells, but I am not sure if I see why. Is there some orthonormality for the spin states at play here?

• I corrected the typesetting and an obvious mistake (missing complex conjugation) without changing the actual meaning. Note specifically that for vectors with subscripts one better include only the label and not the subscript inside \vec{}, i.e. $\vec{r}_{1}$ (\vec{r}_{1}) usually looks better than $\vec{r_1}$ (\vec{r_1}). Oct 28 '15 at 21:09
• Are you sure, by the way, that the second spin orbital has a different spatial part $\psi_2$ and not the same $\psi_1$ as the first spin orbital? I mean, usually for a closed shell case, the restricted formalism is used in which spin orbitals come in pairs with the same spatial part: $\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(s_1)$ and $\chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(s_2)$. Oct 28 '15 at 21:33
• Ohh, you're right. Is that due to that both electrons are occupied in the same orbital, and are thus restricted to the same spatial domain?
– Yoda
Oct 28 '15 at 21:46
• Yes, the usual restricted picture for a closed shell system insists on electron "occupying" the same spatial orbital. I don't think it matters for the very question though, since the exchange integral vanishes due to the defining properties of the spin functions irregardless of this restriction. Oct 28 '15 at 21:49

Let's do the math as explicit as possible. So, the exchange integral is defined as follows, $$\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle := \sum\limits_{m_{s1}=-1/2}^{+1/2} \sum\limits_{m_{s2}=-1/2}^{+1/2} \iint\limits_{-\infty}^{+\infty} \bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2) r_{12}^{-1} \chi_2(\vec{x}_1) \chi_1(\vec{x}_2) \mathrm{d} \vec{r}_1 \mathrm{d} \vec{r}_2 \, ,$$ where I used a more appropriate label for the spin coordinate ($m_{s}$) as well as distinguished between summation over discrete variables $m_{s1}$ and $m_{s2}$ and integration over continuous $\vec{r}_1$ and $\vec{r}_2$ ones.

Then, for the restricted spin orbitals, $$\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(m_{s1}) \, , \\ \chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(m_{s2}) \, ,$$ where $\alpha$ and $\beta$ are the so-called "spin up" and "spin down" spin functions defined as follows, $$\alpha(m_{s}) = \begin{cases} 1, & m_{s} = +1/2 \\ 0, & m_{s} = -1/2 \end{cases} \, , \quad \beta(m_{s}) = \begin{cases} 0, & m_{s} = +1/2 \\ 1, & m_{s} = -1/2 \end{cases} \, ,$$ we get then $$\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle \\ = \sum\limits_{m_{s1}=-1/2}^{+1/2} \sum\limits_{m_{s2}=-1/2}^{+1/2} \iint\limits_{-\infty}^{+\infty} \bar{\psi}_1(\vec{r}_1) \alpha(m_{s1}) \bar{\psi}_1(\vec{r}_2) \beta(m_{s2}) r_{12}^{-1} \psi_1(\vec{r}_1) \beta(m_{s1}) \psi_1(\vec{r}_2) \alpha(m_{s2}) \\ = \sum\limits_{m_{s1}=-1/2}^{+1/2} \sum\limits_{m_{s2}=-1/2}^{+1/2} \alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) \iint\limits_{-\infty}^{+\infty} \bar{\psi}_1(\vec{r}_1) \bar{\psi}_1(\vec{r}_2) r_{12}^{-1} \psi_1(\vec{r}_1) \psi_1(\vec{r}_2) \, ,$$ where we used the fact that spin functions $\alpha$ and $\beta$ are real-valued, so that $\bar{\alpha} = \alpha$ and $\bar{\beta} = \beta$.

At this point we could concentrate our attention exclusively on the coefficient in front of the double integral above, since we could show that it is equal to zero, so that the whole expression vanishes irregardless of the value of the integral. So, the coefficient can be written as follows, $$\sum\limits_{m_{s1}=-1/2}^{+1/2} \sum\limits_{m_{s2}=-1/2}^{+1/2} \alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) \\ = \alpha(-1/2) \beta(-1/2) \beta(-1/2) \alpha(-1/2) + \alpha(-1/2) \beta(+1/2) \beta(-1/2) \alpha(+1/2) \\ + \alpha(+1/2) \beta(-1/2) \beta(+1/2) \alpha(-1/2) + \alpha(+1/2) \beta(+1/2) \beta(+1/2) \alpha(+1/2) \\ = 0 + 0 + 0 + 0 \\ = 0 \, ,$$ where all the terms vanish due to at least either $\alpha(-1/2) = 0$ or $\beta(+1/2) = 0$, so that we indeed get $$\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle = 0 \, .$$

It can be noted that there is a shortcut in proving that the coefficient above is zero that uses the defining properties of spin functions. First, we could rearrange spin functions as follows, $$\sum\limits_{m_{s1}=-1/2}^{+1/2} \sum\limits_{m_{s2}=-1/2}^{+1/2} \alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) = \sum\limits_{m_{s1}=-1/2}^{+1/2} \alpha(m_{s1}) \beta(m_{s1}) \sum\limits_{m_{s2}=-1/2}^{+1/2} \beta(m_{s2}) \alpha(m_{s2}) \, .$$ Secondly we note that by the very definition of $\alpha$ and $\beta$ presented above, $$\sum\limits_{m_{s}=-1/2}^{+1/2} \alpha(m_{s}) \beta(m_{s}) = \alpha(-1/2) \beta(-1/2) + \alpha(+1/2) \beta(+1/2) = 0 \cdot 1 + 1 \cdot 0 = 0 \, ,$$ which is already enough to establish that the whole exchange integral vanishes. The same is, of course, true for the second factor as well, $$\sum\limits_{m_{s}=-1/2}^{+1/2} \beta(m_{s}) \alpha(m_{s}) = \beta(-1/2) \alpha(-1/2) + \beta(+1/2) \alpha(+1/2) = 1 \cdot 0 + 0 \cdot 1 = 0 \, .$$ It is also quite customary to use Dirac bracket notation for such expressions over spin orbitals, so that one could write these findings concisely as $$\langle \alpha \lvert \beta \rangle = \langle \beta \lvert \alpha \rangle = 0 \, ,$$ where $\langle \alpha \lvert \beta \rangle$, for instance, is defined as follows, $$\langle \alpha \lvert \beta \rangle := \sum\limits_{m_{s}=-1/2}^{+1/2} \bar{\alpha}(m_{s}) \beta(m_{s}) \, .$$ In general, for two spin functions $\gamma_1$ and $\gamma_2$, expression $\langle \gamma_1 \lvert \gamma_2 \rangle$ defined as follows, $$\langle \gamma_1 \lvert \gamma_2 \rangle := \sum\limits_{m_{s}=-1/2}^{+1/2} \bar{\gamma_1}(m_{s}) \gamma_2(m_{s}) \, ,$$ is zero if the functions correspond to different spin states and one otherwise which can be expressed as follows, $$\langle \gamma_1 \lvert \gamma_2 \rangle = \delta_{\gamma_1 \gamma_2} \, .$$

Finally, I would like to make a short comment on "integrating" over discrete spin coordinates rather then summing up over it. One could indeed define $$\langle \gamma_1 \lvert \gamma_2 \rangle := \int \bar{\gamma_1}(m_s) \gamma_2(m_s) \mathrm{d} m_s \, ,$$ and then think of integration over discrete $m_s$ variable being reduced to summation. But the integration here is just a "symbolic shorthand" since, strictly speaking, we have a summation over discrete $m_s$ variable from the get go. We could just symbolically write it down as an integration as well, if we think it looks more cute for some reason.

In exact same way the exchange integral can also be symbolically defined as follows, $$\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle := \iint \bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2) r_{12}^{-1} \chi_2(\vec{x}_1) \chi_1(\vec{x}_2) \mathrm{d} \vec{x}_1 \mathrm{d} \vec{x}_2 \, ,$$ where the "integration" is informally done over joint spin-spatial coordinates of the electrons. And again, strictly speaking, a summation is done over the spin coordinates and an integration over spatial ones, as it is written explicitly in the very beginning of the answer.

• Actually, I think \left| and \right| works best. Oct 28 '15 at 21:21
• @orthocresol, MathJax appearance is browser dependent: what looks nicer in one, might looking worse in another. But let me check. Oct 28 '15 at 21:28
• @orthocresol, mmm... Wait! Where exactly do you want to use \left| and \right|? Oct 28 '15 at 21:30
• I see... which looks good to you? meta.chemistry.stackexchange.com/a/2989/16683 My reasoning was, the braket package in $\LaTeX$ defines it that way too. Oct 28 '15 at 21:31
• @AndersMB, there you go then, page 35, lines 10-12: "The spin part defines the electron spin and is labelled $\alpha$ or $\beta$. These spin functions have the value $0$ or $1$ depending on the quantum number $m_s$ of the electron. Thus $\alpha(\frac{1}{2})=1$, $\alpha(-\frac{1}{2})=0$, $\beta(\frac{1}{2})=0$, $\beta(-\frac{1}{2})=1$. Oct 28 '15 at 22:45