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Some context:

I'm having a hard time deriving the results of the Hartree-Fock approximation. Let $H$ have the form $$H = \sum_{i=1}^{n}\left[\frac{p_{i}^{2}}{2 m}+U\left(\vec{r}_{i}\right)\right]+\underbrace{\frac{1}{2} \sum_{i \neq j} \overbrace{\frac{e^{2}}{|\vec{ r}_{i}-\vec{r}_{j} |}}}^{\hspace{1.5cm}=V^{ij}}_{=V_{\rm int}},$$ where $U(\vec{r}_i)$ is some external potentail. We further want $\Psi$ to be of the form $$\Psi=\frac{1}{\sqrt{n !}}\left|\begin{array}{ccc}{\varphi_{\alpha_{1}}(1)} & {\cdots} & {\varphi_{\alpha_{1}}(n)} \\ {\vdots} & {\ddots} & {\vdots} \\ {\varphi_{\alpha_{n}}(1)} & {\cdots} & {\varphi_{\alpha_{n}}(n)}\end{array}\right|.$$ We define the functional $\mathcal{F}$ over $$\mathcal{F}\left[\varphi_{\alpha_{1}}\left(x_{1}\right), \ldots, \varphi_{\alpha_{1}}\left(x_{n}\right)\right]=\langle\Psi|H| \Psi\rangle-\sum_{i} \lambda_{\alpha_{i}}\left\langle\varphi_{\alpha_{i}} | \varphi_{\alpha_{i}}\right\rangle.$$ The result of the derivation was given to us as $$\frac{\delta \mathcal{F}}{\delta\varphi_{\alpha_k}(x)}=0,\quad k\in\{1,\dots,n\},$$ which should be equivalent to $$\begin{aligned}\left[-\frac{\hbar^{2}}{2 m} \vec{\nabla}^{2}+U(\vec{r})+\int d^{3} r^{\prime} \frac{e \rho_{i}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right] \varphi_{\alpha_{i}}(\vec{r}) \\-\sum_{j} \delta_{s_{i} s_{j}} \int d^{3} r^{\prime}\left[\varphi_{\alpha_{j}}^{*}\left(\vec{r}^{\prime}\right) \frac{e^{2}}{\left|\vec{r}-\vec{r}^{\prime}\right|} \varphi_{\alpha_{i}}\left(\vec{r}^{\prime}\right)\right] \varphi_{\alpha_{j}}(\vec{r}) &=\lambda_{\alpha_{i}} \varphi_{\alpha_{i}}(\vec{r}) \end{aligned},$$ where $\rho_i(\vec{r})=\sum_{j}' e\left|\varphi_{\alpha_{j}}\right|^{2}$ and the sum is to be taken ''over every appearing term only once''...


The actual problem:

To get this result one has to calculate $\langle\Psi|H| \Psi\rangle$ and therefore also $\langle\Psi|V_{\rm int}| \Psi\rangle$ and this is where my problem lies. The derivation involves the definitions of $$\begin{align*} \Psi_{\pi}&=\prod_{i} \varphi_{\alpha_{i}}\left(x_{i}\right)\\ \Psi&\equiv\sqrt{N !} a \Psi_{\pi}\equiv 1 / \sqrt{N !} \sum_{p\in S_n}{\rm sgn}(p)\, p\, \Psi_{\pi}, \end{align*}$$ where $S_n$ is the permutation group. For the calculation of $\langle\Psi|V_{\rm int}| \Psi\rangle$ the professor wrote $$\begin{aligned}\left\langle\Psi\left|V_{\mathrm{int}}\right| \Psi\right\rangle &= N !\left\langle\Psi_{\pi} a V_{\mathrm{int}} a \Psi_{\pi}\right\rangle \\ & \downarrow a V_{\mathrm{int}} a=V_{\mathrm{int}} a^{2}=V_{\mathrm{int}} a \\ &=\frac{1}{2} \sum_{j \neq i} \sum_{p}{\rm sgn}(p)\left\langle\Psi_{\pi} V_{\mathrm{int}}^{i j} p \Psi_{\pi}\right\rangle \\ & = \frac{1}{2} \sum_{i \neq j}\left\langle\Psi_{\pi}\left|V_{\mathrm{int}}^{i j}\left(1-p_{i j}\right)\right| \Psi_{\pi}\right\rangle. \end{aligned}$$ Can someone explain to me what is happening here? I literally don't understand any of the steps that where taken here... Why do all elements of $S_n$ vanish besides the identity and $p_{ij}$? Why exactly does the anti-symmetrizer $a$ commute with $V_{\rm int}$?

From this I should be able to conclude that $$\begin{array}{c}{\left\langle V_{\mathrm{int}}\right\rangle=\frac{1}{2} \int d^{3} r d^{3} r^{\prime} \frac{e^{2}}{\left|\vec{r}-\vec{r}^{\prime}\right|} \sum_{i \neq j}\left. [\varphi_{\alpha_{i}}^{*}(\vec{r}) \overbrace{\varphi_{\alpha_{j}}^{*}\left(\vec{r}^{\prime}\right) \varphi_{\alpha_{j}}\left(\vec{r}^{\prime}\right)}^{\rho\left(\vec{r}^{\prime}\right) / e}\right.} \\ {-\delta_{s_{i} s_{j}} \varphi_{\alpha_{i}}^{*}(\vec{r}) \varphi_{\alpha_{j}}^{*}\left(\vec{r}^{\prime}\right) \varphi_{\alpha_{j}}\left(\vec{r}^{\prime}\right) \varphi_{\alpha_{i}}\left(\vec{r}^{\prime}\right) ]},\end{array}$$ but once again I'm completely lost, especially where the $\delta_{s_i,s_j}$ comes from... Assuming this I'm able to derive the desired result.

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