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From Szabo Ostlund 'Modern Quantum Chemistry'

Given a normalized wavefunction of N-electron system $\Phi(\mathbf{x_{1}},...,\mathbf{x_{n}})$, then $\Phi(\mathbf{x_{1}},...,\mathbf{x_{n}})\Phi^{*}(\mathbf{x_{1}},...,\mathbf{x_{n}}) d\mathbf{x_{1}}...d\mathbf{x_{n}}$ is the probability that an electron is in the space-spin volume element $d\mathbf{x_{1}}$ located at $\mathbf{x_{1}}$, while simultaneously another electrons in $d\mathbf{x_{2}}$ located at $\mathbf{x_{2}}$ and so on. If we are intersted only in the probability of finding an electron in $d\mathbf{x_{1}}$ at $\mathbf{x_{1}}$ independent of where the other electrons are, then we must average over all space-spin coordinates of the other electrons, i.e. integrate over $\mathbf{x_{2}},...,\mathbf{x_{N}}$ to obtain $$\rho(\mathbf{x_{1}})=N\int d\mathbf{x_{2}}...d\mathbf{x_{N}} \Phi(\mathbf{x_{1}},...,\mathbf{x_{N}})\Phi^{*}(\mathbf{x_{1}},...,\mathbf{x_{N}}) \quad (Eq.1)$$
where $\rho(\mathbf{x_{1}})$ is called the reduced density function for a single electron in an N-electron system.

I still dont quite understand the meaing of (Eq.1). $\int d\mathbf{x_{1}} |\Phi(\mathbf{x_{1}})|^{2}$ can be interpreted as integral of the charge density at $\mathbf{x_{1}} $ which is the number of electrons at $\mathbf{x_{1}} $. So I think $\int d\mathbf{x_{2}}...d\mathbf{x_{N}} \Phi(\mathbf{x_{1}},...,\mathbf{x_{N}})\Phi^{*}(\mathbf{x_{1}},...,\mathbf{x_{N}})$ is the total number of electrons at $(\mathbf{x_{2}},...,\mathbf{x_{N}})$ as function of $\mathbf{x_{1}}$ and i cant see how my intepretation is connected to the probability of independently finding an electron in $d\mathbf{x_{1}}$ at $\mathbf{x_{1}}$.

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    $\begingroup$ Related to your question: physics.stackexchange.com/questions/497324/… $\endgroup$
    – Tobias Fünke
    Commented Jan 24, 2021 at 13:37
  • $\begingroup$ In addition to the answer linked by Jakob, I'd say your "total number of electrons" approach is somewhat wrong, what you find by any of these integrals is the probability of a set of different electron configurations. Integrating over indices 2 to N simply mean looking at all possible states of these electrons and averaging over them. So $\int d\mathbf{x_{1}} |\Phi(\mathbf{x_{1}})|^{2}$ should give exactly $1$: It's the probability of finding the 1st electron anywhere in the system $\endgroup$
    – acarturk
    Commented Jan 24, 2021 at 14:29
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    $\begingroup$ @Jakob well yeah, $\Phi(\mathbf x_{1,N})$ doesn't have to be separable so defining $\Phi(\mathbf x_1)$ doesn't necessarily make sense. The way I understood it, $\rho(\mathbf x_1)$ in Eq.1 corresponds to the particle density of the entire system, which is made out of N equal $\rho_i(\mathbf x_i)$ terms which correspond to the equal individual electron densities. This density can be defined due to indiscernibility of fermions. Both answers in the link above are better than what I can offer to be honest $\endgroup$
    – acarturk
    Commented Jan 24, 2021 at 17:14
  • $\begingroup$ @ OP: Is your question answered or do you have additional (closely related) questions? $\endgroup$
    – Tobias Fünke
    Commented Jan 24, 2021 at 21:07
  • $\begingroup$ @Jakob Thanks a lot for the link! It does help me to understand the equation. So I would say $\int d^{3} r_{2} \ldots d^{3} r_{N} \psi^{*}\left(r_{1}, r_{2}, \ldots r_{N}\right) \psi\left(r_{1}, r_{2}, \ldots r_{N}\right)$ is the probability density of finding the system within the space created by $d\mathbf{r_{2}},..., d\mathbf{r_{n}}$ (when you integral of that it would be all space apart from that of $d\mathbf{r_{1}}$) averaged over all possible $\mathbf{r_{2}},...,\mathbf{r_{n}}$. $\endgroup$
    – jayjay
    Commented Jan 26, 2021 at 13:54

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