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On wikipedia it says that:

$$\rho(\mathbf{r})= \sum_{{s}_{1}} \cdots \sum_{{s}_{N}} \int \ \mathrm{d}\mathbf{r}_1 \ \cdots \int\ \mathrm{d}\mathbf{r}_N \ \left( \sum_{i=1}^N \delta(\mathbf{r} - \mathbf{r}_i)\right)|\Psi(\mathbf{r}_1,s_{1},\mathbf{r}_{2},s_{2},...,\mathbf{r}_{N},s_{N})|^2 \tag{1}$$

Let's say $N=3$ and spin summations are implicit.

then $$\rho(\mathbf{r})= \int d\mathbf{r}_1 \int d\mathbf{r}_2 \int d\mathbf{r}_3 \left[\delta( \mathbf{r}- \mathbf{r}_1)+ \delta( \mathbf{r}- \mathbf{r}_2) + \delta( \mathbf{r}- \mathbf{r}_3)\right] |\Psi(\mathbf{r}_1,\mathbf{r}_{2},\mathbf{r}_{3})|^2 \tag{2} $$

We will have three terms:

$$\rho(\mathbf{r})= \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 + \int d\mathbf{r}_1 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{1},\mathbf{r}_{3})|^2 + \ \int d\mathbf{r}_1 \int d\mathbf{r}_2 |\Psi(\mathbf{r},\mathbf{r}_{1},\mathbf{r}_{2})|^2 \tag{3}$$

Where I considered the anti-symmetry of the wave functions. This is my final result, I don't see how I can simply it further.

I don't understand why (how) this will be equal to:

$$\rho(\mathbf{r})= 3\int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 \tag{4}$$

or, for the general case:

$$\rho(\mathbf{r})= N \int \ \mathrm{d}\mathbf{r}_2 \ \cdots \int\ \mathrm{d}\mathbf{r}_N \ |\Psi(\mathbf{r},\mathbf{r}_{2},...,\mathbf{r}_{N})|^2 \tag{5}$$

This is basically saying that all the terms in Eq. (3) are equal. Did I do something wrong?

A: Jason Funderberker's comment helped me understand what was my problem.

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    $\begingroup$ I've edited the equations. Feel free to undo if you want. Further, I think that you made a 'mistake' in the third term; there should be no $r_3$ but a $r_1$, no? Anyway, here is a hint: Each of the three terms only depend on $r$, whereas $r_1$, $r_2$ and $r_3$ are dummy variables since you integrate over them. Thus, label them as you want. $\endgroup$ Feb 2, 2022 at 20:40
  • $\begingroup$ Thank you. I corrected the mistake. And yes, your hint made it very clear. $\endgroup$
    – AA10
    Feb 2, 2022 at 21:29
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    $\begingroup$ If you know the answer to your question, you can (if you want to) also write an answer yourself. Actually, this is encouraged, cf. here. This could help potential future readers. $\endgroup$ Feb 2, 2022 at 21:36
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    $\begingroup$ Thanks. I've just done it. $\endgroup$
    – AA10
    Feb 2, 2022 at 22:51

1 Answer 1

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I followed the hint given by @Jason Funderberker in the comments section.

$$\rho(\mathbf{r})= \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 + \int d\mathbf{r}_1 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{1},\mathbf{r}_{3})|^2 + \ \int d\mathbf{r}_1 \int d\mathbf{r}_2 |\Psi(\mathbf{r},\mathbf{r}_{1},\mathbf{r}_{2})|^2 \tag{1}$$

The variables inside the integrals ($d\mathbf{r}_1,d\mathbf{r}_2 ,d\mathbf{r}_3$) are dummy variables. Therefore, it doesn't matter what number we use to label them.

This means that: $$\rho(\mathbf{r})= \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 + \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 + \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 \tag{2}$$

Arriving the final expression:

$$\rho(\mathbf{r})= 3 \int d\mathbf{r}_2 \int d\mathbf{r}_3 |\Psi(\mathbf{r},\mathbf{r}_{2},\mathbf{r}_{3})|^2 $$

Which agrees with the general expression in Eq. (5) on the original post.

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