# Electron density question

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.

• 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. Feb 2, 2022 at 20:40
• Thank you. I corrected the mistake. And yes, your hint made it very clear.
– AA10
Feb 2, 2022 at 21:29
• 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. Feb 2, 2022 at 21:36
• Thanks. I've just done it.
– AA10
Feb 2, 2022 at 22:51

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.