I read that the electron density used for density functional theory in a system of $N$ electrons with wavefunction $\psi$ is defined as

$$\rho(r)=N\int d^3r_2\dots d^3r_N \psi^*(r,r_2,\dots r_N)\psi(r,r_2,\dots r_N) $$

this definition seems weirdly asymmetrical to me. Why is the first coordinate special? How can I interpret this definition physically?


2 Answers 2


In a one particle system, the probability of finding the electron between $\mathbf{r}$ and $\mathbf{r}+d\mathbf{r}$ is $|\Psi(\mathbf{r})|^2d\mathbf{r}$, with the probability density $\rho(\mathbf{r})=|\Psi(\mathbf{r})|^2$.

In a many-electron system, the probability of finding electron 1 in the volume element $d\mathbf{r}_1$ at $\mathbf{r}_1$, electron 2 in the volume element $d\mathbf{r}_2$ at $\mathbf{r}_2$, and so on, is $|\Psi(\mathbf{r}_1,\mathbf{r}_2,\dots,\mathbf{r}_N)|^2d\mathbf{r}_1 d\mathbf{r}_2\dots d\mathbf{r}_N$. Therefore, the probability of finding electron 1 in the volume element $d\mathbf{r}_1$ at $\mathbf{r}_1$ and the other electrons elsewhere is $$P(1)=\left[\int|\Psi(\mathbf{r}_1,\mathbf{r}_2,\dots,\mathbf{r}_N)|^2 d\mathbf{r}_2\dots d\mathbf{r}_N\right]d\mathbf{r}_1=\rho(\mathbf{r}_1)d\mathbf{r}_1$$ This probability is exactly the same for all electrons, because they are indistinguishable, so we don't care about the label. Thus, the probability density of finding an electron in the volume element $d\mathbf{r}$ at $\mathbf{r}$ and the other $N-1$ electrons elsewhere is $N$ times the probability to find one of the electrons at that region. That is, $$\rho(\mathbf{r})=N\int|\Psi(\mathbf{r},\mathbf{r}_2,\dots,\mathbf{r}_N)|^2 d\mathbf{r}_2\dots d\mathbf{r}_N$$


The total density is the sum of each particle's density $\rho(r) = \sum_i^N \rho_i(r)$ with $$ \rho_i (r) = \int d^3r_1...d^3r_{i-1}d^3r_{i+1}...d^3r_n |\psi(r_1, ... r_{i-1}, r, r_{i+1}, ..., r_n)|^2. $$ In DFT, the wave function is a Slater determinant and is thus antisymmetrized, reflecting the indiscernibility of fermions as $\rho_i(r) = \rho_1(r)$. We find $$ \rho(r) = N\rho_1(r) $$ which is another form for your equation of $\rho(r)$.

  • $\begingroup$ ah, so the $N$ terms comes from $\rho(r)=\sum_{i=1}^N \rho_i(r)$ where by $\rho_i(r)$ I mean that we integrate on all the particles except particle $i$, but since exchanging two particles just changes the sign of the wavefunction and the integral only depends on the norm of $\psi$, the $N$ terms are all equal, is this correct? I'm still not sure of the physical meaning, as this sum of probabilities is greater than $1$ it is not a probability $\endgroup$ Aug 17, 2019 at 14:41
  • $\begingroup$ The integral in it-self is the probability of finding the first particle at $r$. The wavefunction is normalized to one, while the integral of the density gives the number of particles in total, that's why you have a $N$. You are right in your first statement. $\endgroup$
    – gingras.ol
    Aug 17, 2019 at 14:50
  • $\begingroup$ I rewrote an answer including your remarks, thanks! $\endgroup$
    – gingras.ol
    Aug 18, 2019 at 2:55
  • $\begingroup$ In DFT, the total electron density is $\rho(\mathbf{r})=\sum_{\sigma}\sum_{i}^{N}n_i|\psi_i(\mathbf{r},\sigma)|^2$, where $\sigma$ represent the spin variable. Also, in Kohn-Sham DFT, the density is $\rho(\mathbf{r})=\sum_{\sigma}\sum_i^N|\psi_i(\mathbf{r},\sigma)|^2$. The latter form (with $n_i=1$ for $N$ orbitals and $0$ for the rest) describes exactly $N$ non-interacting electrons. $\endgroup$
    – Verktaj
    Jan 28, 2020 at 21:58
  • 1
    $\begingroup$ I saw that you joined Matter Modeling SE and for a second thought that you were Michel Gingras!!! Also, hi @Verktaj! $\endgroup$ Jul 29, 2021 at 22:08

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