# Electron density expression

1. May 18, 2014

### arierreF

Suppose we have a system with M nuclei and N electrons

the electron density is defined as:

$\rho(\mathbf{r}) = \sum\limits_{\sigma = \uparrow, \downarrow} \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r\sigma})\psi_{i}(\mathbf{r\sigma})$

i would like to know if it is valid to express it in the following way

$\rho(\mathbf{r}) = 2 \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r})\psi_{i}(\mathbf{r}) = 2 \sum\limits_{i}^N |\psi_{i}(\mathbf{r})|^{2}$

2. May 18, 2014

### Jilang

What happens if N is odd?

3. May 18, 2014

### arierreF

Im not seeing any difference between even or odd N.

here is what i think

the two factor is a consequence of the Pauli Principle, a single electron wave function can be occupied by two electrons (spin up and spin down) . And because the electron density is dependent of the 3 spatial variables and not the spin, i can right psi (r) and not psi (rσ) so the last expression is valid in my opinion.

4. May 19, 2014

### Jilang

I still think it looks wrong. Try a trivial example with 3 electrons.

5. May 20, 2014

### sokrates

To me it looks like all he did was to sum up/down electrons for a spin-free system.

Looks valid if you are interested in "pure" charge density and want to forget about spin.

Nothing wrong with it.

6. May 21, 2014

### arierreF

3 electrons

psi*(r1, down)psi(r1,down) + psi*(r1, up)psi(r1, up) +
psi*(r2, down)psi(r2, down) +psi*(r2, up)psi(r2, up) +
psi*(r3, down)psi(r3, down) +psi*(r3, up)psi(r3,up)

2|psi(r1)|^2 + 2|psi(r2)|^2 + 2|psi(r3)|^2 = 2 sum (from i =1 to 3) |psi(ri)|² consistent with the last expression

7. May 21, 2014

### ChrisVer

1. yes it is valid... otherwise you can think of it as the spinor degrees of freedom...
eg several times you can find that the number density is given as $n= A g \int f(p) d^{3}p$
that g is giving you the dof of the particles you care about- so if you have particles which can take values +1/2 and -1/2, then g=2.
For a dirac s-1/2 particle, the g=4...if you had a massless spin 1 particle, then g=2 as well... if you had massive spin 1, then g=3 etc...
Otherwise you can think of it as you did for the case of N=3

2. N shouldn't be relevant. One reason is that you have the same wavefunction product (you have $\psi_{i}^{*} \psi_{i}$ which you sum its spins over, so N doesn't affect it- equivalently speaking you can interchange the summations)

8. May 23, 2014

### arierreF

Ok! now it is much more clear.

Just more one question

When it is valid to approximate the density as a product of individual functions?

Last edited: May 23, 2014
9. May 23, 2014

### ChrisVer

I don't understand the question....
do you mean to write something like $\psi_{i} \psi_{j}, i \ne j$?

10. May 25, 2014

### arierreF

Forget my last question. What i would like to know is

we have the general equation for electronic density:

$\rho(\mathbf{r})=N \sum_{{s}_{1}}\cdots \sum_{{s}_{N}} \int \mathrm{d}\mathbf{r}_2 \cdots \int \mathrm{d}\mathbf{r}_N$

what are the approximations to simplify the expression to have

$\rho(\mathbf{r}) = 2 \sum\limits_{i}^N \psi^{*}_{i}(\mathbf{r})\psi_{i}(\mathbf{r}) = 2 \sum\limits_{i}^N |\psi_{i}(\mathbf{r})|^{2}$

even with the Hartree product approximation, i can't see how to pass from the first equation to tthe last one.

11. May 25, 2014

### ChrisVer

I still don't get the question...
You don't have infinite numbers of particles, you particles are of number N...
In that case, the number density is given by summing the $\psi_{i}$
You don't make any integral into summation...
But for the integral you've given, your sum appears by the operator of $\rho$, not by some assumption.

Things become clearer if you work for example with momenta. Instead of the number of particle i, to work with momenta $k_{i}$ (go to momentum space).

In that case,a general wf is:

$\psi_{k_{i}} (x) = \frac{1}{sqrt{V}} e^{i k^{0}_{i} x^{0} - i \vec{k}_{i} \vec{x}}$

For finite volume, the momenta are discrete/quantized, because the wf needs to satisfy boundary conditions on the limits...

So the number density is:

$\rho (p)= \sum_{s} \sum_{k_{i}} \delta ^{3} (\vec{p}-\vec{k}_{i}) | \psi_{k_{i}} (x)|^{2}= \frac{2}{V} \sum_{k_{i}} \delta ^{3} (\vec{p}-\vec{k}_{i})$
Now if you want to send the volume to infinity, then your discrete momenta values will come closer and closer to each other, going from a discrete to almost a continuous spectrum.
In that case you can change the sum to integral (Riemann type sum).

Last edited: May 25, 2014