How can the electron density expression be simplified using momenta?

[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

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

 

[Jilang]

What happens if N is odd?

 

[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.

 

[Jilang]

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

 

[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.

 

[arierreF]

3 electronspsi*(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

 

[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)

 

[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?

 

[ChrisVer]

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

 

[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.

 

[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).