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}) <br />

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}<br />

 

[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).