# Getting electron density from probability density function

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1. Dec 28, 2015

### Jan Wo

Hello
Lastly I was thinking a lot about electron density definition. It is not intuitive for me and I'm looking for any mathematical tool that could explain it to me more. My friend told me about idea to derivate it from propability density function using Dirac delta distribution. I'd like to show you this way of thinking and ask you about your opinion about it. Is it right way to think about electron density like this?

First - case of one electron. If we have only one electron, the propability density function should be at the same time its electron density function. We are looking for the operator which gives us as a mean value electron density:
$\rho(r)=\langle \psi | \hat{\rho}|\psi\rangle$
If we assume that $\hat{\rho} \equiv \delta(r-r')$ then we got electron density function for one electron from propability density.

Now let's take a N-electron case:
Let's use sum of Dirac delta operators as a electron density operator:
$\langle \psi | \sum_{i=1}^N \delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \langle \psi |\delta(r-r_i)|\psi\rangle = \sum_{i=1}^N \int \psi^* (r_1, \dots , r_N) \delta (r-r_i)\psi(r_1, \dots , r_N)dr_1 \dots dr_N=*$
As we can see now, Dirac delta 'takes' one position out from this integral, let's assume that it is position number one. Then one summand should looks like: $\int \psi^* (r, r_2, \dots , r_N)\psi (r, r_2, \dots , r_N)dr_2 \dots dr_N$ where $r$ is an position which we choose to 'check' the electron density. There is $N$ summands and the electrons are indistinguishable so we can $N$ times choose the electron number one. Then:
$*=N\int \psi^*\psi dr_2 \dots dr_N \equiv \rho(r)$

Is this right way of thinking about electron density?
Can you tell me if something is wrong in this?
WJ

Last edited: Dec 28, 2015
2. Dec 28, 2015

### Jano L.

What do you mean by electron density? Do you mean the probability density that some electron is at specific location, one specified electron is at specific location, or charge density density at specific location? Or is the density supposed to refer to the whole configuration space?

The common concept of density related to solutions of Schroedinger's equations is probability density of configuration, calculated as $|\psi|^2$. For example, if system consists of two particles, if coordinates of the first are denoted by $\mathbf r_1$ and the coordinates of the second by $\mathbf r_2$, and provided the system is described by a solution of some Schroedinger equation, denoted $\psi(\mathbf r_1,\mathbf r_2)$, then:

the probability that the system is in a configuration belonging to subset $G$ of the configuration space is (one of the Born rules):

$$p = \int_G |\psi(\mathbf r_1, \mathbf r_2)|^2\,d^3\mathbf r_1 d^3\mathbf r_2.$$

The quantity $\rho(\mathbf r_1,\mathbf r_2)=|\psi(\mathbf r_1,\mathbf r_2)|^2$ is therefore called probability density. It gives probability for configurations of the whole system.

If you are interested only in one specific electron, you can calculate probability density for its coordinates from this quantity according to the standard rule of marginal probabilities. For the first electron:

$$\rho_1(\mathbf r_1) = \int\,d^3\rho (\mathbf r_1,\mathbf r_2)$$

3. Dec 28, 2015

### Jan Wo

I forgot to write electron density definition. Sorry about that.When I mentioned about electron density and I meant about the function with electron charge as a value in selected point(argument) of space (it is used for example in x-ray spectroscopy or in DFT studies). If we ignore spin the definition looks like:
$\rho(r_1)=N\int |\psi(r_1, \dots , r_N)|^2d^3r_2, \dots , d^3r_N$
And my question is if it is possible to 'derive' this definition from the propability density function... Now I see that my question is not well posed because i started from operator of electron density mean value definition. Again - sorry for this mess. I not necessarily engaged propability density function.

So, is it possible? Is my way of derivating it using Dirac delta as a electron density operator? Do this derivation goes like that in my first post?
Thanks for patience and attention :)

4. Dec 28, 2015

### A. Neumaier

Your formula correctly gives, up to a constant factor, the measurable (mean) charge density of a system of electrons.. See Section ''How do atoms and molecules look like?'' (and related articles) in Chapter A6 of my theoretical physics FAQ.

o

5. Dec 28, 2015

### Jano L.

You seem to be interested in the probability density function for any particle forming the system (and having its coordinates in the list of arguments of $\psi$) being at given coordinates $\mathbf x$. If the only thing we have is probability density for the system configuration
$$\rho(\mathbf r_1, ...,\mathbf r_N)$$

then the basic rules of probability theory already imply that the probability density of any particle being at $\mathbf x$ is sum of marginal probability densities for all particles:

$$D(\mathbf x) = \sum_{i=1}^{N} \rho_{marginal,i}(\mathbf x).$$
The marginal probability density for $i$-th particle is given by (again a basic rule of probability theory):
$$\rho_{marginal, i}(\mathbf x) = \int d^3\mathbf r_1 ... d^3\mathbf r_N\, \delta(\mathbf r_1 - \mathbf x)\rho(\mathbf r_1, ..., \mathbf r_N).$$

This is what can be said about any probability density in general. If we, in addition, assume the probability density $\rho$ is symmetric with respect to exchange of any two vector arguments $\mathbf r_k, k\in {1,...,N}$ then we arrive at the formula you gave above.

The derivation is purely probability theory, nothing about operators from quantum theory is needed.