Interpreting Electron Density Definition

Click For Summary
The discussion revolves around the interpretation of the electron density definition from the "Handbook of Molecular Physics and Quantum Chemistry." The integral in the equation only includes variables for positions 2 through N because at least one position variable must remain un-integrated to yield a function of position. This allows the electron density, denoted as ρ(r), to depend on a specific point in space, typically represented as ρ(r1). Mathematically, integrating over some variables while leaving others unintegrated is essential for maintaining a functional relationship, as demonstrated in the example with the function f(x,y). The conversation emphasizes the importance of understanding this concept for clarity in quantum chemistry.
Jan Wo
Messages
4
Reaction score
0
Hi
I'd like to ask you about interpretation of electron density definition. According to the "Handbook of Molecular Physics and Quantum Chemistry" it goes like:

##\rho (\textbf{r})=N \sum_{\sigma_1}^{\sigma_N} \idotsint_{\mathbb{R^3}} |\psi_V (\textbf{r}_1, \sigma_1;\dots ;\textbf{r}_N,\sigma_N)|^2 d^3\textbf{r}_2 \dots d^3\textbf{r}_N##

I know that my question may be a little imprecise so please - be patient. I'd like to know how to iterpret this definition. Why in the integral is only ##d^3\textbf{r}_2 \dots d^3\textbf{r}_N## but no ##d^3\textbf{r}_1d^3\textbf{r}_2 \dots d^3\textbf{r}_N##. How to explain this equation using physical viewpoint and how to explain it methematically ?

Feel free to add any helpful information to understand this definition.
Thanks in advance.
 
Physics news on Phys.org
Is your \rho(r) supposed to be \rho(r_1)?

You have to leave at least one position variable un-integrated if you want the result to be a function of position. I think that's why.
 
  • Like
Likes Jan Wo
Ok, thank you for answer.
I think this ##\rho(r)## should be ##\rho(r_1)##
Can you tell me how it works mathematically?
I need to have a function of position, I know that I can't to integrate every position (then it would be only a number). But how exactly it looks. Let's consider simple example of function ##f(x,y)## what would happen if I integrate it like this: ##\int f(x,y)dy=?##. The result for this example would be ##?=g(x)\times <number>##?
Is it true in general case?
Perhaps it is obvious but I really like to simplyfy everything and I'd like to be sure.
 
Thread 'Unexpected irregular reflection signal from a high-finesse cavity'
I am observing an irregular, aperiodic noise pattern in the reflection signal of a high-finesse optical cavity (finesse ≈ 20,000). The cavity is normally operated using a standard Pound–Drever–Hall (PDH) locking configuration, where an EOM provides phase modulation. The signals shown in the attached figures were recorded with the modulation turned off. Under these conditions, when scanning the laser frequency across a cavity resonance, I expected to observe a simple reflection dip. Instead...

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 28 ·
Replies
28
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 175 ·
6
Replies
175
Views
26K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K