# Derivation of two-electron density operator

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• Mart1234
In summary, the conversation discussed the derivation of the second term in 2.11a of the two-electron density. The expert explained that the term eliminates the i=j products through the use of certain identities and integration methods. They also clarified that the two-electron density is not commonly written in this form.
Mart1234
TL;DR Summary
Derivation of two electron density operator using single electron density operator
Hello, I am going over the derivation for two-electron density. I am having a hard time understanding how the second term in 2.11a seen below is derived. I know this term must eliminate the i=j products but can't seem to understand how. Thanks for the help.

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Where is this coming from? The 2nd equality of (2.11a) seems to indicate that you consider a special state of uncorrelated/free particles, but we need more context to make sense of it.

I've never seen the two-electron density written like that. Here are my thoughts but I can't say for sure.

Mart1234 said:
I know this term must eliminate the i=j products but can't seem to understand how.
Considering the terms where ##i=j## $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r' - \mathbf r_i)$$We should be able to use the identities ##\delta(x - y) = \delta(y - x)## and ##\int dy \delta(x-y)\delta(y-x') = \delta(x-x')## So we rewrite the above as $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r_i - \mathbf r')$$ and when we are computing the electron density function we will be integrating over ##\mathbf r_i## so the latter identity suggests $$\sum_i\int d\mathbf r_1\dots d\mathbf r_N \delta(\mathbf r - \mathbf r_i)\delta(\mathbf r_i - \mathbf r')|\Psi(\mathbf r_1,\dots, \mathbf r_N)|^2 = N\int d\mathbf r_2\dots d\mathbf r_N \delta(\mathbf r - \mathbf r')|\Psi(\mathbf r,\dots, \mathbf r_N)|^2$$For the last line. I am assuming $$\int dx\int dy \delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$

Last edited:
vanhees71
Morbert said:
I've never seen the two-electron density written like that. Here are my thoughts but I can't say for sure.Considering the terms where ##i=j## $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r' - \mathbf r_i)$$We should be able to use the identities ##\delta(x - y) = \delta(y - x)## and ##\int dy \delta(x-y)\delta(y-x') = \delta(x-x')## So we rewrite the above as $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r_i - \mathbf r')$$ and when we are computing the electron density function we will be integrating over ##\mathbf r_i## so the latter identity suggests $$\sum_i\int d\mathbf r_1\dots d\mathbf r_N \delta(\mathbf r - \mathbf r_i)\delta(\mathbf r_i - \mathbf r')|\Psi(\mathbf r_1,\dots, \mathbf r_N)|^2 = N\int d\mathbf r_2\dots d\mathbf r_N \delta(\mathbf r - \mathbf r')|\Psi(\mathbf r,\dots, \mathbf r_N)|^2$$For the last line. I am assuming $$\int dx\int dy \delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$

Morbert said:
I've never seen the two-electron density written like that. Here are my thoughts but I can't say for sure.Considering the terms where ##i=j## $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r' - \mathbf r_i)$$We should be able to use the identities ##\delta(x - y) = \delta(y - x)## and ##\int dy \delta(x-y)\delta(y-x') = \delta(x-x')## So we rewrite the above as $$\sum_{i}\delta(\mathbf r-\mathbf r_i)\delta(\mathbf r_i - \mathbf r')$$ and when we are computing the electron density function we will be integrating over ##\mathbf r_i## so the latter identity suggests $$\sum_i\int d\mathbf r_1\dots d\mathbf r_N \delta(\mathbf r - \mathbf r_i)\delta(\mathbf r_i - \mathbf r')|\Psi(\mathbf r_1,\dots, \mathbf r_N)|^2 = N\int d\mathbf r_2\dots d\mathbf r_N \delta(\mathbf r - \mathbf r')|\Psi(\mathbf r,\dots, \mathbf r_N)|^2$$For the last line. I am assuming $$\int dx\int dy \delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$
Got it. I appreciate the help.

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