I Derivation of two-electron density operator

[Mart1234]

TL;DR

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.

 

[vanhees71]

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.

 

[Morbert]

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

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 $$\int dx\int dy \delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$

 

[Mart1234]

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 $$\int dx\int dy \delta(x-y)\delta(y-x')f(y) = \int dx \delta(x-x')f(x)$$

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.