# Diagonal Two-Particle Density Matrix

## PDM Definition

One-particle density matrix

$\langle a_{p\sigma}^\dagger a_{q\tau} \rangle$

Two-particle density matrix

$\langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{r\gamma} a_{s\lambda} \rangle$

Spatial one-particle density matrix

$E_{pq} \equiv \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle$

Spatial two-particle density matrix

$e_{pqrs} \equiv \sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{r\tau} a_{s\sigma} \rangle$

Spatial two-spin density matrix

$s_{pqrs} \equiv \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{r\tau} a_{s\sigma} \rangle$

where

$\begin{split}(-1)^{1+\delta_{\sigma\tau}} = \begin{cases} 1 & \sigma = \tau \\ -1 & \sigma \neq \tau \end{cases}\end{split}$

## NPC Definition

Number of particle correlation (pure spin)

$\langle n_{p\sigma} n_{q\tau} \rangle = \langle a_{p\sigma}^\dagger a_{p\sigma} a_{q\tau}^\dagger a_{q\tau} \rangle$

Number of particle correlation (mixed spin)

$\langle a_{p\sigma}^\dagger a_{p\tau} a_{q\tau}^\dagger a_{q\sigma} \rangle$

## Spin/Charge Correlation

Spin correlation

$S_{pq} = \langle (n_{p\alpha} - n_{p\beta}) (n_{q\alpha} - n_{q\beta}) \rangle = \langle n_{p\alpha} n_{q\alpha} \rangle - \langle n_{p\alpha} n_{q\beta} \rangle - \langle n_{p\beta} n_{q\alpha} \rangle + \langle n_{p\beta} n_{q\beta} \rangle = \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle n_{p\sigma} n_{q\tau} \rangle$

Charge correlation

$C_{pq} = \langle (n_{p\alpha} + n_{p\beta}) (n_{q\alpha} + n_{q\beta}) \rangle = \langle n_{p\alpha} n_{q\alpha} \rangle + \langle n_{p\alpha} n_{q\beta} \rangle + \langle n_{p\beta} n_{q\alpha} \rangle + \langle n_{p\beta} n_{q\beta} \rangle = \sum_{\sigma\tau} \langle n_{p\sigma} n_{q\tau} \rangle$

## Diagonal Spatial Two-Particle Density Matrix (Pure Spin)

Using anticommutation relation

$a_{q\tau}^\dagger a_{p\sigma} = - a_{p\sigma} a_{q\tau}^\dagger + \delta_{pq}\delta_{\sigma\tau}$

We have

$\langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{q\tau} a_{p\sigma} \rangle = -\langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{p\sigma} a_{q\tau} \rangle = \langle a_{p\sigma}^\dagger a_{p\sigma} a_{q\tau}^\dagger a_{q\tau} \rangle - \delta_{pq} \delta_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\tau} \rangle$

Then

$\begin{split}e_{pqqp} \equiv&\ \sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{q\tau} a_{p\sigma} \rangle = -\sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{p\sigma} a_{q\tau} \rangle = \sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{p\sigma} a_{q\tau}^\dagger a_{q\tau} \rangle - \delta_{pq} \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle \\ =&\ \sum_{\sigma\tau} \langle n_{p\sigma} n_{q\tau} \rangle - \delta_{pq} \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle\end{split}$

Therefore,

$\boxed{C_{pq} \equiv \sum_{\sigma\tau} \langle n_{p\sigma} n_{q\tau} \rangle = e_{pqqp} + \delta_{pq} E_{pq}}$

Similarly,

$\begin{split}s_{pqqp} \equiv&\ \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{q\tau} a_{p\sigma} \rangle = -\sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{p\sigma} a_{q\tau} \rangle \\ =&\ \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle a_{p\sigma}^\dagger a_{p\sigma} a_{q\tau}^\dagger a_{q\tau} \rangle - \delta_{pq} \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle \\ =&\ \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle n_{p\sigma} n_{q\tau} \rangle - \delta_{pq} \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle\end{split}$

Therefore,

$\boxed{S_{pq} \equiv \sum_{\sigma\tau} (-1)^{1+\delta_{\sigma\tau}} \langle n_{p\sigma} n_{q\tau} \rangle = s_{pqqp} + \delta_{pq} E_{pq} }$

## Diagonal Spatial Two-Particle Density Matrix (Mixed Spin)

Using anticommutation relation

$a_{q\tau}^\dagger a_{p\tau} = - a_{p\tau} a_{q\tau}^\dagger + \delta_{pq}$

we have

$\begin{split}e_{pqpq} \equiv&\ \sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\tau}^\dagger a_{p\tau} a_{q\sigma} \rangle = -\sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{p\tau} a_{q\tau}^\dagger a_{q\sigma} \rangle + \delta_{pq} \sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle \\ =&\ -\sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{p\tau} a_{q\tau}^\dagger a_{q\sigma} \rangle + 2\delta_{pq} \sum_{\sigma} \langle a_{p\sigma}^\dagger a_{q\sigma} \rangle \\\end{split}$

Therefore,

$\boxed{\sum_{\sigma\tau} \langle a_{p\sigma}^\dagger a_{p\tau} a_{q\tau}^\dagger a_{q\sigma} \rangle = -e_{pqpq} + 2\delta_{pq} E_{pq}}$