I am reading Theory of Simple Liquids by Hansen and McDonald, and they in chapter 3, they describe the density-density correlation for a simple liquid in the grand canonical ensemble. This is how they have defined it: $$H^{(2)}(r,r') = \langle [\rho (r) - \langle \rho (r) \rangle ][ \rho (r') - \langle \rho (r') \rangle ]\rangle $$ $$ = \rho ^{(2)}(r,r') - \rho ^{(1)}(r)\rho ^{(1)}(r') + \rho ^{(1)}(r) \delta (r-r').$$ This is my understanding of where the terms in the RHS comes from. I understand that the two-particle density $\rho^{(2)}$ term arises from: $$\langle \rho (r) \rho (r') \rangle$$ in the RHS.

The second term in the RHS comes from expanding the the product and taking like terms together.

$$-\rho ^{(1)}(r)\rho ^{(1)}(r') = -\langle \rho(r) \langle \rho (r') \rangle \rangle - \langle \rho(r') \langle \rho (r) \rangle \rangle + \rho ^{(1)}(r) \rho ^{(1)}(r'),$$ where $\rho^{(1)}(r)$ is the average single-particle density at $r$, for a homogeneous liquid.

But I still do not get why that $\delta$-function exists there. Is the $\delta$-function just there to say that if the two particles are in the same spot ($r=r'$), the density correlation is... maximized? How does the $\delta$ fall out of the averaging, mathematically?

I would appreciate any advice you have for me!

  • $\begingroup$ Is this a classical or quantum liquid? $\endgroup$
    – march
    Jan 23, 2023 at 4:16
  • $\begingroup$ This is a classical liquid @march $\endgroup$
    – megamence
    Jan 23, 2023 at 4:50

1 Answer 1


This is formal trickery and Hansen & Mc Donald are quite the wizards in this area... It is a matter of definition, whether we want the two point correlation function to include self interaction or not.

The third and last term correspond to $\langle \rho(r)\rho(r') \rangle$. You have: \begin{split} \langle \rho(r)\rho(r') \rangle &=\left\langle \sum_i \sum_j\delta(r_i-r) \delta(r_j-r')\right\rangle\\ &=\left\langle \sum_i \sum_{j\neq i}\delta(r_i-r) \delta(r_j-r')\right\rangle +\left\langle \sum_i \delta(r_i-r) \delta(r_i-r')\right\rangle \\ &=\left\langle \sum_i \sum_{j\neq i}\delta(r_i-r) \delta(r_j-r')\right\rangle +\left\langle \sum_i \delta(r_i-r)\right\rangle\delta(r-r') \\ &=\rho^2(r,r') + \rho^1(r)\delta(r-r') \end{split}

Where we used the identity $\delta(r-r_i)f(r)=\delta(r -r_i)f(r_i)$ and the fact that for $\rho^2$, we don't sum over the same atom twice. The same kind of issues arise for the structure factor for example


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.