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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!

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  • $\begingroup$ 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

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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

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