# Confusion over density-density correlation function

I have a question regarding what may perhaps be quite simple definitions in classical statistical physics. If one considers a set of $n$ random variables with an $n$-dimensional probability density function $p(u_1,\ldots,u_n)$, then the moments of order $k = \sum_{i=1}^n k_i$ are defined via

$\langle u_1^{k_1}\ldots u_n^{k_n} \rangle = \int_{-\infty}^\infty\ldots\int_{-\infty}^\infty u_1^{k_1}\ldots u_n^{k}\,p(u_1,\ldots,u_n)\,du_1\ldots du_n$.

Further, expressions of the form $\langle u(\mathbf{x}_1)u(\mathbf{x}_2) \rangle$ are known as correlation functions. This all seems simple enough, until I came across a concrete example. Solid crystals are in some sense defined by the fact that their density-density correlation functions satisfy the following property:

$\lim_{|\mathbf{x} - \mathbf{x'}|}\langle \rho(\mathbf{x})\rho(\mathbf{x}') \rangle = f(\mathbf{x}-\mathbf{x}')$, where $f$ is periodic in the chosen set of basis vectors. However, based on the above definitions, I propose that

$\langle \rho(\mathbf{x})\rho(\mathbf{x}') \rangle \stackrel{?}{=} \int\int\rho(\mathbf{x})\rho(\mathbf{x}')p(\mathbf{x},\mathbf{x}')'\,d^3\mathbf{x}\,d^3\mathbf{x}'$,

where the integrals are taken over all space.

My questions are therefore

1. Is my understanding of the density-density correlation function correct?
2. If not, what should I understand by $\langle \rho(\mathbf{x})\rho(\mathbf{x}') \rangle$?
3. If so, how can the resultant integral be a function of space; and
4. how does one understand $p$ in this setting?

I'm sorry if this is more straightforward than I am making it, but I am quite new to statistical mechanics and struggling to find my bearings.

## 2 Answers

Density function (normalied) of a $N$-particle classical system with cordinates $\{\vec{r}_{i}^{}\}$ can be defined as : $$\rho(\vec{r})=\frac{1}{N}\sum_{i=1}^{N}\delta_{}^{(3)}(\vec{r}-\vec{r}_{i}^{}).$$ And the density-density correlation function (in canonical ensemble) is defined as : $$\langle\rho(\vec{r}) \rho(\vec{r}')\rangle = \frac{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}\rho(\vec{r}) \rho(\vec{r}')e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}.$$

$\textbf{Addendum (Field theoretic generating function technology):}$ To recast the problem in field theoretic terms (to obtain a statistical field theory) for simplicity consider : $$\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})=\sum_{i=1}^{N}\frac{\vec{p}_{i}^{}\cdot\vec{p}_{i}^{}}{2m}+\frac{1}{2}\sum_{i\neq j=1}^{N}\mathcal{V}(\vec{r}_{i}^{},\vec{r}_{j}^{}).$$ Now define a density moment generating function as : $$\mathcal{Z}[\{\phi(\vec{r})\}] = \frac{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{i\int d^{3}_{}\vec{r}\phi(\vec{r})\rho(\vec{r})}_{}e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}.$$ Using which we can calculated density-density correlation functions to all order as : $$\langle\rho(\vec{r}_{1}^{})\cdots\ \rho(\vec{r}_{n}^{})\rangle=(-i)^{n}_{}\frac{\delta}{\delta \phi(\vec{r}_{1}^{})}\cdots\frac{\delta}{\delta \phi(\vec{r}_{n}^{})}\mathcal{Z}[\{\phi(\vec{r})\}]\Big|_{\{\phi(\vec{r})=0\}}^{}.$$ Recast the moment generating function as (self interaction of particles need to be regularized somehow (denoted as $\Delta$ here)) : $$\mathcal{Z}[\{\phi(\vec{r})\}] = \frac{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{i\int d^{3}_{}\vec{r}\phi(\vec{r})\rho(\vec{r})}_{}e^{-\beta\big[\sum_{i=1}^{N}\frac{\vec{p}_{i}^{}\cdot\vec{p}_{i}^{}}{2m}+\frac{N^{2}_{}}{2}\int\int d^{3}_{}\vec{r} d^{3}_{}\vec{r}'\rho(\vec{r})\mathcal{V}(\vec{r},\vec{r}')\rho(\vec{r}')+\Delta[\rho(\vec{r})]\big]}}{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}.$$ Now we can introduce the functional delta identity : $$\int\mathcal{D}[\sigma(\vec{r})]\delta[\sigma(\vec{r})-\rho(\vec{r})]=1$$ inside the generating function to get : $$\mathcal{Z}[\{\phi(\vec{r})\}] = \frac{\int\mathcal{D}[\sigma(\vec{r})]e^{i\int d^{3}_{}\vec{r}\phi(\vec{r})\sigma(\vec{r})}_{}e^{-\beta\big[\frac{N^{2}_{}}{2}\int\int d^{3}_{}\vec{r} d^{3}_{}\vec{r}'\sigma(\vec{r})\mathcal{V}(\vec{r},\vec{r}')\sigma(\vec{r}')+\Delta[\sigma(\vec{r})]\big]}e^{-\beta\Phi[\sigma(\vec{r})]}}{\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-\beta\mathcal{H}(\{\vec{r}_{i}^{}\},\{\vec{p}_{i}^{}\})}}.$$ Where : $$e^{-\beta\Phi[\sigma(\vec{r})]}=\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}\delta[\sigma(\vec{r})-\rho(\vec{r})]e^{-\beta\big[\sum_{i=1}^{N}\frac{\vec{p}_{i}^{}\cdot\vec{p}_{i}^{}}{2m}\big]}$$ which can be rewritten using functional Fourier representation of functional delta functional as :

$$e^{-\beta\Phi[\sigma(\vec{r})]}=\int\mathcal{D}[\omega(\vec{r})]\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-i\int d_{}^{3}\vec{r}\omega(\vec{r})[\sigma(\vec{r})-\rho(\vec{r})]}e^{-\beta\big[\sum_{i=1}^{N}\frac{\vec{p}_{i}^{}\cdot\vec{p}_{i}^{}}{2m}\big]}.$$ Finally everything can be recast as : $$\mathcal{Z}[\{\phi(\vec{r})\}] = \frac{\int\mathcal{D}[\omega(\vec{r})]\int\mathcal{D}[\sigma(\vec{r})]e^{i\int d^{3}_{}\vec{r}[\phi(\vec{r})-\omega(\vec{r})]\sigma(\vec{r})}_{}e^{-\beta\big[\frac{N^{2}_{}}{2}\int\int d^{3}_{}\vec{r} d^{3}_{}\vec{r}'\sigma(\vec{r})\mathcal{V}(\vec{r},\vec{r}')\sigma(\vec{r}')+\Delta[\sigma(\vec{r})]\big]}e^{-\beta F[\omega(\vec{r})]}}{\mathcal{Z}[\{0\}]}.$$ With : $$e^{-\beta F[\omega(\vec{r})]}=\int\dots\int\prod_{i=1}^{N}d_{}^{3}\vec{r}_{i}^{}d_{}^{3}\vec{p}_{i}^{}e^{-\beta\sum_{i=1}^{N}\big[\frac{\vec{p}_{i}^{}\cdot\vec{p}_{i}^{}}{2m}+i\beta_{}^{-1}\omega(\vec{r}_{i}^{})\big]},$$ is the partition function of non-interacting particles in a external imaginary potential.

$\textbf{Probability functional for densities:}$ Finally, on inverse functional Fourier transforming $\mathcal{Z}[\{\phi(\vec{r})\}]$ we get probability functional for densities as : $$P[\{\sigma(\vec{r})\}]=\frac{e^{-\beta\big[\frac{N^{2}_{}}{2}\int\int d^{3}_{}\vec{r} d^{3}_{}\vec{r}'\sigma(\vec{r})\mathcal{V}(\vec{r},\vec{r}')\sigma(\vec{r}')+\Delta[\sigma(\vec{r})]+\mathbb{V}_{}^{}[\sigma(\vec{r})]\big]}}{\int\mathcal{D}[\sigma(\vec{r})]e^{-\beta\big[\frac{N^{2}_{}}{2}\int\int d^{3}_{}\vec{r} d^{3}_{}\vec{r}'\sigma(\vec{r})\mathcal{V}(\vec{r},\vec{r}')\sigma(\vec{r}')+\Delta[\sigma(\vec{r})]+\mathbb{V}_{}^{}[\sigma(\vec{r})]\big]}}$$ with $$e^{-\beta\mathbb{V}_{}^{}[\sigma(\vec{r})]}=\int\mathcal{D}[\omega(\vec{r})]e^{-i\int d^{3}_{}\vec{r}\omega(\vec{r})\sigma(\vec{r})}_{}e^{-\beta F[\omega(\vec{r})]}.$$

• Thank you for your thorough answer, and in particular for linking your original answer to that provided by "BySymmetry". This has helped me clear up a lot of confusion! – Jakob Lindemann Nov 13 '17 at 11:39
• @JakobLindemann You are welcome. Nice that things are clear for you now. Field theoretic methodology is useful to calculate correlators and partition functions approximately (using mean-field or RPA or loop expansions). There are other methods like Mayer's cluster expansion (described in standard Stat-Mech books) which is generally used to calculate partition function and virial coefficients. – Sunyam Nov 13 '17 at 12:04

You need to pay attention to exactly what the is the random variable you are averaging over. When we measure the density-density correlation function, we choose which points to measure at, so $\mathbf{x}$ is not a random variable, it is a parameter. The randomly fluctuating quantities are $\rho(\mathbf{x})$, that is we have an (infinite) family of correlated random variables, $\rho$ which we parameterise with $\mathbf{x}$.

This means that when we calculate averages we must average over all values of $\rho$, and since all the $\rho$s are generally correlated we must do this for every $\rho$. This means, rather than the integral you wrote down, we must perform a functional integral over all possible functions $\rho(\mathbf{x})$ $$\langle \rho(\mathbf{x})\rho(\mathbf{x}')\rangle = \int \mathcal{D}\rho\; \rho(\mathbf{x})\rho(\mathbf{x}') p[\rho]$$ where the probability $p$ is now a functional of $\rho$

In practice we normally write $p$ in terms of some action $p = \exp(-S[\rho])$ and normally write the entire integral in terms of some underlying field for which we have a microscopic model, such that $\rho = \psi^\dagger \psi$.

• Thanks for your answer. This makes sense in and of itself, and I wish I had found a textbook that states this as clearly as you have here. That said, I'm still not clear on how this relates to Sunyam's answer regarding the more "obvious" explanation in terms of integration against the normalised probability distribution. That is, is there a straightforward way to show that this functional integral is exactly that set forth by Sunyam for the canonical ensemble? – Jakob Lindemann Nov 10 '17 at 16:40
• @JakobLindemann Please see the addendum to my answer for the details you are looking for in BySymmetry's answer. – Sunyam Nov 10 '17 at 19:54