Density density correlations of a simple Brownian particle Suppose, I have a particle satisfying the equation 
\begin{equation}
\frac{dX}{dt}=\eta(t)
\end{equation}
Where $\langle \eta(t)\eta(t')\rangle=\delta(t-t')$. I can now define a density like $\rho(x,t)=\delta(x-X(t))$. I am confused how do I compute $$\langle \rho(x,t)\rho(x',t')\rangle~?$$ Any help would be greatly appreciated.  
 A: Welcome to Physics StackExchange! This question can be related to the calculation of dynamic density correlations in liquids, in the diffusion limit. Because of this, and the familiarity of the results in that physical application, I'm going to explicitly include the diffusion coefficient $D$ in the right-hand side of the Brownian dynamics equation (Langevin equation without inertia) and write
$$
\langle \eta(t)\eta(t') \rangle = 2D\,\delta(t-t')
$$
but obviously the result for your equation will follow from setting 
$D=\frac{1}{2}$.
It is also standard to assume that $\langle \eta(t)\rangle=0$ and
that 
$\eta(t)$ is a Gaussian random process, so all its higher moments are
determined by the first two: you don't explicitly state this, 
but I assume that it is true.
Direct integration of the equation gives the usual results $\langle \Delta X\rangle = 0$ and $\langle \Delta X^2\rangle = 2D |t-t'|$,
where $\Delta X=X(t)-X(t')$.
Your interest lies in the density-density correlation function
$$
G(x,t,x',t') = G(x-x',t-t') = \left\langle \rho(x,t) \, \rho(x',t') \right\rangle
$$
where $\rho(x,t)=\delta(x-X(t))$.
I've incorporated here the fact that the system is translationally invariant
in both the space and time coordinates.
This in turn means that the time correlation function
of the Fourier transformed density 
$$\tilde{\rho}(k,t)=
\int \rho(x,t) \exp(-ikx) dx
=\exp [-ikX(t)]$$
which we write
$$
F(k,t-t') = \left\langle \tilde{\rho}(k,t) \tilde{\rho}(-k,t') \right\rangle
$$
is itself the Fourier transform of $G$:
$$
F(k,t) = \int G(x,t) \exp(-ikx) dx
$$
More details on these kind of manipulations may be found, for example,
in Theory of Simple Liquids by J-P Hansen and IR McDonald.
In our case, from the definition of $\tilde{\rho}(k,t)$,
$$
F(k,t-t') = \left\langle \exp [-ik\{X(t)-X(t')\}] \right\rangle
= \left\langle \exp [-ik \Delta X] \right\rangle
$$
Now we use the Gaussian nature of the random term $\eta$,
which implies that $\Delta X$ is also distributed as a Gaussian,
and satisfies the identity
$$
\left\langle \exp [-ik \Delta X] \right\rangle = 
\exp \left(-\frac{1}{2}k^2 \langle \Delta X^2\rangle \right)
=
\exp \left(-Dk^2 |t-t'| \right)
$$
This is a familiar diffusional form,
a Gaussian function of $k$,
and we can immediately deduce the form of $G$,
which is a Gaussian in $x$:
$$
G(x-x',t-t') = \frac{1}{\sqrt{4\pi D\,|t-t'|}} \exp\left(-\frac{(x-x')^2}{4D\,|t-t'|}\right)
$$
You may need to check in case I have made any slips along the way.
This problem may also be approached by interpreting the density-density correlation function in the sense of a conditional probability
which satisfies the partial differential equation corresponding to the
original Langevin-like equation. That PDE is, of course, the diffusion equation,
and $G$ is the solution of it which you get when you set up the initial probability distribution as a delta-function, in other words, it is the propagator. But I've deliberately approached this as a problem involving dynamical variables based on $X(t)$, because that seemed to match the way the question is phrased.
