# The Two-Particle Density function

1. Jul 14, 2005

### MalleusScientiarum

I have a question regarding the two-particle density function, in particular its Fourier transform. I know that in a liquid or gas the function $$n_2(\mathbf{R}_1, \mathbf{R}_2)$$ is the probability that two particles will be found at $$\mathbf{R}_1$$ and $$\mathbf{R}_2$$. But what is the significance of its Fourier transform,
$$G(\mathbf{k}) = \frac{1}{2} \int d^{3N}\mathbf{R}_1 d^{3N}\mathbf{R}_2 e^{\imath \mathbf{k}\cdot(\mathbf{R}_1 - \mathbf{R}_2)} n_2(\mathbf{R}_1, \mathbf{R}_2)$$

My guess is that it is some sort of momentum distribution, but that's only a guess.

2. Jul 26, 2005

### DaTario

It seems to me that, in the classical context, where there is no de Broglie relation between position and momentum, this formalism is not expected to yield any significative result concerning what we classically call as linear momentum probability distribution.
I would interpretate this calculation as purely a description in terms of weights of wave vectors of plane waves such that, when accordingly superposed yield the surface you transformed.

Another point: With no uncertanty principle, deltas of Dirac in position representation have no reason to yield plane wave in "momentum space" and therefore, minimum knowledge about momentum.

3. Jul 28, 2005

### alien308

In the general case:
$$G(\mathbf{k}_1,\mathbf{k}_2 ) = \int d^{3N}\mathbf{R}_1 d^{3N}\mathbf{R}_2 e^{\imath \cdot(\mathbf{k}_1 \mathbf{R}_1+\mathbf{k}_2 \mathbf{R}_2)} n_2(\mathbf{R}_1, \mathbf{R}_2)$$

Last edited by a moderator: Jul 28, 2005
4. Jul 28, 2005

### MalleusScientiarum

Well obviously, but not if the interaction potential is a central force.