Question about a passage in Blundell's QFT

[WWCY]

TL;DR Summary

On the second quantisation of potential term associated with inter-particle interactions

In Bundell's QFT book, the second-quantised, pairwise interaction potential was defined as:

in a later step, this was re-written in a momentum-space representation

1. I don't understand why it goes from $V(x, y)$ to $V(x-y)$, why do we consider only this form of interaction potentials?

2. Also, could someone show how the expression for $\hat{V}$ derive from the general form of the second-quantised two particle operator below?

Thanks in advance!

 

[vanhees71]

ad 1) The usual expression is $V(\vec{x},\vec{y})=V(\vec{x}-\vec{y})$, because you want to fulfill Newton's 3rd Law, i.e., if there's an interaction and particle 2 excerts a force $\vec{F}_{12}=-\vec{\nabla}_1 V(\vec{x}_1,\vec{x}_2)$ then necessarily particle 1 excerts a force on particle 2, $\vec{F}_{21}=-\vec{F}_{12}=-\vec{\nabla}_2 V(\vec{x}_1,\vec{x}_2).$
So you have
$$\vec{\nabla}_1 V(\vec{x}_1,\vec{x}_2)=-\vec{\nabla}_2 V(\vec{x}_1,\vec{x}_2).$$
Now introduce center-of-mass and relative coordinates for the particle pair,
$$\vec{R}=\frac{1}{2} (\vec{x}_1+\vec{x}_2), \quad \vec{r}=\vec{x}_1-\vec{x}_2.$$
and consider $V$ as a function of $\vec{R}$ and $\vec{r}$. Then you have
$$\partial_{1j} V= \frac{\partial R_k}{\partial x_{1j}} \frac{\partial}{\partial R_k} V + \frac{\partial r_k}{\partial x_{1j}} \frac{\partial}{\partial r_k} V=\frac{1}{2} \frac{\partial}{\partial R_k} V +\frac{\partial}{\partial r_k} V$$
and in the same way
$$\partial_{2j} V=\frac{1}{2} \frac{\partial}{\partial R_j} V -\frac{\partial}{\partial r_j} V.$$
Now Newton's 3rd law tells us that
$$\partial_{1j} V + \partial_{2j} V=0 \; \Rightarrow \; \frac{\partial}{\partial R_j} V=0 \; \Rightarrow \; V=V(\vec{r})=V(\vec{x}_1-\vec{x}_2).$$
And that's what you plug into your Hamiltonian. Note that the factor 1/2 comes from overcounting the sum over all particles (in the field formalism that's the integration over $\vec{x}$ and $\vec{y}$) since the total potential energy of one pair has to count only once and not twice. So summing over all particle indices gives a factor 2 too large contribution of the two-body contributions. And that's why you have to compensate by the factor $1/2$.

ad 2) Just go on with the calculation as it stands! Obviously the author wisely uses first a quantization volume (with single-particle wave functions periodic, leading to discrete momenta $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3$, where $L$ is the length of the cube making up the quantization volume). Then you get
$$\hat{V}=\frac{1}{2 \mathcal{V}} \sum_{\vec{p}_1,\ldots \vec{p}_4} \hat{a}_1^{\dagger} \hat{a}_2^{\dagger} \hat{a}_3 \hat{a}_4 \int_{\mathbb{R}^3} \mathrm{d}^3 x \int_{\mathbb{R}^3} \mathrm{d}^3 y \exp[\mathrm{i} \vec{x}\cdot (\vec{p}_4-\vec{p}_1) + \mathrm{i} \vec{y} \cdot (\vec{p}_3-\vec{p}_2)] V(\vec{x}-\vec{y}).$$
From this you get
$$V_{1234}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \int_{\mathbb{R}^3} \mathrm{d}^3 y \exp[\mathrm{i} \vec{x}\cdot (\vec{p}_4-\vec{p}_1) + \mathrm{i} \vec{y} \cdot (\vec{p}_3-\vec{p}_2)] V(\vec{x}-\vec{y}).$$
Now you should go on and use that $V$ depends only on the relative coordinates $\vec{r}=\vec{x}-\vec{y}$ to simplify the matrix elements.

Hint: Introduce $\vec{R}$ and $\vec{r}$ as integration variables (note that you need the Jacobian of the corresponding coordinate transformation!).

 

[WWCY]

Cheers, I think i get it now. I'll try and work through the details!