# Cluster decomposition $\stackrel{?}{=}$ Translation invariance

In Weinberg Volume 1 (section 4.4), Weinberg argues for a certain structure of the interaction Hamiltonian by demanding that it produce an $$S$$-matrix satisfying cluster decomposition. The proposed structure is the following (flavor/spin indices suppressed) $$H = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m)$$ where $$\widetilde h$$ is a smooth function.

It seems to me that one could more easily arrive at the above structure simply by postulating that the Hamiltonian is translation invariant. Indeed, since the translation group acts on the ladder operators as $$U^\dagger(\vec b) a(\vec p) U(\vec b) = e^{-i\vec p \cdot \vec b} a(\vec p)$$ (for translation by $$\vec b \in \mathbb{R}^3$$), we find that the above Hamiltonian is translation invariant: \begin{align} U^\dagger(\vec b)H U(\vec b) & = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) \, U^\dagger(\vec b) a^\dagger(\vec p’_1) U(\vec b)\cdots U^\dagger(\vec b)a^\dagger(\vec p’_n) U(\vec b) \, U^\dagger(\vec b)a(\vec p_1)U(\vec b) \cdots U^\dagger(\vec b)a(\vec p_m)U(\vec b) \\ & = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) \exp\left[i \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \cdot\vec b\right] a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m) \\ & = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m) \\ & = H \end{align} where in the second-to-last equality we used the fact that the delta function forces $$\exp\left[i \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \cdot\vec b\right] = 1$$.

Conversely, starting with the requirement of translation invariance, I think we can deduce the above form of Hamiltonian.

Given that the translation group is a subgroup of Poincare, why does Weinberg bother introducing cluster decomposition as an additional postulate if it can be deduced already from translation invariance?

The following interaction is translation invariant but nonlocal

$$$$H_{\rm int} = \lambda \int d^3 x \int d^3 y \phi^2(\vec{x}) \phi^2(\vec{y})$$$$ An $$S$$-matrix computed with this interaction Hamiltonian should fail to satisfy the cluster decomposition principle.

The point of the cluster decomposition principle is that it implies that the Hamiltonian should be a local function of the fields.

• Cluster decomposition doesn't actually require the Hamiltonian to be local. There exist non-local theories that satisfy both microcausality and cluster property, for example generalized free theories. Sep 22, 2022 at 21:09

Cluster decomposition $$\stackrel{?}{=}$$ Translation invariance

No, these are not the same.

Conversely, starting with the requirement of translation invariance, I think we can deduce the above form of Hamiltonian.

But, why do you think the converse is true? You have not shown it. You have not provided support for such a statement.

Given that the translation group is a subgroup of Poincare, why does Weinberg bother introducing cluster decomposition as an additional postulate if it can be deduced already from translation invariance? (Emphasis added)

Because it can not be deduced from translation invariance.

I completely agree with the answers already provided by hft and Andrew. Thanks.

Here is an explicit counter-example (closely related to Andrew’s) but expressed in the language of ladder operators.

Consider

$$H = \int {\rm d}\vec p’_1 \, {\rm d}\vec p’_2 \, {\rm d}\vec p_1\, {\rm d}\vec p_2 \, \delta^3 \left(\vec p’_1 - \vec p_1\right) \delta^3\left(\vec p’_2 - \vec p_2\right) \, f(\vec p’_1,\vec p’_2, \vec p_1,\vec p_2) \, a^\dagger(\vec p’_1) \, a^\dagger(\vec p’_2) \, a(\vec p_1) \, a(\vec p_2)$$ with $$f$$ a smooth function.

Acting with translation operator,

\begin{align} U^\dagger H U & = \int {\rm d}\vec p’_1 \, {\rm d}\vec p’_2 \, {\rm d}\vec p_1\, {\rm d}\vec p_2 \, \delta^3 \left(\vec p’_1 - \vec p_1\right) \delta^3\left(\vec p’_2 - \vec p_2\right) \, f(\vec p’_1,\vec p’_2, \vec p_1,\vec p_2) \, \exp\left[i\big(\vec p’_1+\vec p’_2-\vec p_1 - \vec p_2\big)\cdot \vec b\right] a^\dagger(\vec p’_1) \, a^\dagger(\vec p’_2) \, a(\vec p_1) \, a(\vec p_2) \end{align} The presence of two delta functions again forces $$\exp\left[i\big(\vec p’_1+\vec p’_2-\vec p_1 - \vec p_2\big)\cdot \vec b\right] =1$$.