18
$\begingroup$

In Weinberg's "The Quantum Theory of Fields, Vol. 1", Section 4.4, page 182, the author says:

We now ask, what sort of Hamiltonian will yield an $S$-matrix that satisfies the cluster decomposition principle? It is here that the formalism of creation and annihilation operators come into its own. The answer is contained in the theorem that the $S$-matrix satisfies the cluster decomposition principle if (and as far as I know, only if) the Hamiltonian can be expressed as in

$$H=\sum_{N=0}^\infty\sum_{M=0}^\infty \int dq_1'\cdots dq_N' dq_1\cdots dq_M a^\dagger(q_1')\cdots a^\dagger(q_N')a(q_M)\cdots a(q_1)h_{NM}(q',q)$$

with coefficient functions $h_{NM}$ that contain just a single three-dimensional momentum-conservation delta function (returning here here briefly to a more explicit notation):

$$h_{NM}(\mathbf{p}_1'\sigma_1'n_1',\cdots \mathbf{p}_N'\sigma_N'n_N',\mathbf{p}_1\sigma_1n_1,\cdots,\mathbf{p}_M\sigma_Mn_M)=\delta^{(3)}(\mathbf{p}_1'+\cdots+\mathbf{p}_N'-\mathbf{p}_1-\cdots -\mathbf{p}_M)\tilde{h}_{NM}(\mathbf{p}_1'\sigma_1'n_1',\cdots \mathbf{p}_N'\sigma_N'n_N',\mathbf{p}_1\sigma_1n_1,\cdots,\mathbf{p}_M\sigma_Mn_M)$$

where $\tilde{h}_{NM}$ contains no delta function factors.

Now, Weinberg says that as far as he knows the converse holds: if the $S$-matrix satisfies the cluster decomposition principle the interaction has this form.

My question here is, out of curiosity, is it known today a proof of this statement? Or Weinberg's belief was wrong and indeed it has been found $S$-matrices satisfying the cluster decomposition principle without the interaction having this form?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.