# Product of VEVs vs. VEV of product

How can we prove the following cluster decomposition formula

$$\langle \phi_1 \phi_2 \rangle ~=~ \langle \phi_1 \rangle \langle \phi_2 \rangle,$$

where brackets denote vacuum expectation value (VEV) on some vacuum state, and $\phi_i$ are constant fields in some quantum field theory?

The only justification for this I'm aware of comes from Weinberg (QFT 2, p. 166).

-
vev? I do not know this term. –  Bernhard Apr 20 '13 at 18:29
But if the field is constant, then, isn't this statement obvious? –  Bernhard Apr 20 '13 at 18:30
@Bernhard vev = vaccum expectation value, the mean value of the field taken for a specific vaccum state –  Learning is a mess Apr 20 '13 at 19:00
There formula you're writing down is nonsense. If the $\phi_i$ are constant then it's trivial, but it's not even true for free scalar particles with zero vev. The cluster decomposition principle is treated in Ch. 4 of the first book from Weinberg. Please read that chapter first and come back with precise questions, if you have any : ) –  Vibert Apr 20 '13 at 19:31
You are mistaken, Weinberg's condition (v. 2, p. 166) differs from this, it's $\langle u| A(\mathbf{x})B(0)|v\rangle \to \sum_{w}\langle u|A(0)|w\rangle\langle w|B(0)|v\rangle$ as $|\mathbf{x}|\to\infty$ ... which Weinberg notes (v. 2, p. 167) is related to "the cluster decomposition condition (see Chapter 4)". –  Alex Nelson Apr 20 '13 at 23:22

How about that way of looking at it. Starting from path-integral definitions: $$\langle\phi_1\phi_2\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1\phi_2e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2]e^{iS[\phi_1,\phi_2]}}$$ $$\langle\phi_1\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2] e^{iS[\phi_1,\phi_2]}},\quad \langle\phi_2\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_2 e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2] e^{iS[\phi_1,\phi_2]}}$$ We immediately see that your equality doesn't generally holds, but...
Suppose that $S[\phi_1,\phi_2] = S[\phi_1] + S[\phi_2]$, then: $$\langle\phi_1\rangle\langle\phi_2\rangle=\frac{\int [D\phi_1]\phi_1e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]\phi_2e^{iS[\phi_2]}}{\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}}$$ $$=\frac{\int [D\phi_1]\phi_1e^{iS[\phi_1]} \int [D\phi_2]\phi_2e^{iS[\phi_2]}}{\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}}=\frac{\int [D\phi_1][D\phi_2] \phi_1\phi_2e^{iS[\phi_1]+S[\phi_2]}}{\int [D\phi_1][D\phi_2]e^{iS[\phi_1]+S[\phi_2]}}$$ $$=\langle\phi_1\phi_2\rangle$$
This works in finite dimensional integrals, but your assumption $S[\phi_1, \phi_2] = S[\phi_1] + S[\phi_2]$ does not work for functional integrals. This is the so-called multiplicative anomaly, reviewed in e.g. Elizalde or E. Elizalde and M. Tierz –  Alex Nelson Apr 20 '13 at 23:19
Hi Kostya, I think you're reading the question wrong. If $\phi_1$ and $\phi_2$ are different, non-interacting fields, then the factorisation is indeed trivial. The cluster decomposition formula holds for S-matrices that factorise "in space-time," i.e. when cluster of incoming and outgoing particles are very far apart. –  Vibert Apr 21 '13 at 7:32