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).
 A: 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:
$$\langle\phi_1\rangle\langle\phi_2\rangle=\frac{\int [D\phi_1]\phi_1e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]\phi_2e^{iS[\phi_2]}}{\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}} $$
$$=\frac{\int [D\phi_1]\phi_1e^{iS[\phi_1]} \int [D\phi_2]\phi_2e^{iS[\phi_2]}}{\int [D\phi_1]e^{iS[\phi_1]}\int [D\phi_2]e^{iS[\phi_2]}}=\frac{\int [D\phi_1][D\phi_2] \phi_1\phi_2e^{iS[\phi_1]+S[\phi_2]}}{\int [D\phi_1][D\phi_2]e^{iS[\phi_1]+S[\phi_2]}}$$
$$=\langle\phi_1\phi_2\rangle$$
