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In general, the product of two Hermitian operators $\phi$ will not be Hermitian, unless the two operators commute. Question: is $X = T \phi(t_1) \phi(t_2)$ Hermitian? It doesn't seem to be if $T \phi_1 \phi_2 = \phi_1\phi_2\theta(t_1-t_2) + \phi_2\phi_1\theta(t_2-t_1).$ (Weinberg) If not, how can we interpret $\langle 0 |X|0\rangle$ as the vacuum correlation function, if X is not an observable? |
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$X$ isn't a Hermitian operator, as you've discerned, and $\langle 0 | X |0\rangle$ isn't the expectation value of an observable. It's traditional to call the quantum mechanical quantity $\langle 0| T(\phi(t_2) \phi(t_1) )| 0 \rangle$ a "correlation function", but this is actually an abuse of terminology. It is actually an amplitude! Or, if you prefer, it is the matrix element of an operator which is not hermitian. It is related to the correlation functions of classical observables in the statistical field theory that you get when you look at the Wick rotation of your quantum theory. It's a boundary value of the analytic continuation of the correlation function of two Euclidean field values. |
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Here we are just rephrasing user1504's correct answer using slightly different words. Off-shell correlation functions $$\tag{1}\langle 0| T\varphi(x_1)\ldots \varphi(x_m)\varphi(y_1)\ldots \varphi(y_n) | 0 \rangle$$ are related via the LSZ reduction formula to on-shell $S$-matrix elements $$\tag{2}\langle p_1, \ldots p_n~{\rm out} | p_1, \ldots p_m~ {\rm in}\rangle.$$ In other words, the correlation function (1) is related to the probability amplitude a.k.a. the overlap between the 'in' and the 'out' Hilbert space of states. In particular, eq. (1) should not be viewed as an expectation value of some Hermitian operator. |
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