Given the theory

$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}m_{\phi}^{2}\phi^{2}+\partial_{\mu}\chi^{*}\partial^{\mu}\chi-m_{\chi}^{2}\chi^{*}\chi+\mathcal{L}_{\text{int}},\qquad \mathcal{L}_{\text{int}}=-g\phi\chi^{*}\chi,$

the time-correlation function $\langle \Omega | \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})|\Omega\rangle$ is given by

$$\langle \Omega | \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})|\Omega\rangle = \langle 0| \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})|0\rangle -ig \int d^{4}x\ \langle 0 | \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})\phi(x)\chi^{*}(x)\chi(x)|0\rangle + \mathcal{O}(g^{2})$$

Is $\langle 0| \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})|0\rangle = 0$?


1 Answer 1


Yes $\langle 0| \phi(x_{1})\chi^{*}(x_{2})\chi(x_{3})|0\rangle = 0$ is zero since the expectation value of $\phi$ in the vacuum state is zero.


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