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In the derivation of the LSZ reduction formula equation (5.21) Srednicki claims that in case of an interaction term in the Lagrangian density $a^{\dagger}(\textbf{k})$ will no longer be time dependent. Now does that mean that the only modification to the formula of $\varphi(x)$ will be a time dependence of $a$'s and $a^{\dagger}$'s? So is that like an ansatz we use for $\varphi(x)$? Now even if we do assume this form of $\varphi(x)$, in line two of the derivation he makes the substitution $a^{\dagger}(t,\textbf{k})= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]$.$$a^{\dagger}(\textbf{k},t)= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]\tag{5.2}.$$ Now in the derivation of this formula if we remember we explicitly had to use the time independence of the $a$'s. So what am I missing in this derivation that justifies the substitution in the case the $a$'s are time dependent?

In the derivation of the LSZ reduction formula equation (5.21) Srednicki claims that in case of an interaction term in the Lagrangian density $a^{\dagger}(\textbf{k})$ will no longer be time dependent. Now does that mean that the only modification to the formula of $\varphi(x)$ will be a time dependence of $a$'s and $a^{\dagger}$'s? So is that like an ansatz we use for $\varphi(x)$? Now even if we do assume this form of $\varphi(x)$, in line two of the derivation he makes the substitution $a^{\dagger}(t,\textbf{k})= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]$. Now in the derivation of this formula if we remember we explicitly had to use the time independence of the $a$'s. So what am I missing in this derivation that justifies the substitution in the case the $a$'s are time dependent?

In the derivation of the LSZ reduction formula equation (5.21) Srednicki claims that in case of an interaction term in the Lagrangian density $a^{\dagger}(\textbf{k})$ will no longer be time dependent. Now does that mean that the only modification to the formula of $\varphi(x)$ will be a time dependence of $a$'s and $a^{\dagger}$'s? So is that like an ansatz we use for $\varphi(x)$? Now even if we do assume this form of $\varphi(x)$, in line two of the derivation he makes the substitution $$a^{\dagger}(\textbf{k},t)= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]\tag{5.2}.$$ Now in the derivation of this formula if we remember we explicitly had to use the time independence of the $a$'s. So what am I missing in this derivation that justifies the substitution in the case the $a$'s are time dependent?

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LSZ reduction derivation Srednicki

In the derivation of the LSZ reduction formula equation (5.21) Srednicki claims that in case of an interaction term in the Lagrangian density $a^{\dagger}(\textbf{k})$ will no longer be time dependent. Now does that mean that the only modification to the formula of $\varphi(x)$ will be a time dependence of $a$'s and $a^{\dagger}$'s? So is that like an ansatz we use for $\varphi(x)$? Now even if we do assume this form of $\varphi(x)$, in line two of the derivation he makes the substitution $a^{\dagger}(t,\textbf{k})= -i \int\:d^3x \:[e^{ikx} \partial_0\varphi(x) - e^{ikx} ik_0 \varphi(x)]$. Now in the derivation of this formula if we remember we explicitly had to use the time independence of the $a$'s. So what am I missing in this derivation that justifies the substitution in the case the $a$'s are time dependent?