Vacuum to vacuum transition amplitude confusion Quantum field theory texts often use the expression $\langle 0,\infty|0,-\infty\rangle_J\,$ denoting the vacuum-to-vacuum transition amplitude.  


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*What is the meaning of $|0,-\infty\rangle_J?$ Am I to take the Hamiltonian with the source present, ie. $H(t)=H-J(t)q(t)\,,$ find the ground state at time $0\,,$ ie. $H(0)=E_0|0\rangle$ and evolve it backwards to $-\infty?$ 

*Why isn't the correct expression $_J\langle 0,\infty|0,-\infty\rangle_J$ instead? I also see $\langle q',t'|q,t\rangle_J$ and have the same question about that, ie. what is $|q,t\rangle_J$ and why aren't we considering $_J\langle q',t'|q,t\rangle_J$ instead? 
Reference
 A: I think this is a good question and need to be addressed because this is very often not explained well in books. 
The notation $|0,\pm\rangle_j$ has no meaning. But the expression, $(\langle 0,+\infty|0,-\infty\rangle)_j\equiv \langle 0,+\infty|0,-\infty\rangle_j$ is perfectly meaningful. Let me explain what does it represent.
Consider a system being subjected to an externally applied time-dependent "perturbation" $j(t)$. We look for the effect of this
perturbation on the transition amplitude $\langle q_f,t_f|q_i,t_i\rangle$. I put the word perturbation in quotations. Because at the moment,
it's an arbitrary function of time, and I don’t assume it to be small. Sometimes $j(t)$ is called a source term  which can be thought of as a external field like an electric field $\textbf{E}(t)$. To ensure that $j(t)$ enters the equation of motion as an inhomogeneous source, a term $q(t)j(t)$ has to be added to the Lagrangian so that:
$$L\to L + q(t)j(t).$$ For concreteness, assume that the system is a linear harmonic oscillator. Suppose that I prepare the system in the
ground state at $t=-\infty$. I’ll call denote it by $|0_-\rangle\equiv |0,-\infty\rangle$. At some finite past $t =-\tau$, I turn on a time-dependent "perturbation" $j(t)$, let it act for sometime, and then switch it off at time $t\to +\tau$. Then let the system evolve asymptotically to $t\to +\infty$ on its own. Whatever state the system evolves to will be
denoted by $|0_+\rangle\equiv |0,+\infty\rangle$. If you wish, you can call it the ground state at $t=+\infty$.
Now here is the word of caution. The state $|0_+\rangle$ need not be same as $|0_-\rangle$. At $t=+\infty$, the system has a finite probability to be found in any of the excited states. Therefore, the overlap $\langle 0_+|0_-\rangle_j\leq 1$ in presence of $j$. If $j$ never acted
this overlap would have been equal to $1$.
