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

  1. 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 $-\infty?$

  2. 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

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    $\begingroup$ I believe this is sloppy notation. It is not the dot product of $|0,-\infty\rangle_J$ and $|0,\infty\rangle$. Rather, it is a modification of the standard partition function $\langle 0,\infty|0,-\infty\rangle$ by a term involving $J$. Thus, for example, $(\langle 0,\infty|0,-\infty\rangle)_J$ would be better. $\endgroup$ – Ryan Unger Jul 14 '15 at 17:54
  • $\begingroup$ @0celo7 What about here? They say $|0,t\rangle^J$ is the vacuum in the presence of a source: books.google.ca/… $\endgroup$ – JLA Jul 14 '15 at 19:35
  • $\begingroup$ I'm not completely sure, but they might just be lazy and are writing $^J\langle 0,\infty |0,-\infty\rangle ^J$ and omitting the left $J$. $\endgroup$ – Ryan Unger Jul 14 '15 at 19:43
  • $\begingroup$ @0celo7 I thought about that, but in the first link I posted, inside the integral they put precisely one J and it seems to be on purpose. $\endgroup$ – JLA Jul 14 '15 at 19:44
  • $\begingroup$ The first link is a standard treatment. Giving a special meaning to $|0,t\rangle^J$ is not standard. (Zee, Weinberg, Srednicki and Peskin & Schroeder do not give it special meaning, for what it's worth.) $\endgroup$ – Ryan Unger Jul 14 '15 at 19:50
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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 $1$.

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  • $\begingroup$ How negative time; i.e., $t \to - \infty$ is possible? In other words, what does it physically mean when time $t \to - \infty$? $\endgroup$ – rainman Jul 30 '17 at 9:36

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