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I'm reviewing Srednicki's chapter on path integrals and am having trouble understanding how he arrives at formula 6.21:

$$\left<0|0\right>_{f,h}= \int \mathcal{D}q \,\mathcal{D}p \, \exp \left[i \int_{-\infty}^{+\infty} dt \Big( p\dot{q}-(1-i\epsilon)H+fq+hp \Big) \right]. \tag{6.21}$$

He gives a wordy explanation that sounded like it made sense at first before I thought about it too hard, but on further thought, I just don't see his logic. He begins with the following equation, which I can accept:

$$\left<0|0\right>_{f,h}=\lim\limits_{t'\to-\infty}\lim\limits_{t''\to+\infty}\int dq'' dq' \psi_0^*(q'') \left<q'',t''|q',t'\right>_{f,h} \,\,\psi_0(q')\,, \tag{6.19}$$

where the $\psi_0$'s represent the ground state of the Hamiltonian (without the $-i\epsilon$). Then he introduces the trick of changing $H$ to $(1-i\epsilon)H$, which I am also willing to accept. With this change, he shows that

$$\lim\limits_{t'\to-\infty} \left| \, q',t'\right> = \psi_0^*(q') \left|0\right>$$ $$\lim\limits_{t''\to+\infty} \left< \, q'',t''\right| = \psi_0(q'') \left<0\right|.$$

This is the end of his mathematical derivation—he then uses two paragraphs to explain how this all means that "we can be cavalier about the boundary conditions on the endpoints of the path" and jump to Eq. (6.21).

Can someone please fill in the mathematical gaps here? The issue I'm having is that the above fact seems to only bring us in a circle:

\begin{align} \left<0|0\right>_{f,h}&=\lim\limits_{t'\to-\infty}\lim\limits_{t''\to+\infty}\int dq'' dq' \psi_0^*(q'') \left<q'',t''|q',t'\right>_{f,h} \,\,\psi_0(q') \tag{6.19}\\ &=\int dq'' dq' \psi_0^*(q'') \bigg[ \psi_0(q'') \left<0\right| \bigg] \bigg[ \psi_0^*(q') \left|0\right> \bigg] \,\,\psi_0(q') \\ &=\left<0|0\right> \int dq'' \left|\psi_0(q'')\right|^2 \int dq' \left|\psi_0(q')\right|^2 \\ &=\left<0|0\right> \end{align}

Qmechanic
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WillG
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    It is really frustrating to see statements like this, so I fully empathise with you. Of course the boundary terms are important if you want to actually know what you are doing. The best place that I know of that discusses these contributions with great clarity and extremely carefully is Fadeev's lecture notes from the Les Houches summer school 1975, http://inspirehep.net/record/116350?ln=en or https://www.worldscientific.com/worldscibooks/10.1142/0004 – Wakabaloola Apr 12 '19 at 16:20
  • Can you also write the path integral definition of $\langle q',t' | q,t \rangle_{f,h}$? – MannyC Apr 12 '19 at 16:25
  • Thanks for the references. I'm not even sure what he means by boundary terms: The $\psi_o(q)$'s? Or the whole ket vectors $|q,t>$? – WillG Apr 12 '19 at 17:25
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    I think you should think of it in reverse. The point is to show that $\langle q'',t'' | q', t'\rangle_{f,h,\epsilon} = \langle 0 | 0 \rangle_{f,h}$ up to normalization. The discussion with limits, etc is what demonstrates this. What he means by boundary conditions on the endpoints is that with $H-i\epsilon$, is that in fact the infinite past and future position eigenstates for the path integral become the ground states up to normalization. – Aaron Apr 13 '19 at 02:55
  • Essentially a duplicate of https://physics.stackexchange.com/q/409907/2451 – Qmechanic Apr 13 '19 at 12:10

1 Answers1

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The goal is to show that \begin{align} \lim_{t''\rightarrow \infty}\lim_{t'\rightarrow -\infty} \langle q'',t''|q',t'\rangle_{f,h,\epsilon} \stackrel{?}{\propto} \langle 0|0\rangle_{f,h} \end{align} where the $\epsilon$ denotes that the we've replaced $H\rightarrow (1-i\epsilon)H$. We know that the LHS gives the path integral, so if we can demonstrate this equality, we are done.

\begin{align} \lim_{\epsilon \rightarrow 0}\lim_{t''\rightarrow \infty} \lim_{t'\rightarrow -\infty} \langle q'',t''|q',t'\rangle_{f,h,\epsilon} &= \psi^*(q'')\psi(q')\langle 0|0\rangle_{f,h} \\ &\propto \langle 0|0\rangle_{f,h} \end{align}

Note that the addition of $i\epsilon$ essentially converts the initial and final position eigenstates to be the ground state, which is what is meant by "$i\epsilon$ fixes the boundary condition."

Aaron
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    Ok I think I'm starting to understand: So we deduce $(6.19)$ (see post) based on general principles, but the $\psi_0$'s don't let us immediately turn this expression into the path integral $(6.21)$. However, the argument involving $1-i\epsilon$ and time limits lets us conclude that $\lim ... \int dq'' dq' \left<q'',t''|q't'\right> $ is also equal to $\left<0|0\right>$ (up to a constant), and this is what gets us to $(6.21)$ by invoking the path integral expression. Is that accurate? – WillG Apr 13 '19 at 22:23
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    If so, then my only remaining source of confusion is how to handle the integrals in $q'$ and $q''$. Srednicki derives the general expression $\left<q'',t''|q',t'\right>=$ (Eq. 6.8). So when you say above, "the LHS gives the path integral," how do we arrive at this (in light of those $q$ integrals)? – WillG Apr 13 '19 at 22:29
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    I guess since this path integral is taken from $t=-\infty$ to $+\infty$, it would be overly restrictive to nail down a starting and ending position—so the $Dq$ in the path integral formalism, if time goes from $-\infty$ to $+\infty$, should be treated as including the $\int dq'' dq'$? – WillG Apr 13 '19 at 22:38
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    Sorry, I just had a typo in the answer. The $dq$ integrals were unnecessary and I had accidentally added them in. However, it would've basically gave the same final result because it would have been equivalent to integrating over the initial and final positions, but the choice of initial and final position is irrelevant to the answer. – Aaron Apr 14 '19 at 04:53