Transition from an initial/final position state to the ground state in the path integral

Question

I am reading chapter 6 of M.Srednicki's book. On page 47 he argues why it is possible to choose for the initial/final state the ground state instead of a position state. Actually I don't understand his argument. He starts his argument off with

$$<0|0>_{f,h} = \lim_{t'\longrightarrow -\infty, t''\longrightarrow+\infty} \int dq'' dq' \psi^{*}_0(q'') <q'', t''|q' t'>_{f,h} \psi_0(q')\tag{6.19}.$$

Then he argues that if $|n>$ is an eigenstate of the Hamilton operator with the eigenvalue $E_n$ with the assumption $E_0=0$ the position state $|q't'>$ can be developed in the following way:

$$|q', t'>=e^{iHt'}|q'>=\sum^{\infty}_{n=0} e^{iHt'}|n><n| q'> =\sum^{\infty}_{n=0} \psi_n^*(q') e^{iE_n t'} |n>,\tag{6.20}$$

and if $H$ is substituted by $(1-i\epsilon)H$ the result would be

$$\lim_{t'\longrightarrow -\infty}|q', t'>=\psi_0^*(q')|0>.\tag{1}$$

A similar conclusion could also be drawn for $<q'',t''|$:

$$\lim_{t''\longrightarrow +\infty}<q'', t''|=\psi_0(q'')<0|.\tag{2}$$

Actually if I plug both results in the first formula, I would get a kind of tautology since I would find $$<0|0>_{f,h} = <0|0>.\tag{3}$$

under the assumption that the function $\psi_0(q)$ is normalized: $$\int|\psi_0|^2 dq =1.\tag{4}$$

Or should I better conclude that

$$<0|0>=\lim_{t'\longrightarrow -\infty, t''\longrightarrow+\infty} <q'',t''|q',t'>\frac{1}{\psi_0(q'')\psi_0^*(q') }\sim Z~ ?\tag{5}$$

Srednicki proposes also to multiply this expression by a function $\chi(q')$ (with $<0|\chi>\neq 0$) and, I guess, with $\xi(q'')$ too and integrate, but then the expression would look like:

$$\int dq' dq'' \chi(q') \xi(q'') <0|0> $$ $$= \int dq' dq'' \lim_{t'\longrightarrow -\infty, t''\longrightarrow+\infty} <q'',t''|q',t'>\frac{\chi(q')\xi(q'')}{\psi_0(q'')\psi_0^*(q') }.\tag{6}$$

Finally I could bring the integral $\int dq' dq'' \chi \xi$ on the rhs of the equation which would lead to an expression which even more complicated than the expression Srednicki started off. So all that does not look very reasonable for me. I would be grateful if someone who has more experience with the path integral than me would explain it.

Answer 1

1. Let us first of all mention that Ref.1 is considering QM as opposed to QFT, i.e. no particle creation & annihilation are allowed. Eq. (6.19) can alternatively be written as $$\begin{align} \langle 0 \text{ out} |0 \text{ in}\rangle_{f,h} ~=~~& \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty} \langle 0|\hat{U}(t^{\prime\prime},t^{\prime})|0\rangle \cr \stackrel{\begin{matrix}\text{Compl.}\cr\text{Rel.}\end{matrix}}{=}& \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty} \int \! dq^{\prime}dq^{\prime\prime} \langle 0|q^{\prime\prime}\rangle \langle q^{\prime\prime}| \hat{U}(t^{\prime\prime},t^{\prime})|q^{\prime} \rangle \langle q^{\prime} |0\rangle \cr ~=~~&\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty} \int \! dq^{\prime}dq^{\prime\prime} \langle 0|q^{\prime\prime}\rangle \langle q^{\prime\prime}, t^{\prime\prime}| q^{\prime}, t^{\prime} \rangle_{f,h} \langle q^{\prime} |0\rangle, \end{align}\tag{6.19}$$ where we in last equality introduced instantaneous eigenstates in the Heisenberg picture, cf. the text above eq. (6.2) in Ref. 1. The overlap in the operator formalism can formally be written as a phase space path integral $$\begin{align} \langle q^{\prime\prime}| \hat{U}(t^{\prime\prime},t^{\prime})|q^{\prime} \rangle ~=~&\langle q^{\prime\prime}, t^{\prime\prime}| q^{\prime}, t^{\prime} \rangle_{f,h}\cr ~=~&\int_{q(t^{\prime})=q^{\prime}}^{q(t^{\prime\prime})=q^{\prime\prime}}\!{\cal D}q~{\cal D}p~\exp\left[\frac{i}{\hbar}\int_{t^{\prime}}^{t^{\prime\prime}} \! dt \left(p\dot{q}-H+fq +hp \right) \right].\end{align} \tag{6.16} $$ The Dirichlet boundary conditions (DBC) are mentioned in the text below eq. (6.8) in Ref. 1. Note that unlike the operator formalism, there is a $p\dot{q}$ term in the path integral. In eq. (6.19) the time-evolution operator is $$ \hat{U}(t^{\prime\prime},t^{\prime})~:=~T\exp\left[-\frac{i}{\hbar}\int_{t^{\prime}}^{t^{\prime\prime}} \! dt \left(\hat{H}-f\hat{q} -h\hat{p} \right) \right],$$ cf. e.g. this related Phys.SE post.

2. Hidden assumption: It is implicitly assumed in Ref. 1 that the sources $f,h$ vanish outside a compact time interval $[t_1,t_2]\subseteq[t^{\prime},t^{\prime\prime}]$.

3. We next introduce the Feynman $\color{red}{i\epsilon}$-prescription $$ \hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t^{\prime})~:=~T\exp\left[-\frac{i}{\hbar}\int_{t^{\prime}}^{t^{\prime\prime}} \! dt \left(\color{red}{(1-i\epsilon)}\hat{H}-f\hat{q} -h\hat{p} \right) \right].$$

4. Ref. 1 tries to make the point that $$\begin{align} \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}&\int \! dq^{\prime}dq^{\prime\prime}\frac{\langle \chi^{\prime\prime}|q^{\prime\prime}\rangle}{\langle \chi^{\prime\prime}|0\rangle} \langle q^{\prime\prime}| \hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t^{\prime})|q^{\prime} \rangle \frac{\langle q^{\prime} |\chi^{\prime}\rangle}{\langle 0|\chi^{\prime}\rangle}\cr \stackrel{\begin{matrix}\text{Compl.}\cr\text{Rel.}\end{matrix}}{=}\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \sum_{n^{\prime}=0}^{\infty}\sum_{n^{\prime\prime}=0}^{\infty}\frac{\langle \chi^{\prime\prime}|n^{\prime\prime}\rangle}{\langle \chi^{\prime\prime}|0\rangle} \langle n^{\prime\prime}| \hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t_2) \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1) \hat{U}_{\color{red}{i\epsilon}}(t_1,t^{\prime}) |n^{\prime} \rangle \frac{\langle n^{\prime} |\chi^{\prime}\rangle}{\langle 0|\chi^{\prime}\rangle}\cr \stackrel{\begin{matrix}\text{Hidden}\cr\text{assum.}\end{matrix}}{=} \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \sum_{n^{\prime}=0}^{\infty}\sum_{n^{\prime\prime}=0}^{\infty}\frac{\langle \chi^{\prime\prime}|n^{\prime\prime}\rangle}{\langle \chi^{\prime\prime}|0\rangle} \exp\left[-\frac{i}{\hbar}\color{red}{(1-i\epsilon)} E_{n^{\prime\prime}}(t^{\prime\prime}\!-\!t_2) \right]\cr &\times~ \langle n^{\prime\prime}| \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1)|n^{\prime} \rangle \exp\left[-\frac{i}{\hbar}\color{red}{(1-i\epsilon)} E_{n^{\prime}}(t_1\!-\!t^{\prime}) \right] \frac{\langle n^{\prime} |\chi^{\prime}\rangle}{\langle 0|\chi^{\prime}\rangle}\cr ~=~\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \exp\left[-\frac{i}{\hbar}\color{red}{(1-i\epsilon)} E_0(t^{\prime\prime}\!-\!t_2) \right] \langle 0| \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1)|0 \rangle\cr &\times~\exp\left[-\frac{i}{\hbar}\color{red}{(1-i\epsilon)} E_0(t_1\!-\!t^{\prime}) \right]\cr \stackrel{\begin{matrix}\text{Hidden}\cr\text{assum.}\end{matrix}}{=} \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \langle 0|\hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t_2)|0\rangle\langle 0| \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1)|0\rangle\langle 0| \hat{U}_{\color{red}{i\epsilon}}(t_1,t^{\prime})|0 \rangle \cr \stackrel{\begin{matrix}\text{Ortho.}\cr\text{Rel.}\end{matrix}}{=} \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \sum_{n^{\prime}=0}^{\infty}\sum_{n^{\prime\prime}=0}^{\infty}\langle 0|\hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t_2)|n^{\prime\prime}\rangle\langle n^{\prime\prime}| \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1)|n^{\prime}\rangle\langle n^{\prime}| \hat{U}_{\color{red}{i\epsilon}}(t_1,t^{\prime})|0 \rangle \cr \stackrel{\begin{matrix}\text{Compl.}\cr\text{Rel.}\end{matrix}}{=} \lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \langle 0|\hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t_2) \hat{U}_{\color{red}{i\epsilon}}(t_2,t_1) \hat{U}_{\color{red}{i\epsilon}}(t_1,t^{\prime})|0 \rangle\cr ~=~\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \langle 0|\hat{U}_{\color{red}{i\epsilon}}(t^{\prime\prime},t^{\prime})|0 \rangle ~=~\langle 0 \text{ out} |0 \text{ in}\rangle_{f,h}, \end{align} $$ independently of the 2 arbitrary states $|\chi^{\prime}\rangle$ and $|\chi^{\prime\prime}\rangle$ as long as the denominators don't vanish. In the third equality we used that the Feynman $\color{red}{i\epsilon}$-prescription washes out higher energy states.

5. Altogether the above conveniently implies that we don't have to impose boundary conditions for the phase variables $q,p$ if we replace the Hamiltonian $H$ with $\color{red}{(1-i\epsilon)}H$: $$\begin{align} &\langle 0 \text{ out} |0 \text{ in}\rangle_{f,h} \cr ~=~\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \int\! dq^{\prime}dq^{\prime\prime}\frac{\langle \chi^{\prime\prime}|q^{\prime\prime}\rangle}{\langle \chi^{\prime\prime}|0\rangle} \frac{\langle q^{\prime} |\chi^{\prime}\rangle}{\langle 0|\chi^{\prime}\rangle}\cr \int_{q(t^{\prime})=q^{\prime}}^{q(t^{\prime\prime})=q^{\prime\prime}}\!&{\cal D}q~{\cal D}p~\exp\left[\frac{i}{\hbar}\int_{t^{\prime}}^{t^{\prime\prime}} \! dt \left(p\dot{q}-\color{red}{(1-i\epsilon)}H+fq +hp \right) \right] \cr ~=~\lim_{t^{\prime}\to -\infty, t^{\prime\prime}\to+\infty}& \int\!{\cal D}q~{\cal D}p~\exp\left[\frac{i}{\hbar}\int_{t^{\prime}}^{t^{\prime\prime}} \! dt \left(p\dot{q}-\color{red}{(1-i\epsilon)}H+fq +hp \right) \right].\end{align} \tag{6.21} $$

6. OP's eq. (3) is not true.

References:

1. M. Srednicki, QFT, 2007; Chapter 6, p.47. A prepublication draft PDF file is available here.