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I'm studying qft from Srednicki's book. If one writes down the full $i\epsilon$ terms, passing from Eq. (6.21) (non-perturbative definition) to Eq. (6.22) (perturbative definition) yields $$\left<0|0\right>_{f,h} = \int\mathcal{D}p\mathcal{D}q \exp\left[i\int_{-\infty}^{+\infty}dt\left(p\dot{q} - (1 - i\epsilon )H(p,q) + fq + hp\right)\right] \\ =\exp\left[-i(1 - i\epsilon )\int_{-\infty}^{+\infty}dt H_{1}\left(\frac{1}{i}\frac{\delta}{\delta h(t)},\frac{1}{i}\frac{\delta}{\delta f(t)}\right)\right]\\ \times\int\mathcal{D}p\mathcal{D}q \exp\left[i\int_{-\infty}^{+\infty}dt\left(p\dot{q} - (1 - i\epsilon )H_{0} + fq + hp\right)\right] \tag{6.22}$$ where $f\left( t\right) $, $g\left( t\right) $ are external sources with $% \lim_{t\rightarrow \pm \infty }f\left( t\right) =\lim_{t\rightarrow \pm \infty }g\left( t\right) =0$, $\left\langle 0|0\right\rangle _{f,h}$ is the vacum-vacum probability amplitude in the presence of sources, $H=H_{0}+H_{1}$ , and $\left\vert 0\right\rangle $ is the $\textit{ground state ket}$ (assumed non-degenerate) of the $\textit{total}$ Hamiltonian operator $\hat{H}=H(\hat{p},\hat{q})$.

However, in the physics literature the $i\epsilon$ factor multiplying $H_{1}\left(\frac{1}{i}\frac{\delta}{\delta h(t)},\frac{1}{i}\frac{\delta}{\delta f(t)}\right)$ is absent (see, e.g., Eq. (9.6) from Srednicki), even though the $i\epsilon$ factor multiplying $H_{0}$ is still present during the calculations giving the famous $i\epsilon$ prescription for the free propagator, and $\epsilon$ is taken to the limit $0^{+}$ only in the very end for $H_{0}$.

Question: How to justify getting rid of the $i\epsilon$ that multiplies $H_{1}$, while keeping to the very end the $i\epsilon$ that multiplies $H_{0}$?


Update 06.06.2019: I've tried to reason about it as follows. Let $$ Z_{0}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \equiv\int \mathcal{D}p\mathcal{D}q\exp\left[ i\int_{-\infty }^{+\infty }dt\ \left( p\dot{q}-\left( 1-i\varepsilon \right) H_{0}+fq+hp\right) \right] $$ and $$ x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \equiv \frac{\left( -1\right) ^{n}}{n!}\left[ \int_{-\infty }^{+\infty }dt\text{ }% H_{1}\left( \frac{1}{i}\frac{\delta }{\delta h\left( t\right) },\frac{1}{i}% \frac{\delta }{\delta f\left( t\right) }\right) \right] ^{n}Z_{0}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] $$

Since $$ \exp \left[ -i\left( 1-i\varepsilon \right) \int_{-\infty }^{+\infty }dt \text{ }H_{1}\left( \frac{1}{i}\frac{\delta }{\delta h\left( t\right) }, \frac{1}{i}\frac{\delta }{\delta f\left( t\right) }\right) \right] \\ =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n!}\left( i+\varepsilon \right) ^{n}\left[ \int_{-\infty }^{+\infty }dt\text{ }H_{1}\left( \frac{1}{i }\frac{\delta }{\delta h\left( t\right) },\frac{1}{i}\frac{\delta }{\delta f\left( t\right) }\right) \right] ^{n} $$ it follows that \begin{eqnarray*} \left\langle 0|0\right\rangle _{f,h} &=&\lim_{\varepsilon \rightarrow 0^{+}}\left\{ \sum_{n=0}^{\infty }\left( i+\varepsilon \right) ^{n}x_{n} \left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \right\} \\ &=&\lim_{\varepsilon \rightarrow 0^{+}}\left\{ \sum_{n=0}^{\infty }i^{n}x_{n} \left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \right\} \\ &&-i\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \left\{ \sum_{n=0}^{\infty }n\cdot i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \right\} +\lim_{\varepsilon \rightarrow 0^{+}}O\left( \varepsilon ^{2}\right) \end{eqnarray*}

I assume that the series $S_{1}\left( \varepsilon \right) \equiv \sum_{n=0}^{\infty }i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] $ is convergent, i.e., that $S_{1}\left( \varepsilon \right) $ is finite. However, the series $$ S_{2}\left( \varepsilon \right) \equiv \sum_{n=0}^{\infty }n\cdot i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] $$ is not necessarily convergent due to the multiplication of $i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] $ by $n$.

Therefore, the limit $$ \lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \left\{ \sum_{n=0}^{\infty }n\cdot i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \right\} =\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \cdot S_{2}\left( \varepsilon \right) $$ may not even exist.

Only in the case in which $\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \cdot S_{2}\left( \varepsilon \right) =0$ can one write \begin{eqnarray*} \left\langle 0|0\right\rangle _{f,h} &=&\lim_{\varepsilon \rightarrow 0^{+}}\left\{ \sum_{n=0}^{\infty }i^{n}x_{n}\left[ f\left( t\right) ,g\left( t\right) ;\varepsilon \right] \right\} \\ &=&\lim_{\varepsilon \rightarrow 0^{+}}\exp \left[ -i\int_{-\infty }^{+\infty }dt\text{ }H_{1}\left( \frac{1}{i}\frac{\delta }{\delta h\left( t\right) },\frac{1}{i}\frac{\delta }{\delta f\left( t\right) }\right) \right] \\ &&\times \int \mathcal{D}p\mathcal{D}q\exp \left[ i\int_{-\infty }^{+\infty }dt\text{ }\left( p\dot{q}-\left( 1-i\varepsilon \right) H_{0}+fq+hp\right) % \right] \end{eqnarray*} and the $i\varepsilon $ that multiplies $H_{1}$ can be taken as $0$, while the $i\varepsilon $ that multiplies $H_{0}$ is beeing kept until the very end of calculations.

Question: Is there any other way to justify this replacement? Is the perturbative definition for path integrals ill-defined due to possible divergence of the series $S_{2}\left( \varepsilon \right) $?


I would be most thankful if you could help me with this question concerning the perturbative definition of Green functions via path integrals in ordinary qm and in non-relativistic and relativistic qft.

Following the epoch making papers by Osterwalder and Schrader, it has become common practice to work in Euclidean time with Euclidean Green functions and then to analytically continue them to real time. Concerning path integrals, one usually defines them perturbatively.

My elementary analysis shows as to what happens when one makes an infinitesimal Wick rotation $t\rightarrow (1 - i\epsilon)t$ (much used in the literature). When the limit $\epsilon\rightarrow 0^{+}$ is taken in the end, in the perturbative definition of the path integral, I show that one encounters a nonsensical result of the form $\epsilon\times (divergent\ series)$ which doesn't have a well-defined limit when $\epsilon\rightarrow 0^{+}$. It seems that for an infinitesimal Wick rotation, when one analytically continues back to the real time (i.e., takes the limit $\epsilon\rightarrow 0^{+}$),one obtains a nonsensical result.

An analytic continuation is not a mere replacement $t \rightarrow it$, but can be thought of as a series of infinitesimal Wick rotations.


The problem of analytic continuation from real time to Euclidean time is very nicely presented in Ch. 13 of Giorgio Parisi's famous book on statistical field theory. This is the essence of the \epsilon trick in Srednicki's book.

In Parisi's analysis everything works very well since it is a non-perturbative analysis, based on the total Hamiltonian. The problem shows up in the perturbative approach as I have shown above. The perturbative approach for the analysis of analytic continuation $t\rightarrow it$ is not presented in any book that I know of. In order for the theory to be consistent, the perturbation series must be convergent for any $t\exp(i\theta)$ with $0 < \theta < \pi/2$. For the theory to exist, it is NOT sufficient that the case $\theta = \pi/2$ is convergent.

I think that the problem is very serious since the perturbative definition of path integrals is the most common tool in physics. The question I'm raising is not merely an academic one.

I have no proof that the series $S_2$ is divergent, since this is model dependent. However, there are no theorems in complex or real analysis that discuss the convergence of a series $S_2 = \sum_{n=0}^{\infty} nx_n$, if it is known that the series $S_1 = \sum_{n=0}^{\infty} x_n$ is convergent.

If one continues the series expansion in $\epsilon$ to higher order terms, the divergences are even worse.

Questions: Is the perturbative definition of path integrals wrong? Is there a version of the O-S axioms for non-relativistic qm and condensed matter? Is the perturbative definition of path integrals in contradiction with the O-S axioms?

I would be most thankful if you could send me your opinion on these questions.

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  • $\begingroup$ Is it not assumed that $H_1$ is just a small perturbation from the non-interacting theory? In that case the $\epsilon H_0$ term would be of $\mathcal{O}(2)$. $\endgroup$ – Akerai May 31 at 18:16
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    $\begingroup$ $H_{1}$ may be "small" but is finite and doesn't depend on $\epsilon$. Therefore cannot be $\mathcal{O}(2)$. $\endgroup$ – pathint May 31 at 18:36
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I do not think the understanding is to get rid of the $i \epsilon$ multiplying the perturbation Hamiltonian $H_1$, but rather to assume it implicitly multiplying $H_1$ even if not shown for notational simplicity. This is confirmed by the fact that in eq. (6.22) in Srednicki "Quantum field theory", also the basic Hamiltonian $H_0$ is not given the $i \epsilon$ either.

Looking through Srednicki's book, in many demonstrations the $i \epsilon$ is suppressed, even if when required it is shown again.

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  • $\begingroup$ What I'm most concerned about is the very perturbative definition of path integrals, which doesn't seem to make sense due to divergence of $S_2 (\epsilon )$. Srednicki has no comment on that, and neither could I find it in the literature. $\endgroup$ – pathint Jun 8 at 9:10
  • $\begingroup$ If you add a perturbation to the basic hamiltonian, you assume the expansion is convergent and the first terms are enough to describe the physical occurrence. Of course, only the experimental evidence can validate the procedure. $\endgroup$ – Michele Grosso Jun 9 at 19:34
  • $\begingroup$ Sometimes one needs nonperturbative results and for that sums infinite special classes of Feynman diagrams, or if possible all of them. There is more to the world than QED. See, e.g., Mattuck's book about Feynman diagrams. I've thought that if $S_1$ is finite one has solved the problem exactly, but $S_2$ being most likely divergent spoils everything. $\endgroup$ – pathint Jun 9 at 19:57

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