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  Is the perturbative definition of path integrals ill-defined and contradicts the O-S axioms?

+ 1 like - 0 dislike

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 $H_{0}$?


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]
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
\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)

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
\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) %
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 of 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 a 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 back to the 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.

asked Jun 8, 2019 in Theoretical Physics by pathint (5 points) [ revision history ]
reshown Jun 26, 2019 by pathint

The perturbative definition is strictly speaking meaningless in the absence of a renormalization prescription which removes the divergences. Therefore one expects problems when trying to give it precise mathematical meanings. Manipulations are appropriate only in as much they lead to finite results, in which case they are usually useful.

@Arnold Neumaier I have assumed that $S_1$ is convergent. This means that regularization and renormalization have been enforced and I even have assumed that the series converges. The problem is one of analytic continuation from real time to Euclidean time in the perturbative definition of path integrals, the most commonly one used in the literature

So your regularized Dp and Dq are finite-dimensional measures?? Or how is your perturbative path integral defined? This is not what Srednicki is doing. He uses the ill-defined infinite-dimensional measure, and get divergences in (9.21). 

Note also that the OS axioms only hold for the correctly renormalized but unregularized n-point functions (i.e., after the removal of the cutoff). There you can analytically continue from imaginary positive time to any time in the upper half plane. 

@Arnold Neumaier I repeat that my problem is a about analytic continuation in time of a perturbative expression whose terms are all finite. Take a simple qm example. If it doesn't work in 0 space dimensions, it doesn't work in d dimensions either. In d dimensions there are regularization and renormalization problems, which can be dealt with in some models, e.g. superrenormalizable theories. Please, if you can, answer the question about TIME and forget about space.

1 Answer

+ 1 like - 0 dislike

The perturbative expansion doesn't have poles and hence reflects the analytic structure very incompletely. Thus it should not surprise you that you get nonsense when you try to analytically continue it. You need to resum enough before you can expect analytic continuation to be approximately faithful.

answered Jun 19, 2019 by Arnold Neumaier (15,787 points) [ no revision ]

Thank you very much for your explanation.

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