A very common inverse problem in mathematical physics is trying to understand the potential of a quantum mechanical system given its scattering data. Such problems, although very interesting, are very challenging and usually ill-posed. I'm trying to understand one inverse problem that I guess is particularly simple.

Some quantum mechanical models are ill-defined because their potential is unbounded from below (e.g., $\phi^3$). Still, perturbation theory is typically well-defined, as least in the sense of formal power series. As far as Feynman diagrams is concerned, the theories $\phi^3$ and $\phi^4$ are not fundamentally different, although only the second one represents an approximation to a meaningful non-perturbative theory. The same thing can be said about standard quantum mechanical models such as an anharmonic oscillator with cubic and quartic terms.

Assume we are given an arbitrarily large but finite number of terms in the perturbative expansion of, say, the partition function of a certain unknown system. (Recall that the logarithm of partition function represents the ground state energy, so it is in principle observable). Question: Can we predict whether the underlying potential of such a series is bounded from below?

In other words, a single term in the perturbative series is qualitatively identical in the bounded and unbounded case. But what if we zoom-out and look at many terms? Does the (truncated) series contain any information that allows us to tell them apart? Or is the series truly oblivious to the behaviour of the potential far from the equilibrium position?

It seems clear to me that from a truncated series we can at best predict the probability that the potential is bounded. The more terms, the better the prediction (in the sense of a confidence interval). As long as the number is finite, we can never be sure the potential is bounded or not. But is there such a probabilistic estimator at all? or is it really impossible to even predict a probability?

  • $\begingroup$ This seems to be a mathematical question about how well a truncated series can represent a function which requires an infinite series of terms. $\endgroup$ Mar 28, 2018 at 20:03
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    $\begingroup$ @sammygerbil No, I'm not interested in how well the series represents an observable. Im interested in the information contained in the series, regardless of any notion of convergence. The series has no numerical value; it's the sequence of its coefficients what matters. In other words, a formal power series in its rigorous mathematical sense (not the vague notion of formal series we physicists typically use). But yes: my question is mostly mathematical (although that doesn't make it offtopic: its about mathematical physics, which is ontopic) $\endgroup$ Mar 28, 2018 at 20:08
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    $\begingroup$ Your question seems hard in general but one can make a trivial observation. Take $\phi^4$ with the wrong sign for the coupling, which makes the potential unbounded below. This will be visible in the perturbation series: all the terms will have the same sign. $\endgroup$ Mar 28, 2018 at 22:11

1 Answer 1


I hope I'm not misreading; I think the answer is certainly no. Let's say we have a potential $V(\phi)$.

  • Perturbation theory can never see nonanalytic things like $e^{-1/x^2}$. So if the potential has a piece like $\phi\, e^{-1/\phi^2}$ it isn't bounded and you can never tell from any perturbation series. An even simpler example would be, say, $\delta(\phi - \phi_0)$ for any nonzero $\phi_0$.
  • Even assuming the potential is analytic, any finite truncation can say nothing. Let $$V(\phi) = \sum_{n=0}^\infty c_n g^n \phi^n$$ where $g$ is our expansion parameter. If we expand any quantity up to order $g^m$ we only know the coefficients up to $c_m$. But it doesn't matter whether or not $$V_m(\phi) = \sum_{n=0}^{m} c_n g^n \phi^n$$ is bounded from below because it is completely overwhelmed by even just $c_{m+1}$ alone. For any probability distribution on the $c_n$ whatsoever, as long as all the $c_n$ can be both positive and negative and the $c_n$ are independent, knowing any finite truncation gives exactly zero information.
  • Alternatively, we could use a much more special prior on the $c_n$. For example, if we somehow knew the full summed form of $V(\phi)$ was, say, $$V(\phi) = (\text{finite order polynomial with } O(1) \text{ coefficients}) \times (\text{exponential})$$ then we would indeed pick up a bit of information at each order of perturbation theory on whether the exponential was growing or decaying. However I think it's meaningless to assign a prior outside of any physical context, to give us a hint towards what the UV completion is. If you do specify the context your question basically reduces to 'all of particle physics' (i.e. still impossible) since finding UV completions from physical data is what the field is all about.
  • $\begingroup$ While I agree that finite order perturbation theory can never uniquely specify the potential, the statement that "Perturbation theory can never see nonanalytic things" seems perhaps too strong. Indeed there has been active research recently on "resurgence" in QFT, and (in the cases that are tractable to calculate) there are nontrivial cancellations between divergences in the resummed perturbation series and instantons which are required for the theory to be semiclassically well-defined. You can do the same thing in simple QM but as far as I know you basically just rediscover Borel summation. $\endgroup$
    – Logan M
    Mar 28, 2018 at 21:57
  • $\begingroup$ Could you help me see your first point better? I feel that such a term like $g\phi e^{-1/\phi^2}$ will still contribute to perturbation theory just as much as a $g\phi^3$ term would. I could add a term $g x e^{-1/x^2}$ term to the simple harmonic oscillator and get a perturbation series in $g$. Further, $g\phi e^{-1/\phi^2}$ would still give a bounded-below potential, assuming a $\phi^2$ term exists with the right sign. $\endgroup$
    – user196574
    Mar 18, 2021 at 2:32

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