It is well known that perturbation theory in quantum field theory leads to a series that is (at best) asymptotic. Dyson's famous argument for quantum electrodynamics is a good justification for this.

What about in non relativistic quantum mechanics - are there examples where (time independent, say) perturbation theory leads to a convergent series / a divergent series / an asymptotic series?

As an example, say one were to study the anharmonic oscillator, with some perturbation $\mu x^{4}$. Perturbation theory for the ground state energy would lead to a power series in $\mu$. I don't think Dyson applies here as the sign of $\mu$, for sufficiently small $\mu$ should not lead to such dramatic change in the physics as in negating the electric charge. Are there examples were perturbative series such as this diverge, converge, are asymptotic?

Thanks for any input.

  • $\begingroup$ QFT is a subset of QM, so the question isn't particularly clear. You should edit your post to make it clear what other characteristics you're looking for in such an asymptotic series, and why your QFT example doesn't already fit the bill. If you're looking for an exhaustive list of such series, that question is too broad for our format. $\endgroup$ Commented May 29, 2017 at 16:18
  • $\begingroup$ Edited: I'm interested in non relativistic quantum theory (no fields) $\endgroup$
    – nox
    Commented May 29, 2017 at 16:30
  • $\begingroup$ That's not a particularly well-defined disjunctive. There's plenty of situations where the relevant framework is QFT over a non-relativistic hamiltonian, including in particular condensed matter and cold quantum gases. So, again - the QFT example still applies. $\endgroup$ Commented May 29, 2017 at 16:34
  • $\begingroup$ Agreed but I am not interested in such situations since, as you already point out, I already know what happens when QFT applies. I have edited my question to give a simple, specific, clear example of where a perturbative series could arise in QM. Hopefully this explains what I'm looking to find out. $\endgroup$
    – nox
    Commented May 29, 2017 at 16:42
  • 2
    $\begingroup$ Dyson's argument of applies to quantum mechanics. If the fourth order term is negative, then the ground state is unstable and tunneling takes place. This means that perturbation theory cannot be convergent. $\endgroup$
    – Thomas
    Commented May 29, 2017 at 16:47

2 Answers 2


For certain systems, there indeed are convergent perturbation series. In the case of the anharmonic oscillator described by a Hamiltonian,

$$H = \frac12 p^2 + \frac12 m^2x^2 + \frac14 gx^4$$

one may construct a convergent series which is convergent for any $g>0$ and arbitrary harmonic term, valid in both the weak and strong coupling limits.

Similarly, using a procedure involving splitting the Hamiltonian in a particular way and applying perturbation theory, a system of coupled harmonic oscillators admitted a convergent series.

Furthermore, there exists a generalisation of a convergent perturbation series for a $q$-deformed anharmonic oscillator, which is to say a system based on a $q$-deformed Heisenberg algebra, with a Hamiltonian based on operators with modified commutation relations.

In a more general setting, it has been shown that for a class of hyperbolic differential equations, there exists convergent perturbation series for particular families under certain conditions. (As an aside this is used to show the Einstein field equations in a particular scheme yield a divergent perturbation series as opposed to asymptotic.)

  • $\begingroup$ Thank you for a detailed and fully​ referenced answer. I will explore your links and let you know if everything is settled. :-) $\endgroup$
    – nox
    Commented May 29, 2017 at 18:08
  • $\begingroup$ But these series are not constructed by perturbation theory, rather in some more advanced way! $\endgroup$ Commented Oct 10, 2017 at 16:30
  • $\begingroup$ @Arnold Neumaier I don't quite understand your comment. Perturbation theory assumes a formal power series expansion. Then it uses constrains to find the coefficients. If there exist a set of coefficients that give a convergence power series. Then the perturbation theory procedure must give those coefficients (since power series expansion is unique). Can you tell me how my argument is flawed? $\endgroup$
    – Bohan Xu
    Commented Jan 24, 2020 at 15:58
  • 1
    $\begingroup$ @BohanXu: There is no flaw if the series is convergent. But for the quartic oscillator, the convergence radius of the standard perturbation series is zero. The link given in the answer does not use the standard procedure, hence, though useful, doesn't answer the original question. $\endgroup$ Commented Jan 31, 2020 at 16:24

You should take a look at Francisco Fernandez "Introduction to Perturbation Theory in Quantum Mechanics", in chapter 6 he notes that almost all perturbation series are divergent, including fan-favorites Stark and Zeeman effect, so that having a convergent series is an exception. This has nothing to do with the infinite degrees of freedom of QFT, is just a fact of QM.

As an addendum, let me just note that although almost all perturbation series are indeed divergent the problem only arises (or arised) in QFT. That's because in finite degrees of freedom QM we can define rigorously the hamiltonian, for example the Zeeman effect one, we just can't compute exactly the eigenvalues.

In the early days of QFT people only knew how to define free fields, there was no satisfactory definition of a interacting QFT. For some time there was some hope of defining QFT to BE the results of perturbation series. That's why Dyson's argument was significant, it shows that one cannot define interacting QFT via perturbation expansion.

(The last couple of paragraphs we're my extrapolation of why you we're curious about convergence in QM, and so I thought appropriate to include. I'll be happy to delete them if people find it not pertaining to the questions as such)

  • $\begingroup$ An interesting collection of thoughts: thank you for writing and for the reference. $\endgroup$
    – nox
    Commented May 30, 2017 at 3:11

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