Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Asymptotic series in field theory and quantum mechanics

+ 3 like - 0 dislike
3214 views

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.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user lux
asked May 29, 2017 in Theoretical Physics by lux (15 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user Emilio Pisanty
Edited: I'm interested in non relativistic quantum theory (no fields)

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user lux
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.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user Emilio Pisanty
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.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user lux
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.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user Thomas

2 Answers

+ 4 like - 0 dislike

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.)

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user JamalS
answered May 29, 2017 by JamalS (895 points) [ no revision ]
Thank you for a detailed and fully​ referenced answer. I will explore your links and let you know if everything is settled. :-)

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user lux
But these series are not constructed by perturbation theory, rather in some more advanced way!

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user Arnold Neumaier
+ 4 like - 0 dislike

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)

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user cesaruliana
answered May 30, 2017 by cesaruliana (215 points) [ no revision ]
An interesting collection of thoughts: thank you for writing and for the reference.

This post imported from StackExchange Physics at 2017-10-16 12:24 (UTC), posted by SE-user lux

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...