4
$\begingroup$

Consider some $1d$ system $H_0$ with energy levels $E_n^{(0)}$ that has been perturbed by $\lambda V$, i.e. $H = H_0 + \lambda V$ for some $\lambda >0$. Consider the ratio of the first order perturbation $E_n^{(1)}$ to the unperturbed $E_n^{(0)}$ as a function of $n$.

If this ratio $E_n^{(1)}/E_n^{(0)}$ diverges as $n\to \infty$, must the perturbation series fail to converge in $\lambda$?


Here's what a counterexample might look like: For example, suppose the true energy levels of the system were $n^q e^{-\lambda n^p}$ for $q, p>0$. Then the first order perturbation would be $-\lambda n^{p+q}$, which would suffer $E_n^{(1)}/E_n^{(0)} \to \infty$ as $n \to \infty$. Nevertheless, the form of the energy levels is analytic in $\lambda$ and hence one would expect the perturbation series to converge.

My trouble is that I'm unsure of how to construct a counterexample Hamiltonian with such energy levels. Indeed, many examples where $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$ in fact do have perturbation series that fail to converge for any nontrivial $\lambda$! For example, the Harmonic oscillator perturbed by $\lambda x^k$ for even integers $k>2$ suffers a first order correction growing strongly in $n$ and for which $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$, and these perturbed systems are known to have energies that are not analytic in $\lambda$ about $\lambda = 0$ by Dyson's argument (and in turn have merely asymptotic and not convergent perturbation series).


I'm curious if such a counterexample is known/can be constructed or if there's some friendly argument why $\lim_{n\to \infty} E_n^{(1)}/E_n^{(0)} = \infty$ completely ensures the perturbation series cannot converge in $\lambda$. To avoid any subtleties with negative energy bound states approaching an energy of $0$, let's restrict to the case of $E_n^{(0)} \to \infty$ as $n \to \infty$.

$\endgroup$
0

4 Answers 4

5
$\begingroup$

Here is a counterexample:

  1. Let $\hat{H}_0=\hbar\omega (\hat{n}+\frac{1}{2})$ be the 1D quantum harmonic oscillator, and let the full Hamiltonian be $\hat{H}=\hat{H}_0+\lambda\hat{V}$ with perturbation $\hat{V}=\hbar\omega \hat{n}^2$.

  2. The energy eigenstates $$|n\rangle~=~|n^{(0)}\rangle, \qquad |n^{(\geq 1)}\rangle~=~0,$$ are undeformed.

  3. Since the perturbation series $$E_n~=~\underbrace{E^{(0)}_{n}}_{=\hbar\omega (n+\frac{1}{2})}+\lambda \underbrace{E^{(1)}_{n}}_{=\hbar\omega n^2}, \qquad E^{(\geq 2)}_{n}~=~0,$$ only has 2 terms, it is obviously convergent and analytic in $\lambda$, but $$\lim_{n\to\infty}\frac{E^{(1)}_{n}}{E^{(0)}_{n}} ~=~\infty.$$

  4. If $\lambda\geq 0$ ($\lambda<0$), then the full Hamiltonian $\hat{H}$ is unbounded from above (below) but bounded from below (above) with a ground state (highest energy state), respectively.

    Viewed as an isolated QM system following the TDSE it is in both cases mathematically well-defined, as the overall sign of the energy is a matter of convention.

    (In a more realistic model, one should include the interactions with the environment.)

$\endgroup$
4
  • $\begingroup$ i.e. any perturbation which doesn't change the eigenvectors of the Hamiltonian, and the perturbation grows with $n$ is a counterexample. $\endgroup$
    – AXensen
    Jun 27, 2023 at 11:23
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Jun 27, 2023 at 11:49
  • $\begingroup$ +1 Thank you for the discussion. 1. Do you have a sense for why this perturbation enjoys a convergent series while $\lambda x^4$ suffers a divergent series in $\lambda$, despite the fact that both perturbations lead to an energetic catastrophe for $\lambda<0$? 2. @AXensen The perturbation should grow rapidly enough in $n$, not just growing in $n$ $\endgroup$
    – user196574
    Jun 27, 2023 at 18:02
  • $\begingroup$ 1. The $\lambda x^4$ potential also deforms the energy eigenstates. $\endgroup$
    – Qmechanic
    Jun 28, 2023 at 7:14
1
$\begingroup$

The first order correction is the diagnonal element of pertirbation: $$ E_n^{(1)}=\langle n|\lambda V|n\rangle, $$ which is why it is usually absorbed in the unperturbed Hamiltonian $$ E_n^{(0)}=\langle n|H_0|n\rangle \longrightarrow \tilde{E}_n^{(0)}=\langle n|H_0|n\rangle + \langle n|\lambda V|n\rangle $$

The rest of the perturbation terms contains denominators inversely proportional to energy differences $$ \sim\frac{1}{E_n^{(0)}-E_m^{(0)}}. $$

The condition $$E_n^{(1)}/E_n^{(0)}$$ might make the diagonal part of the Hamiltonian somehow irregular - e.g., by introducing degeneracies that would break perturbation theory, $\tilde{E}_n^{(0)}\approx \tilde{E}_m^{(0)}$ - it depends on the details of the limit $n\rightarrow 0$, and no general answer can be given. However, as should be clear from the above, this is more an issue of a poorly defined unperturbed Hamiltonian, than an actual pattern in perturbation theory.

Example
Let us perturb a harmonic oscillator with unperturbed energies $$ E_n = \hbar\omega_0 n $$ We now take a perturbation with diagonal elements $$ \langle n|V|n\rangle = V_0n^2 $$ If $V_0>0$ the distances between the levels would only grow with increasing $n$, and the perturbation theory works fine, as long as the non-diagonal elements of the perturbation is smaller than the distance between the levels. However, if $V_0<0$, we might get level crossings, i.e., situations where the levels are nearly degenerate and the PT breaks down.

$\endgroup$
1
  • 1
    $\begingroup$ +1 I feel I must have missed this discussion somehow (and maybe Qmechanic did too, since their answer has some nice similarities to yours). I'm going to mull this over a bit. $\endgroup$
    – user196574
    Jun 27, 2023 at 18:00
1
$\begingroup$

After thinking about it a bit, I think the two conditions (the perturbative series converging and the perturbation growing with $n$) actually have nothing to do with each other. There's no reason to think one implies the other, and it's pretty easy to make counterexamples. See for example qmechanic's trivial counterexample. I'm only writing another answer in case of the rebuttal "but maybe there are only counterexamples with this property that they don't change the eigenfunctions." I want to make it clear that these two properties of a perturbative series really have nothing to do with one another.

Consider an example which does satisfy both conditions. $V_0=m\omega^2x^2/2$, $V_1=cx^4$. Often the non-convergence of perturbation theory in this example is proven by pointing out that if $c$ is negative, then the potential goes to $-\infty$ as $x$ goes to $\pm\infty$, so we no longer have any bound states. So the groundstate energy cannot be an analytic function of $c$. But perturbation theory to a finite number of orders will still give a good approximation of how positive $c$ values affect the energy levels. And the perturbation grows with $n$. But this is really just a coincidence - not the cause of perturbation theory's failure.

Also, as other answers have noted, the condition $E_n^{(1)}/E_n^{(0)}\rightarrow \infty$ cannot possibly be the condition for convergence, because $E_n^{(0)}$ can be shifted arbitrarily by adding a constant to the potential. Typically, the validity of perturbation theory is expressed in terms of the correction to the energy level divided by energy level differences. The validity meaning that the first order is reasonably accurate - nothing to do with whether or not it converges after infinitely many orders. Likewise the validity of the adiabatic theorem is expressed in terms of energy level differences. I understand that you restricted your attention to situations where $E_n^{(0)}>0$ trying to avoid this issue, but the fact that $E_n^{(0)}$ is arbitrary just proves that $E_n^{(1)}/E_n^{(0)}\rightarrow \infty$ will never come up in a proof that (for example) perturbation theory doesn't converge.

As OP pointed out, the following example doesn't satisfy their condition $E_n^{(1)}/E_n^{(0)}\rightarrow \infty$, but it does have the property that the correction divided by energy level differences goes to infinity.

If I just remove the property of the potential which makes it non-convergent and still have the perturbation increasing with $n$ I will make a counterexample. Take the same harmonic oscillator from before and use $V_1=\lambda m\omega_1^2 x^2$. The energy levels are now $$ E_n=(n+1/2)\hbar\sqrt{\omega^2+\lambda\omega_1^2} $$ If you used perturbation theory it would reveal, obviously, the taylor series of that expression for small $\lambda$. And it's not an asymptotic series, it has a radius of convergence given by $|\lambda\omega_1^2|=\omega^2$. But the perturbation also obviously grows with $n$, it's proportional to $n+1/2$.

$\endgroup$
2
  • $\begingroup$ +1 Thanks for the discussion. Note that it's not quite enough for the counterexample to have an energy correction growing in $n$; it needs to be growing rapidly enough in $n$, since I'm considering the case that $E_n^{(1)}/E_n^{(0)}$ diverges $\endgroup$
    – user196574
    Jun 27, 2023 at 17:52
  • $\begingroup$ @user196574 yeah I see now that this didn't satisfy your divergence condition. I edited the answer because I still think it has some useful ideas, and I added some discussion about how $E_n^{(1)}/E_n^{(0)}\rightarrow \infty$ is probably not a well-defined condition. But I couldn't easily fix my "counterexample" to make it satisfy your condition. $\endgroup$
    – AXensen
    Jun 28, 2023 at 8:29
-1
$\begingroup$

The limit $n\rightarrow \infty$ is unnecessary (as is looking for a divergence in this ratio). A condition that is typically necessary for perturbation theory to be valid is that \begin{equation} \frac{E^{(1)}_n}{E^{(0)}_n} \ll 1 \end{equation} for all $n$. (At least, all $n$ where you trust your theory is valid).

An exception is if $E_n^{(0)}=0$, then the leading contribution to the energy can the first order correction. But then you would expect the second order correction to be smaller than the first.

$\endgroup$
3
  • $\begingroup$ Hi, do you have a specific case where the ratio diverges in $n$ but the perturbation theory converges in $\lambda$ for any fixed $n$? Or a proof that that cannot happen? $\endgroup$
    – user196574
    May 8, 2022 at 22:22
  • $\begingroup$ @user196574 If the ratio diverges in $n$, then $E^{(1)}_n/E^{(0)}_n > 1$ for all $n>N$ for some $N$. This means you would expect that perturbation theory is very badly breaking down. It's not a proof, since as I said in my answer it could be that $E^{(0)}_n$ is simply $0$ (or very small); what really matters is the behavior of $E^{(p+1)}_n / E^{(p)}_n$ for large $p$. But, it's evidence that something odd is happening. The limit as $n\rightarrow \infty$ is not really relevant for asking about perturbation theory, it is enough to look at one $n$. $\endgroup$
    – Andrew
    May 8, 2022 at 22:57
  • $\begingroup$ I think this answer really doesn't get at what the question is asking. Perturbation theory can be "valid" given your definition but still not converge (a harmonic oscillator with a small $x^4$ perturbation). In fact that's usually the case - usually perturbation theory creates an "asymptotic series." And to fix your issue with $E_n^{(0)}=0$, usually people express the validity of perturbation theory with the ratio of the correction to energy level differences. I can also add a constant potential to make $E_n^{(0)}$ huge and erroneously make it look like perturbation theory is always valid. $\endgroup$
    – AXensen
    Jun 27, 2023 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.