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My course instructor mentioned that the Perturbative Series are not convergent but diverge as we consider more and more terms in the expansion. He then briefly mentioned that the Perturbative Series are Asymptotic Series. I have some idea about Asymptotic series, such as Stirling's approximation $\left (n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n} \right )$, which gets better as $n$ increases.

So, does it mean that the perturbative expansion of $n^{th}$ state energy ($E_n$) is gets better as $n$ increases?

Qmechanic
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Dev
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2 Answers2

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The difference between a convergent and asymptotic series comes from reversing the order of two quantifiers.

  • Convergent: For every $x$, there exists a large enough number of terms such that the error is less than $\epsilon$.
  • Asymptotic: For every number of terms, there exists a small enough $x$ such that the error is less than $\epsilon$.

The utility of the latter can be summed up by Carrier's rule: "Divergent series converge faster than convergent series because they don't have to converge". In other words, the approximation will get better for awhile and then start to get worse at a point that depends on $x$.

In the case of Stirling's approximation, the small number $x$ would be $1/n$. For perturbation theory in quantum mechanics though, it is some coefficient of an interaction Hamiltonian. I'm not aware of any result stating that in systems with a discrete spectrum, excited state energies are more easily approximated than the ground state energy. The eigenstate thermalization hypothesis more or less tells us to expect the opposite.

J.G.
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Connor Behan
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Just because I don't think there's a clear enough answer to your proposal that $n$ might be the energy level (as in $E_n$). No that's the wrong parameter to be increasing to get better precision. The equivalent to the $n$ in the Stirling approximation is the number of terms in the perturbative expansion. I.e. first order energy perturbation ($n=1$), second order ($n=2$), etc. Although most people don't go beyond second order energy perturbation in QM anyway because even if it was convergent the terms get complicated and difficult to compute.

So if you're using $n$'th order perturbation theory to compute the perturbed energy levels of a system, as you increase $n$, at some point the terms will start getting bigger rather than smaller, and you'll know that your answer is getting less, not more precise.

As was discussed in the other answer, the optimal choice for $n$ will depend on how big your perturbation is. Roughly speaking, I'd guess you can use first order perturbation theory to see how big the perturbations are, then the nominal energy level differences over the size of the perturbation is the order of magnitude of the optimal $n$.

AXensen
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