Time independent perturbation theory

  • #1
This isn't explained anywhere so it must be super basic and I'll probably kick myself for not getting it, but on the wiki page for time independent perturbation theory, section 3.1:
https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)
It talks about first order corrections and says if the perturbation is weak you can write E and ##|n\langle## as a Maclaurin series in ##\lambda##. Why can we and why would we do that? I've tried a number of sources including Griffiths and online lecture notes but I still don't get why that step is done, isn't E just a constant? Why can it be written as a power series of that particular form?
 

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  • #2
hilbert2
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Well, if we have something like a nonlinearly perturbed quantum oscillator with Hamiltonian

##\hat{H} = \frac{\hat{p}^2}{2m}+\frac{1}{2}k\hat{x}^2 + \beta \hat{x}^4##,

with ##\beta## some parameter with correct dimensions and ##k## fixed to some value, then the sequences of eigenstates ##\left|\right.n\left.\right>## and eigenenergies ##E_n## are both functions of the parameter ##\beta##. The perturbation theory assumes that they depend on ##\beta## like an analytical function, i.e. something that can be written as a Maclaurin series. The assumption is not correct in general, but can produce quite good results for small perturbations even if the perturbation series does not converge in an exact sense.
 
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  • #3
Vanadium 50
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s if the perturbation is weak you can write E and |n⟨|n\langle as a Maclaurin series in λ\lambda. Why can we
Why not? You can always* write a function as a Maclaurin series. The question is how many terms you need to get a result that is "close enough" to what you want.

* Provided it is sufficiently well-behaved.
 
  • #4
Keith_McClary
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Provided it is sufficiently well-behaved.
In Hilbert2's example the Hamiltonian is unbounded below for negative β so eigenvalues do not exist. In that example it is known that the series diverges.
 

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