Time-independent perturbation theory

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SUMMARY

Time-independent perturbation theory in quantum mechanics assumes that perturbed energies and wavefunctions can be expressed as power series in a small perturbation parameter, λ. The standard form is E_n = E_n^{(0)} + λE_n^{(1)} + λ^2E_n^{(2)} + ..., and |n⟩ = |n^{(0)}⟩ + λ|n^{(1)}⟩ + λ^2|n^{(2)}⟩ + .... This assumption holds under the condition that the perturbation Hamiltonian is small, allowing for a smooth transition to unperturbed results. However, this approach fails in cases such as the Coulomb potential in hydrogen, where the energy expansion diverges, indicating that the perturbation cannot be treated as small.

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ehrenfest
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Homework Statement


In each of my QM books, they always say something like "we can write the perturbed energies and wavefunctions as"

E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots

|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots

without any justification. This is really not obvious to me and although it seems reasonable, it do not see why the perturbed energies and wavefunctions might not be something completely different like

E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 \log \left(E_n^{(2)}\right)^{-1}\sin E_n^{(2)} + \cdots

|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots

Homework Equations


The Attempt at a Solution

 
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Expanding the energies and wave functions as a function of the perturbation parameter is an iterative manner which is well behaved. The function you wrote for the energy has poles and in multivalued.
 
ehrenfest said:

Homework Statement


In each of my QM books, they always say something like "we can write the perturbed energies and wavefunctions as"

E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots

|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots

without any justification. This is really not obvious to me and although it seems reasonable, it do not see why the perturbed energies and wavefunctions might not be something completely different like

E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 \log \left(E_n^{(2)}\right)^{-1}\sin E_n^{(2)} + \cdots

|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots

Homework Equations


You are right that it's an assumption that is being made. The assumption is that the perturbation Hamiltonian is a small perturbation, which means that as lambda goes to zero, the energies and wavefunctions smoothly approach the unperturbed results. This assumption does fail in some cases (for example, trying to treat the Couloumb potential in Hydrogen as a perturbation would fail completely because the energy expansion would diverge in lambda which would signal the fact that the Coulomb potential has to be included to all orders so it cannot really be treated as a perturbation)
 

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