Degenerated perturbation theory

[Lucas-]

TL;DR

I don't understand why we cannot obtain perturbative correction to a specific eigenstate in the degenerated case (ie. |2s>)

Hello,
In the case of Stark effect for example, one may find the correction for the |1s> state easily by applying non degenerated perturbation theory. However in the degenerated case it's seems as though we can only treat the whole n=2 level for example and not individual eigen states. That, I don't understand as, in reality we could make 2s hydrogen atom and place them in an electric field. What would happen ? Why can't we get let's say the first order |2s> correction ?
When we do the usual setup, we find the common eigenkets in the degenerate subspace of the original hamiltonian and the perturbation and their associated eigen values.
We observe linear combination of eigenkets yet, they don't seem to depend on the field intensity, so this would mean that when I turn the perturbation off, If, let's say I started with a |2s> hydrogen I would keep the superposition ? Shouldn't it go back to |2s> ?

I'm a bit confuse and I would appreciate it if someone could enlighten me !

 

[Mordred]

Look into parity and the linear Stark effect in regards to degenerated perturbation

See here
https://bohr.physics.berkeley.edu/classes/221/1112/notes/stark.pdf

Section 19 goes into detail on 2s

 

[Lucas-]

Look into parity and the linear Stark effect in regards to degenerated perturbation

See here
https://bohr.physics.berkeley.edu/classes/221/1112/notes/stark.pdf

Section 19 goes into detail on 2s

So if I understand correctly, we find a new basis in which the perturbation is diagonal and when we want a specific ket we can write it in this new basis and we'll get the energy shift we want when we turn on the perturbation ?

 

[Mordred]

In essence yes but extra care must be dealt with. You may or may not have noted that the Stark effect is described by Old Quantum Mechanics using the Bohr model. There is a further detail to add to your statement above. "With standard first-order degenerate perturbation theory the Hamilton operator is diagonal in parabolic coordinates but not in spherical coordinates between the perturbed and unperturbed relations. "

However numerous problems with the Stark effect gets resolved using the Schrödinger equations and wave mechanics.
A good coverage of these details can be found here.
https://arxiv.org/abs/1404.5333

The above quotation is directly from said article.

Key note the two Stark effects are linear (first order) and quadratic (parabolic)