Doubt in understanding degenerate perturbation theory

[Kashmir]

McIntyre, quantum mechanics,pg360

Suppose states $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ are degenerate eigenstates of unperturbed Hamiltonian $H$


Author writes:
"The first-order perturbation equation we want to solve is
$\left.\left(H_{0}-E_{n}^{(0)}\right)\left|n^{(1)}\right\rangle=\left(E_{n}^{(1)}-H^{\prime}\right) \mid n^{(0)} \right >)$"
"But the energy degeneracy of these two states creates an ambiguity. Both $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ satisfy the zeroth-order energy eigenvalue equation for the energy $E_{2}^{(0)}$, but so does any linear combination of the two states. If we are trying to find the energy correction to the state with zeroth-order energy $E_{2}^{(0)}$, how do we know whether to use the state $\left|2^{(0)}\right\rangle$ or the state $\left|3^{(0)}\right\rangle$ in the perturbation equation? "
Why can't I use $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$individually in the perturbation equation and carry on?

 

[Orodruin]

Because they are degenerate, the perturbation to the energies will be larger than the difference in energy. Therefore, off-diagonal contributions become important.

 

[Kashmir]

Because they are degenerate, the perturbation to the energies will be larger than the difference in energy. Therefore, off-diagonal contributions become important.

Whose difference in energies?

 

[Orodruin]

Whose difference in energies?

The zeroth order energies, which have difference zero

 

[Kashmir]

Also is it that when we apply the perturbation $\lambda H'$ to the original Hamiltonian $H_0$ which has degenerate eigenstates, only a select few eigenstates change smoothly as we increase $\lambda$ from 0 to 1 ,to be the eigenstate of the new total Hamiltonian. These eigenstates of $H_0$ are called "good states"?

 

[Kashmir]

So the states which change abruptly aren't physically realizable so is it correct to say that after applying the perturbation if we make a measurement, the system can only be in the "good states"?

 

[Orodruin]

The thing is that the degenerate states span some subspace of the full state space where any vector will be an eigenvector of the unperturbed Hamiltonian. Just looking at this degenerate subspace and the unperturbed Hamiltonian, you can pick any set of orthonormal vectors as your basis. However, this changes once you introduce a perturbation because such a perturbation will generally break the degeneracy and no longer can you pick any state within the degenerate space and have an eigenvector with the same eigenvalue. Instead, the (approximate) eigenvectors of the perturbed Hamiltonian within the subspace will be given by eigenstates of the perturbation restricted to the degenerate subspace. No states change in a discontinuous way, but the introduction of the perturbation restricts what states are eigenstates of the full Hamiltonian.

 

[FuzzySphere]

In ordinary, non-degenerate perturbation theory, the first order correction to the state is given by $$|n^{(1)}\rangle = -\sum_{k \not = n}\frac {|k^{(0)}\rangle H'_{kn}}{E^{(0)}_k -E^{(0)}_n}$$ where $H'_kn =\langle k^{(0)}|H'|n^{(0)}\rangle$. You can see that if two or more eigenvalues are equal, then the sum blows up. The reason why you cannot just consider perturbation expansions of the original eigenstates is as follows: the sum blows up if the state has a component along the degenerate subspace, and the degenerate subspace takes up a sizable portion of the Hilbert space, so one cannot a priori assume that the corrections have no component along that subspace. In nondegenerate perturbation theory, one can always choose the $n$th order corrections to have no component along the original, unperturbed state. It is one thing for there to be no component along a single vector, but it is a more substantial thing if there is no component along an entire subspace, and the latter turns out to fail in general. If you want a better understanding of degenerate perturbation theory, going all the way to degeneracy lifted at second order, see the following lecture videos and notes by Barton Zweibach: https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/

 

[Orodruin]

In nondegenerate perturbation theory, one can always choose the nth order corrections to have no component along the original, unperturbed state. It is one thing for there to be no component along a single vector, but it is a more substantiala thing if there is no component along an entire subspace, and the latter turns out to fail in general.

To have no component along the original eigenstate in first-order non-degenerate perturbation theory is not a choice. It is directly implied by the normalisation of the perturbed eigenstate to first order in the perturbation. The second order correction, however, must include a component in that direction.

Furthermore, a single vector spans a one-dimensional subspace so that is an "entire subspace" as well. I have already described in this thread what happens in degenerate perturbation theory (the degenerate subspace is spanned by any (sufficiently large) set of linearly independent vectors in that space. The reason that you cannot apply non-degenerate perturbation theory is that what is the eigenstates in that space after perturbation is not restricted to be a small rotation away from whatever basis you had chosen. That things still work out if you are more careful with your choice of basis in the degenerate space is apparent from the fact that the problematic terms never appear in the first order correction if you choose a basis such that $H’$ is diagonal when restricted to the degenerate subspace. This occurs because then again the correction is necessarily a small perturbation.

 

[FuzzySphere]

To have no component along the original eigenstate in first-order non-degenerate perturbation theory is not a choice. It is directly implied by the normalisation of the perturbed eigenstate to first order in the perturbation. The second order correction, however, must include a component in that direction.

Furthermore, a single vector spans a one-dimensional subspace so that is an "entire subspace" as well. I have already described in this thread what happens in degenerate perturbation theory (the degenerate subspace is spanned by any (sufficiently large) set of linearly independent vectors in that space. The reason that you cannot apply non-degenerate perturbation theory is that what is the eigenstates in that space after perturbation is not restricted to be a small rotation away from whatever basis you had chosen. That things still work out if you are more careful with your choice of basis in the degenerate space is apparent from the fact that the problematic terms never appear in the first order correction if you choose a basis such that $H’$ is diagonal when restricted to the degenerate subspace. This occurs because then again the correction is necessarily a small perturbation.

Okay, an $N$ dimensional space is a lot bigger than a one dimensional space, to be clear ($N$ is the number of degenerate eigenstates). That was my point in the post, I apologize if that was unclear. I was answering the question of OP's, which was "Why can't I use |2(0)⟩ and |3(0)⟩individually in the perturbation equation and carry on?" The answer to this is because the first order correction can have components along the entire degenerate subspace ($N$ dimensional). The only thing the OP can get without considering this is the component of the correction along the non-degenerate subspace, which does not make up the entire correction, because there is an $N$ dimensional subspace that we haven't yet considered. I don't exactly understand your "small rotation" requirement, but I suspect it to be related to the "smallness" of the energy differences in relation to the perturbation. I clearly explained, that when one tries to apply non-degenerate perturbation theory to the original eigenstates, the sum blows up, i.e., you get a 1/0 term, so that the term $\frac 1 {E_n^{(0)} - E_k^{(0)}}$ is no longer small in regards to anything, much less the perturbation. I'm not giving the OP a full course on non-degenerate perturbation theory, I gave him resources for more information to understand the topic better. As for whether you can choose the correction to have no component along the original eigenstate, in my studies of perturbation theory, we have not started out by assuming the full state is normalized, only that it is normalizable, see the following notes, particularly page 5 in regards to this "choice": https://ocw.mit.edu/courses/8-06-qu...c5ca8a479c3e56c544d646fb770_MIT8_06S18ch1.pdf. Also, according to this same source, the $n$th order correction need not have a component along the original state, but it may be that when we impose normalization, we get components along the original eigenstate.

 

[Orodruin]

Okay, an N dimensional space is a lot bigger than a one dimensional space, to be clear (N is the number of degenerate eigenstates)

In the sense of being a part of a higher dimensional space, not really. Both would have measure zero in an M-dimensional space where N < M. The probability to have zero component in either ”by chance” for a random vector is zero. But perturbation theory is not about random vectors.

I was answering the question of OP's, which was "Why can't I use |2(0)⟩ and |3(0)⟩individually in the perturbation equation and carry on?" The answer to this is because the first order correction can have components along the entire degenerate subspace (N dimensional).

No. The answer is what I said in the previous post. That the degenerate subspace has no sense of differentiating between what will become the non-defenerate eigenvectors after perturbation. This is because the perturbation is not really a small correction in the degenerate subspace as ”small” has to be considered relative to the difference in eigenvalues (if you have a zeroth order Hamiltonian that has a non-degenerate subspace where eigenvalues differ by less than the typical entry in the perturbation Hamiltonian you also cannot use non-degenerate perturbation theory and expect a reasonable answer).

The ”corrections along the entire degenerate subspace” you refer to are not really corrections in the sense of being small. They are there to pick out the eigenstates of the perturbation Hamiltonian restricted to the subspace.

I don't exactly understand your "small rotation" requirement, but I suspect it to be related to the "smallness" of the energy differences in relation to the perturbation.

This is the entire point of perturbation theory. To consider what a small change in the Hamiltonian does to eigenvalues and eigenstates. If you make a small change, then the corrections are small. The entire problem with degenerate zeroth order Hamiltonian is that the perturbation cannot be considered small within that subspace, which is why the perturbation Hamiltonian itself dictates the good states for perturbation.

I clearly explained, that when one tries to apply non-degenerate perturbation theory to the original eigenstates, the sum blows up, i.e., you get a 1/0 term, so that the term 1En(0)−Ek(0) is no longer small in regards to anything, much less the perturbation.

This is true only if you make the mistake of thinking that any eigenstates of the zeroth order Hamiltonian goes. This again results from the perturbation not actually being small in that subspace. As I also stated previously, the non-degenerate perturbation theory breaks down before the states actually become degenerate. It is sufficient that the perturbation Hamiltonian is not small conpared to the eigenvalue differences within a particular subspace.

As for whether you can choose the correction to have no component along the original eigenstate, in my studies of perturbation theory, we have not started out by assuming the full state is normalized, only that it is normalizable

If you do not consider normalised states, then the correction will generally have a component in the original direction. ”Choosing” the first order correction not to have that is equivalent to giving the perturbed state the same normalization as the original state up to second order in the perturbation. It does not have anything to do with not ”hitting” the one-dimensional subspace with the correction.