# Degenerate Perturbation Theory

• I

## Main Question or Discussion Point

Hello! I am reading Griffiths and I reached the Degenerate Time Independent Perturbation Theory. When calculating the first correction to the energy, he talks about "good" states, which are the orthogonal degenerate states to which the system returns, once the perturbation is gone. I understand mathematically how it works as you need to find simultaneous eigenfunction for the unperturbed Hamiltonian and for the perturbation itself. I am confused from a physics point of view, why would nature favor one orthonormal combination over the other. As in the non-perturbed state all the combinations have the same energy, I would be tempted to believe that once the perturbation is gone, all the states are equally likely. Why isn't this the case (I would prefer some physics insight rather than math, but any answer would be really appreciated)?

Related Quantum Physics News on Phys.org
kuruman
Homework Helper
Gold Member
Perhaps you missed that if you have two degenerate eigenstates of the unperturbed Hamiltonian $H_0$, $|\alpha>$ and $|\beta>$, then any normalized linear combination $|\psi>=a|\alpha>+b|\beta>$ is also an eigenstate of $H_0$ for the same eigenvalue. In short, Nature has no way of "knowing" which state to favor.

Perhaps you missed that if you have two degenerate eigenstates of the unperturbed Hamiltonian $H_0$, $|\alpha>$ and $|\beta>$, then any normalized linear combination $|\psi>=a|\alpha>+b|\beta>$ is also an eigenstate of $H_0$ for the same eigenvalue. In short, Nature has no way of "knowing" which state to favor.
On the contrary, I am aware of that. This is why I don't understand why when you do the math you obtain a particular combination and not any of them

kuruman
Homework Helper
Gold Member
I am trying to understand what you mean by the following two statements.
As in the non-perturbed state all the combinations have the same energy, I would be tempted to believe that once the perturbation is gone, all the states are equally likely. Why isn't this the case (I would prefer some physics insight rather than math, but any answer would be really appreciated)?
Let's consider the four states in the excited Hydrogen atom (no spin), $R_{20}Y_{00}$, $R_{21}Y_{11}$, $R_{21}Y_{10}$, $R_{21}Y_{1\bar 1}$. Let's call this set of basis states "Basis 1". The unperturbed Hamiltonian using t"Basis 1" will be diagonal with $E_2$ along the diagonal. If you measured the energy of this hydrogen atom, you will get $E_2$. If someone asked you "what state is this hydrogen atom after the measurement?", the answer is "I don't know."

Now suppose you choose 4 orthonormal linear combinations of the four states and call your choice "Basis 2". Calculating the energies and constructing the Hamiltonian is trivial. Measuring the energy still gives $E_2$ as a result and as for the state of the atom after the measurement, the answer is still "I don't know". I'm sure you agree with all of this so far.

Now here comes the perturbation. We put a uniform magnetic field in some direction relative to our lab and go about our business of measuring energy or orbital angular momentum or both. This is where Griffiths's "good states" come in. Basis 1 are the good states if (and this is an important if) we identify the direction of the magnetic field as the axis of quantization of the spherical harmonics. Then the Hamiltonian is already diagonal and we have nothing more to do. The energies are $E_2$ for $R_{20}Y_{00}$, $E_2+\epsilon$ for $R_{21}Y_{11}$, $E_2$ for $R_{21}Y_{10}$ and $E_2-\epsilon$ for $R_{21}Y_{1\bar 1}$. If you measured the energy, you would get $E_2$, $E_2+\epsilon$ and $E_2-\epsilon$. If you get $E_2$, you would still not know what state the hydrogen atom is in after the measurement, but that would no longer be the case for the non-degenerate states. If the measurement returns $E_2+\epsilon$, then you know that the atom is in state $R_{21}Y_{11}$.

If, however, we chose "Basis 2" and chose the magnetic field in the x-direction relative to the quantization axis, we will end up with a Hamiltonian with a whole lot of off diagonal elements that we will have to diagonalize. Say that we do that and then proceed to get the eigenvectors. We will end up with linear combinations of the "Basis 2" states which. lo and behold, are nothing but $R_{20}Y_{00}$, $R_{21}Y_{11}$, $R_{21}Y_{10}$ and $R_{21}Y_{1\bar 1}$.

When we are done and turn the magnetic field off, we are back to where we started. So can you explain to me what you mean when you say that all the states are not equally likely when the perturbation is gone? In what sense are they not equally likely?

stevendaryl
Staff Emeritus
When we are done and turn the magnetic field off, we are back to where we started. So can you explain to me what you mean when you say that all the states are not equally likely when the perturbation is gone? In what sense are they not equally likely?
It's no good asking the original poster what is meant, because he's trying to understand what Griffiths meant. Maybe the page from Griffiths should be quoted verbatim, because it's hard to tell from the paraphrase.

I am trying to understand what you mean by the following two statements.

Let's consider the four states in the excited Hydrogen atom (no spin), $R_{20}Y_{00}$, $R_{21}Y_{11}$, $R_{21}Y_{10}$, $R_{21}Y_{1\bar 1}$. Let's call this set of basis states "Basis 1". The unperturbed Hamiltonian using t"Basis 1" will be diagonal with $E_2$ along the diagonal. If you measured the energy of this hydrogen atom, you will get $E_2$. If someone asked you "what state is this hydrogen atom after the measurement?", the answer is "I don't know."

Now suppose you choose 4 orthonormal linear combinations of the four states and call your choice "Basis 2". Calculating the energies and constructing the Hamiltonian is trivial. Measuring the energy still gives $E_2$ as a result and as for the state of the atom after the measurement, the answer is still "I don't know". I'm sure you agree with all of this so far.

Now here comes the perturbation. We put a uniform magnetic field in some direction relative to our lab and go about our business of measuring energy or orbital angular momentum or both. This is where Griffiths's "good states" come in. Basis 1 are the good states if (and this is an important if) we identify the direction of the magnetic field as the axis of quantization of the spherical harmonics. Then the Hamiltonian is already diagonal and we have nothing more to do. The energies are $E_2$ for $R_{20}Y_{00}$, $E_2+\epsilon$ for $R_{21}Y_{11}$, $E_2$ for $R_{21}Y_{10}$ and $E_2-\epsilon$ for $R_{21}Y_{1\bar 1}$. If you measured the energy, you would get $E_2$, $E_2+\epsilon$ and $E_2-\epsilon$. If you get $E_2$, you would still not know what state the hydrogen atom is in after the measurement, but that would no longer be the case for the non-degenerate states. If the measurement returns $E_2+\epsilon$, then you know that the atom is in state $R_{21}Y_{11}$.

If, however, we chose "Basis 2" and chose the magnetic field in the x-direction relative to the quantization axis, we will end up with a Hamiltonian with a whole lot of off diagonal elements that we will have to diagonalize. Say that we do that and then proceed to get the eigenvectors. We will end up with linear combinations of the "Basis 2" states which. lo and behold, are nothing but $R_{20}Y_{00}$, $R_{21}Y_{11}$, $R_{21}Y_{10}$ and $R_{21}Y_{1\bar 1}$.

When we are done and turn the magnetic field off, we are back to where we started. So can you explain to me what you mean when you say that all the states are not equally likely when the perturbation is gone? In what sense are they not equally likely?
Thank you for your response. Here you can find the pdf version of the book I am reading and the paragraph (and section) I am talking about is at page 228 (or 242 in pdf numbering), section 6.2. There Griffiths says "Typically, the perturbation will break the degeneracy, the common unperturbed energy $E^0$ splits into two (he talks about double degenerate case). Going the other direction, when we turn off the perturbation, the upper state reduces down to one linear combination of $\psi_a^0$ and $\psi_b^0$ and the lower sate reduces to some orthogonal linear combination, but we don't know a priori what these "good" linear combinations will be. For this reason we can't even calculate the first-order energy, we don't know what unperturbed states to use."
The way I understand this, is that there is a privileged, "good" initial orthogonal combination that we should use, where the system returns when the perturbation is turned off. Mathematically it makes perfect sense, you just follow the formalism, find the eigenstates of the perturbed hamiltonian and done. But again, initially all the states should be on equal footing. So physically, why are there good and bad states? Or I just miss understand what Griffiths is saying?

stevendaryl
Staff Emeritus
The way I understand this, is that there is a privileged, "good" initial orthogonal combination that we should use, where the system returns when the perturbation is turned off. Mathematically it makes perfect sense, you just follow the formalism, find the eigenstates of the perturbed hamiltonian and done. But again, initially all the states should be on equal footing. So physically, why are there good and bad states? Or I just miss understand what Griffiths is saying?
I think he's just saying this: In order to do perturbation theory, you start off with a complete set of eigenstates for the unperturbed hamiltonian, $H^0$. If there is degeneracy, then there isn't a unique set of unperturbed states to serve as a starting basis. So this "good" versus "bad" is about which states to use in forming your initial complete basis.

kuruman
When I posted #4, I assumed you had read Griffiths's "Moral" on page 230. Post #4 was an attempt to illustrate what he's saying using the first excited state of the hydrogen atom. My "Basis 1" are the "good" states because they are also eigenstates of $L$ and $L_z$. Griffiths is saying that if you pick good states, then you might be able to avoid degeneracies and make your calculations easier. If you can eliminate degeneracies by choosing good states, then you will not have to worry about the annoying zeroes in the denominator when you do second order PT and you might even avoid doing second order PT altogether. That's Griffiths's moral as I see it.
Choosing a basis set is like choosing a coordinate system of axes. A projectile will reach a certain height with a certain speed regardless of whether you use the standard vertical-horizontal set of axes or one rotated by $\theta$ in the vertical plane. Likewise, the energy levels of a hydrogen atom, when perturbed or left alone, will split in a particular way regardless of whether you use the standard $R_{nl}Y_{lm}$ basis set or one rotated by a unitary transformation that gives you linear combinations thereof. Nature will do its thing regardless of your choice.