How are good quantum numbers related to perturbation theory?

[mathingenue]

TL;DR Summary

Meaning of good quantum numbers and relationship to perturbation theory and matrix diagonalization

Hello folks, I am currently studying from Griffiths' Introduction to Quantum Mechanics and I've got a doubt about good quantum numbers that the text has been unable to solve.

As I understand it, good quantum numbers are the eigenvalues of the eigenvectors of an operator O that remain eigenvectors of such an operator with the same eigenvalues as time evolves. At least that's the Wikipedia definition: https://en.wikipedia.org/wiki/Good_quantum_number. This is equivalent to saying that the operator O commutes with the Hamiltonian H of the problem in question.

But what does this have to do with degenerate perturbation theory? As I see it, degeneracy introduces two problems in the use of nondegenerate perturbation theory: 1) you don't know which states to use, since any linear combination of the degenerate eigenstates is also an eigenstate with the same energy, and 2) the expressions derived in the nondegenerate case involve dividing by zero when there is degeneracy. How do good quantum numbers help solve these problems?

And what does all of this have to do with diagonalizing the Hamiltonian matrix?

Any help would be greatly appreciated, and sorry for the (very) broad questions, I couldn't find any other way to put it.

 

[vanhees71]

It's all about solving the eigenvalue-eigenstate problem for the Hamiltonian. The importance comes from the fact that the eigenstates of the Hamiltonian are the stationary states of the system.

For each eigenvalue $E$ of $\hat{H}$ there's an eigenspace of $\hat{H}$ related to this eigenvalue, $\text{Eig}(E,\hat{H})$. It's the subspace of Hilbert space of eigenvectors to the eigenvalue $E$, i.e., for all $|\psi \rangle \in \text{Eig}(E,\hat{H})$ you have $\hat{H} |\psi \rangle=E|\psi \rangle$. Since $\hat{H}$ is self-adjoint there's an orthonormal basis $|E,\alpha \rangle$ of eigenvectors that spans $\text{Eig}(E,\hat{H})$, where $\alpha$ is some label denoting these orthonormal eigenvectors. All the eigenspaces of $\hat{H}$ together make up the entire Hilbert space (completeness), i.e., you have
$$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|,$$
where for simplicity I assume that the eigenvalues of $\hat{H}$ as well as the labels $\alpha$ are given by a discrete set. If you have continuous eigenvalues you have to put integrals instead of sums and Dirac $\delta$-distributions instead of Kronecker $\delta$-symbols.

If not all eigenspaces are one-dimensional, i.e., if $\mathrm{dim} \text{Eig}(E,\hat{H})>1$ for some eigenvalues of $\hat{H}$, then usually there are one or more additional compatible observables $\hat{A}_k$ ($k \in \{1,\ldots,n\}$) which commute among each other and with $\hat{H}$. Usually we can choose a complete set of such compatible independent observables, which is a set for which the common eigenspaces $\text{Eig}(E,a_1,\ldots,a_n;\hat{H},\hat{A}_1,\ldots,\hat{A}_n)$ are all one-dimensional. This implies that when preparing the system such that $\hat{H}$ and the $\hat{A}_k$ take simultaneously determined values $E$ and $a_k$ it is in the unique pure state $\hat{\rho}(E,a_1,\ldots,a_n)=|E,a_1,\ldots,a_n \rangle \langle E,a_1,\ldots,a_n|$.

Now suppose the Hamiltonian splits as
$$\hat{H}=\hat{H}_0+\hat{H}_1.$$
Now you may have a complete set $\hat{H}_0$, $\hat{A}_k$ of independent compatible observables. In general, however, the $\hat{A}_k$ won't be also compatible with $\hat{H}_1$. If you now want to apply perturbation theory with $\hat{H}_1$ as the perturbation, it's not a priori clear, if the chosen basis $|E_0,a_1,\ldots,a_n \rangle_0$ of $\hat{H}_0$ is appropriate to do perturbation theory. That's why in the perturbative solution of the eigenproblem you have to find the right choice of basis vectors in each subspace $\mathrm{Eig}(E_0,a_1,\ldots,a_n)$ at any order of perturbation theory, and usually the perturbation will "lift the degeneracy", i.e., for each of these eigenspaces of $\hat{H}_0$ the perturbative solutions of the eigenvalue problem of $\hat{H}$ will lead to different approximate eigenvalues of $\hat{H}$, though also the eigenvalue problem of $\hat{H}$ may also be degenerate. In this case you'd need another complete set of independent observable $\hat{A}_1',\ldots,\hat{A}_{n'}'$ which commutes with $\hat{H}$, but these need not be compatible with the orignal set $\hat{A}_1,\ldots, \hat{A}_n$.

One way to find the most convenient basis of $\hat{H}_0$ is to look for a complete compatible set of observables $\hat{A}_k$, of which as many as possible are also compatible with $\hat{H}_1$ and thus with $\hat{H}$. Using the corresponding basis of eigenvectors of $\hat{H}_0$ solves the problem of degenerate perturbation theory. Usually that's however not feasible, because then you could solve the eigenvalue problem of $\hat{H}$ exactly and wouldn't need perturbation theory. That's why usually you have to do the techniques described in the textbooks to find the proper basis of the degenerate eigenspaces of $\hat{H}_0$ order by order in perturbation theory.

The most elegant formulation I know of, using projection techniques, can be found in the textbook

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics, Addison Wesley (1993).

which is much more carefully written and better to understand than Griffiths.

 

[mathingenue]

It's all about solving the eigenvalue-eigenstate problem for the Hamiltonian. The importance comes from the fact that the eigenstates of the Hamiltonian are the stationary states of the system.

For each eigenvalue $E$ of $\hat{H}$ there's an eigenspace of $\hat{H}$ related to this eigenvalue, $\text{Eig}(E,\hat{H})$. It's the subspace of Hilbert space of eigenvectors to the eigenvalue $E$, i.e., for all $|\psi \rangle \in \text{Eig}(E,\hat{H})$ you have $\hat{H} |\psi \rangle=E|\psi \rangle$. Since $\hat{H}$ is self-adjoint there's an orthonormal basis $|E,\alpha \rangle$ of eigenvectors that spans $\text{Eig}(E,\hat{H})$, where $\alpha$ is some label denoting these orthonormal eigenvectors. All the eigenspaces of $\hat{H}$ together make up the entire Hilbert space (completeness), i.e., you have
$$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|,$$
where for simplicity I assume that the eigenvalues of $\hat{H}$ as well as the labels $\alpha$ are given by a discrete set. If you have continuous eigenvalues you have to put integrals instead of sums and Dirac $\delta$-distributions instead of Kronecker $\delta$-symbols.

If not all eigenspaces are one-dimensional, i.e., if $\mathrm{dim} \text{Eig}(E,\hat{H})>1$ for some eigenvalues of $\hat{H}$, then usually there are one or more additional compatible observables $\hat{A}_k$ ($k \in \{1,\ldots,n\}$) which commute among each other and with $\hat{H}$. Usually we can choose a complete set of such compatible independent observables, which is a set for which the common eigenspaces $\text{Eig}(E,a_1,\ldots,a_n;\hat{H},\hat{A}_1,\ldots,\hat{A}_n)$ are all one-dimensional. This implies that when preparing the system such that $\hat{H}$ and the $\hat{A}_k$ take simultaneously determined values $E$ and $a_k$ it is in the unique pure state $\hat{\rho}(E,a_1,\ldots,a_n)=|E,a_1,\ldots,a_n \rangle \langle E,a_1,\ldots,a_n|$.

Now suppose the Hamiltonian splits as
$$\hat{H}=\hat{H}_0+\hat{H}_1.$$
Now you may have a complete set $\hat{H}_0$, $\hat{A}_k$ of independent compatible observables. In general, however, the $\hat{A}_k$ won't be also compatible with $\hat{H}_1$. If you now want to apply perturbation theory with $\hat{H}_1$ as the perturbation, it's not a priori clear, if the chosen basis $|E_0,a_1,\ldots,a_n \rangle_0$ of $\hat{H}_0$ is appropriate to do perturbation theory. That's why in the perturbative solution of the eigenproblem you have to find the right choice of basis vectors in each subspace $\mathrm{Eig}(E_0,a_1,\ldots,a_n)$ at any order of perturbation theory, and usually the perturbation will "lift the degeneracy", i.e., for each of these eigenspaces of $\hat{H}_0$ the perturbative solutions of the eigenvalue problem of $\hat{H}$ will lead to different approximate eigenvalues of $\hat{H}$, though also the eigenvalue problem of $\hat{H}$ may also be degenerate. In this case you'd need another complete set of independent observable $\hat{A}_1',\ldots,\hat{A}_{n'}'$ which commutes with $\hat{H}$, but these need not be compatible with the orignal set $\hat{A}_1,\ldots, \hat{A}_n$.

One way to find the most convenient basis of $\hat{H}_0$ is to look for a complete compatible set of observables $\hat{A}_k$, of which as many as possible are also compatible with $\hat{H}_1$ and thus with $\hat{H}$. Using the corresponding basis of eigenvectors of $\hat{H}_0$ solves the problem of degenerate perturbation theory. Usually that's however not feasible, because then you could solve the eigenvalue problem of $\hat{H}$ exactly and wouldn't need perturbation theory. That's why usually you have to do the techniques described in the textbooks to find the proper basis of the degenerate eigenspaces of $\hat{H}_0$ order by order in perturbation theory.

The most elegant formulation I know of, using projection techniques, can be found in the textbook

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics, Addison Wesley (1993).

which is much more carefully written and better to understand than Griffiths.

Thanks for taking the time to write a long answer vanhees71, and thank you for the Sakurai reference, but I'm afraid your answer is simply too complicated for me. I did try going through it in detail, but basically you lost me after the equations $$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|,$$. If anyone else would like to take a shot at it, it would be greatly appreciated.

 

[gentzen]

... I did try going through it in detail, but basically you lost me after the equations $$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|,$$. If anyone else would like to take a shot at it, it would be greatly appreciated.

That is simply a typo, it should be $$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|=1,$$
The first equation says that the $|E, \alpha \rangle$ are orthonormal, and the second equation says that they are complete.

 

[mathingenue]

That is simply a typo, it should be $$\langle E',\alpha' |E,\alpha \rangle=\delta_{EE'} \delta_{\alpha \alpha'}, \quad \sum_{E,\alpha} |E, \alpha \rangle \langle E,\alpha|=1,$$
The first equation says that the $|E, \alpha \rangle$ are orthonormal, and the second equation says that they are complete.

Hello gentzen, I'm afraid the equations aren't the problem, the text is. I merely pointed to them as a reference of where in vanhees71's text I got lost.

 

[gentzen]

Good, then you already understood quite a lot. Especially, there is an eigenspace $\text{Eig}(E,\hat{H})$ for each eigenvalue $E$, and the $|E,\alpha \rangle$ are an orthonormal basis of that subspace (where $\alpha$ indexes the different basis vectors, and $E$ is fixed).

 

[mathingenue]

Good, then you already understood quite a lot. Especially, there is an eigenspace $\text{Eig}(E,\hat{H})$ for each eigenvalue $E$, and the $|E,\alpha \rangle$ are an orthonormal basis of that subspace (where $\alpha$ indexes the different basis vectors, and $E$ is fixed).

Yes, but I can't seem to move forward in understanding his answer.

 

[A. Neumaier]

If $H$ commutes with $A$ and $\psi$ is an eigenvector of $H$ with nonzero $A\psi$ then the latter is also an eigenvector of $H$ with the same eigenvalue. Proving this is an easy exercise. Thus the existence of such $A$ implies that most eigenvalues of $H$ have multiplicity greater than 1, hence call for degenerate perturbation theory.
This can be exploited computationally by the methods of representation theory, applied to the group or Lie algebra generated by the operators commuting with $H$, such as SO(3) for angular momentum. This provides the link to the reply given by @vanhees71.

 

[mathingenue]

If $H$ commutes with $A$ and $\psi$ is an eigenvector of $H$ with nonzero $A\psi$ then the latter is also an eigenvector of $H$ with the same eigenvalue. Proving this is an easy exercise. Thus the existence of such $A$ implies that most eigenvalues of $H$ have multiplicity greater than 1, hence call for degenerate perturbation theory.
This can be exploited computationally by the methods of representation theory, applied to the group or Lie algebra generated by the operators commuting with $H$, such as SO(3) for angular momentum. This provides the link to the reply given by @vanhees71.

Yes A. Neumaier, unfortunately I'm not well versed in representation theory...

 

[mathingenue]

Anyone else would like to help?

 

[A. Neumaier]

Yes A. Neumaier, unfortunately I'm not well versed in representation theory...

The first paragraph of my answer already resolves your question and can be understood without representation theory. If you don't understand that you need to brush up your linear algebra.

The second paragraph just pointed to where you could go on learning about the consequences...

 

[DrClaude]

Hello folks, I am currently studying from Griffiths' Introduction to Quantum Mechanics and I've got a doubt about good quantum numbers that the text has been unable to solve.

Could you please be more specific as to what part of Griffiths is giving you problems?

 

[mathingenue]

Could you please be more specific as to what part of Griffiths is giving you problems?

I think the part that has been giving me the most trouble is the end of page 259 of the second edition, when Griffiths talks about the "good states". He says that if we could guess the good states then we could go ahead and use nondegenerate perturbation theory, and he introduces a theorem that allows us to find such good states. I fail to see why these states are special, so that we can choose them over all other linear combinations of degenerate eigenfunctions, and I also don't see how choosing these good states solves the problem of dividing by zero in the equations derived in the nondegenerate case.

 

[vanhees71]

Griffiths seems to be always confusing with his strange formulations. The point of degenerate time-independent perturbation theory precisely is to figure out what he calls "the good states". As I said the best treatment of this indeed not so simple subject is in Sakurai's textbook, Modern Quantum Mechanics (my favorite edition is the socalled "revised edition", coauthored/completed by Tuan; although as far as I know this particular section is also the same in the later editions coauthored by Napolitano).

 

[mathingenue]

Could you please be more specific as to what part of Griffiths is giving you problems?

I do hope my doubts are clearer now.

 

[hutchphd]

I do hope my doubts are clearer now.

The purpose here is not to justify your confusion.
Here is the cheap and dirty version: the fact that an energy level is n-fold degerate gives us the leeway to arbitrarilly choose linear combinations of those states as our new starting point.

I fail to see why these states are special, so that we can choose them over all other linear combinations of degenerate eigenfunctions,

They are special because they keep the perturbation expansion from "blowing up". In terms of the original Hamiltonian they are not "special". We choose them because we can and we are clever humans.
The way to choose them is diagonalize the perturbed formerly degenerate subspace. Voila!
.

 

[mathingenue]

The purpose here is not to justify your confusion.
Here is the cheap and dirty version: the fact that an energy level is n-fold degerate gives us the leeway to arbitrarilly choose linear combinations of those states as our new starting point.

They are special because they keep the perturbation expansion from "blowing up". In terms of the original Hamiltonian they are not "special". We choose them because we can and we are clever humans.
The way to choose them is diagonalize the perturbed formerly degenerate subspace. Voila!
.

I understand that we can choose any linear combination of these states as our starting point, and that we are trying to diagonalize the perturbation in the degenerate subspace, but why exactly are we trying to diagonalize the perturbation in the degenerate subspace?

 

[hutchphd]

To keep the perturbation expansion from blowing up. The matrix elements which would cause trouble are made manifestly zero. The details are in every textbook for undergraduate quantum. You should look at several...this is bread and butter (or better: nuts and bolts) quantum mechanics. At each order of the expansion you can keep the lid on and as I recall (help requested here as required!) to each order the perturbation expansion is unique..

 

[DrClaude]

I think the part that has been giving me the most trouble is the end of page 259 of the second edition, when Griffiths talks about the "good states". He says that if we could guess the good states then we could go ahead and use nondegenerate perturbation theory, and he introduces a theorem that allows us to find such good states. I fail to see why these states are special, so that we can choose them over all other linear combinations of degenerate eigenfunctions, and I also don't see how choosing these good states solves the problem of dividing by zero in the equations derived in the nondegenerate case.

Griffiths is not talking about good quantum numbers here, but rather using the adjective "good" to indicate the linear combinations that are eigenstates of both the base Hamiltonian (with the degeneracy) and another operator A, such that these states are also eigenstates of this operator and the Hamiltonian when the perturbation is included.