Definition of a Degenerate Subspace (QM)

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Summary:

I was learning about Degenerate Perturbation Theory and I encountered the term 'Degenerate Subspace', I didn't really understand what it meant so I came here to ask - what does it mean? will it matter if i'll say 'Degenerate space' instead of 'Degenerate Subspace'? and subspace of what? ( subspace of the space composed from all eigenfunctions of the Hamiltonian maybe? )

Main Question or Discussion Point

I was learning about Degenerate Perturbation Theory and I encountered the term 'Degenerate Subspace', I didn't really understand what it meant so I came here to ask - what does it mean? will it matter if i'll say 'Degenerate space' instead of 'Degenerate Subspace'? and subspace of what? ( subspace of the space composed from all eigenfunctions of the Hamiltonian maybe? )

From what I understood, the definition is: Set of all eigenfunctions of the Hamiltonian which have the same eigenvalue. ( so essentially, Degenerate subspace is the set of all degenerate eigenfunctions corresponding to a specific Hamiltonian ).

But I would like another opinion and knowledge about the matter.
Thanks for the help in advance.
 

Answers and Replies

  • #2
PeroK
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Let's take any (Hermitian) operator and a given eigenvalue. There are two possibilities. First, there may be a single eigenfunction corresponding to that eigenvalue. In this case, of course, the eigenspace spanned by that eigenfunction is one dimensional. Or, there may be more than one (linearly independent) eigenfunction corresponding that that eigenvalue. In this case, the eigenspace is multi-dimensional and we say that the eigenvalue (and loosely the eigenspace) is degenerate.
 
  • #3
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Let's take any (Hermitian) operator and a given eigenvalue. There are two possibilities. First, there may be a single eigenfunction corresponding to that eigenvalue. In this case, of course, the eigenspace spanned by that eigenfunction is one dimensional. Or, there may be more than one (linearly independent) eigenfunction corresponding that that eigenvalue. In this case, the eigenspace is multi-dimensional and we say that the eigenvalue (and loosely the eigenspace) is degenerate.
From what I understood, when you say "eigenspace" , you mean the set of eigenvalues corresponding to the set of eigenfunctions belonging to the Hermitian operator?
 
  • #4
PeroK
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From what I understood, when you say "eigenspace" , you mean the set of eigenvalues corresponding to the set of eigenfunctions belonging to the Hermitian operator?
No. An eigenspace is a subspace of the space of all functions. It consists of all eigenfunctions that correspond to a given eigenvalue for the given operator. Eigenvalues are in general complex numbers, but are real for Hermitian operators.

The set of eigenvalues is called the spectrum of the operator.

In general, the terminology you used in these posts seems confused. This is all standard linear algebra terminology. You can't learn QM without linear algebra!
 
  • #5
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Oh, I have confused the terminology between set and space o0).
So Degenerate subspace is a subspace of the eigenspace of a certain Hermitian operator?
In that case, Degenerate subspace is the space of degenerate eigenfunctions corresponding to the same certain eigenvalue ?
 
  • #6
PeroK
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Oh, I have confused the terminology between set and space o0).
So Degenerate subspace is a subspace of the eigenspace of a certain Hermitian operator?
In that case, Degenerate subspace is the space of degenerate eigenfunctions corresponding to the same certain eigenvalue ?
All your terminology is either wrong or confused. You'll have to learn some linear algebra.
 
  • #7
vanhees71
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Oh, I have confused the terminology between set and space o0).
So Degenerate subspace is a subspace of the eigenspace of a certain Hermitian operator?
In that case, Degenerate subspace is the space of degenerate eigenfunctions corresponding to the same certain eigenvalue ?
I've never heard the expression "degenerate subspace". I guess what's meant is the eigenspace of an operator of an eigenvalue.

Let ##\hat{A}## be a linear operator. Then a eigenvector ##|u_{\lambda} \rangle \neq 0## of ##\hat{A}## with eigenvalue ##\lambda## fulfills the equation
$$\hat{A} |u_{\lambda} \rangle=\lambda |u_{\lambda} \rangle.$$
Now obviously the set of vectors ##|u \rangle## with eigenvector ##\lambda## of the operator ##\hat{A}## form a subspace, ##\text{Eig}(\hat{A},\lambda)##, because for any two vectors ##|u_1 \rangle## and ##|u_2 \rangle## in this subspace you have for all ##\alpha,\beta \in \mathbb{C}##
$$\hat{A} (\alpha |u_1 \rangle + \beta |u_2 \rangle)=\alpha \hat{A} |u_1 \rangle + \beta \hat{A} |u_2 \rangle=\alpha \lambda |u_1 \rangle + \beta \lambda |u_2 \rangle = \lambda (\alpha |u_1 \rangle + \beta |u_2 \rangle).$$
Thus indeed ##\text{Eig}(\hat{A},\lambda)## is a subspace of ##\mathcal{H}##.

The eigenvalue ##\lambda## is called non-degenerate if and only if ##\mathrm{dim} \text{Eig}(\hat{A},\lambda)=1##. Otherwise it's called degenerate.
 
  • #8
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I've never heard the expression "degenerate subspace". I guess what's meant is the eigenspace of an operator of an eigenvalue.

Let ##\hat{A}## be a linear operator. Then a eigenvector ##|u_{\lambda} \rangle \neq 0## of ##\hat{A}## with eigenvalue ##\lambda## fulfills the equation
$$\hat{A} |u_{\lambda} \rangle=\lambda |u_{\lambda} \rangle.$$
Now obviously the set of vectors ##|u \rangle## with eigenvector ##\lambda## of the operator ##\hat{A}## form a subspace, ##\text{Eig}(\hat{A},\lambda)##, because for any two vectors ##|u_1 \rangle## and ##|u_2 \rangle## in this subspace you have for all ##\alpha,\beta \in \mathbb{C}##
$$\hat{A} (\alpha |u_1 \rangle + \beta |u_2 \rangle)=\alpha \hat{A} |u_1 \rangle + \beta \hat{A} |u_2 \rangle=\alpha \lambda |u_1 \rangle + \beta \lambda |u_2 \rangle = \lambda (\alpha |u_1 \rangle + \beta |u_2 \rangle).$$
Thus indeed ##\text{Eig}(\hat{A},\lambda)## is a subspace of ##\mathcal{H}##.

The eigenvalue ##\lambda## is called non-degenerate if and only if ##\mathrm{dim} \text{Eig}(\hat{A},\lambda)=1##. Otherwise it's called degenerate.

Thanks, I understand now. I never encountered the term "eigenspace" in my Linear Algebra studies, It seems to have a little different definition than conventional vector spaces, in which you have a set of vectors and an operator associated with them.

So now with your assistance, I have the answers to what I asked above:
Q: Degenerate subspace is a subspace of the eigenspace of a certain Hermitian operator?
A: Yes, because the whole eigenspace consists of all eigenfunctions associated with different corresponding eigenvalues, and Degenerate subspace is a subspace of the whole eigenspace since it ONLY contains eigenfunctions associated with one same eigenvalue.

Q: Degenerate subspace is the space of degenerate eigenfunctions corresponding to the same certain eigenvalue ?
A: Yes, as explained above.


All your terminology is either wrong or confused. You'll have to learn some linear algebra.
My terminology is fine - you learn new things as you go along a subject and there is always more to learn on a subject you think you've already covered.
When you study hard and worthy subjects like math and physics ( which not many people in the world know - after all, how many math and physics graduates are out there compared to 7 billion people? ) it is okay to ask stupid questions, do mistakes and sometimes even trivial ones. So there's no need to condescend to those who are not rigorous and exact enough if what they said is understood and just fine.
 
  • #9
PeterDonis
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My terminology is fine
You don't know that, and neither do we. First of all, we don't know where you got it from; you said in your OP:

I was learning about Degenerate Perturbation Theory and I encountered the term 'Degenerate Subspace'
But you didn't say what source you were learning from and encountered the term "Degenerate Subspace" in. It would be very helpful if we knew specifically what source that was (and also the exact context in which the term "Degenerate Subspace" occurs).

there's no need to condescend
He's not condescending: he's giving you feedback based on what you post. If what you post is confused, that's the feedback you're going to get.
 
  • #10
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Nothing's confused. The term is from the basics of degenerate perturbation theory in QM , if someone else doesn't understand what it is then I don't tell him he's confused/wrong entirely and surely don't patronize as to tell him he has no clue of the subject as a-whole.
So this my feedback and if he/she is still in the mindset "wrong/confused", he can go over all the subject again, but since we learn as we proceed ( at least any logical person I've met) some sources are: MIT quantum mechanics 3 ( spring 2018, chapter 1, perturbation theory ) , " The Oxford Solid State Basics" - Steven H. Simon , Chapter 15 ( Nearly Free Electron Model ) and my professor's lectures.
 
  • #11
PeterDonis
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Nothing's confused. The term is from the basics of degenerate perturbation theory in QM
Then it's up to you to provide a specific reference for the term when asked. See below.

some sources are
Links would be helpful, particularly to whichever source you saw the term "Degenerate Subspace" in. We're not going to go through an entire semester's worth of course material or an entire textbook chapter to find the term, and how are we supposed to have access to your professor's lectures?
 
  • #12
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The term is from the basics of degenerate perturbation theory in QM
I don't understand your attitude. Mind you that people that are replying to your thread are people who know QM very well (to say the least). So if they say that terminology that you are using is non-standard/confused then it is non-standardand/confused and you should provide reference as soon as they ask for it.
 
  • #13
vanhees71
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I've never heard the term "degenerate subspace". Apparently what's meant is the eigenspace of a given eigenvalue of some linear operator in Hilbert space. A very good treatment of old-fashioned perturbation theory in the case of degenerate eigenvalues is in the textbook

J. J. Sakurai, Modern Quantum Mechanics, Revised Edition.
 

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