# Degeneracy of a quantum state

1. Feb 6, 2009

### Niles

Hi all.

When a quantum state is said to be degenerate, then it means that two states $\psi_1$ and $\psi_2$ result in the same energy and that $|\psi_1|2\neq|\psi_2|2$, am I correct? Now in my book we have a wavefunction given by:

$$\psi_n =\frac{1}{\sqrt{L}}\exp(2\pi i nx/L),$$

where n is a whole integer (i.e. n can also be negative) and L is some constant. The energies are $E_n\propto n^2$. Lets look at n=1 and n=-1. These states result in the same energy, but the absolute square of the wavefunctions are equal. Now according to my book, this is a degenerate state, but according to: Where is my reasoning wrong?

2. Feb 6, 2009

### jensa

I have never heard statements about absolute square of wave-function in relation to degeneracy. I would just say that two quantum states are degenerate if they have the same energy. The fact that the two quantum states are different implies that they are orthogonal to each other. So I would state the two conditions $E_1=E_2$ and $\langle \psi_1|\psi_2\rangle=0$ as the condition for degeneracy. Clearly the two eigenstates n,-n in your example are orthogonal to each other so the book is correct.

Hope this helps

3. Feb 6, 2009

### Niles

Yes, using your conditions it works out. There must be some error with my conditions then.

I can't find anywhere on WWW what the exact conditions have to be in order for two states to be degenerate.

4. Feb 6, 2009

### jensa

http://en.wikipedia.org/wiki/Degeneracy_(quantum_mechanics [Broken])

Last edited by a moderator: May 4, 2017
5. Feb 6, 2009

### Niles

"In physics two or more different physical states are said to be degenerate if they are all at the same energy level. Physical states differ if and only if they are linearly independent. An energy level is ..."

Ok, then we can conclude two things: Apparently I haven't search very thorough, and you are correct.

Thanks

Last edited by a moderator: May 4, 2017
6. Feb 6, 2009

### xepma

There's also some other method to check this, namely such degenerate states might not be degenerate with respect to some other Hermitian operator (which necessarily commutes with the Hamiltonian). In your case the eigenvalues of the momentum operator differ.

7. Jun 30, 2009

### yogeshbua

Physical states can differ even if they are not linearly independent... For example... consider
$$\psi_a$$ and $$\psi_b$$ to be linearly independent. Then $$\frac{1}{\sqrt{2}}\psi_a+\frac{1}{\sqrt{2}}\psi_b$$ is a different state than either $$\psi_a$$ or $$\psi_b$$! Without being linearly independent!

In fact the only condition for $$\psi_a$$ and $$\psi_b$$ to be degenerate states (i.e. apart from having the same eigen-value) is that $$\psi_a\not=\psi_b$$

Sorry... The last part should read $$\psi_a\not=c\psi_b$$, where c is a scalar in the scalar-field assumed for the vector-space/ state-space under consideration! I.e., sometimes, the values of c are restricted to real numbers, while most of the times, c can be complex... (In mathematics, Linear Algebra, a scalar field can be any set with some definite properties... Wiki it!)

Last edited: Jul 1, 2009
8. Jun 30, 2009

### Fredrik

Staff Emeritus
That Wikipedia article is pretty bad. Yogeshbua pointed out the worst thing about it. The bolded text (in his quote) should be something like "state vectors represent different states if and only if they belong to different one-dimensional subspaces". Another way of thinking about that is that it's the one-dimensional subspaces that represent states, not the state vectors.

Now consider the set of eigenvectors with eigenvalue $\lambda$ of some hermitian operator A. It's very easy to verify that this set is a subspace. If this subspace isn't one-dimensional, then we're dealing with "degeneracy". Note that the operator A doesn't have to be the Hamiltonian, so degeneracy doensn't really have anything to do with energy.

I don't know exactly what the standard usage of the word "degenerate" is. I would say that the spectrum of A is degenerate, or that $\lambda$ is a degenerate eigenvalue of A, but I wouldn't call one of the corresponding eigenvectors degenerate. Maybe that's just me though.

9. Jul 1, 2009

### Fredrik

Staff Emeritus
Regarding the terminology...My books on functional analysis don't have the words "degenerate" or "degeneracy" in the index. The QM books by Isham and Ballentine don't either. The only one I've found it in is Sakurai.

He defines the word by saying that if two linearly independent eigenvectors of an operator have the same eigenvalue, then the eigenvalues of these eigenvectors are degenerate. Later in the book, he used the phrase "these are two states with the same energy, that is, they are degenerate", so apparently he's using it about the state vectors as well.

10. Jul 1, 2009

### Bob_for_short

Yes, it is like moving forward and backward with the same energy E but with different (opposite) momenta p.

Bob_for_short.

11. Jul 1, 2009

### jensa

I was surprised to see this old thread resurrected. Eventhough neither I or the wikipedia article was very exhaustive, I believe the OP was happy with the original response.

Personally I don't see what is wrong with (the cited part of) the Wikipedia article. When we talk about two different "physical states" we mean physically distinguishable states, i.e. there exists some observable for which the two states have different eigenvalues. This implies a) that two state vectors $$\psi$$ and $$c\psi$$ correspond to the same physical state and b) that two different physical states must be orthogonal.

12. Jul 1, 2009

### yogeshbua

But you see, the wiki article talks about, to quote, "In physics two or more different physical states are said to be degenerate if they are all at the same energy level"

If we say that, to quote, "When we talk about two different "physical states" we mean physically distinguishable states", then there cannot be two different physical states which are at the same energy level! The latter quote is in contradiction with the former... (Given that we are talking about the same observable!; and given that we are discussing degeneracy, we are!)

Last edited: Jul 1, 2009
13. Jul 1, 2009

### yogeshbua

I hope that the next time you see an old post opened for discussion again, you'll not be surprised. Here's why:
Phy Forums keeps old threads? Why? For people in the future to come and see if their doubts are solved. If a future person has to add to the discussion, (s)he should do so, right? Because, people from even further future are going to have a look at the thread!

Cheers,
Y

14. Jul 1, 2009

### jensa

Usually when we have a degeneracy there exist at least one other observable the corresponding operator of which commutes with the Hamiltonian (symmetry of the Hamiltonian). The eigenvalue of this operator distinguishes the two degenerate (wrt Hamiltonian) states. See xepmas reply. Examples include momentum when the Hamiltonian is translationally invariant and angular momentum when the Hamiltonian is rotationally invariant. This is actually also described in the Wiki.

/Jens

15. Jul 1, 2009

### yogeshbua

Yes... Thank you. I had not taken that perspective.

Edit 1: I agree with you completely, given the interpretation of 'physical states' as you have.
Thank you...

Last edited: Jul 1, 2009
16. Jul 1, 2009

### jensa

No problem, and I hope you did not take my response as a discouragement to taking up old threads. If you find something you think is wrong or you feel the need to add something then you should definitely do that. I was just surprised that people objected to the Wiki (and obviously not just you but other people like Fredrik whom I respect very much. I am still interested to hear if he agrees with me or not).

17. Jul 1, 2009

### Fredrik

Staff Emeritus
I didn't even notice that it was an old one.

I don't either. I'm not sure what I was thinking yesterday. Now I think the bolded part says the same thing as my "correction".

I don't think this is a good way to define "different" and "the same", because if we do it this way, we can find two states that are neither "different" nor "the same". Shouldn't any two states that aren't "the same" be "different"?

18. Jul 2, 2009

### yogeshbua

Yes. We can 'find' (have in existence, though not 'observe') states that are neither 'different' nor the 'same'. But we cannot 'find' 'physical states' (and 'find' for 'physical states' means 'observe') that are neither 'different' nor the 'same'!

19. Jul 2, 2009

### jensa

Yes this is quite loose and confusing language. Physical state in this sense corresponds to the equivalence class $[\psi]$ of states which are physically undestinguishable. Two different vectors can still belong to the same equivalence class and thus correspond to the same physical state. So if we define the equivalence class to be the set of all eigenvectors of a set of commuting observables with the same eigenvalues then the following relations should hold:
$\psi\sim c\psi$ (or $[\psi]=[c\psi]$, i.e. they belong to the same equivalence class) and $\langle \psi_a|\psi_b\rangle = 0$ if $[\psi_a]\neq [\psi_b]$ (i.e. they belong to different equivalence classes).

I think you can always find an observable with a non-degenerate spectrum so the equivalence class would reduce to the one dimensional subspace (or ray) spanned by one of the eigenvector $[\psi_a]=\text{span}_\mathbb{C}\{\psi_a\}$ which is what you mentioned in one of your posts.

Really I should probably not be so confident in this because I have not been able to find a good reference that defines the term "physical state" in this way. Although rays are often used.

EDIT: Just to avoid confusion, when I used the term "degenerate spectrum" I was referring to the spectrum of a general observable, i.e. it doesn't necessarily have anything to do with the energy spectrum :)

EDIT2: What the... PF must be messing with me :) post keep coming up wrong.

Last edited: Jul 2, 2009
20. Jul 2, 2009

### yogeshbua

Here's something we can think about. I was convinced by Jensa that different 'physical states' would mean ones that can be 'observed' distinctly and hence, states, I quote (recollected version) 'that can be physically distinguished from one another'.

Suppose |1,1> and |1,2> are the eigen vectors with eigenvalue 1 for operator O and eigen values 1 and 2 for operator T, respectively.
Are the states |1,1> and $$\frac{1}{\sqrt{2}}\left(|1,1\rangle+|1,2\rangle\right)$$ capable of being distinguished from one another?
I thought that since $$\frac{1}{\sqrt{2}}\left(|1,1\rangle+|1,2\rangle\right)$$ could not be 'observed', not being an eigen vector (say), one cannot distinguish the 2 states.
But if a system were in that state, we can observe, with repeated experiments, the probability of 'observing' |1,1> and |1,2>. And from this, conclude that this state is different from either |1,1> or |1,2>. So they are 'physically distinguishable', after all! And this state also has evalue 1 wrt the operator O.

Sorry for being so fickle, but I then revert to say that even 'physical states' which are different need not be orthogonal. Earlier, I interpreted that 'physical state' was a state with a particular set of measurement values. Now, we interpret it to be 'a possible state of existence of the system' , distinguishable from the a state which gives a particular set of measurement values, with the additional knowledge of the probability of observing a particular set of measurement values.

In summary, we used only the measurement values to distinguish 'physical states'. Now, we use the values of probability of obtaining a particular set of measurement values as well, to distinguish 'physical states'. (In fact, the Stern-Gerlach experiment, the earliest QM one envisaged, and then verified, employs the measurement of probability of obtaining a particular eigen-state heavily to distinguish between physical states! Though there, these states are not degenerate...)

Edit 1: Even more simply put, the essence of what I said is that 'physical states' need not be those which yield only a particular set of eigen-values (measurements). It is simply 'a possible state of existence of a system'. And these states are distinguishable using the added knowledge of the probabilities concerned...

Last edited: Jul 2, 2009
21. Jul 2, 2009

### yogeshbua

We should ask someone the conditions for this implication. That is, that of,
'physically distinguishable states' $$\Rightarrow$$ 'there exists some observable for which the two states have different eigenvalues'
It would be worth knowing...

Sorry, but tell me, are all of us students, or do professors also participate in PF? (I'm new to this...)

Last edited: Jul 2, 2009
22. Jul 2, 2009

### jensa

Yogeshbua,
You ask very relevant questions, but I must warn you: If you are an undergraduate student studying QM I would be careful to read too much into what I say as much of it goes outside of standard textbook quantum mechanics (and some people may even disagree with me) so much of this may just confuse you. Also I am definitely not an expert in these matters and am only conveying my very vague understanding of these matters. Most of these ideas and concepts are taken from the algebraic approach to quantum theory. I would be glad if some experts would join in and comment.

That being said:

Let's talk about observables, imagine we could list all the possible observables we could (in principle) measure (in a single shot measurement, i.e. we are not allowed to make repeated measurements). Now the way "I" defined a physical state it has to be an eigenstate of a subset of mutually commuting observables. For most systems the set of observables would be very large and the set of eigenstates to these observables would be larger still. The set of all eigenstates of all observables is equivalent to the union of all equivalence classes (or the set of all physical states), so it describes what you might call "the physical Hilbert space". Now one interesting question would be: does the "physical Hilbert space" cover the whole Hilbert space or is it just a subset of the Hilbert space? I.e. are there vectors in the Hilbert space which do not correspond to physical states, i.e. that are not eigenstates of some observable? An interesting property of a vector in Hilbert space that does not correspond to the eigenstate of an observable is that we could never measure it directly in a single-shot measurement. The only information we can obtain about it are probabilites, acquired by repeated measurements. This has the implication that we could never distinguish such a pure state from a so called mixed state (if you are familiar with those). Let me use your example to demonstrate:

In your example you had only two operators O and T (which commute), in reality there could be many more observables. So there may very well be a third operator A (which commutes with O) which has the property:

$$A\left(\frac{1}{\sqrt{2}}|1,1\rangle+\frac{1}{\sqrt{2}}|1,2\rangle\right)=a\left(\frac{1}{\sqrt{2}}|1,1\rangle+\frac{1}{\sqrt{2}}|1,2\rangle\right)$$

Now, the observable associated with operator A may be quite difficult to access in the sense that an experiment measuring the quantum number 'a' may be difficult to perform or even conceive. Still it is possible "in principle" to distinguish this state physically.

However, If we, for the sake of the argument, assume that the only observables are O and T, then one reaches the interesting conclusion (as you pointed out) that the (more general) state $\alpha|1,1\rangle+\beta|1,2\rangle$ when $\alpha,\beta\neq 0$, is not a physical state in the sense that I just described since it is not an eigenstate of T.
So what does this mean exactly? Well since O and T are the only observables of the system and the state we are looking at is not an eigenstate of T we have the problem that we can only obtain probabilities $|\alpha|^2,|\beta|^2$. We can never measure the phase difference! It then does not make any difference if we describe this state, not as a superposition, but as a statistical mixture of states $|1,1\rangle$ and $|1,2\rangle$.

So even if the state $\alpha|1,1\rangle+\beta|1,2\rangle$ exists in the Hilbert space it could never be measured as a pure state. We then have two choices: 1) we accept that there are certain pure states in nature that we cannot directly measure, or 2) we say that such states do not exist! This is what is called a super selection rule.

Regardless which choice you make, for all practical purposes it is completely sufficient to treat such vectors in Hilbert space as mixed states!

23. Jul 2, 2009

### Fredrik

Staff Emeritus
I'm with you so far.

I think you're just saying the same thing as before in a more complicated way. It's much simpler to just define the equivalence classes to be the one-dimensional subspaces (i.e. the "rays"). You're making a mistake when you say that two vectors in different equivalence classes must be orthogonal. Think of the plane $\mathbb R^2$ for a moment. Any straight line through the origin is a one-dimensional subspace, but (1,1) isn't orthogonal (1,0).

24. Jul 2, 2009

### jensa

Well the orthogonality should follow from the definition of the equivalence class. Two vectors which belong to two different equivalence classes have different eigenvalues of some hermitian operator and therefore have to be orthogonal. Right?

EDIT: BTW I agree with you that if this equivalence class is just a one-dimensional subspace then it is definitely easier to talk about that. However, I am not sure if this is always the case.

25. Jul 2, 2009

### jensa

Some additional thought makes me think Fredrik may be right. The way I tried to construct this thing is not exactly waterproof. There is probably a reason why they use a much more sophisticated construction in algebraic quantum mechanics. Take what I wrote with handfull of salt.

EDIT:
Since I am not sure about the way I constructed things let me throw out what i think is the correct way to define the physical states:
My understanding is that the physical states are defined by the irreducible representation of the algebra of observables. These define the relevant quantum numbers that can be used to label the physical states. Vectors belonging to different irreps are naturally orthogonal.

Again, would be happy if an expert would comment as this is somewhat over my head. (Can't believe this thread went this deep :) , my own fault I guess)

/jens

Last edited: Jul 2, 2009