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While I acknowledge that this topic has been discussed extensively, and I've read numerous similar questions along with their respective answers, I am still struggling to comprehend why all the eigenvectors of the harmonic oscillator are non-degenerate.

My confusion stems from a particular proof presented in the book "QUANTUM MECHANICS Volume I Basic Concepts, Tools, and Applications" by Claude Cohen-Tannoudji, Bernard Diu, and Franck Laloë, specifically in Chapter five on page 510.

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I grasp that the proof establishes the non-degeneracy of $|\phi_n\rangle$, and subsequently, proving that $|\phi_{n+1}\rangle$ is proportional to it implies its non-degeneracy. If this were my sole perspective, everything would make sense. However, I'm struggling to understand why having all vectors proportional to $|\phi_n\rangle$ doesn't lead to a problem of degeneracy.

In a nutshell why proportionality does not lead to degeneracy?

These are some of the questions that I have read

Degeneracy of the ground state of harmonic oscillator with non-zero spin

Proof that the one-dimensional simple harmonic oscillator is non-degenerate?

my2cts
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user353399
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    For the proof to be complete you need to add (B-13) and (B-41) maybe you can add all details in your own words (pictures can go broken very easily)? – Mauricio Jan 12 '24 at 11:49
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    If I have two physical systems A and B and I describe them with wave functions $\psi$ and $15 \psi$, is there any physical difference between their states? – Marius Ladegård Meyer Jan 12 '24 at 11:51
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    To hint in a different way: if I google "degenerate eigenvalue" then the first quote I find reads: "An eigenvalue is degenerate if there is more than one linearly independent eigenstate belonging to the same eigenvalue." Emphasis is mine. – Marius Ladegård Meyer Jan 12 '24 at 11:52
  • Yes I understand what you mean, if they are not linearly independent therefore the eigenvalue is not degenerate. But that means that the construction of all the ${|\phi\rangle}$ representation needs just one vector to create all the other vectors of the space. – user353399 Jan 12 '24 at 12:11
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    Possible duplicate: Proof that the one-dimensional simple harmonic oscillator is non-degenerate? – Qmechanic Jan 12 '24 at 12:12
  • Please don't use images for that. Instead, use MathJax and the quote function. – Tobias Fünke Jan 12 '24 at 12:15
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    @ManuelBorra yes, which is why we call the $a, a^\dagger$ ladder operators; If you have somewhere to start, and the ladder operators, you can reach any other state. What this section you have quoted proves, on the other hand, is that you won't "miss" any eigenstates by only climbing the ladder up and down. – Marius Ladegård Meyer Jan 12 '24 at 12:20
  • @Tobias Fünke Whenever I have more time I will write the equations by myself – user353399 Jan 12 '24 at 13:01
  • Well, thx Marius, it is clearer for me now.I will think more about what implies that you can create everything with the ladder operators – user353399 Jan 12 '24 at 13:07

2 Answers2

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The idea is to identify all the distinct eigenstates $|\phi_n\rangle$ of the number operator $\hat{n}$. These eigenstates are normalized $\langle\phi_n|\phi_n\rangle=1$. In the process, we find that, given an eigenstate $|\phi_n\rangle$, the state $\hat{a}^{\dagger}|\phi_n\rangle$ is not the same state as $|\phi_n\rangle$. In fact, we find that it is proportional to $|\phi_{n+1}\rangle$. There may seem to be several different candidates for $|\phi_{n+1}\rangle$ all associated with the eigenvalue $(1+n)$, but they are all proportional to $\hat{a}^{\dagger}|\phi_n\rangle$. Now clearly, all the states that differ only by a constant proportionality factor must be identically the same state. All that remains is to normalize these states to remove the proportionality constants, as required for the eigenstates. Then they would all be equal and there would not be any degeneracy. Does that make sense?

flippiefanus
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  • Yes does makes more sense, and with the answers of Marius i start to understand – user353399 Jan 12 '24 at 13:07
  • Thinking about that. I found it strange that all my space is constructed with the same vector. I know that it works well and fulfills all the requirements to be a representation. But I feel that I cant make an analogy with an ordinary coordinate system and its canonical representation. – user353399 Jan 13 '24 at 12:11
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Since any states are proportional they are identical after normalisation. Hence there is only one state and there is no degeneracy.

my2cts
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