How to prove that all eigenstate $|0\rangle,|1 \rangle,...| n\rangle $ are non-degenerate where $a^{\dagger}a|n \rangle =n|n \rangle $
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Degeneracy for 1D systems is impossible. Look under degeneracy in 1D in https://en.wikipedia.org/wiki/Degenerate_energy_levels – Superfast Jellyfish Feb 05 '20 at 15:59
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It would be better to start with definitions. What is $|1\rangle$? It is an eigenstate of a Hamiltonian, I guess. Which Hamiltonan? etc. I think the answer will become clear once you fix the definitions. – Cryo Feb 05 '20 at 16:29
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2Degeneracy or non-degeneracy is not a property of states, but of an operator. The operator $a^{\dagger}a$ as you defined above is non-degenerate in the space that is spanned by $|0\rangle, |1\rangle, \ldots$. Can you clarify your question accordingly? – Feb 05 '20 at 16:53
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Possible duplicates: https://physics.stackexchange.com/q/23028/2451 and links therein. – Qmechanic Feb 05 '20 at 17:40