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Griffiths states that the "ladder" of stationary states for a harmonic oscillator should be unique. That should mean that for one particular energy level, there exists only one energy state. So if I have an energy state $\psi$ with energy $E$, $a_+ \psi$ should take me to a state with energy $E + \hbar \omega$, and $a_- a_+ \psi$ should have energy $E$. And similarly, $a_+ a_- \psi$ should have energy $E$. Shouldn't this imply that $a_+ a_- \psi = a_- a_+ \psi$, since they both correspond to the same energy? What is wrong with the logic here?

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No. The up-down $\sim$ down-up property implies only that that $a_+a_- \psi= \lambda a_-a_+ \psi$ for some number $\lambda$. Here $\lambda$ can be (and indeed is) a number other than 1.

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  • $\begingroup$ But then what does it mean for the ladder to be unique? $\endgroup$
    – Math boi
    Jan 28, 2023 at 13:59
  • $\begingroup$ We have $a_+|n\rangle = \sqrt{n+1} |n+1\rangle$ and $a_-|n\rangle = \sqrt{n} |n-1\rangle$ so $a_+a_- |n\rangle =(n+1) a_-a_+ |n\rangle$. No problem with uniqueness. $\endgroup$
    – mike stone
    Jan 28, 2023 at 14:10

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