# Why is the commutator of ladder operators non-zero?

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?

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.
• 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. Jan 28, 2023 at 14:10