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This question already has an answer here:

If $\psi$ is acted upon by both the operators one by one, it should return the same wave function. Thus order in which you increase or decrease the energy shouldn't matter. Then why is it so that the order of operators matters?

$$a_\pm = \frac {1}{\sqrt{(\hbar2m\omega)}} (\mp ip + m\omega x).$$

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marked as duplicate by Emilio Pisanty quantum-mechanics Oct 6 '18 at 6:32

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    $\begingroup$ Why "return the same wave function"? Or by same, did you mean proportional to? $\endgroup$ – Alfred Centauri Oct 5 '18 at 21:45
  • $\begingroup$ I meant the exact same wave function. If we operated H on $\psi$ and H on a+a-$\psi$, we should get the same eigen state. So, It should return the same wave function. $\endgroup$ – Astik Oct 9 '18 at 8:32
  • $\begingroup$ If the state is a state of definite energy, an energy eigenstate, then operating on the energy eigenstate with H gives that same state multiplied by the energy eigenvalue: $H|E\rangle =E|E\rangle$. So, to be clear, do you consider the state $E|E\rangle$ to be same as or proportional to $|E\rangle$? $\endgroup$ – Alfred Centauri Oct 9 '18 at 11:39
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Your assumption is incorrect. The commutator is not zero because the state function returned from $a_-a_+$ is not the same as that returned from $a_+a_-$. One may end up in the original energy eigenstate, but there are scalar constants which are part of the result, too. The scalar constants depend on the particular state you operate on and whether the operator is $a_-$ or $a_+$.

The primary basis of this is that $p$ and $x$ don't commute: $\left[ p,x\right]=-i\hbar.$ $$\left[ a_-,a_+\right]=\dfrac{1}{2m\omega\hbar}\left([p,p]+m^2\omega^2[x,x]+i2m\omega[p,x]\right)$$

(within a sign). That last term is not zero

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  • $\begingroup$ Mathematically, I understand that [p,x] don't commute and so [ a-,a+] don't commute. But, intuitively, how will you explain the result? $\endgroup$ – Astik Oct 9 '18 at 8:26
  • $\begingroup$ Read my answer closely: "One may end up in the original energy eigenstate, but there are scalar constants which are part of the result, too. " It makes a difference if you go down-then-up (in energy) versus up-then-down. You can't ignore constants generated by the operations, and they aren't symmetric. $\endgroup$ – Bill N Oct 10 '18 at 15:17

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