# If $\psi$ acting on $a_+$ and $a_-$ operator just moves up and down the ladder, why is $[a_-, a_+] = 1$ and not 0? [duplicate]

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?

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

• Why "return the same wave function"? Or by same, did you mean proportional to? – Alfred Centauri Oct 5 '18 at 21:45
• 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. – Astik Oct 9 '18 at 8:32
• 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 $|E\rangle$? – Alfred Centauri Oct 9 '18 at 11:39

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: $$\left[ p,x\right]=-i\hbar.$$ $$\left[ a_-,a_+\right]=\dfrac{1}{2m\omega\hbar}\left([p,p]+m^2\omega^2[x,x]+i2m\omega[p,x]\right)$$