Why use of creation and then annihilation operator on n state gives n+1 state when they are supposed to reverse each others effect?

Question

I read this in Griffith in harmonic oscillator topic(page 45 equation 2.66) . Why when we use creation operator then annihilation operator on w.f of state n it gives wavefunction of state n+1? Isn't the state should remain n as one operator reverse the effect of another which is the case when we use annihilation operator then creation operator. I know they do not commute so what I read also seems correct but according to what I know about these operators and how they access the ladder of states my argument also seems fine. What i am missing here?

Answer 1

You must have misread something. The raising $(\hat a^\dagger)$ and lowering $(\hat a)$ operators of the quantum harmonic oscillator act as follows on the $n^\text{th}$ state:

$$\hat a|n\rangle=\sqrt{n}|n-1\rangle,\quad \hat a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle \tag{1}.$$

Your suggestion that acting with one and then the other should return us to our original state is correct, let's do it explicitly:

$$\hat a^\dagger (\hat a|n\rangle)=\hat a^\dagger(\sqrt{n}|n-1\rangle)=\sqrt{n}\sqrt{n}|n\rangle=(n)|n\rangle \tag{2.1}$$

$$\hat a (\hat a^\dagger|n\rangle)=\hat a(\sqrt{n+1}|n+1\rangle)=\sqrt{n+1}\sqrt{n+1}|n\rangle=(n+1)|n\rangle \tag{2.1}$$

For completeness we now have everything we need to derive the commutator of these two operators, as you say they do not commute:

$$[\hat a,\hat a^\dagger]|n\rangle=(\hat a\hat a^\dagger-\hat a^\dagger\hat a)|n\rangle=\hat a(\hat a^\dagger|n\rangle)-\hat a^\dagger(\hat a|n\rangle)$$

$$=(n+1)|n\rangle-(n)|n\rangle=(n+1-n)|n\rangle=|n\rangle\tag{3}$$ This means that: $$[\hat a,\hat a^\dagger]=\Bbb{\hat 1}\tag {4}.$$