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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?

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  • $\begingroup$ What page of Griffiths are you asking about? It will help people greatly if you mention that, also one should spell out acronyms as it's not always immediately obvious what abbreviations might mean. $\endgroup$
    – Triatticus
    Aug 19, 2020 at 15:41
  • $\begingroup$ Sure I did the changes. $\endgroup$
    – Pirateking
    Aug 19, 2020 at 16:23

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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: $$[\hat a,\hat a^\dagger]=\Bbb{\hat 1}\tag {4}.$$

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  • $\begingroup$ Sorry you were right I misread the state was same just the coefficient got changed. $\endgroup$
    – Pirateking
    Aug 19, 2020 at 16:28
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    $\begingroup$ @PiyushGalav It's fine, the raising and lowering operators can be deceptively tricky especially since they're usually one of the first operators you actually use in quantum mechanics. Just remember that the lowering operator has a factor of $\sqrt{n}$ not $\sqrt{n-1}$, it's not simply the opposite of the raising operator unfortunately. $\endgroup$
    – Charlie
    Aug 19, 2020 at 16:36

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