I was watching the following video on the harmonic oscillator using ladder operators :


Clicking on the video above will take you to the exact point where my questions are based off (30:09).

At that point, he has circled in black the step up and step down operators and how they act on the wavefunction $\psi$. He does not explain how he obtains the normalisation constants $\sqrt n$ and $\sqrt{n+1}$.

Here's my attempt starting with the step down operator (C is my normalisation constant) :

$$\int{(a^-\psi)^*a^-\psi}.dx=1$$ $$\int{ C^*\psi^*_{n-1}a^-\psi}.dx=1$$ $$\int{C^*\psi^*_{n-1}C\psi_{n-1}}.dx=1$$ $$\mid C\mid^2\int{\psi^*_{n-1}\psi_{n-1}}.dx=1$$

Where do I go from here ?

EDIT : I would also like to know whether the equation for the energy of the harmonic oscillator is derived from pure observation of pattern or is there a general method to derive it ?

  • 1
    $\begingroup$ This is explained in Griffiths Quantum Mechanics. $\endgroup$
    – jinawee
    Apr 5, 2014 at 8:27
  • $\begingroup$ I don't have that textbook unfortunately nor can I access it anytime soon. Could you help me out please ? $\endgroup$ Apr 5, 2014 at 8:32
  • $\begingroup$ Related: physics.stackexchange.com/questions/82691/… $\endgroup$
    – Kyle Kanos
    Apr 5, 2014 at 17:29
  • $\begingroup$ @KyleKanos I'm still a tad confused as to why PhotonicBoom and my textbook equate this to $\mid C\mid$^2 (see below) ? $\endgroup$ Apr 5, 2014 at 17:56
  • $\begingroup$ Search your book for the phrase "completeness relation." Recall also from my previous comment that $C$ is not necessarily real, hence the modulus-squared. $\endgroup$
    – Kyle Kanos
    Apr 5, 2014 at 18:07

1 Answer 1


For the raising operator case: We know that $\hat{a}_+^\dagger = \hat{a}_-$

(Don't forget that you operator acting on your conjugate is daggered)

Therefore $\langle\psi_n$|$\hat{a}_- \hat{a}_+$|$\psi_n\rangle$ = $|c_n|^2$ $\langle\psi_{n+1}|\psi_{n+1}\rangle$ = $|c_n|^2$

But $[\hat{a}_-, \hat{a}_+]=1$ and when you expand the commutator out you get $\hat{a}_- \hat{a}_+ = \hat{a}_+\hat{a}_- + 1$

So substituting in we get:

$\langle\psi_n$|$\hat{a}_+ \hat{a}_-$ + 1|$\psi_n\rangle$ = $|c_n|^2$

We also know that $\hat{a}_+ \hat{a}_-=\hat{N}$ which is the number operator with eigenvalues equal to $n$.

Therefore $\langle\psi_n$|$\hat{N}+1$|$\psi_n\rangle$ = $|c_n|^2$ which leads to:

$n + 1$ = $|c_n|^2$ and finally we get: $\sqrt{n+1} = c_n$

With a very similar manner you derive the lowering operator eigenvalue, and also the Hamiltonian! Just remember that $\hat{H} = \hbar\omega(\hat{a}_+ \hat{a}_- + \frac{1}{2})$

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    $\begingroup$ I understand and follow your proof perfectly, but I don't understand : $<\psi_n$|$\hat{a}_- \hat{a}_+$|$\psi_n$> = $|c_n|^2$ Why have you chosen to set it equal to $|c_n|^2$ ? Why not 1 ? $\endgroup$ Apr 5, 2014 at 13:02
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    $\begingroup$ That is just the arbitrary eigenvalue which I named $c_n$. $\endgroup$
    – PhotonBoom
    Apr 5, 2014 at 13:10
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    $\begingroup$ So that $\hat{a}_+|\psi_n> =c_n|\psi_n>$ $\endgroup$
    – PhotonBoom
    Apr 5, 2014 at 13:11
  • $\begingroup$ So we're not actually 'normalising' this wavefunction are we ? Because if we were we'd be setting this equal to 1 ? I'm getting a bit confused with what we're doing here conceptually... Could you take me through it please ? $\endgroup$ Apr 5, 2014 at 13:52
  • 1
    $\begingroup$ I'm doing exactly what you have done, but the integral will not be equal to one because we are applying an operator to it. If you like it will be equal to $1$ x $|c_n|^2$ which is the eigenvalue when you apply the operator twice. Conceptually this is the number of particles in that state $\psi_n$ $\endgroup$
    – PhotonBoom
    Apr 5, 2014 at 18:21

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