Normalisation of Linear Harmonic Oscillator - Ladder Operator Method

Question

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

http://youtu.be/gRdCV9p8sAU?t=30m9s

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 ?

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 $\hat{H} = \hbar\omega(\hat{a}_+ \hat{a}_- + \frac{1}{2})$