Operation of Hamiltonian roots on wave functions

[SherLOCKed]

How come a₊a_- ψ_n = nψ_n ? This is eq. 2.65 of Griffith, Introduction to Quantum Mechanics, 2e. I followed the previous operation from the following analysis but I cannot get anywhere with this statement. Kindly help me with it. Thank you for your time.

 

[BvU]

HI,

I only have 1st ed PDF and there, in the text "Algebraic method" he shows that the Schroedinger eqn

[ ..² + .. ²] $\psi = E\psi$​

can be factored into

$[a_+ a_- ] + {1\over 2 } \hbar \omega \psi = E\psi$
​

Or $[a_- a_+ ] - {1\over 2 } \hbar \omega \psi = E\psi$.Then he proves that if $\psi$ satisfies the Schroedinger eqn, with energy $E$, then $a_+\psi$ satisfies the Schroedinger eqn, with energy $E+\hbar\omega$
and - same way - $a_-\psi$ satisfies the Schroedinger eqn, with energy $E-\hbar\omega$

(Walking down the ladder there is a ground state ($\ {1\over 2 } \hbar \omega\$) and in n steps down you have subtracted n times $\hbar\omega$.)

The eigenfunction part has now been shown. The normalization coefficient is left as an exercise. That's where the $i\sqrt{\left (n+1\right) \hbar\omega\,}$ and $-i\sqrt{n \hbar\omega\,}$ pop up.