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.