# A Operation of Hamiltonian roots on wave functions

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1. May 16, 2017

### SherLOCKed

How come a+a- ψn = nψn ? 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.

2. May 16, 2017

### 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.