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As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?

$$a_{\pm} \equiv \frac 1 {\sqrt{2m}}\bigg( \frac {\hbar}{i} \frac {d}{dx} \pm im \omega x\bigg)$$

$$a_{\pm} \equiv \frac 1 {\sqrt{2\hbar m \omega}}\bigg(\mp ip + m \omega x\bigg)$$

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Your second equation is the standard definition for $a$ ($a_-$) and $a^\dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.

Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is $$ [a_-, a_+] = \hbar\omega , $$ compare with the standard $[a, a^\dagger] = 1$.

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In the first definition, $a_{+}$ and $a_{-}$ have been defined such that $[a_{-},a_{+}]=\hbar\omega$. Here the position and momentum have been scaled as $X=\sqrt{m}x$ & $P=\sqrt{\frac{1}{m}}p$ and $a_{\pm}= \frac{1}{\sqrt{2}}(X {\pm}iP)$.

Whereas, in the second definition, $a_{+}$ and $a_{-}$ have been defined such that $[a_{-},a_{+}]=1$. Here the position and momentum have been scaled as $X=\sqrt{\frac{m\omega}{\hbar}}x$ & $P=\sqrt{\frac{1}{m\omega\hbar}}p$ and $a_{\pm}=\frac{1}{\sqrt{2}}(X{\pm}iP)$.

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