Solving Schrodinger for harmonic oscillator(griffiths analytic method) I was just getting into quantum mechanics. But I'm having a bit of trouble following Griffiths for the analytic method. It goes like so:
The Schrodinger equation is:
$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{d x^2} + \frac{1}{2} m \omega^2 x^2 \psi = E \psi $$
Griffiths expresses $\xi$ as:
$$\xi = \sqrt{\frac{m \omega}{\hbar}}x$$
and $K$ as:
$$K=\frac{2E}{\hbar \omega}$$
Ultimately, leading to the equation:
$$\frac{d^2\psi}{d\xi^2}=(\xi^2-K) \psi$$
I've tried to rearrange on my own, but:


*

*I do not understand why $\xi$ equals the square root, except for $x$.  $\xi$ is eventually squared. 

 A: This follows from non-dimensionalisation or scaling - the removal of units from an equation involving physical quantities. In the case of the wavefunction, this is done through the following substitution:
\begin{equation}
    x = x_{c} \tilde{x}\\
    \frac{d}{dx} =  \frac{d\tilde{x}}{dx} \frac{d}{d\tilde{x}} = \frac{1}{x_c} \frac{d}{d\tilde{x}}
\end{equation}
where $x_{c}$ is an intrinsic characteristic unit (length) of the system and $\tilde{x}$ is the nondimensionalized counterpart of $x$. Additionally, $\psi(x) = \psi( x_{c} \tilde{x} ) = \tilde{\psi}(\tilde{x})$. After a substitution the Schrödinger equation becomes:
\begin{equation}
 \left(-\frac{\hbar^2}{2m}\frac{1}{x_c^2}\frac{d^2}{d\tilde{x}^2} + m\omega^2 x_{c}^{2} \tilde{x}^{2} \right) \tilde{\psi}(\tilde{x}) = E \tilde{\psi}(\tilde{x})
\end{equation}
After dividing by the coefficients in front of the largest term, we get:
\begin{equation}
 \left(-\frac{d^2}{d\tilde{x}^2} + \frac{m^{2} \omega^{2} x_{c}^{4}}{\hbar^2} \tilde{x}^{2} \right) \tilde{\psi}(\tilde{x}) = \frac{2mE x_{c}^{2}}{\hbar^2} \tilde{x}^{2} \tilde{\psi}(\tilde{x})
\end{equation}
To make the terms dimensionless we equate the coefficients to unity:
\begin{equation}
 \frac{m^{2} \omega^{2} x_{c}^{4}}{\hbar^2} = 1 \implies  x_c = \sqrt{\frac{\hbar}{m\omega}}\\
 \frac{2mE x_{c}^{2}}{\hbar^2} =  \frac{2mE \hbar}{\hbar^2 \omega} = 1\implies E= \frac{\hbar \omega}{2} \tilde{E} 
\end{equation}
From the above equations we find that the characteristic units of energy and length of the Harmonic oscillator are, $\hbar \omega$ and $\sqrt{\frac{\hbar}{m \omega}}$ , respectively. The unit free version of the time-independent Schrödinger equation for the quantum harmonic oscillator becomes:
\begin{equation}
 \left(-\frac{d^2}{d\tilde{x}^2} + \tilde{x}^2\right) \tilde{\psi}(\tilde{x}) = \tilde{E} \tilde{\psi}(\tilde{x}) \text{ or } \frac{d^2}{d\tilde{x}^2} = (\tilde{x}^2 - \tilde{E}) \tilde{\psi}(\tilde{x}) 
\end{equation}
