# 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:

$$K=\frac{2E}{\hbar \omega}$$

$$\frac{d^2\psi}{d\xi^2}=(\xi^2-K) \psi$$

I've tried to rearrange on my own, but:

1. I do not understand why $$\xi$$ equals the square root, except for $$x$$. $$\xi$$ is eventually squared.
• It's just done to non-dimensionalize the equation. It's annoying keeping track of the constants when trying to solve this ODE. Instead we non-dimensionalize, solve that ODE and then put the units back in at the end. – Cameron Williams Jul 16 '13 at 2:12
• as Cameron Williams points out, this is done for convenience. You could just as well not do it and suffer through the next couple pages in Griffith's QM with constants galore... – James S. Cook Jul 16 '13 at 2:22
• @Dustin incidentally, be sure to look at some of the related Q and As such as: math.stackexchange.com/questions/87655/… – James S. Cook Jul 16 '13 at 3:28

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:

$$\begin{equation} \left(-\frac{d^2}{d\tilde{x}^2} + \tilde{x}^2\right) \tilde{\psi}(\tilde{x}) = \tilde{E} \tilde{\psi}(\tilde{x}) \text{ or } \frac{d^2}{d\tilde{x}^2} = (\tilde{x}^2 - \tilde{E}) \tilde{\psi}(\tilde{x}) \end{equation}$$