If we have a function: $\psi(x)$ and $x=\rho\gamma$ where $\rho$ is a constant
Then is the following statement correct:
$1)~~$ $\psi(x) =\psi(\rho \gamma)=\rho \psi(\rho)$
What is the equation for : $\frac{d\psi(\gamma)}{d\gamma}$ and $\frac{d^2\psi(\gamma)}{d\gamma^2}$
Below is my attempt at finding it:
This question is the context of findings the solution of simple harmonic oscillator
I am having trouble understanding how $9.19$ is derived from $9.18$ .
Question: I am having trouble understanding how $9.19$ is derived from $9.18$ .
Answer: There are some mistakes (may be printing mistake) in their calculation. You can see those in my answer which I have already fixed (in red color). Here we go:
$$-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi(x)}{dx^2}+\dfrac{mw^2x^2}{2}\psi(x)=E~\psi(x)\tag{9.18}$$Putting $~x=\rho\gamma\implies \psi(x) =\psi(\rho \gamma)=\rho \psi(\color{red}\gamma)~$ and the partial derivatives became $$\dfrac{d\psi(x)}{dx}=\dfrac{d\psi(\rho\gamma)}{d\gamma}\cdot\dfrac{d\gamma}{dx}=\dfrac{1}{\rho}\cdot\rho\dfrac{d\psi(\gamma)}{d\gamma}=\dfrac{d\psi(\gamma)}{d\gamma}~~~~~~~~\text{and}$$ $$\frac{d^2\psi(x)}{dx^2}=\dfrac{d}{dx}\left(\dfrac{d\psi(x)}{dx}\right)=\dfrac{d}{d\gamma}\left(\dfrac{d\psi(\gamma)}{d\gamma}\right)\cdot\dfrac{d\gamma}{dx}=\frac{d^2\psi(\gamma)}{d\gamma^2}\cdot\dfrac{d\gamma}{dx}=\color{red}{\dfrac 1\rho}\cdot\frac{d^2\psi(\gamma)}{d\gamma^2}~.$$
Now putting the above values in equation $(9.18)$ we have, $$-\dfrac{\hbar^2}{2m}\dfrac 1\rho\dfrac{d^2\psi(\gamma)}{d\gamma^2}+\dfrac{mw^2\rho^2 \gamma^2}{2}\rho\psi(\gamma)=E~\rho\psi(\gamma)$$ $$\implies \dfrac{d^2\psi(\gamma)}{d\gamma^2}-\dfrac{2m\rho}{\hbar^2}\dfrac{mw^2\rho^2 \gamma^2}{2}\rho\psi(\gamma)=-\dfrac{2m\rho}{\hbar^2}E~\rho\psi(\gamma)$$ $$\implies \dfrac{d^2\psi(\gamma)}{d\gamma^2}-\dfrac{ m^2w^2\rho^4}{\hbar^2}\gamma^2\psi(\gamma)=-\dfrac{2mE\rho^2}{\hbar^2}\psi(\gamma)\tag{9.19}$$