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I am following a quantum mechanics text book which uses a simple looking substitution in a derivative.

The substitution is $$\xi=\gamma x\tag1$$

It then says that $$\frac{d\psi}{dx}=\frac{d\psi}{d\xi}\frac{d\xi}{dx}=\gamma\frac{d\psi}{d\xi}\tag2$$

So far so good. Now comes the part I don't follow. It says: $$\frac{d^2\psi}{dx^2}=\gamma^2\frac{d^2\psi}{d\xi^2}\tag3$$

I don't know how they get this. I tried $$\frac{d^2\psi}{dx^2}=\frac{d^2\psi}{d\xi^2}\frac{d^2\xi}{dx^2}\tag4$$ but I get $$\frac{d^2\xi}{dx^2}=0$$ by using the substitution in equation (1). I'd appreciate it if someone could explain this to me.


Another thing I tried is (suggested by comment below): $$\frac{d}{dx}\left(\frac{d\psi}{dx}\right)=\frac{d}{dx}\left(\frac{d\psi}{d\xi}\frac{d\xi}{dx}\right)=\frac{d^2\psi}{dxd\xi}\frac{d\xi}{dx}+\frac{d\psi}{d\xi}\frac{d^2\xi}{dx^2}\tag5$$

Now since $$\frac{d^2\xi}{dx^2}=0$$ equation (5) reduces to:$$\frac{d}{dx}\left(\frac{d\psi}{dx}\right)=\frac{d}{dx}\left(\frac{d\psi}{d\xi}\frac{d\xi}{dx}\right)=\gamma\frac{d^2\psi}{dxd\xi}$$

Once again, I'm stuck..

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(4) is wrong. You need to use (2) again, but with $d\psi/dx$ in $\psi$'s place. –  Hans Lundmark Nov 28 '12 at 10:33
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1 Answer

Ok, I got it.

$$\gamma\frac{d^2\psi}{dxd\xi}=\gamma\frac{d^2\psi}{d\xi dx}=\gamma \frac{d}{d\xi}\frac{d\psi}{dx}$$

$$\gamma \frac{d}{d\xi}\frac{d\psi}{dx}=\gamma \frac{d}{d\xi}\left(\gamma\frac{d\psi}{d\xi}\right)=\gamma^2\frac{d^2\psi}{d\xi^2}$$

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