# Quantum harmonic oscillator 1d

Ok, so im trying understand how to derive the following version of the Schrodinger Equation for QHO:

$$\frac{d^2u}{dz^2} + (2\epsilon-z^2)u=0$$

where

$$\ 1. z=(\frac{m\omega}{hbar})^{1/2}x$$ and

$$\ 2. \epsilon= \frac{E}{hbar\omega}$$

I've started with the TISE, used a potential of V(x)=1/2mw^2x^2, and with a little rearranging have the following:

$$\ 3. \frac{d^2u}{dx^2} + \frac{2mE}{hbar^2}u - \frac{m^2\omega^2x^2}{hbar^2}u = 0$$

Using equation 1 we have:

$$\ 4. \frac{d^2u}{dx^2} = \frac{m\omega}{hbar}\frac{d^2u}{dz^2}$$

and

$$\ 5.x^2=\frac{hbar}{m\omega}z^2$$

Plugging 4 and 5 into equation 3 and rearranging gives the required answer.

I can follow all of the steps, but get stuck on how to acquire the equation 4. Any help would be much appreciated.

SpectraCat
Ok, so im trying understand how to derive the following version of the Schrodinger Equation for QHO:

$$\frac{d^2u}{dz^2} + (2\epsilon-z^2)u=0$$

where

$$\ 1. z=(\frac{m\omega}{hbar})^{1/2}x$$ and

$$\ 2. \epsilon= \frac{E}{hbar\omega}$$

I've started with the TISE, used a potential of V(x)=1/2mw^2x^2, and with a little rearranging have the following:

$$\ 3. \frac{d^2u}{dx^2} + \frac{2mE}{hbar^2}u - \frac{m^2\omega^2x^2}{hbar^2}u = 0$$

Using equation 1 we have:

$$\ 4. \frac{d^2u}{dx^2} = \frac{m\omega}{hbar}\frac{d^2u}{dz^2}$$

and

$$\ 5.x^2=\frac{hbar}{m\omega}z^2$$

Plugging 4 and 5 into equation 3 and rearranging gives the required answer.

I can follow all of the steps, but get stuck on how to acquire the equation 4. Any help would be much appreciated.

Chain rule for derivatives: for f(v) and v(y), df/dy=(df/dv)(dv/dy) (the variables have been renamed to protect the innocent ).

Is that enough of a hint, or do you need more help?