# Quantum harmonic oscillator 1d

## Main Question or Discussion Point

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.

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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?