- #1
rooney123
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Ok, so I am trying understand how to derive the following version of the Schrödinger 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.
\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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