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I was wondering if anyone knows the origins of how Schrodinger arrived to his equation? And can it be derived from Newtonian mechanics? How did Schrodinger form the equation out of his MIND?

I also I was asking because I was playing around with Newtonian mechanics, I believe I derived the Schrodinger equation...

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  • $\begingroup$ Try searching for "derivation schrodinger equation" and you'll get rather a lot of information. You'll even find some answers on StackExchange listed. $\endgroup$ – StephenG Nov 14 '18 at 19:20
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    $\begingroup$ In the future, it would be helpful to us who volunteer our time here to have a more descriptive title. $\endgroup$ – garyp Nov 14 '18 at 19:34
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    $\begingroup$ physics.stackexchange.com/q/69982/25851 $\endgroup$ – bolbteppa Nov 14 '18 at 19:36
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    $\begingroup$ Just read Schrödinger's original papers (4). $\endgroup$ – DanielC Nov 14 '18 at 21:00
  • $\begingroup$ no check again. I'm pretty sure I derived it from Newtonian mechanics, but I wasn't trying to. I was just trying to make my HW easier to work with... $\endgroup$ – Only Lauren Nov 15 '18 at 17:25
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It can be derived from the classical wave equation for photons and the de Broglie wavelength. Suppose you have an electromagnetic wave $$\psi = Ae^{i (\mathbf{k} \cdot \mathbf{r} - \omega t)}.$$ Taking the spatial derivative yields $$\nabla^2 \psi = - k^2 \psi.$$ Since $\hbar k = p$, $$-\frac{\hbar^2}{2m} \nabla^2 \psi = \frac{p^2}{2m}\psi.$$ Taking the time derivative yields $$\frac{\partial \psi}{\partial t} = -i \omega \psi.$$ Since $\hbar \omega = E$, $$ i \hbar \frac{\partial \psi}{\partial t} = E \psi.$$ The energy is $$E = \frac{p^2}{2m} + V.$$ Hence $$i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + V\psi.$$

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    $\begingroup$ Saying "it can be derived from the classical wave equation for photons" is misleading; it can be motivated via the route above, but the Schrodinger equation cannot truly be derived from classical mechanics. Quantum mechanics is not contained within classical mechanics. $\endgroup$ – Grayscale Nov 15 '18 at 0:36
  • $\begingroup$ @Grayscale fair point. You need to use the de Broglie wavelength to arrive at the final result. $\endgroup$ – user110971 Nov 15 '18 at 0:48

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