jdstokes
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Hi all,
I was trying to explain to my girlfriend how you can `derive' Schrodinger's equation using the Planck relation E = \hbar \omega, the de Broglie relation p= \hbar k and conservation of energy.
If you assume that the fundamental wavefunction is of the form \psi = e^{i(kx - \omega t)}, then it follows easily that the momentum and energy operators have the representation \hat{p} = -i \hbar d/dx and \hat{E} = i \hbar d/dt. Substituting into the operator relation \hat{E} \psi = \hat{K} \psi + \hat{V} \psi then gives the time-dependent Schrodinger equation
i \frac{\partial \psi }{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi.
She was getting stuck on the assumption that \psi = e^{i(kx - \omega t)}. She is aware that a classical wave can be represented by \psi = \sin (kx - \omega t + \phi) = A \sin (kx - \omega t) + B\cos(kx -\omega t). I was not able to convince her of the leap to the oscillating exponential.
Any ideas?
Thanks
I was trying to explain to my girlfriend how you can `derive' Schrodinger's equation using the Planck relation E = \hbar \omega, the de Broglie relation p= \hbar k and conservation of energy.
If you assume that the fundamental wavefunction is of the form \psi = e^{i(kx - \omega t)}, then it follows easily that the momentum and energy operators have the representation \hat{p} = -i \hbar d/dx and \hat{E} = i \hbar d/dt. Substituting into the operator relation \hat{E} \psi = \hat{K} \psi + \hat{V} \psi then gives the time-dependent Schrodinger equation
i \frac{\partial \psi }{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi.
She was getting stuck on the assumption that \psi = e^{i(kx - \omega t)}. She is aware that a classical wave can be represented by \psi = \sin (kx - \omega t + \phi) = A \sin (kx - \omega t) + B\cos(kx -\omega t). I was not able to convince her of the leap to the oscillating exponential.
Any ideas?
Thanks
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