- #1
jdstokes
- 523
- 1
Hi all,
I was trying to explain to my girlfriend how you can `derive' Schrodinger's equation using the Planck relation [itex]E = \hbar \omega[/itex], the de Broglie relation [itex]p= \hbar k[/itex] and conservation of energy.
If you assume that the fundamental wavefunction is of the form [itex]\psi = e^{i(kx - \omega t)}[/itex], then it follows easily that the momentum and energy operators have the representation [itex]\hat{p} = -i \hbar d/dx[/itex] and [itex]\hat{E} = i \hbar d/dt[/itex]. Substituting into the operator relation [itex]\hat{E} \psi = \hat{K} \psi + \hat{V} \psi[/itex] then gives the time-dependent Schrodinger equation
[itex] i \frac{\partial \psi }{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi [/itex].
She was getting stuck on the assumption that [itex]\psi = e^{i(kx - \omega t)}[/itex]. She is aware that a classical wave can be represented by [itex]\psi = \sin (kx - \omega t + \phi) = A \sin (kx - \omega t) + B\cos(kx -\omega t)[/itex]. 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 [itex]E = \hbar \omega[/itex], the de Broglie relation [itex]p= \hbar k[/itex] and conservation of energy.
If you assume that the fundamental wavefunction is of the form [itex]\psi = e^{i(kx - \omega t)}[/itex], then it follows easily that the momentum and energy operators have the representation [itex]\hat{p} = -i \hbar d/dx[/itex] and [itex]\hat{E} = i \hbar d/dt[/itex]. Substituting into the operator relation [itex]\hat{E} \psi = \hat{K} \psi + \hat{V} \psi[/itex] then gives the time-dependent Schrodinger equation
[itex] i \frac{\partial \psi }{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi [/itex].
She was getting stuck on the assumption that [itex]\psi = e^{i(kx - \omega t)}[/itex]. She is aware that a classical wave can be represented by [itex]\psi = \sin (kx - \omega t + \phi) = A \sin (kx - \omega t) + B\cos(kx -\omega t)[/itex]. I was not able to convince her of the leap to the oscillating exponential.
Any ideas?
Thanks
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