I have almost no background in physics and I had a question related to Schrodinger's Equation. I think, it is not really research level so feel free to close it, but I would request you to kindly suggest some existing literature which can help me develop a better understanding for the same.
While reading up about it from an introductory text on Quantum Physics, I wondered for a little while how does one derive this equation. Very soon, the author (HC Verma) added the detail that he just took it on faith when he was himself a student. He goes on to say Schrodinger equation rightly predicts the behavior of atomic transitions etc and people believe that it is a fundamental law of nature for quantum systems.
Then he raises the question himself which I wanted to know the answer to. Namely
What made Schrodinger write such an equation which became a fundamental equation, not to be derived from more fundamental equations?
He then adds that it will be an interesting topic for students of history of science which does not answer my question. Could you please try answering this question (or is this question really useless) ?
Thanks for your time
Answer
Schrodinger was following Hamilton, deBroglie and Einstein. DeBroglie had noted that matter waves obeyed a relation between momentum and wavenumber, and energy and frequency,
$$ E = \hbar \omega $$ $$ p = \hbar k $$
For plane waves of the form $\psi(x) = e^{ikx - i \omega t}$, you learn that the $\omega$ and the $k$ of the wave are the energy and the momentum, up to a unit-conversion factor of $\hbar$. Einstein then noted that the DeBrodlie waves will obey the Hamilton Jacobi equation in a semi-classical approximation, and Schrodinger just went about looking for a real wave equation which would reproduce the Hamilton Jacobi equation when you use phases.
But the end result is easier than the Hamilton Jacobi equation. For pure sinusoidal waves, the energy and wavenumber are related by
$$ E = {p^2\over 2m}$$
Which means that the plane wave satisfies the free Schrodinger equation
$$ i\hbar {\partial\over \partial t} \psi = -{\hbar^2 \over 2m} \nabla^2\psi $$
You can check that for a sinusoid, this reproduced the energy/momentum relation.
If there is an additional potential, when the wavelength is short, the wavefronts should follow the changing potential to reproduce Newton's laws. The way this is done is to add the potential in the most obvious way
$$ i\hbar {\partial\over \partial t} \psi = -{\hbar^2\over 2m} \nabla^2 \psi + V(x) \psi $$
When $V(x) = A - F\cdot x$, where A is a constant offset and B is a constant force vector, the local frequency is slowed down in the direction of bigger potential, curving the wavefronts downward according to Newton's laws.
One way of seeing that the equation reproduces Newton's laws comes from Fourier transforms. There is a group-velocity formula for the motion of wavepackets centered at a certain frequency and wavenumber:
$$ {dx\over dt} = {\partial \omega \over \partial k} = {p\over m} $$
This equation comes from the idea of beating--- waves with a common frequency move together, but the location of constructive interference changes according to the derivative of the frequency with respect to the wavenumber. Identifying the freqency with the energy and the wavenumber with the momentum, this relation reproduces one of Hamilton's equations of motion as a law of motion for the wavepacket solutions of Schrodinger's equation (in the limit of short wavelengths).
The other Hamilton equation can be found by Fourier transform, which makes the wavepacket in k become a wavepacket in x, and the group velocity relation becomes the equation for the changing k as a function of time.
$$ {dp\over dt} = - {\partial \omega \over \partial x} = -{\partial V\over \partial x}$$
Schrodinger's equation is really is the first thing you would guess, and there is no need to make Schrodinger's straightforward ideas look intimidating or axiomatic. It is much more transparent than Heisenberg's reasoning of the time, or for that matter, Einstein's.
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