What inspired Schrödinger to derive his equation?

Question

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 1

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.

Answer 2

Schrödinger's derivation is a good guess since from the well-known double-slit experiment we know that electrons behave just like photons that can behave like waves and particles. We also know that the distribution of photons is proportional to the intensity of the electromagnetic field and the superposition principle explains the intensity pattern. For photons we have $E(x)=Ae^{iT(x)/\omega}$, where $E(x)$ is the electric field at $x$, $A$ is an amplitude, $\omega$ is a constant with dimension of time and $T(x)$ is the time for a wave to propagate from the source to point $x$. Intensity of the electric field is $|E(x)|^2$.

From geometric optic we know that ray of light is the path that can be traversed in the least time $T$. From mechanics we know that particle travels the path of least action $S$. With these ideas we can assume that the field of electron is $\psi(x,t)=Ae^{iS(x,t)/\hbar}$, where $\psi(x,t)$ is the electron field at $x$, $A$ is an amplitude, $\hbar$ is a constant with dimension of action and $S(x,t)$ is the action for a path from the source to point $x$. Intensity of the electron field is $|\psi(x,t)|^2$ and it is the probability density for the location of the electron. We also know that the action obeys the Hamilton–Jacobi equation that in this case takes the form $S_t+S^2_x/2m+V=0$. When we insert $\psi(x,t)$ into the previous equation we have - voilà - the Schrödinger equation with the statistical interpretation.

The very basis for previous "derivation" is the correspondence of electrons and photons in the double-slit experiment.

Answer 3

As Wigner pointed out quantum mechanics is about Fourier transforms and relativity is about Lorentz transforms. When a plane wave function is used in the Schrödinger equation it is a Fourier basis function and you therefore get the Fourier transform of the differential equation. If the nonrelativistic energy conservation equation is multiplied by a function of the Fourier variables Φ(E,p) and then inverse transformed you get the Schrödinger equation returned..

Also if you look at fiber-optic equations you see that They are also equivalent to the paraxial Schrödinger equations for optics (photons). The paraxial equations allow for transverse coupling between waveguides and the same thing happens between atoms and molecules. The atomic and molecular orbitals are much more sophisticated than fiber optics or quantum dots and point to the sophistication of molecular electronics and explaining living structures. See the paper by Man’ko: Analogs of time-dependent quantum phenomena in optical fibers_Margarita A Man’ko 2007, IOP Science. Also see the Schrödinger wave equation derivation in section 4.4 of George and David Beard's QUANTUM MECHANICS WITH APPLICATIONS; actually a 1970 edition.

Answer 4

Schroedinger equation is derivde from HAMILTON JACOBI EQUATION , i thinks that Hamilton had a similar idea comparing his equation to the Eikonal equation

$$ |\nabla S | =n(r) $$