Seeking a derivation of Schrödinger's wave equation

  • #1
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Main Question or Discussion Point

I am interested in the derivation of Schrödinger’s wave equation from the Klein Gordon equation. I have looked in Penfold’s ‘The Road to Reality’, the open University’s Quantum Mechanics books, Feynman’s lectures, the internet, but not found what I want. Everyone seems to take it as a given, or work back to it from quantum theory. Presumably at least in the early days people looked at how to justify the equation.

I think I have succeeded in constructing a derivation, but I would like to check my work. I think the derivation gives insights that are not obvious in the equation fait accompli. Can anyone point me at an ‘approved’ derivation?
 

Answers and Replies

  • #2
phyzguy
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Isn't it simply the non-relativistic form of the Klein-Gordon equation.? If you replace the momentum
[itex] p_i [/itex] with [itex] i \hbar \frac{\partial}{\partial x_i}[/itex],
then the relativistic energy equation [itex] E^2 = p^2 c^2 + m^2 c^4 [/itex] gives you the Klein Gordon equation, and the energy equation [itex] E = \frac{p^2}{2 m} [/itex] gives you the Schrodinger equation, no?

Edit: Edited out the square in the non-relativistic energy equation.
 
Last edited:
  • #3
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Thanks. But that has a number of steps which take some faith. As an aside, how do I use Latex in the Physics forum? I've poked about but failed to see how.
 
  • #4
stevendaryl
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The nonrelativistic wave function is not the limit of the relativistic wave function:

Let ##\phi## be a solution to ##\hbar^2 (-\frac{1}{c^2} \frac{\partial^2}{\partial t^2} + \nabla^2) \phi = m^2 c^2 \phi##.
Let ##\psi = \phi e^{+i \frac{mc^2}{\hbar} t}##

Then ##\psi## obeys the equation:

##\frac{-\hbar^2}{2m} \nabla^2 \psi = i \hbar \frac{\partial}{\partial t} \psi - \frac{\hbar^2}{2 mc^2} \frac{\partial^2}{\partial t^2} \psi##

That's exact. To get Schrodinger's equation, you just assume that

##|\frac{\partial^2}{\partial t^2} \psi| \ll |2 mc^2 \psi|##

so that the last term on the right is negligible.
 
  • #5
DrClaude
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  • #6
phyzguy
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Thanks. But that has a number of steps which take some faith. As an aside, how do I use Latex in the Physics forum? I've poked about but failed to see how.
But I don't think they did anything as rigorous as what @stevendaryl showed in the early days. I think they weren't having much success with relativistic wave equations, so Schrodinger decided to try a non-relativistic version (based on [itex] E = \frac{p^2}{2m}[/itex]), and found that it worked brilliantly. I have not looked up the original papers, but in Vol3 Ch 16 of "The Feynman Lectures on Physics", Feynman claims:

"When Schrödinger first wrote it down, he gave a kind of derivation based on some heuristic arguments and some brilliant intuitive guesses. Some of the arguments he used were even false, but that does not matter; the only important thing is that the ultimate equation gives a correct description of nature. "
 
  • #8
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But I don't think they did anything as rigorous as what @stevendaryl showed in the early days. I think they weren't having much success with relativistic wave equations, so Schrodinger decided to try a non-relativistic version (based on [itex] E = \frac{p^2}{2m}[/itex]), and found that it worked brilliantly. I have not looked up the original papers, but in Vol3 Ch 16 of "The Feynman Lectures on Physics", Feynman claims:

"When Schrödinger first wrote it down, he gave a kind of derivation based on some heuristic arguments and some brilliant intuitive guesses. Some of the arguments he used were even false, but that does not matter; the only important thing is that the ultimate equation gives a correct description of nature. "
Thanks. That's the impression I've got: particularly the Feynman 'brilliant guesses'. I think it is worth putting it on a solid basis, even though as Feynman wrote, it does not matter because it works.

Thanks for the replies. I'll just press on with my attempt, but continue to monitor here in case anyone can point me to a rigorous derivation.
 
  • #9
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I see this thread has been over from relativity to quantum theory. One of my objections to the ad hock approach to Schrödinger's equation is the common assertion that it does not take account of relativity. As Stevendaryl posts(if I understand him correctly), Schrödingers equation is not non-relativistic, it just contains a low velocity approximation. The notion that the rate of vibration of anything can vary with momentum (and hence with speed) either relies on special relativity, or on an easily disproved assumption of a special frame of reference.
 
  • #10
DrClaude
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Schrödingers equation is not non-relativistic, it just contains a low velocity approximation.
Schrödinger's equation is non-relativistic in the same sense that Newton's equation of motion are non-relativistic. The fact that it can be obtained as a certain limit of a relativistic equation does not make it relativistic in any sense. One expects a good theory not to be superseded by a more general theory, but encompassed by it.

The notion that the rate of vibration of anything can vary with momentum (and hence with speed) either relies on special relativity, or on an easily disproved assumption of a special frame of reference.
What are you referring to here?
 
  • #11
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Schrödinger's equation is non-relativistic in the same sense that Newton's equation of motion are non-relativistic. The fact that it can be obtained as a certain limit of a relativistic equation does not make it relativistic in any sense. One expects a good theory not to be superseded by a more general theory, but encompassed by it.


What are you referring to here?
Sorry, I should have referred to wavelength rather than frequency, then the point I wish to make is more obvious. The de Broglie wavelength of a particle is proportional to the momentum of the particle. This implies special relativity. The frequency follows from de Broglie plus special relativity. The same arguments apply to the Schrödinger's probability waves, and I doubt that was just a happy accident.

Still, I think my original question has been answered: there is no accepted derivation of Schrödinger's equation, it is just presented as a very successful intuition.
 
  • #12
DrClaude
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You might be interested in reading this:
Derivation of the Schrödinger equation from the Hamilton–Jacobi equation in Feynman's path integral formulation of quantum mechanics
J H Field, Eur. J. Phys. 32, 63 (2011) preprint: https://arxiv.org/abs/1204.0653
Abstract said:
It is shown how the time-dependent Schrödinger equation may be simply derived from the dynamical postulate of Feynman's path integral formulation of quantum mechanics and the Hamilton–Jacobi equation of classical mechanics. Schrödinger's own published derivations of quantum wave equations, the first of which was also based on the Hamilton–Jacobi equation, are also reviewed. The derivation of the time-dependent equation is based on an a priori assumption equivalent to Feynman's dynamical postulate. de Broglie's concepts of 'matter waves' and their phase and group velocities are also critically discussed.
 
  • #13
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Thanks.
 

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