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My derivation of the Shrodinger Equation

  1. Jul 1, 2011 #1
    I wasn't sure where the best place to put it was so if this is not it, feel free to move it.
    Anyway, I came up with my own derivation for Shrodinger's equation. Unlike the normal way it is derived, assuming E=h-bar * omega, and the solutions are complex plane waves, I made a few different assumptions.

    I assumed the solutions were waves (ie. they solve the wave equaiton), although I do not specify what form of wave they are. I also assumed time was imaginary. The motivation here was Minkowski space where you can interpret time to be i*c*t. The last assumption was that all angular momentum was quantized as the reduced plank constant.

    Writing out the whole derivation here would be cumbersome so I threw it into latex and exported it as a pdf:
    View attachment shrodinger_only.pdf

    I am wondering if my derivation is novel, or is it equivalent to how it is already derived (like on wikipedia), or even correct? I can't seem to find something wrong with it, nor have I seen the equation derived this way before. I suppose you could argue the assumptions are hand wavy.
     
  2. jcsd
  3. Jul 1, 2011 #2
    Re: My derivation of the Schrödinger Equation

    I do not fully understand why you did not deal directly with the Dirac equation ?
    It is relativistic from the origin.
    So it gives the right intrinsic frequencies, spatial and temporal.
    It gives the negative energy and negative frequency components, so it is natural to understand how an electron which has travelled by both sides of a solenoid (Aharanov-Bohm experments, or so) can converge on a final absorber much smaller than the solenoid.
    The Schrödinger-Dirac equidistance is exactly the one required for the Bragg law in a Compton diffusion : http://deonto-ethique.eu/quantic/index.php?title=Calcul_diffusion_Compton_et_Zitterbewegung" [Broken]

    So why sticking on a non-relativistic approximation ? Where any frequency is arbitrary... It is unphysical.
     
    Last edited by a moderator: May 5, 2017
  4. Jul 1, 2011 #3

    SpectraCat

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    I think your "derivation" is incorrect for the following reasons.

    1) The Schrodinger equation cannot be derived within non-relativistic QM. It is a postulate of non-relativistic QM.

    2) You seem to be missing the fact that angular momentum is an OPERATOR in QM ... so you cannot simply substitute L=hbar in any meaningful physical context.

    3) even in classical physics, the expression p=L/r is meaningless. P, L and r are VECTOR quantities.
     
  5. Jul 1, 2011 #4
    Certainly not.
    An angular momentum has none of the vectorial symmetries. It has the symmetries of an antisymmetric tensor.
     
  6. Jul 1, 2011 #5

    SpectraCat

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    Sure, but my point was that it isn't a scalar. Furthermore, the relation [itex]\vec{L}=\vec{r}\times\vec{p}[/itex] (i.e. vector cross product) holds in non-relativistic CM, and that was really what I was getting at in my criticism of the OP's "derivation", since his derivation required that [itex]\hat{p}=\frac{L}{r}[/itex] (hjis notation).
     
  7. Jul 1, 2011 #6

    Matterwave

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    This is just being pedantic. Sure L is technically a "pseudo-vector"...but how does that help the discussion in any way?

    Additionally, if we are being pedantic here, I wouldn't say that L has "none" of the vectorial symmetries, it certainly transforms correctly under rotations. It only behaves differently under a reflection.
     
  8. Jul 1, 2011 #7
    If I answer, it will overflow the thread.
    I could redirect you where the lecture is written, but it is in french.
    Of course, to throw loudly "antisymmetric tensor of rank two" in a workshop is a good way to find yourself hanged by the collar at the cloakstand... It is necessary to have a short and clear word for that, usable in the workshops, so that they can reason efficiently with their hands. In english, this task is not yet done, is still to do. If you are a good philologist, please do.

    If you wish to deepen this point, please open a new thread in the appropriate forum.
     
    Last edited: Jul 1, 2011
  9. Jul 1, 2011 #8
    You all are correct about how I go about working with the momentum, I have overlooked the cross product for the angular momentum [itex]\vec{L}=\vec{r}\times \vec{p}[/itex] but in the Shrodinger Equation, the momentum is squared so we might as well concern ourselves with the norm of these vectors so:

    [itex]|L|=|r||p|sin(\theta)[/itex] where [itex]\theta=\frac{\pi}{2} [/itex] to confine everything to a circle so [itex]|L|=|r||p|[/itex]. Square everything to get [itex]L^2=r^2 p^2[/itex] and rearrange to get [itex]p^2=\frac{L^2}{r^2}[/itex] so the end result is the same.
     
  10. Jul 1, 2011 #9

    Matterwave

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    You cannot arbitrarily choose your angle to be pi/2. That's like saying the momentum and the position of a particle is always perpendicular to each other. That's not true in general. Particles don't always move in circles...
     
  11. Jul 1, 2011 #10
    Wouldn't the fact that angular momentum is quantized imply something is moving in a circle thus [itex]\theta={\pi \over 2}[/itex]?
     
  12. Jul 1, 2011 #11

    Matterwave

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    What about particles in free space? Why would they tend to move in circles?

    You want to use some sort of classical reasoning, and then "quantize" to get your Q.M. correspondence right?
     
  13. Jul 1, 2011 #12
    No. There, you are projecting our macrophysical world onto microphysics, and this is viciously wrong. Please accept that new things are to be learned, far from your familiar macroscopic experience.
     
  14. Jul 1, 2011 #13
    Your derivation amounts to this:

    let's assume some arbitrary facts and mix it with some physics equations.

    Schrödinger was a master of hamilton-jacobi theory, you are a master of meaningless arrangements of some equations.
     
  15. Jul 1, 2011 #14
    To be a little more clear. I chose [itex]\theta={\pi \over 2}[/itex] because in order for [itex]\hbar[/itex] to be constant, the "thing" that is rotating must be going in a circle at a constant velocity. The velocity vector would be perpendicular to the position vector r, thus [itex]\theta={\pi \over 2}[/itex].
     
  16. Jul 1, 2011 #15

    Ken G

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    To my knowledge, Schroedinger himself never "derived" the Schroedinger equation, he just "arrived at" it, using classical analogs of the behavior he knew must be a solution (a plane wave), and replacing omega by a time derivative to generate a dynamical equation. The only test of the equation was not how it was "derived", it was simply that it worked, once it was arrived at. This means that triclon's approach would have been perfectly successful had he done it a week before Schroedinger, but remember, Schroedinger did not have access to hindsight-- he wasn't looking for the path to his equation, he was looking for his equation. So it's not really possible to compare the feats, since neither one is actually a derivation.
     
  17. Jul 1, 2011 #16

    Matterwave

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    I don't understand your reasoning here. Angular momentum conservation certainly does not require circular motion. Elliptical orbits, and even parabolic and hyperbolic orbits conserve angular momentum. In fact, a straight line "orbit" can even conserve angular momentum. A free particle traveling in a straight line certainly conserves angular momentum (around any arbitrary origin).

    For an impact parameter b, and a particle traveling in a straight line with constant momentum:

    [tex]\vec{L}=\vec{r}\times\vec{p} \rightarrow |\vec{L}|=|\vec{r}||\vec{p}|\sin\theta[/tex]

    Now for simplicity, I denote magnitudes by neglecting the vector sign (instead of having to put the bars around them all the time):
    [tex]b=r\sin\theta \rightarrow L=pb[/tex]

    Now note that even though theta varied through the whole "orbit", the angular momentum is nevertheless conserved because the impact parameter is a constant.

    Therefore, I see no merit in your assertion that theta=pi/2 always.
     
  18. Jul 2, 2011 #17
    An interesting side note is that if you replace E in the general wave function with c then you get a nice simple 2nd order non-complex wave equation which provides accurate results with the exception that you must work in per-unit mass since mass is dropped from the function.

    Of course E does not equal c so what kind of wave this is, I don't know. However the result of plugging the new wave function into the non-complex 2nd order wave equation does satisfy the per unit mass solution of E = c^2

    Just a thought.
     
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