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Derivation of Schrodingers Equation

  1. Oct 27, 2009 #1
    I followed -Concepts of Modern Physics --by Arthur Beiser --- and in the Derivation of Schrödinger Equation, The author says---> (chapter 5.3)--> At speeds small compared to the speed of light, the Total Energy of a Particle is sum of its Kinetic Energy K.E. and Potential Energy U, And Shows that
    Total Energy E = p2/2m + U

    But according to the Theory of Relativity, The total Energy of a particle is sum of its kinetic Energy and Rest mass Energy + (potential Energy, if present)
    So the Relation had to be (for what I think) --->
    Total Energy E = p2/2m + U + m0c2.

    Why is the Author not considering the rest Energy???

    The derivation in the book, is almost same as here --

    Oops, just noticed that the link don't directly take you there. For now, First Click Quantum Mechanics, then Schrödinger's Equation
    Last edited: Oct 27, 2009
  2. jcsd
  3. Oct 27, 2009 #2
    Because he's just stated that he is not concerned with the Relativistic case. For the Relativistic equivalent, you may be interested in the derivation of the Dirac equation.
  4. Oct 28, 2009 #3
    Yes, he considers it as a constant part of the potential energy. A constant part is not influential.
  5. Oct 28, 2009 #4
    Even for non-relativistic (v<<c) cases, The theory of Relativity Implies that every mass have equivalent Rest Energy given by E = mc2 Regardless of the velocity of the Body.
    So I don't see any point that rest Energy Can be neglected for low velocities.
  6. Oct 28, 2009 #5
    Do you mean to say that the Function U(x,t) is actually defined as
    U(x,t) = m0c2 + U'(x,t)

    If yes, then in later derivations of cases - like particle in finite potential well -- He supposes U = 0 in the well. Since m0c2 can't me 0, how can U be?
  7. Oct 28, 2009 #6
    U is a potential energy, so you can add any arbitrary constant to it. In physical processes you will only encounter energy differences anyway.
  8. Oct 28, 2009 #7

    Vanadium 50

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    Look at what you wrote. The Schroedinger Equation is non-relativistic. That means it does not incorporate relativity.
  9. Oct 28, 2009 #8


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    I think the point of the OP is that if you take the small-velocity approximation of the relativistic equation, then there is still the m0 c^2 term, which is absent in the Newtonian formula for total energy.

    The point to understand is, as some pointed out already, that you can add any arbitrary constant to total energy, or to potential energy, and that this will not affect the physically measurable results of your calculations. Normally, this property is assigned to the function "potential energy", but in fact you can put it anywhere you want.

    This is so in classical physics.

    Consider the simple problem of a stone in the gravity field near the earth's surface:

    U = m g h

    KE = m v^2/2

    E = KE + U = m v^2/2 + m g h

    h is the "height" of the stone....

    But measured from what reference surface ? The sea surface ? The ocean bottom ? The roof of your house ? The basement of your house ?

    Answer: it doesn't matter, because if U is calculated with an h, which is measured from the roof of your house, and U' is calculated with h', which is measured from the basement of your house, we will have that h' = h + 10 m (say), which means that U' = U + m g (10 m).

    The last term is constant (doesn't change when the stone changes position). We can always do that, with any potential energy, and hence with any total energy.

    So adding a term m0 c^2 or not to U (or to E) won't change the physics.

    In classical physics, all trajectories will be identical, whether you do your calculation with U or with U'. As all physically measurable results are derived from the trajectories, all physically measurable quantities will be the same, whether you've used U or U'.

    In quantum mechanics too, the physically measurable reasults will be the same.

    HOWEVER. It WILL change the exact mathematical form of your wave function ! It will change the mathematical evolution of the wave function you will find!

    In fact, you will find an extra factor exp( - i t m0 c^2 / hbar) in front of your wave function.

    But every quantity that you will calculate, will contain a psi and a psi-dagger, and this factor will cancel. All physically measurable quantities will hence be the same.
  10. Oct 28, 2009 #9
    Thats what I was trying to tell.

    That's what I was wondering on. Since the mathematics involved in Every Cases will be different if we use U'(x,t) = m0c2 + U(x,t); I was wondering How this could not make any changes.


    That was a Reliefer.
    But as you said the mathematical form of wave function will be different, don't it make difference (though no difference if we calculate observables from this wave function), when the wave function superpose in say, interference.? Does it make no difference everywhere?

    Of course the Answer would be no, (since this is what physicists) have been doing, But I would love to learn why?
  11. Oct 28, 2009 #10
    The effect of adding mc^2 is to superimpose a high frequency oscillation with the frequency [tex]mc^2/\hbar[/tex] (~[tex]8*10^{20}[/tex] Hz) on top of the original solution.

    It makes no difference because rest masses of all electrons are the same and there's no interference.
  12. Oct 28, 2009 #11
    I am talking about the interference of two particles, that we have assigned each, this new type of wave function (in your terms, high frequency superimposed wave function).
  13. Oct 28, 2009 #12


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    Nevertheless, when you calculate the expected result of any physical measurement, the extra constant energy does not make any difference. For example, if you calculate the average position of the particle:

    [tex]<x> = \int {x \Psi^*\Psi dx}[/tex]

    without the extra energy you have a wave function [itex]\Psi_0(x,t)[/itex] and get

    [tex]<x> = \int {x \Psi_0^* \Psi_0 dx}[/tex]

    With the extra energy the wave function is [itex]\Psi_0(x,t)e^{-im_0c^2t/\hbar}[/itex] and you get

    [tex]<x> = \int {x \Psi_0^* e^{+im_0c^2t/\hbar} \Psi_0 e^{-im_0c^2t/\hbar} dx}[/tex]

    [tex]<x> = \int {x \Psi_0^*\Psi_0 dx}[/tex]

    the same as before.
    Last edited: Oct 29, 2009
  14. Oct 28, 2009 #13
    and both wavefunctions would be multiplied by exactly the same high frequency factor, so there won't be any interesting interference effects.
  15. Oct 29, 2009 #14


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    I think you mean

    [tex]<x> = \int {x\Psi^*\Psi dx}[/tex]

    right? ;)
  16. Oct 29, 2009 #15
    So, I have learned that, by taking
    U(x,t) = m0c2 + U'(x,t)
    It makes a lot of difference to equations thereafter but won't make even the slightest difference to answers for observables. It do change the geometry of the wave function, but such that the observables remain the same.

    Thanks to all of you for the help.

    But isn't it unjustis, to ignore m0c2 (even though it won't make any difference to observables) and just say it out loud that
    the total Energy E = K.E. + P.E.
  17. Oct 29, 2009 #16


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    Argh, you're right. :blushing:

    I've fixed it. Fortunately it doesn't affect the main point of that exercise because any expectation value, as well as the plain old integrated probability, works out the same way.
  18. Oct 30, 2009 #17


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    Well, this is a very important lesson, and actually, I would say that it will concern about half of everything you need to learn in physics. It is also one of the hardest things to grasp, and what gives you often the deepest insight once you understand it. You can call it "symmetries".

    What are "symmetries" ? They are changes of the description that don't change the outcome. And as I said, half of physics is about symmetries.

    They are of many different kinds.

    Probably the first symmetry you encountered, was in Euclidean geometry. When applied to physics, it comes down to this: when you solve a problem, you have a free choice of "coordinate system". You can shift it or rotate it. This will of course change the mathematical description of your problem, but it won't change "the outcomes". One of the important arts in problem solving, is to pick the coordinate system that will make the solution mathematically the easiest.

    Another, already more difficult, symmetry to grasp, is that you can pick any reference frame, with any velocity. You can do the calculation in the reference frame of a train, of the rails, of the sun, of the earth.... if you do it correctly, the mathematics will be different, but the outcomes will be the same.

    However, these symmetries are somehow easy to understand, because they correspond to different "observers". But they are not the only kinds of symmetries. Another one we've met is that we can pick our "reference of potential energy" where we want. That's already more abstract, because it doesn't correspond to another observer, another little assistant during experimentation, but rather another little assistant during *calculation*.

    It is connected to the previous "potential energy" shift, but one of the important symmetries in quantum theory is the choice of "phase". People call it also "gauge invariance", but we've only scratched the surface with our constant phase factor. Particle physics is full of these kinds of symmetries.

    General relativity is also based upon symmetries, that is, changes of "calculation" while "keeping the physical outcomes the same".

    As I said, half of physics is the study of symmetries.
  19. Oct 30, 2009 #18
    Thanks again for the wonderful lesson, vanesch!!!
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