Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Dirac Equation vs. Schrodinger Equation

  1. Jul 18, 2016 #1
    The Dirac equation is the more generalized form of the Schrodinger equation and accounts for relativistic effects of particle motion (say an electron) by using a second order derivative for the energy operator. If you have an electron that is moving slowly relative to the speed of light, then you can get away with the Schrodinger equation, which is first order in time.

    This makes no sense to me, Why is it that you need to bump the energy operator up to a second order once you get moving at close to the speed of light? Conversely, why does the energy shift to a first order derivative when the electron slows down. I haven't been able to find an explanation for this.
     
  2. jcsd
  3. Jul 18, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The second order derivative is necessary for Lorentz invariance.

    When you go to the non relativistic limit from the relativistic case (I suggest looking at the Klein-Gordon equation first), you can get rid of an overall phase related to the rest energy (ie, mass). The leading correction is a first order derivative wrt time.
     
  4. Jul 18, 2016 #3
    Thanks for the response, Orodruin, but that pithy explanation doesn't advance my understanding much of the situation. Admittedly, I'm a bit of an amateur in these affairs, hence the "B" rating in the title, so perhaps you or someone else can unpack your response some so I can integrate it into my level of understanding.

    What is that level? Well, here's what I at least think I understand. We have three basic or main wave equations in QM. The famous one is the Schrodinger equation which essentially mimics the classical-physics energy-mass-momentum relationships. Total energy=kinetic energy + potential energy. In the Hamiltonian formulation I think it's H=P^2/2m + V, or something like that.

    So, in the classical formulation, H is first order, and P is second order. So Schrodinger thought to mimic that by introducing the Energy and Momentum operators and, viola, you have the Schrodinger equation.

    What about the relativistic version where we have E^2=P^2+M^2 (setting c to 1)? Well, at first glance we could just replace E and P with the quantum operators that represent both and all should be good. This is the Klein-Gordon equation but it doesn't work for fermions. Why? Who knows, but Dirac's solution was to try and find a solution to Einstein's relativistic formula, E^2=P^2+M^2, but by formulating an equation that saw E as a first order term as in the classical relation rather than the Einsteinian relativistic second order relation. The result of his efforts were the Dirac matrices, of course, and I understand the logic behind those fairly well.

    What I missed, though, was the part where "The second order derivative is necessary for Lorentz invariance." I don't know what this means. I think I understand pretty well SR and Lorentz invariance/co-cariance/transforms, at least as it relates to the 4-vector and d(tau)^2=dt^2-dx^2 variety, but I'm not sure what you mean by "The second order derivative is necessary for Lorentz invariance." Again, I'm an amateur at this so I need it explained in relation to the amateurish understanding of the situation as I've outlined above. Any help would be fantastic!
     
  5. Jul 19, 2016 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    If the wavefunction is a scalar with respect to the Lorentz group, then the derivative operator acting on it should be a scalar, too. Hence it can be only 0th, 2nd, 4th, 6th, etc. order. 0th order is the mass term, 2nd is the D'Alembertian, 4th and etc. are non-physical.
     
  6. Jul 20, 2016 #5

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Umm. The Dirac equation only involves first order derivatives. Are you thinking of the Klein-Gordon equation?
     
  7. Jul 23, 2016 #6
    Thanks for all the replies, but these one sentence responses are not really helping me answer my question. These seem to be more of dictionary definitions of the differences between the various wave functions rather than an explanation of why they are different per the specific concern I had. Which is all fine and good, and I'm glad everyone is giving each other in this thread virtual "high fives" with their "likes," but I want to appeal to someone who can demonstrate their didactic prowess and go beyond the phrase "it has to do with the Lorentz group" to try to explain this to a "B" level inquirer.

    Most of the resources I've looked at to try to glean an answer to this equation do not mention the "Lorentz" group or invariance when they discuss the motivation for the Dirac equation. They just say that Dirac was looking for a first order solution to Einstein's relativistic equation that the Klein-Gordon solution didn't seem to solve. So I have no idea how the "Lorentz group" solves this problem and by simply stating that it does does not help me. So someone please unpack this.




    FF to 16 minutes in:





    Nowhere in these discussions are the statement that Dirac developed his equations to satisfy some Lorentz relationship. So please explain.

    Maybe I'm wrong, but my question seems fairly straightforward and legitimate. Why is an electron moving at a slow speed modeled accurately by a first order time derivative but a fast moving electron can only be modeled by a second order time derivative?
     
  8. Jul 23, 2016 #7
    Ok, so we don't incur a further hiccup with this misinterpretation of my quandary, let me clarify that Dirac's equations involve a first order derivative solution to E^2=p^2+m^2 through the process outlined in the first video I posted above. The Klein-Gordon equation left all the quantities squared. But the Schrodinger equation from the outset unapologetically used a first order time derivative while simultaneously setting the space derivative to p^2/2m. Where was the justification for this? SImply because it fit with the classical energy-momentum relationship? Please, someone give me a brief history of this, it seems so basic I don't understand why someone can't explain it to me.
     
  9. Jul 23, 2016 #8
    Apologies if this is a non answer as i am unqualified but how Dirac came to the final solution is moot.

    The thing to ask is; is it correct in known cases and can it predict stuff.

    Physics is the most creative act of the human mind, the op is asking for someone to explain the creative process, nobody has ever been able to explain creativity.
     
  10. Jul 23, 2016 #9

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    Yes.

    And the relativistic classical energy-momentum relationship leads in the same way to the Klein-Gordon equation and its variants, such as the Dirac equation. From the outset, unapologetically.
     
  11. Jul 24, 2016 #10
    Think of it this way - the Schrödinger equation is of first order with respect to time, but second order with respect to spatial derivatives. This is a problem once you go into the realm of relativistic physics, quite simply because in relativity time and space are on equal footing, so the wave equation must reflect this. Klein-Gordon fixes this ( it is fully Lorentz covariant ), but it does not account for spin. Hence the need for the various other relativistic wave equations.
     
  12. Jul 24, 2016 #11
    Thank you Markus, I think you're the first person who has actually understood what my quandary is. I also agree with your statements above. However, regarding those statements, this again is my central question: Why is time and space on an equal footing in relativistic physics but not in non-relativisitic physics? (To use your phrasing) Perhaps this is a question that doesn't have an answer at the moment, which is fine, but if that's the case, then I wish people would just say that instead of throwing out terms like "Lorentz group/invariance," and my favorite, D'Alembertian :redface:, and not elaborating on it.
     
  13. Jul 24, 2016 #12

    jtbell

    User Avatar

    Staff: Mentor

    Because it works, i.e. leads to predictions that agree with old and new experiments to date. If there were serious experimental evidence otherwise, then physicists (as a community) would eventually change their minds. And there are theorists who consider such possibilities, and experimentalists who look for them. Do a Google search for "Lorentz violations." If something like that turns out to be true, it will be a sure route to a Nobel prize.
     
    Last edited: Jul 24, 2016
  14. Jul 24, 2016 #13

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    The first-order version is an approximation to the second-order version, which should be valid for low energies. I don't think there is anything more complicated to it than that.

    Start with the relativistic energy-momentum relation:

    [itex]E = \sqrt{p^2 c^2 + m^2 c^4}[/itex]

    You can solve for this as a power series in [itex]p[/itex]:

    [itex]E = mc^2 + \frac{1}{2} \frac{p^2}{m} - \frac{1}{8c^2} \frac{p^4}{m^4} + ...[/itex]

    Now, go to quantum mechanics by letting this be an operator equation acting a wave function [itex]\Psi[/itex]:

    [itex]H = mc^2 + \frac{1}{2} \frac{p^2}{m} - \frac{1}{8c^2} \frac{p^4}{m^4} + ...[/itex]

    The full Schrodinger equation is:

    [itex]H \Psi = i \hbar \frac{\partial \Psi}{\partial t}[/itex]

    Then you can define a nonrelativistic hamilton [itex]\mathcal{H}[/itex] to be [itex]H - mc^2[/itex], and similarly an auxiliary nonrelativistic wave function [itex]\psi[/itex] to be [itex]\Psi e^{\frac{i mc^2 t}{\hbar}}[/itex]. Then [itex]\psi[/itex] obeys the equation:

    [itex]\mathcal{H} \psi = i \hbar \frac{\partial \psi}{\partial t}[/itex]

    where [itex]\mathcal{H} = \frac{1}{2} \frac{p^2}{m} - \frac{1}{8c^2} \frac{p^4}{m^4} + ...[/itex].

    If you only keep the lowest-order terms in [itex]\frac{1}{c^2}[/itex], then you have the nonrelativistic Schrodinger equation.
     
  15. Jul 24, 2016 #14

    kith

    User Avatar
    Science Advisor

    In non-relativistic physics, the laws are invariant under Galilean transformations. For example, if a law like [itex]F = ma [/itex] is valid in a certain coordinate system and you rotate your axes in space, the law is also valid in the new coordinate system. The important thing is that no matter how you change your space coordinates, you can always choose your time coordinate independently.

    In relativistic physics, the laws are invariant under Lorentz transformations. These include not only rotations in space but also something similar to rotations in spacetime. So time and space are no longer independent. An equation which involves derivatives of different orders for time and space can't be invariant under such transformations.
     
    Last edited: Jul 24, 2016
  16. Jul 24, 2016 #15

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Another very simple way to rephrase kith's second paragraph is: because special relativity has an invariant speed (which is also the maximum value attainable by matter particles), both space and time become relative, hence one can say they are on equal footing.
     
  17. Jul 24, 2016 #16
    Ok, I get the concept of the "invariant interval" when it comes to world-lines in special relativity. Actually, in a bit of lost history, Einstein initially wanted to call his model the "Invariante" theory, or the theory of invariance, seeing as the minus sign between space and time always kept proper time a constant and the the 4-velocity at M^2C^2, etc:

    http://www.asa3.org/ASA/education/views/invariance.htm

    But, still, this doesn't answer my question. You can make a statement that, at high speeds, both space and time become relative, but that doesn't address the question as to why this is so, and that, at the limit of non-relativistic speeds, a first order time derivative works just fine.

    Maybe stevendaryl has the right answer, maybe it has to do with a Taylor series expansion kind of thing. I'm terrible at math so I'll have to unpack that to see if it satisfies me. It wouldn't be the first time. I was highly rewarded when I worked out the Taylor series expansion of E=MC^2 to find the second order (or was it first?) kinetic energy term and the higher-order adjustments (way cool). Perhaps this is the same deal..
     
  18. Jul 24, 2016 #17
    Because the laws of physics - including your wave equation - are the same in all inertial frames of reference. Hence, the line element must be preserved when you go from one frame into another; this is possible only if you treat time and space equally.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dirac Equation vs. Schrodinger Equation
  1. Spin vs Dirac Equation (Replies: 11)

Loading...