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

Klein Gordon equation and particles with spin

  1. Dec 31, 2011 #1
    I am a newbie to QM. Why can't the Klein Gordon equation be used to describe particles with spin?

    Thanks
     
  2. jcsd
  3. Dec 31, 2011 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The Klein-Gordan equation [itex] (\Box -m^2) \phi (\vec{x},t)= 0[/itex] involves a single function on spacetime [itex]\phi(\vec{x},t)[/itex]. A particle with nonzero spin is actually described by a collection of such functions [itex]\phi_i(\vec{x},t)[/itex]. It turns out that each of the [itex]\phi_i[/itex] must actually satisfy the KG equation on its own, but the requirement that the collection describes a particle of definite spin under the Lorentz group amounts to additional, more restrictive conditions, such as the Dirac equation for spin 1/2 particles.
     
  4. Dec 31, 2011 #3

    tom.stoer

    User Avatar
    Science Advisor

    Spin is something that is related to SU(2) matrices acting on two-spinors in non-rel. QM (Pauli equation). Based on SR and the symmetries of spacetime one can relate spin to SO(3,1) ~ Spin(4) (Dirac equation). So essentially spin always requires a spinor (a multi-component object) whereas the Klein-Gordon equation deals with a single wave function describng a scalar.
     
  5. Dec 31, 2011 #4
    This statement may be too strong. Indeed, in a general case, the Dirac equation is equivalent to a fourth-order partial differential equation for just one complex function. Furthermore, the latter can be made real by a gauge transform (at least locally). Source: http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in the Journal of Mathematical Physics).
     
  6. Dec 31, 2011 #5

    tom.stoer

    User Avatar
    Science Advisor

    This seems to be rather exotic is is definitly confusing for a beginner.

    How does one represent spin and rotations? How does one represent helicity and chirality?
     
  7. Dec 31, 2011 #6
    Why is it exotic? Indeed, the physics is the same as for the Dirac equation. It's just a new form of the Dirac equation, which may have its strong and weak points. But it is usually good to know different forms of the same theory, as some of them may be better suited for a specific problem.
    I don't know what I can do about that. It is definitely confusing for me as well, although I'm not exactly a beginner. However, OP asked:
    And it seems that the correct answer is "Actually, after some significant modifications, it can." I believe that might be interesting for OP, judging by the question (s)he asked. And it may be interesting for other readers, and maybe even for you:-) So I believe my information is relevant and appropriate here.

    I don' have a clear idea right now. But as the equation is generally equivalent to the Dirac equation, there is no doubt it does adequately describe spin.
     
  8. Jan 1, 2012 #7
    it may be worth mentioning that just as the 1-component non relativistic schrodinger equation can be tweaked to accommodate spin by adding on a [itex] \sigma . {\bf B} [/itex] type term, so too can the klein-gordon be modified with a [itex] \sigma . ({\bf B} + \imath{\bf E} ) [/itex] term i.e:
    [tex] (\Box + \sigma . ({\bf B} + \imath{\bf E} ) )\psi = m^2 \psi [/tex]
    which is actually just the dirac equation massaged into 2-component quadratic form (assuming that is, the e-m field is constant)

    i am only too well aware i am not explaining this well! and i cannot find a handy on-line reference either. pp101-2 of schweber introduction to relativistic quantum field theory does discusses this though, but perhaps someone else can jump in and explain it a lot better?
     
  9. Jan 2, 2012 #8

    tom.stoer

    User Avatar
    Science Advisor

    I hope I am not too harsh, but the original question was, why the Klein-Gordon-equation cannot explain spin. I think that has been answered in posts #2 and #3.

    The Klein-Gordon-equation does not contain enough degrees of freedom to describe spin. The paper presented by akhmeteli is rather exotic and I doubt that along these lines something like QFT can be constructed; and even this paper has to start with the Dirac equation. Of course one can introduce spin in some way into scalar equations (the well known Pauli-equation is the best example) but this means a) modifying the equation and b) guesswork b/c not everything can be constructed rigorously (g-factor) w/o referring to the original Dirac equation.

    So it is correct that other equations are able to describe certain aspects of spin, but this is not an explanation for its origin rooted in the spacetime symmetry.
     
  10. Jan 2, 2012 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The question is ill-posed. It's incomplete. <Why can't the Klein Gordon equation be used to describe particles with spin?> should have been put as <Why can't the Klein Gordon equation for a scalar field be used to describe particles with spin?> and the answer would be: 'It can. The scalar field describes (non-composite) particles with spin 0'.
     
  11. Jan 2, 2012 #10
    In my post 4 in this thread, I just commented on one specific statement of post 3. For reasons outlined in post 6, I still believe the comment was appropriate.
     
  12. Jan 2, 2012 #11

    tom.stoer

    User Avatar
    Science Advisor

    the reduction described in the article still relates to the original Dirac spinor; the effective 4th order equation alone is not able to describe rotations in spinor-space as far as I can see; it's like the Kepler problem: a certain solution can be described using a special ellipsis, but in order to describe the whole SO(3) symmetry you still need the full setup in R³

    anyway, I doubt that anybody here will agree that for spin 1/2, 1, ... a single wave equation for a scalar field is sufficient
     
  13. Jan 2, 2012 #12
    No matter how true or false this may be, as far as I can see, this has nothing to do with my statement: "spin does not always require a spinor (a multi-component object)", because in a general case there is a one-to-one mapping between the entire set of solutions of the Dirac equation and the set of solutions of the fourth-order PDE of the article. So these two equations contain pretty much the same information about spin. Then why does spin always require a multi-component object?

    Again, I fail to see how this is relevant, as, on the one hand, there is a one-to-one mapping between the entire set of solutions of the Dirac equation and the set of solutions of the fourth-order PDE in the article, so it is not just about "a certain solution", on the other hand, I said nothing about describing any symmetry. Again, I commented just on one statement. For example, I specifically wrote in the article: "Presenting the above results in a more symmetric form is beyond the scope of this work."

    I am afraid I have several issues with this phrase, but for now let me just say that it is not quite relevant, as, strictly speaking, I did not say that "for spin 1/2, 1, ... a single wave equation for a scalar field is sufficient", as I did not discuss "a scalar field" at all, I discussed "a one-component field", and there is a difference in this case, as this "one-component field" has more complex transformation properties than the scalar field.
     
  14. Jan 3, 2012 #13

    tom.stoer

    User Avatar
    Science Advisor

    Please show me how your single wave function transforms w.r.t a general Lorentz transformation; what is the transformation corresponding to

    [tex]\Lambda_\mu^\nu \to S(\Lambda)[/tex]

    [tex]x^\prime = \Lambda x;\;\psi^\prime(x^\prime) = S(\Lambda)\,\psi(x)[/tex]
     
  15. Jan 3, 2012 #14
    I have not written it out, and I am not going to do it now, but the transformation couples the wave function with its derivatives. To get the formula, you just need to substitute the expressions (via the first component) from my article for the three algebraically eliminated components into the above standard transformation formula for the Dirac spinor. This does not change the fact that the Dirac equation is generally equivalent to the fourth order PDE for just one component, so the latter PDE has all the physical content of the Dirac equation.
     
  16. Jan 3, 2012 #15

    tom.stoer

    User Avatar
    Science Advisor

    Your claim is that a single wave function is able to fully describe spin 1/2 fields including its Lorentz transformation properties, so there is a burden of proof on you - not me ;-)
     
  17. Jan 3, 2012 #16
    I respectfully disagree. I did not claim that. I did not even mention Lorentz transformation properties until recently. All I claimed was that "spin does not always require a spinor (a multi-component object)" and that the equation "does adequately describe spin", and this is correct. Otherwise one can say that, e.g., equations of electrodynamics become inadequate as soon as one fixes the gauge, as gauge symmetry disappears in that case. However, all the physical contents is still there.
     
  18. Jan 3, 2012 #17

    tom.stoer

    User Avatar
    Science Advisor

    Sorry, this is totally confusing!

    You claim that "spin does not always require a spinor and that the equation does adequately describe spin". I interpreted this as "a single wave function is able to fully describe spin 1/2 fields including its Lorentz transformation properties". Now you disagree? So your equation does not describe spin 1/2 fields including Lorentz transformation? What else do you then mean by "the equation does adequately describe spin"? How can your equation describe spin if it can't describe the Lorentz transformation of spin?

    btw.: gauge symmetry is an 'unphysical symmetry' whereas Lorentz invariance (in SR) is a physical symmetry; therefore fixing a gauge is not the same as reducing e.g. rotational symmetry; fixing a gauge does not reduce the physical content of a theory whereas reducing e.g. rotational symmetry explicitly excludes physical solutions.
     
  19. Jan 3, 2012 #18
    With all due respect, I reject any responsibility for your arbitrary interpretations.

    I am not ready to answer this question, as the answer depends on definitions, and I don't want to start any flame. I stand by my claims, but that does not mean I am under obligation to take any unrelated challenge.

    As I explained, the equation adequately describes spin as it shares all the contents of the Dirac equation (I mentioned the one-to-one correspondence between the sets of solutions). I believe that if I can correctly describe spin in just one reference frame, that means I do adequately describe it.

    Furthermore, strictly speaking, transforms to different reference frames are not defined unambiguously, for example, in principle, you can describe these transforms using stationary aether and Lorentz contractions, as it was done before Einstein - note that all experiments can be adequately described in this way.

    Another example: you can assume that the Dirac spinor does not change at all under Lorentz transforms, but the gamma-matrices do transform (it would take me quite some time to find the reference though).

    Yet another example: the Pauli equation can't describe the Lorentz transformation of spin, but it does describe spin to a limited extent.

    But generally no physical solutions are excluded in the fourth-order PDE (remember, there is a one-to-one correspondence between the sets of solutions).
     
  20. Jan 3, 2012 #19

    tom.stoer

    User Avatar
    Science Advisor

    b/c you are
    "so your equation does not describe spin 1/2 fields including Lorentz transformation?"

    I think that your work is of little relevance for the explanation of the nature spin in the context of spinor or Dirac theory. I still think that it is rather exotic and confusing for a beginner.

    You (or we ;-) should try to find how to relate your equations to the standard Lorentz transformation for spinors. That's definitly required in order to support your claims.
     
    Last edited: Jan 3, 2012
  21. Jan 3, 2012 #20
    With all due respect, I reject the statement about "definitely required" as arbitrary and baseless.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Klein Gordon equation and particles with spin
  1. Klein-Gordon equation (Replies: 6)

Loading...