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Klein-Gordon field is spin-0: really?

  1. Jul 11, 2011 #1
    Usually in the first sentence of the definition of the Klein Gordon equation is the statement that it describes spin-0 particles.

    Similarly, in the first sentence of the definition of the Dirac equation is the statement that it describes spin-1/2 particles.

    But then comes the bit that got me to write this post -- a few sentences later in that latter definition, you're bound to run into the fact that the Dirac equation is the formal square root of the Klein-Gordon equation, and that therefore any solution to the Dirac equation is also solution of the Klein-Gordon equation (though obviously not vice-versa; [itex]x = y \rightarrow x^{2} = y^{2}[/itex], but [itex]x^{2} = y^{2} !\rightarrow x = y[/itex]).

    So...what gives? How can you say that "the Klein-Gordon equation describes spin-0 particles" when every spin-1/2 particle (or at least all of those described by the Dirac equation) is also a solution?

    Looking at the Klein-Gordon equation, it seems like it should be valid for *any* relativistic particle -- [itex]E^{2} = p^{2} + m^{2}[/itex] is generally true, right?

    My gut is telling me that the answer is that while the Klein-Gordon has many non-scalar-particle solutions, it only provides a *complete* description of spin-0 particles...but that seems like a big enough distinction to warrant at least an asterisk in that ever-present first sentence.

    Also, feel free to school me on spin, field quantization, or anything else that I obviously have only a really shallow understanding of.

  2. jcsd
  3. Jul 12, 2011 #2


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    I think there was a similar topic here on this issue in June, IIRC. Anyways, there's a general theory of <relativistic field equations> first set up in the 1930's (Dirac, Fierz and Pauli). One of its conclusions is that essentially p^2 = m^2 and p^2 = 0, the classical contraints for massive & massless particles, translate into second order partial diff. equations for fields. So essentially in the PDE

    [tex] \left(\partial_{\mu}\partial^{\mu} + m ^2\right) \mbox{object}(x,t) = 0 [/tex]

    the <object> can be any spin field (0,1/2,1,3/2) for a fundamental particle of mass m.

    Thus, in the statement you mention below there's a mistake by omission. Other fields, too, obey the K-G equation.

    <Usually in the first sentence of the definition of the Klein Gordon equation is the statement that it describes spin-0 particles.>
  4. Jul 12, 2011 #3


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    If the "object" in the dextercioby´s equation is a one-component object, THEN this Klein-Gordon equation describes a spin-0 particle. Originally, Klein and Gordon assumed that this object IS a one-component object, which is why it is said that their equation describes a spin-0 particle.
  5. Jul 12, 2011 #4
    Both very helpful! I get it. So it's the number of components of the wavefunction.

    Now since you got that one so quickly :devil:, I guess that brings up another question -- is there a first-order equation that describes spin-0 particles? It seems equally weird to me that the formal square root of the K-G equation describes only multi-component fields. I totally get that the way the Dirac equation achieves that square root necessitates that it operate on a multi-component field, but it still seems like there should be some other way.

    And while we're at it, by analogy with putting multi-component wavefunctions into the K-G equation, does the Dirac equation also then describe higher-spin fields, e.g. with a higher-dimensional representation of the Dirac matrices? I realize that there's no lower-dimensional representation that would let you do scalars (since scalars commute), so I guess this is sort of a companion to the prior question.
  6. Jul 12, 2011 #5


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    I find it difficult to summerize, but the issues raised in your post find an answer in the wonderful book by Fushchych & Nikitin <The symmetries of equations of Quantum Mechanics>, especially Chapter 2, Sections 6 & 8, anyway from page 75 onwards.

    Ammendment to post 2: Even spin 2 can be brought to KG form (but necessarily with mass 0), as the field equations for gravitational waves prove it.
  7. Jul 12, 2011 #6
    Awesome, thanks for the recommend. I just got some Amazon credit, so if it's as interesting as it sounds I'll probably be flipping through it soon.

    Correction: Looks like it's out-of-print, and no one is selling for any price on Amazon. sigh.

    The way I read the post (and my gut feeling since it seems to be a simple re-statement of the relativistic kinetic energy relation) I took it to mean that anything would need to satisfy the Klein-Gordon equation, since not doing so would mean violation of relativity (off-shell virtual quanta or whatever obviously excluded -- though the reasons for that are well over my head).
    Last edited: Jul 12, 2011
  8. Jul 13, 2011 #7


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    That is the right understanding, indeed. Poincare invariant field equations obey special relativity principles.
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