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Dirac wave function

  1. Apr 16, 2015 #1
    Hi, is the wave function that couples to the Dirac equation the same as that which couples to the Schrodinger equation? Thanks.
  2. jcsd
  3. Apr 16, 2015 #2
    No, it's different, the biggest difference is that the Dirac equation has a bispinor as a solution. But generally the Dirac wavefunctions follow relativistic transformation rules, the Schroedinger ones are Galileian.
  4. Apr 16, 2015 #3


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    In addition, it has a completely different physical meaning. It cannot be interpreted easily as a "wave function" like in nonrelativistic physics. The reason is that at relativistic energies, you always can create and destroy particles in scattering processes. The Dirac equations solutions are Dirac-spinor fields. They are best interpreted in their quantized form, leading to relativistic quantum-field theory, because this is the most elegant way to describe particle creation and destruction or, more generally, many-body systems in quantum theory.
  5. Apr 16, 2015 #4
    Yes. Though I didn't want to stress that because single-particle relativistic QM, inconsistent as it is in the fundamental interactions context, has consequences like Klein's paradox that actually show up in solid state physics iirc.
  6. Apr 16, 2015 #5


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    Well, the Klein paradox occurs precisely because of the problems when enforcing a single-particle interpretation!
  7. Apr 16, 2015 #6
  8. Apr 16, 2015 #7
    I don't quite understand why if particles can be created and destroyed in scattering processes at relativistic energies we can't easily interpret it as a wave function? Also by scattering processes are you referring to particle collisions?
  9. Apr 16, 2015 #8


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    If particle numbers are not fixed you have a Fock Space and the QFT formalism.

    Its inevitable when you combine it with relativity that particle numbers are not fixed - see for example section 8.3 of:

    Last edited by a moderator: May 7, 2017
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