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Wavefunction of relativistic free particle

  1. Apr 20, 2008 #1
    Could anyone please help with the following, rather unusual, query?

    I know that for spin 0 bosons, the Klein Gordon equation gives solutions that are similar to the solutions of the Schrodinger equation for a non-relativistic free particle, the only difference being that the energy used when calculating the wave frequency (E/h-bar) is the relativistic energy m*c-square*gamma.

    I don't know how the wavefunctions look for spin 1 or spin 1/2 particles. What I would like to know is if there is any kind of particle for which one gets a wavefunction where the frequency is different then m*c-square*gamma/h-bar? Also, is it possible to verify these things experimentaly? I mean, can an experiment get an eigenstate and measure both the energy and the wave frequency? Has that been done for relativistic particles? Have any discrepancies been found? Where could I find such data?

    Thanks!
     
  2. jcsd
  3. Apr 20, 2008 #2
    As far as I know, for any relativistic free particle, the frequency [tex]\omega[/tex] that appears in [tex]e^{-i \omega t}[/tex] when regarding the evolution of a steady state (i.e. an eigenstate of the Hamiltonian operator) must be equal to [tex] \sqrt{m^2c^4+p^2c^4}/\hbar [/tex] because the hamiltonian is the generator of time translations.
    However, the relation between [tex]\omega[/tex] and the wave vector [tex]\vec{k}[/tex] will be spin dependent.

    P.S1 : Unfortunately the word "wave function" is no more appropriate to qualify a state in relativistic quantum theory. Results can be obtained with this point of vue only for the Klein-Gordon equation and for the Dirac equation (i.e. for a spin zero particle and a spin 1/2 particle) but they lead to fondamental problems that necessite a new approach of quantum mechanics ; that is quantum field theory.

    P.S2 : Excuse me for the english, I live in France and as everybody knows french people are pretty bad in english :) .
     
  4. Apr 20, 2008 #3
     
  5. Apr 20, 2008 #4
    Actually wathever the formalism (i mean the "classical" one or the "field" one) any states in QM has to satisfy the equation of evolution :
    [tex]i \hbar \frac{d}{dt} |\psi(t) \rangle = \hat{H} |\psi(t)\rangle [/tex]
    You know that, if [tex]|\psi(t)\rangle [/tex] is an eigenket of [tex]\hat{H}[/tex] so that :
    [tex]\hat{H} |\psi(t)\rangle = \hbar \omega |\psi(t) \rangle [/tex]
    you get obviously :
    [tex] |\psi(t) \rangle = e^{-i \omega t} |\phi \rangle [/tex]
    Thus [tex]\omega[/tex] is directly linked to the eigenvalue of the hamiltonian in QM, in non relativistic QM but also in realtivistic QM.

    So, if there exists an experiment which result is that [tex] \omega [/tex] is different from the energy, then QM is false and we have to find something else...
     
  6. Apr 20, 2008 #5
    For what it's worth, http://www.ensmp.fr/aflb/AFLB-301/aflb301m416.pdf . The experiment was performed with weakly relativistic electrons (80 MeV). The article is unusual, as it was published in 2005 in a rather obscure journal, whereas the experiment seems to have been performed in 1988, and at that time the authors published some of its results in Phys. Rev. B (I have not read that article). So use your judgement.
     
  7. Apr 20, 2008 #6
    For somewhat unrelated reasons, I invite you to write the Klein-Gordon equation in lightcone coordinates. The result is interesting
     
  8. Apr 21, 2008 #7
    Thanks to everyone!

    I_wonder...
     
  9. Apr 22, 2008 #8
    Peskin and Schroeder, An Introduction to Quantum Field Theory, Chap. 3, give a very accessible description of the free-particle solutions of the Dirac equation (the governing equation for relativistic spin-1/2 particles). It's one of the classic texts for relativistic quantum mechanics and field theory, very well written, and I recommend it wholeheartedly!

    One of the subtleties of the Dirac equation (and other relativistic equations) is that it admits both positive and negative-frequency solutions, which correspond to electron and positron wavefunctions respectively.
     
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