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Homework Help: Harmonic Oscillator in Dirac Theory

  1. Apr 25, 2012 #1
    Hello everyone,

    i'm looking for anypaper or such kind of thing that explain the resolution of the harmonic oscillator in the Dirac Theory.

    I have worked with the exact spin symmetry. I feel like a fish out the water and i'm sure that there are lot of bibliography about this area, but i promise i didn't find anything.

    Thanks since now,
    and may the force be with you!
  2. jcsd
  3. Apr 25, 2012 #2
    What do you mean by resolution? Also, you may have more luck getting replies in the Advanced Physics section (if this is related HW), or in the Quantum Physics section (if you simply want a reference).
  4. Apr 25, 2012 #3
    The problem with the term in the title is that "harmonic oscillator" refers to the non-relativistic concept of a potential energy as a function of coordinates, which is manifestly Lorentz non-covariant, and Dirac's equation, which is a fully Lorentz covariant quantity.

    Also, the proper meaning of Dirac's equation is that is non-quantum equation of motion for a Dirac spinor field operator in second quantization. This reinterprets the negative energy solutions as antiparticles. But, the problem essentially becomes a many-particle one.

    If you are only interested in single-particle solutions, then you must project out the antiparticle component from the Dirac 4-spinor, to get the Pauli 2-spinor. The equation it obeys is Pauli equation, which already incorporates the interaction with an external electromagnetic field of a charged fermion.

    Just take a scalar potential:
    \varphi(x) = \frac{\mu^3 \, c^4}{2\, q \, \hbar^3} x^2
    where [itex]q[/itex] is thecharge of the particle, and [itex]\mu[/itex] has a dimension of mass. Of course, this non-relativistic equation is only valid when:
    q \varphi(x) \ll m c^2 \Rightarrow \vert x \vert \ll \sqrt{\frac{2 m}{\mu}} \, \frac{\hbar}{\mu \, c}
    Last edited: Apr 25, 2012
  5. Apr 26, 2012 #4
    Well, this isnt exactly what im looking for, but its given me an idea anyway, so thank you very much for your time and effort.
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