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Breaking Hamiltonian of a particle

  1. Apr 18, 2012 #1
    Dear all,

    I have a fundamental question about breaking the Hamiltonian. Here is the description:

    Suppose a particle, [itex]\lambda^{0}[/itex], is produced in a high energy nuclear collision with proton beam. It is produced by strong interaction, and it has fixed energy (can be obtained from its momentum and mass). Its propagation with time is given by unitary operator.

    [itex]\hat{U}[/itex](t, 0) = exp(-i[itex]\hat{H}[/itex]t)

    The momentum that it got is created by physics involving strong interaction during the collision. So, the Hamiltonian is Strong Hamiltonian, [itex]\hat{H}_{S}[/itex].

    This [itex]\lambda^{0}[/itex] decays via weak interaction. I have seen in books that its Hamiltonian can be break into strong, electromagnetic and weak part, [itex]\hat{H} = \hat{H}_{S} + \hat{H}_{EM} + \hat{H}_{W}[/itex]. How other Hamiltonians come here?

  2. jcsd
  3. Apr 18, 2012 #2


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    The Hamiltonian appearing in your expression is the general Hamiltonian describing all the interactions that influence the evolution of the particle. While it might be created via a strong interaction, it subsequently evolves according to all allowed interactions; for example, it might radiate a Z_0 or photon.
  4. Apr 18, 2012 #3
    Major decay mode of [itex]\lambda[/itex] are [itex]\lambda \rightarrow p \pi^{-} \text{and} ~\lambda \rightarrow n \pi^{0}[/itex]. These are weak decays, and [itex]\hat{H}_{EM}[/itex] and [itex]\hat{H}_{W}[/itex] are reffered in these cases.
  5. Apr 18, 2012 #4


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    Right. And they're included in the full Hamiltonian, correct? I think I'm misunderstanding your question.
  6. Apr 18, 2012 #5


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    Staff: Mentor

    This is just an approximation. The other parts of the full hamiltonian are present during the production, too, but their influence on the production is small as the strong force is much stronger than the electroweak stuff.
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