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S-matrix structure of the lightest particle

  1. Nov 22, 2014 #1
    If you have a particle that is the lightest particle, then it cannot decay into other particles. As a consequence of this, its T-matrix amplitude on-shell should be zero, since the S-matrix is S=1+iT, and the amplitude for the particle to be found with the same quantum numbers and momentum is 1 since it can't decay, so T=0. The T-matrix in this case should be the self-energy of the particle. So you should always have [itex] \Pi(p^2=m^2)=0 [/itex]. But what if you choose a different renormalization scheme so that [itex] \Pi(p^2=m^2) \neq 0 [/itex]? Then the S-matrix doesn't seem like it can be equal to 1, since this is the lightest particle, so the T-matrix should always be real on-shell, so that S=1+imaginary number, and the amplitude of this is greater than 1.

    Unitarity requires:

    $$
    - \text{Im} \, \Pi (i \leftarrow i)=\sum_n (2 \pi)^4 \delta^4(P_{ni})(\Pi_n \frac{1}{\rho^2})F (n \leftarrow i)^* F (n \leftarrow i)
    $$

    In this formula, [itex]n[/itex] are all on-shell intermediate states of all possible particles, and [itex]\rho [/itex] are normalization factors of these particles, and [itex]F [/itex] are Feynman amplitudes. Since this is the lightest particle and can't decay, only when [itex]n=i[/itex], [itex]F \rightarrow -i \Pi(p^2=m^2)[/itex] does the RHS contribute. This seems to imply that if the self-energy is not equal to zero on-shell, then the RHS is non-zero, so the imaginary part of the self-energy on the LHS can be non-zero on-shell. Is this correct?
     
  2. jcsd
  3. Nov 22, 2014 #2

    vanhees71

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    Of course, you also have to renormalize the wave function. The physical mass of the particle is always defined by the pole of the corresponding single-particle propagator (one-particle Green's function).
     
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