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Pseudoscalar matrix elements

  1. Jun 26, 2013 #1
    Hi all. I have a question. What is the behaviour of the polarization vector of a pseudoscalar particle under a parity transformation??
    Let me explain my problem. I know for sure that the effective matrix element which links a [itex]D^*[/itex] and a [itex]\pi[/itex] can be written as:

    $$
    \langle \pi(p)D^*(q,\lambda) | D^*(k,\eta)\rangle=\frac{g}{M_{D^*}}\epsilon_{\alpha\beta\gamma\delta} \lambda^\alpha \eta^\beta p^\gamma q^\delta,
    $$
    where $g$ is an effective coupling.

    What I am trying to prove is that such a matrix element is (as it must be) a scalar. Now if, for example, we put ourselves in the rest frame of the [itex]\pi[/itex] we have just [itex](\vec{\lambda}\times\vec{\eta})\cdot \vec{q}[/itex]. Is that a scalar function?

    Thank you very much
     
  2. jcsd
  3. Jun 26, 2013 #2

    Bill_K

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    Science Advisor

    Pseudoscalar particles don't have polarization vectors, Einj.
     
  4. Jun 26, 2013 #3
    I am sorry that's clearly a typo. I am talking about the [itex]D^*[/itex], a pseudovector. Sorry again.
     
  5. Jun 26, 2013 #4
    why it should strictly have this form.
     
  6. Jun 26, 2013 #5
    It is a consequence of Heavy Quark Effective Theory. You can take a look, for example, at arXiv:hep-ph/9605342 [hep-ph]. In this article there should be such a result.
     
  7. Jun 26, 2013 #6
  8. Jun 26, 2013 #7
    I am sorry but I wasn't able to find such a lagrangian in the reference, where is it?. However, let me just underline that we are talking of a matrix element between a [itex]\pi[/itex] and two [itex]D^*[/itex], not a [itex]D[/itex] and a [itex]D^*[/itex]. Does your point still hold also for such matrix element??
     
  9. Jun 26, 2013 #8

    Hepth

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    Gold Member

    ##D^* ## is a vector with parity ##P = -1 ##
    ##\pi ## is a pseudoscalar ##P = -1##

    So you have ##(-1)^3 = -1## for your matrix in and out states.

    You have two vectors (three momentum, but not independent) and two pseudo-vectors(polarizations) ##P = +1 ## that you can decompose this matrix element into.

    It should be proportional to the two polarizations for sure, and have no free indices.

    ##\eta_1^{\alpha} \eta_2^{* \beta}##

    now you could just multiply this by the metric to get an invariant but this is ## P = +1 ##. You need a negative parity thing somewhere. If you introduce a momentum you'll have a free index if the polarizations are contracted. So you have a ## P = -1 ## thing but its not invariant.

    ##\eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta}##

    So you use another momenta (assuming some metric multiplications here, any permutation)

    ##\eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta} p_2^{\sigma} ##

    BUT now its positive parity again. Luckily we have a 4-index negative parity pseudo-tensor, the antisymmetric Levi-Civita tensor giving us an object:

    ##\epsilon_{\alpha \beta \delta \sigma} \eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta} p_2^{\sigma} ##

    That has both the parity and lorentz-invariant properties of the amplitude. Throw in a constant out front for good measure.

    For the Lagrangian look into Heavy Meson Chiral Perturbation Theory. the ##\frac{1}{M}## is from the expansion in the heavy quark mass.

    In heavy meson chiral pt the vector and pseudoscalar mesons (D,DSTAR) are grouped into a spin-doublet "H" due to transformation properties.

    You'll find the lagrangian as Eq 40 in http://arxiv.org/abs/hep-ph/9605342

    which is a great overview of HMxPT
     
  10. Jun 27, 2013 #9
    Thank you very much. You have been very clear! :biggrin:
     
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