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Projection operators and Weyl spinors

  1. Nov 11, 2015 #1

    CAF123

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    I am working through some course notes where the aim is to derive the equations of motion satisfied by the left handed and right handed components of the Dirac spinor ##\psi##. From the Dirac lagrangian, we have $$\mathcal L = \bar \psi (i \not \partial P_L - m P_L)\psi_L + \bar \psi (i \not \partial P_R - m P_R)\psi_R$$ The next line is $$\mathcal L = \bar \psi_L i \not \partial \psi_L - \bar \psi_R m \psi_L + \bar \psi_R i \not \partial \psi_R - \bar \psi_L m \psi_R$$ I understand where the terms involving m come from but I am not sure about the other two. Can anyone help?

    Another question is to do with the acting of the projection operators onto Dirac spinors. Since $$P_L \psi = P_L \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \frac{1}{2}(1-\gamma^5)\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \frac{1}{2} \left( \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \right) \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$$ which does not project out the left handed component. If I use the Weyl representation of ##\gamma^5##, it works but I am trying to understand why not in the Dirac representation of the ##\gamma^5##.

    Thanks!
     
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  3. Nov 11, 2015 #2

    Orodruin

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    What is your problem with them? They follow directly from the form of the derivative terms.

    Because in Dirac representation the field is not represented as
    $$
    \begin{pmatrix}
    \psi_L \\ \psi_R
    \end{pmatrix}
    $$
    with ##\psi_L## and ##\psi_R## being the left- and right-handed components of the Dirac spinor.
     
  4. Nov 11, 2015 #3

    CAF123

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    Hi Orodruin,
    If I write out the leftmost term explicitly then I get, $$i\psi^{\dagger} \gamma^0 \gamma^{\mu} \partial_{\mu} P_L \psi_L$$ I don't see the manipulations that made the adjoint spinor go to a left handed adjoint spinor.

    I see, how do we write the field in Dirac representation? Just like ##\psi = \psi_L + \psi_R##?

    Thanks!
     
  5. Nov 11, 2015 #4

    Orodruin

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    What is the commutation relation of ##\gamma^5## with the other gamma matrices?

    That is true in any representation. In Dirac representation, the components just do not have the same interpretation in terms of left and right handed fields.
     
  6. Nov 12, 2015 #5

    CAF123

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    I think I see: $$\psi^{\dagger} \gamma^0 \gamma^{\mu} \partial_{\mu} (P_L \psi_L)= \psi^{\dagger} \gamma^0 \gamma^{\mu} P_L \partial_{\mu} \psi_L = \frac{1}{2} \psi^{\dagger} \gamma^0 \gamma^{\mu} (1-\gamma^5) \partial_{\mu} \psi_L = \frac{1}{2}\psi^{\dagger} (\gamma^0 \gamma^{\mu} + \gamma^0 \gamma^5 \gamma^{\mu}) \partial_{\mu} \psi_L = \frac{1}{2} \psi^{\dagger} \gamma^0 (1+\gamma^{5}) \gamma^{\mu} \partial_{\mu} \psi_L = \psi^{\dagger} \gamma^0 P_R \gamma^{\mu} \partial_{\mu} \psi_L = \psi^{\dagger} P_L \gamma^0 \gamma^{\mu} \partial_{\mu} \psi_L \Rightarrow \text{result}$$ Is that ok?

    Ok, so in Dirac representation, the spinors are those labelled by ##u(p,s)## and ##v(p,s)##, where e.g $$u(p,s) = N\begin{pmatrix} \phi^s \\ \frac{\sigma \cdot p}{E+m} \phi^s \end{pmatrix}$$ The components are not written in terms of LH and RH fields. Is that what you meant? So when I (incorrectly) applied the Dirac representation of ##\gamma^5## onto the Weyl spinor ##\langle \psi_L, \psi_R \rangle## I was going to get, as a result, a state with the left and right handed fields mixed and thus no notion of left and right handed states in the Dirac representation?

    Thanks!
     
  7. Nov 12, 2015 #6

    Orodruin

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    Yes.

    Yes, this is no stranger than using a rotated representation of SO(2) or any other group. The representations are unitary equivalent, but look different in terms of components.
     
  8. Nov 14, 2015 #7

    CAF123

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    Ok so with the manipulations in the OP understood I get, in the zero mass limit, a lagrangian like $$\mathcal L = i \bar \psi_L \not \partial \psi_L + i \bar \psi_R \not \partial \psi_R.$$ The first term here is supposed to describe a massless fermion with negative helicity. (or equivalently a massless antifermion with positive helicity). I want to see why this is the case.

    The first term gives, for the equation of motion for the LH component of the field, ##\not \partial \psi_L = 0##. The generic solution to the Dirac equation is ##\psi = \omega(p,s) e^{-ip\cdot x}##, where ##\omega## is four component spinor describing particle (u) or antiparticle (v) state, both of positive energy. A negative helicity means that the momenta is always in direction opposite to that of the spin? which by definition is what left handed means. Are these statements correct? Thanks :)
     
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