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Dirac and Majorana spinors for neutrinos

  1. Mar 14, 2015 #1
    Dirac description
    If I well understood a Dirac description for fermions is :
    ##\Psi_{D}=\Psi_{L}+\Psi_{R}## where ##\Psi_{L}## is the left-chiral spinor and ##\Psi_{R}## the right-chiral spinor.
    Each spinor, ##\Psi_{L} ## and ##\Psi_{R}## has 2 components cotrresponding to the particle and antiparticle :
    Q1 : Can we write ##\Psi_{L}=(\nu_{L},\bar{\nu}_{R}) ##? and ##\Psi_{R}=(\nu_{R},\bar{\nu}_{L})## ?

    Majorana description
    The Majorana condition is ##\Psi_{L}=\Psi_{L}^{c}## and ##\Psi_{R}=\Psi_{R}^{c}##.
    Q2: is it right ?
    If yes, ##\Psi_{M}=\Psi_{L}+\Psi_{R}## can be written as ##\Psi_{M}=\Psi_{L}+\Psi_{R}^{c}=\Psi_{L}+(\Psi_{L})^{c}##
    with ##\Psi_{L}=(\nu_{L},\bar{\nu}_{R}) ## and ##(\Psi_{L})^{c}=((\nu_{L})^{c},(\bar{\nu}_{R})^{c}) =(\nu_{R}^{c},\bar{\nu}_{L}^{c})=(\nu_{R},\bar{\nu}_{L})##
    So the Majorana field describes the 4 states of the neutrino (##\nu_{L},\bar{\nu}_{R},\nu_{R},\bar{\nu}_{L}##)
    Q3: In such notation what is the difference between ##\nu_{L}^{c}## and ##\bar{\nu}_{L} ## ? Majorana condition ##\nu_{L}^{c} = \nu_{L}## can be also written as : ##\bar{\nu}_{L} = \nu_{L}## ???

    I realise I am lost between antiparticle notation ##\bar{\nu}## and charge conjugate ##\nu^{c}##
    Can you help me ?
     
    Last edited: Mar 14, 2015
  2. jcsd
  3. Mar 16, 2015 #2
  4. Mar 16, 2015 #3

    Avodyne

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    Your questions cannot be answered without careful explanation of the notation. Are you using a particular text?

    I recommend Srednicki's text (draft version available free from his web page) for a detailed explanation of Weyl, Dirac, and Majorana fields.
     
  5. Mar 17, 2015 #4
    Thank you, this text is very useful.
    But I am still not sure to understand the difference between the fields component and the particle.
    For example for the Dirac field where ##\Psi_{D}=\Psi_{L}+\Psi_{R}##, the four components of the field ##\Psi_{D}## are ##\nu_{L},\bar{\nu}_{R},\nu_{R},\bar{\nu}_{L}## or is it something completely different ?
    Thank you...
     
  6. Mar 17, 2015 #5

    samalkhaiat

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    No, [itex]\Psi_{L}[/itex] and [itex]\Psi_{R}[/itex] are 4-component spinors just like [itex]\Psi_{D}[/itex]. Why don’t you work it out yourself? In Dirac representation, you have [tex]\gamma_{5} = \left( \begin{array}{cc} 0_{2 \times 2} & I_{2 \times 2} \\ I_{2 \times 2} & 0_{2 \times 2} \end{array} \right) .[/tex] So, if you write [itex]\Psi_{D} = \left( \chi , \phi \right)^{T}[/itex], with [itex]\chi = \left( \psi_{1} , \psi_{2} \right)^{T}[/itex] and [itex]\phi = \left( \psi_{3} , \psi_{4} \right)^{T}[/itex], you find [tex]\Psi_{L} = \frac{1}{2} \left( \begin{array}{c} \chi - \phi \\ \phi - \chi \end{array} \right) , \ \ \ \Psi_{R} = \frac{1}{2} \left( \begin{array}{c} \chi + \phi \\ \phi + \chi \end{array} \right) .[/tex]
    Wrong again, Dirac equation has 4 independent solutions, each (massive) solution is given by 4-component spinor. Two of these solutions (spin up: [itex]u^{(1)} (p)[/itex] & spin down: [itex]u^{(2)} (p)[/itex]) represent a (massive) fermion and the other two solutions ( [itex]v^{( 1 , 2 )} (p)[/itex] ) describe the corresponding anti-fermion. The same is true for massless fermions except that the Dirac equation splits into two decoupled equations for the 2-component spinors [itex]\chi[/itex] and [itex]\phi[/itex] : [tex]E \chi = - ( \sigma \cdot p ) \chi ,[/tex] [tex]E \phi = ( \sigma \cdot p ) \phi .[/tex] Each one of these equations has 2 independent solutions: one (2-spinor) for [itex]E = | p |[/itex] ( [itex]\nu_{L}[/itex], if you like), and the other 2-spinor solution is for [itex]E = - | p |[/itex] ( i.e., [itex]\bar{\nu}_{R}[/itex] ) . So, when we say that [itex]\chi[/itex] describes [itex]\nu_{L}[/itex] and [itex]\bar{\nu}_{R}[/itex], this DOSE NOT mean that [itex]\nu_{L}[/itex] is the first component of [itex]\chi[/itex] and [itex]\bar{\nu}_{R}[/itex] is the second component. No, [itex]\nu_{L}[/itex] and [itex]\bar{\nu}_{R}[/itex] are 2 independent solutions of the SAME equation and each (massless particle) is described by 2-component spinor. Indeed, in the Weyl (chiral) representation, you find [itex]\Psi_{L} = ( \chi , 0 )^{T}[/itex] and [itex]\Psi_{R} = ( 0 , \phi )^{T}[/itex].
    What does it mean to pair together 2 independent solutions of the same equation? Is [itex]\Psi_{D} ( p ) = \left( u ( p ) , v ( p ) \right)[/itex]? Does the object [itex]( e^{-} , e^{+} )[/itex] make any mathematical sense? No, [itex]\Psi_{D} = u^{(1)} , u^{(2)} , v^{(1)} , v^{(2)}[/itex], these are four independent 4-component spinors solutions of the Dirac equation.

    Sam
     
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