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Spinor fields and spinor wave functions

  1. Aug 9, 2005 #1
    Hi everyone, I've confused myself trying to understand Weyl spinors... here's my best attempt at well-posed questions: (by the way, a nice--if incomplete--reference on computations using Weyl spinors can be found here (.ps), from Harvard's Phys 253B page)


    1. Fermions are spinor fields which obey anticommutation relations (Zee's motivation for this is that a pair of electrons have to be antisymmetric under exchange of quantum numbers). These spinor fields can be fourier transformed as a linear combination of spinor wavefunctions. In the above link, the author (Nima Arkani-Hamed, apparently) writes this as:

      [tex]\xi_\alpha (x) = \sum_s \int \frac{d^3p}{(2\pi)^{3/2}(2E_p)^{1/2}}[x_\alpha a e^{-ip\cdot x} + y_\alpha a^\dagger e^{ip\cdot x}][/tex] (NAH, eq.3)

      Where now [tex]x[/tex] and [tex]y[/tex] are the wavefunctions (fourier components). [tex]x, y, a, a^\dagger[/tex] are all functions of 3-momentum and spin, and the [tex]a^\dagger[/tex] and [tex]a[/tex] are raising/lowering operators.

      That being said, one thing that is not clear to me is why the wavefunctions [tex]x[/tex] and [tex]y[/tex] obey commutation relations while the fields [tex]\xi[/tex] obery anticommutation relations. I understand that the creation operators obey anticommutation relations, but why would the wavefunctions not?


      [*]I've noticed that I should treat bars and daggers very carefully. For example, the reference above relates barred and daggered (where a dagger means hermetian conjugation) by [tex]\bar{x}_{\dot{\alpha}} \equiv x^\dagger_\beta[/tex].

      Is it fair for me to interpret this as [tex]\bar{x}_{\dot{\beta}} = x^*_\beta[/tex], that is to say that one takes the complex conjugate of each component (without "transposing" the column spinor into a row spinor)? (This is consistent, I think, with the Peskin and Schroeder definition of [tex]\bar{u(p)} = u^\dagger(p)\gamma^0[/tex].) I understand that indices are important and are somewhat subtle--is it correct that hermetian conjugation would raise or lower the spinor index?

      Now I become a little more uneasy when, in the context of the Standard Model, authors will use the bar notation to represent a right-handed particle. For example, [tex]\bar{e}^\dagger = e_R[/tex]. Is this meaning for the bar still consistent with my previous understanding?


      [*]Anyway, in addition to the bar, I've noticed that indices are very important and subtle. In particular, since indices are raised and lowered by the totally antisymmetric [tex]\epsilon[/tex] tensor,

      [tex]x^\alpha y_\alpha = -x_\alpha y^\alpha[/tex]

      but

      [tex]x^\alpha y_\alpha = y^\alpha x_\alpha[/tex].

      (I suppose this is a justification for why the wavefunctions commute?) Anyway, it becomes important that one picks a convention for the contraction of dotted and undotted indicies, and the above document suggests having undotted indices go from upper left to lower right and dotted going from lower left to upper right. That is to say that we can drop the indices for terms that look like these: [tex]x^\alpha y_\alpha = xy[/tex], or [tex]\bar{x}^{\dot{\alpha}}\bar{y}_{\dot{\alpha}} = \bar{x}\bar{y}[/tex].

      So computing Feynman diagrams amounts to following fermion lines and writing down terms with some consistent convention on raised and lowered indicies. (for example, following a fermion line: starting external fermion is raised if it's undotted and lowered if dotted, ending external fermion is the opposite) Once the amplitude is written this way, one needs to square it to get something proportional to cross section. [tex]|M|^2 = MM^*[/tex].

      However, now I'm unsure what the index structure should look like. For example, if I have a term (writing indices explicitly):

      [tex]x^\alpha y_\alpha[/tex]

      is the complex conjugate equal to [tex]\bar{x}_{\dot{\alpha}}\bar{y}^{\dot{\alpha}}[/tex] or [tex]\bar{x}^{\dot{\alpha}}\bar{y}_{\dot{\alpha}}[/tex]? Or is it equal to something in terms of daggers?

      The appendix to the above reference claims that: [tex](xy)^\dagger = \bar{y}\bar{x}[/tex], which, using the above index notation is:

      [tex](x^\alpha y_\alpha)^\dagger = \bar{x}_{\dot{\alpha}}\bar{y}^{\dot{\alpha}}[/tex].

      The claim is made that this is applicable to both anti-commuting and commuting spinors. (Why?) Well, then... why does the author use a dagger when the term in parenthesis should be a scalar? (i.e. a complex conjugate should do) Further, is this equation arbitary, or is there any motivation for why the indices must end up this way? Earlier, we picked an arbitrary convention for when indices can be dropped. It was "understood" that unbarred and barred spinors were contracted with the indices a certain way... but it's not clear that the complex conjugate of a contraction of the unbarred spinors would give a contraction of the barred spinors in with the appropriate convention. That is to say, why is it that undotted indices are contracted in one way (upper then lower) while dotted indices are contracted in the opposite way (lower then upper)?



    Any insight would be much appreciated!

    -Flip
     
  2. jcsd
  3. Aug 19, 2005 #2
    Hi tomato,

    Let me try.


    I have three answers:
    1) What makes you think that x and y should anticommute? - There is NO principle that says that for fermions all comutators are blindly replaced by anticommutators.
    2) We have the same in Dirac fermions: x and y (which are usually denoted by u and v) commute.
    3) If x and y were anticommuting, their product with the raising and lowering operators would be a commuting object, so the resulting field would be a commuting object, which would not make you happy.

    Probably you meant here [tex]\bar{x}_{\dot{\beta}} \equiv x^\dagger_\beta[/tex]

    For Weyl spinors, the situation is more complicated than in the more simple cases where we have only covariant and contravariant vectors (so that they can be represented by row and column vectors). Weyl spinors include not 2 types of objects, but 4:
    For a Lorentz transformation with rotation angle [tex]\theta[/tex] and rapidity [tex]\beta[/tex], we have the matrix
    [tex]M=\exp\left(-\frac{1}{2}(i\theta+\beta)\sigma\right)[/tex]
    How do the spinors transform? There are 4 cases: [tex]\psi_\alpha[/tex] transforms by the matrix [tex]M[/tex], [tex]\bar\psi_{\dot\alpha}[/tex] by the matrix [tex]M^*[/tex], [tex]\psi^\alpha[/tex] by the matrix [tex]M^{-1}[/tex], and [tex]\bar\psi^{\dot\alpha}[/tex] by the matrix [tex]M^{-1*}[/tex]. This is where the subtlety is coming from.

    You should not switch the order of x and y on the RHS.

    Maybe the author wanted to include the case of a spinor field (like the expression for [tex]\xi[/tex] you wrote at the beginning of your post), and the dagger means that also [tex]a[/tex] should be relaced by [tex]a^\dagger[/tex], and vice versa, which is more than just complex conjugation. Is this the right thing to do? - I am confused about that at the moment.
     
    Last edited: Aug 19, 2005
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