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Help! How to understand classical Fermion field from anticommuting relations?

  1. Apr 4, 2006 #1
    Since we have anticommuting relations for the quantum Dirac fields, this
    will bring us to the similar classical correspondance but result in Grassmann
    spinor field function instead. (such as path integral)

    So when we consider an arbitrary interaction term that like (\bar{\psi} \psi)^n, if the field is really a Grassmann field, then it may be zero when two identical fields meet (at same spacetime point). So why such a term can still be possible?

    thanks for clarify this concept.
  2. jcsd
  3. Apr 5, 2006 #2
    [tex]\bar{\psi}\psi[/tex] is a real number and raising a real number to an integer power gives another real number.

    Also, [tex](\bar{\psi}\psi)^{n} \not= \bar{\psi}^{n}\psi^{n}[/tex]
  4. Apr 5, 2006 #3
    [tex](\bar{\psi}\psi)^{n} \not= \bar{\psi}^{n}\psi^{n}[/tex]
  5. Apr 5, 2006 #4
    "[tex](\bar{\psi}\psi)^{n} \not= \bar{\psi}^{n}\psi^{n}[/tex]"
    Is this writing formate popular in The West?
  6. Apr 5, 2006 #5
    It's Latex, the standard format that most maths and physics papers (and even many books) are written in. Gives for nice looking equations and easy to read articles. A damn sight better than the equation editor that comes with Microsoft Word.
  7. Apr 5, 2006 #6
    To AlphaNumeric :

    [tex]\bar{\psi}\psi[/tex] is an operator under quantization, classically there
    won't be a problem, it is just a real number. But when you consider its classical
    correspondence from quantization, or you can write
    it in component form, such as
    can be written as [tex]\bar{\psi}_1\psi_1\bar{\psi}_1\psi_1[/tex] in component form(which is one component of this term, 1 is a component index). So when you consider Grassmann field function this component would be zero.
    Last edited: Apr 5, 2006
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