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Fermion annihilation operators from position and momentum

  1. May 30, 2010 #1
    Is it possible to express fermion annihilation operator as a function of position and momentum?

    I've seen on Wikipedia the formula for boson annihilation operator:
    [tex]
    \begin{matrix} a &=& \sqrt{m\omega \over 2\hbar} \left(x + {i \over m \omega} p \right) \\ a^{\dagger} &=& \sqrt{m \omega \over 2\hbar} \left( x - {i \over m \omega} p \right) \end{matrix}
    [/tex]

    But what about fermions? Is it possible to get anticommutation relations from canonical relations alone, or is it necessary to postulate something else?
     
  2. jcsd
  3. May 30, 2010 #2

    tom.stoer

    User Avatar
    Science Advisor

    You have to use Grassmann numbers or something like that. This can be seen by inspection of the (fermionic) anti-commutators

    [tex]\{b, b\} = 2b^2 = 0[/tex]
    [tex]\{b^\dagger, b^\dagger\} = 2{b^\dagger}^2 = 0[/tex]

    which cannot be derived from commuting objects.
     
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