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Creation operator in k-space

  1. Nov 4, 2013 #1
    Hi,

    Could anyone tell if there exists an identity [tex]a_k^\dagger = a_{-k}[/tex] because intuitively there should be no difference between creating a particle with momentum k and destroying a particle with momentum -k.
    If true is it possible to show that from the definition [tex]a_k = \frac{1}{√V}∫e^{ikx} a(x)[/tex]?
     
  2. jcsd
  3. Nov 5, 2013 #2
    Hello Trave11er,

    Given the value of p, there are two solutions for the energy, one possitive the other negative.

    The vacuum state is defined as having all the negative eigenstates full (no particles with
    negative energy) and all the possitive eigenstates empty (no particles). In QED negative energy particles are interpreted as particles with opposite charge travelling backwards in time (positrons), so the destruction of a hole (creation as a positron) is not the same that the creation of an electron.
     
  4. Nov 5, 2013 #3

    DrDu

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    That depends on the kind of particle. I think uncharged particles like Majorana Fermions and bosons can act as their own anti-particles. For charged particles, there is no such relation between creation and anihilation operators with k and -k respectively.
     
  5. Nov 5, 2013 #4
    Thanks for the replies,

    The original question actually arose in the context of spin waves which have bosonic excitations on chain of spins - they are not charged so to me it seems that the relation in k-space should hold and it should be possible to prove starting from definition.
     
  6. Nov 5, 2013 #5
    Thanks for the replies,

    The original question actually arose in the context of spin waves which have bosonic excitations on chain of spins - they are not charged so to me it seems that the relation in k-space should hold and it should be possible to prove starting from definition.
     
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