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Fermionic creation and annihilation

  1. Jan 3, 2014 #1
    Hi guys,
    I consider fermionic holes and I know the creation and annihilation operators of them.

    I have shown that [itex]\vec{S}= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{c}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{c}_{k \nu} =...= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{-k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ -k \nu}^{ }[/itex].
    But I have to show that [itex]\vec{S}=\sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ }[/itex].

    It would be nice if somebody has an idea how I can get rid of the minus sign.


    Is the sum maybe from minus infinity to plus infinity? Then I think I got it.
    Last edited: Jan 3, 2014
  2. jcsd
  3. Jan 3, 2014 #2

    George Jones

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    Yes, if there is no hard cut-off on the momenta.
  4. Jan 3, 2014 #3
    So because k is from infinity to infinity I can write:
    [itex]\vec{S}= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{c}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{c}_{k \nu} =...= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{-k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ -k \nu}^{ }

    = \sum\limits_{-k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ } \stackrel{(1)}{=}
    \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ }[/itex].

    I would say (1) is true because if I regard the k from "-"infinity to "+"infinity:

    [itex]\sum\limits_{-k}^{}f(k)=\sum\limits_{k}^{}f(-k)= \sum\limits_{k}^{}f(k) [/itex]

    Could somebody check my thoughts? Would be nice!!!
    THX Abby
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