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Can we sum out the vacuum state ##|0\rangle\langle 0|## ?

  1. Oct 6, 2014 #1
    For example, when we write down the operator definition of quark fragmentation matrix element:
    ##\Phi_{ij} = \sum_X \int d^4 x e^{ikx}\langle 0|\psi_i(x)|P,X\rangle\langle P,X|\bar{\psi}_j(0)|0\rangle##.
    Can we rewrite is as:
    ##\Phi_{ij} = \sum_X \int d^4 x e^{ikx}\langle P,X|\bar{\psi}_j(0)|0\rangle\langle 0|\psi_i(x)|P,X\rangle = \sum_X \int d^4 x e^{ikx}\langle P,X|\bar{\psi}_j(0)\psi_i(x)|P,X\rangle##?
     
  2. jcsd
  3. Oct 6, 2014 #2

    Orodruin

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    No, this is not allowed. The entity ##\left|0\rangle\langle 0\right|## is a projection operator onto the vacuum state and not the unit operator.
     
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