Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can we sum out the vacuum state ##|0\rangle\langle 0|## ?
  1. Bird 1, LHC 0 (Replies: 2)

  2. Spin-0 antiparticle (Replies: 6)

  3. D 0 →K+ K− (Replies: 1)

Loading...