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

1. Oct 6, 2014

### ccnu

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. Oct 6, 2014

### Orodruin

Staff Emeritus
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.