- #1
LayMuon
- 149
- 1
I am trying to write down the LSZ formula for e- e+ -> 2 gamma:
$$ S_{fi} \propto \langle 0| T \{\bar{\Psi}(x_1) \vec{A}(x_2) \Psi (x_4) \vec{A}(x_3) \left\lbrack \frac{(-i)^2}{2} \int d^4x A_\mu (x) \bar{\Psi}(x) \gamma^\mu \Psi (x) \int d^4y A_\nu (y) \bar{\Psi}(y) \gamma^\nu \Psi (y) \right\rbrack \} |0\rangle $$
So I have electron at ##x_1## and positron ##x_4##.
But now if i try to contract I can contract ##\bar{\Psi}(x_1)## with one at x or at y, and same with ##\vec{A}(x_2)## and this would produce a factor of 4 which would cancel with 1/2 from exponential expansion giving a factor 2, but this is wrong, there should not be a factor of 2.
Any ideas? thanks.
$$ S_{fi} \propto \langle 0| T \{\bar{\Psi}(x_1) \vec{A}(x_2) \Psi (x_4) \vec{A}(x_3) \left\lbrack \frac{(-i)^2}{2} \int d^4x A_\mu (x) \bar{\Psi}(x) \gamma^\mu \Psi (x) \int d^4y A_\nu (y) \bar{\Psi}(y) \gamma^\nu \Psi (y) \right\rbrack \} |0\rangle $$
So I have electron at ##x_1## and positron ##x_4##.
But now if i try to contract I can contract ##\bar{\Psi}(x_1)## with one at x or at y, and same with ##\vec{A}(x_2)## and this would produce a factor of 4 which would cancel with 1/2 from exponential expansion giving a factor 2, but this is wrong, there should not be a factor of 2.
Any ideas? thanks.
