- #1
CAF123
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Consider a real scalar field described through the following lagrangian $$\mathcal L = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}m^2 \phi^2 - \frac{g}{3!}\phi^3$$ The second order term in the S matrix expansion produces the diagrams in which we have a ##2 \rightarrow 2## scattering event and such diagrams are given by evaluating $$S^{(2)} = -\frac{9 g^2}{2 \cdot 6^2} \int d^4 x_1 \int d^4 x_2 i \Delta_F(x_1-x_2) (2 \langle p_3p_4| \phi_2^- \phi_2^- \phi_1^+ \phi_1^+ | p_2p_1\rangle + 4 \langle p_3p_4 | \phi_1^- \phi_2^- \phi_1^+ \phi_2^+| p_1p_2 \rangle) $$
My questions are: From this, apparently it is clear that the ##s,t,u## diagrams can be obtained but I'm not sure how this is the case. I can see that in the first term, say, ##p_1## and ##p_2## are destroyed at ##x_1## and then ##p_3## and ##p_4## created at ##x_2## which would give the ##s## diagram and similarly depending on whether ##\phi_2^+## annihilates ##p_1## or ##p_2## at ##x_2## in the second term I can get the ##t## and ##u## channel diagrams. There should also be a factor of 4 alongside the first term there and a factor of 2 for the second term (in order to cancel the combinatorial prefactor already present), but I don't see where this comes from.
Should I write $$\langle p_3 p_4 | \phi_1^- \phi_2^- \phi_1^+ \phi_2^+ | p_1 p_2 \rangle = e^{-i (p_1-p_3) \cdot x_1} e^{-i (p_2-p_4) \cdot x_2} + e^{-i(p_2-p_3) \cdot x_1} e^{-i(p_1-p_4) \cdot x_2} + \text{perms}$$ where perms is sending ##x_1 \leftrightarrow x_2##. Under the integration over ##x_1## and ##x_2## each perm matches exactly one of the terms written explicitly above so that is why we have factor of 2? Maybe? But using this logic does not explain the factor of 4 in the first term (as far as I can see).
Thanks!
My questions are: From this, apparently it is clear that the ##s,t,u## diagrams can be obtained but I'm not sure how this is the case. I can see that in the first term, say, ##p_1## and ##p_2## are destroyed at ##x_1## and then ##p_3## and ##p_4## created at ##x_2## which would give the ##s## diagram and similarly depending on whether ##\phi_2^+## annihilates ##p_1## or ##p_2## at ##x_2## in the second term I can get the ##t## and ##u## channel diagrams. There should also be a factor of 4 alongside the first term there and a factor of 2 for the second term (in order to cancel the combinatorial prefactor already present), but I don't see where this comes from.
Should I write $$\langle p_3 p_4 | \phi_1^- \phi_2^- \phi_1^+ \phi_2^+ | p_1 p_2 \rangle = e^{-i (p_1-p_3) \cdot x_1} e^{-i (p_2-p_4) \cdot x_2} + e^{-i(p_2-p_3) \cdot x_1} e^{-i(p_1-p_4) \cdot x_2} + \text{perms}$$ where perms is sending ##x_1 \leftrightarrow x_2##. Under the integration over ##x_1## and ##x_2## each perm matches exactly one of the terms written explicitly above so that is why we have factor of 2? Maybe? But using this logic does not explain the factor of 4 in the first term (as far as I can see).
Thanks!