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Let's say we want to calculate the two-point Green's function for a fermion to a given order for a two particle interaction of the form ##U(x,y)=U(y,x)##. For the first order calculation we have to do all contractions related to
$$\mathcal{T}[\psi^\dagger_\mu(x_1)\psi^\dagger_\lambda(x_1')\psi_{\lambda'}(x_1')\psi_{\mu'}(x_1)\psi_\alpha(x)\psi^\dagger_\beta(y) ]$$
we have three daggered operators and three-undaggered, so we have 3!=6 possible contractions However there are 2! possible unconnected diagrams. That leaves us with 4 diagrams, but they are repeated, so we have to divide by two to avoid double counting. That means that there are only connected 2 diagrams for order 1.
What about order 2? I was proceeding the same way, we have 5! possible diagrams, when the two interactions are not connected then we have to remove 4!. Also when a single interaction is disconnected, we have 4 possible configurations (per first order), multiplied by 2 to account for the two possible disconnected diagrams. That leaves 80 diagrams, however if we account for double counting twice (there are two interactions), we have 20 connected diagrams.
However when drawing them altogether (and confirming with Fetter&Walecka) there are just 10 distinct connected diagrams. So where did I go wrong here? In general, is there an easy way to count them all for a given order?
$$\mathcal{T}[\psi^\dagger_\mu(x_1)\psi^\dagger_\lambda(x_1')\psi_{\lambda'}(x_1')\psi_{\mu'}(x_1)\psi_\alpha(x)\psi^\dagger_\beta(y) ]$$
we have three daggered operators and three-undaggered, so we have 3!=6 possible contractions However there are 2! possible unconnected diagrams. That leaves us with 4 diagrams, but they are repeated, so we have to divide by two to avoid double counting. That means that there are only connected 2 diagrams for order 1.
What about order 2? I was proceeding the same way, we have 5! possible diagrams, when the two interactions are not connected then we have to remove 4!. Also when a single interaction is disconnected, we have 4 possible configurations (per first order), multiplied by 2 to account for the two possible disconnected diagrams. That leaves 80 diagrams, however if we account for double counting twice (there are two interactions), we have 20 connected diagrams.
However when drawing them altogether (and confirming with Fetter&Walecka) there are just 10 distinct connected diagrams. So where did I go wrong here? In general, is there an easy way to count them all for a given order?
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