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I have a two point Green's Function with a phi^4 interacting Hamiltonian and am considering the second order of pertubation, and must work out the following:
[tex]T[\phi(x_1) \phi(x_2) \phi(y_1)^4 \phi(y_2)^4] = T[\phi(x_1) \phi(x_2) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_2) \phi(y_2) \phi(y_2) \phi(y_2)][/tex]
where [tex]\phi(x_1)[/tex]and [tex]\phi(x_2)[/tex] are two interacting fields.
Now I am trying to work out the weight of a diagram, but I am not sure if I am doing it right?
This is our diagram:
x1___y1____y2____x2
with one loop at y1 and one loop at y2.
-For the line joining x1 to y1, there are 4 ways of contracting x1 with y1 (there are three y1's left after doing this)
-For the loop at y1, there are 2 ways of contracting the next y1 with another y1 (there is one y1 left after doing this)
-For the line joining y1 and y2, there are 4 ways of contracting the remaining y1 with a y2
-For the loop at y2, there are 2 ways of contracting the next y2 with another y2 (there is one y2 left after doing this)
-Finally there is one way of contracting the remaining y2 with x2.
Hence the weight for this diagram is 4x2x4x2=64. Is this right?
Thanks.
[tex]T[\phi(x_1) \phi(x_2) \phi(y_1)^4 \phi(y_2)^4] = T[\phi(x_1) \phi(x_2) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_2) \phi(y_2) \phi(y_2) \phi(y_2)][/tex]
where [tex]\phi(x_1)[/tex]and [tex]\phi(x_2)[/tex] are two interacting fields.
Now I am trying to work out the weight of a diagram, but I am not sure if I am doing it right?
This is our diagram:
x1___y1____y2____x2
with one loop at y1 and one loop at y2.
-For the line joining x1 to y1, there are 4 ways of contracting x1 with y1 (there are three y1's left after doing this)
-For the loop at y1, there are 2 ways of contracting the next y1 with another y1 (there is one y1 left after doing this)
-For the line joining y1 and y2, there are 4 ways of contracting the remaining y1 with a y2
-For the loop at y2, there are 2 ways of contracting the next y2 with another y2 (there is one y2 left after doing this)
-Finally there is one way of contracting the remaining y2 with x2.
Hence the weight for this diagram is 4x2x4x2=64. Is this right?
Thanks.