- #1
ChrisVer
Gold Member
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Can someone check the symmetry factor I've found in the following diagram and verify it?
Do you know if there is any way to determine the SF of a diagram fast?
So I have [itex]\phi_x[/itex] which can be contracted with 4 [itex]\phi_w[/itex] : 4
Then [itex]\phi_y[/itex] which can be contracted with 3 [itex]\phi_w[/itex] : 3
Then I have 2 [itex]\phi_w[/itex] which can be contracted with 4 [itex] \phi_u [/itex]: 2x4
Then I have 3 [itex]\phi_u[/itex] which can be contracted with 4 [itex]\phi_z[/itex]: 3x4
The rest [itex]\phi_u[/itex] get contracted together (only 2 left), whereas the [itex]\phi_z[/itex] we get a factor of 2 since there are 2 possibilities to contract 3 fields.
Finally the last contraction gives just a factor 1 (no possible alternative choices).
So the result from contractions is 4x3x2 x4 x3 x2 x 4 = 4! x 4! x 4
The diagram is of order 3 (3 vertices) so there is a factor 1/3! and also the 1/4! from the coupling constant for [itex]\phi^4[/itex] theory.
So is it correct to say that the symmetry factor is afterall SF=16?
Do you know if there is any way to determine the SF of a diagram fast?
So I have [itex]\phi_x[/itex] which can be contracted with 4 [itex]\phi_w[/itex] : 4
Then [itex]\phi_y[/itex] which can be contracted with 3 [itex]\phi_w[/itex] : 3
Then I have 2 [itex]\phi_w[/itex] which can be contracted with 4 [itex] \phi_u [/itex]: 2x4
Then I have 3 [itex]\phi_u[/itex] which can be contracted with 4 [itex]\phi_z[/itex]: 3x4
The rest [itex]\phi_u[/itex] get contracted together (only 2 left), whereas the [itex]\phi_z[/itex] we get a factor of 2 since there are 2 possibilities to contract 3 fields.
Finally the last contraction gives just a factor 1 (no possible alternative choices).
So the result from contractions is 4x3x2 x4 x3 x2 x 4 = 4! x 4! x 4
The diagram is of order 3 (3 vertices) so there is a factor 1/3! and also the 1/4! from the coupling constant for [itex]\phi^4[/itex] theory.
So is it correct to say that the symmetry factor is afterall SF=16?