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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 \phi_x which can be contracted with 4 \phi_w : 4
Then \phi_y which can be contracted with 3 \phi_w : 3
Then I have 2 \phi_w which can be contracted with 4 \phi_u: 2x4
Then I have 3 \phi_u which can be contracted with 4 \phi_z: 3x4
The rest \phi_u get contracted together (only 2 left), whereas the \phi_z 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 \phi^4 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 \phi_x which can be contracted with 4 \phi_w : 4
Then \phi_y which can be contracted with 3 \phi_w : 3
Then I have 2 \phi_w which can be contracted with 4 \phi_u: 2x4
Then I have 3 \phi_u which can be contracted with 4 \phi_z: 3x4
The rest \phi_u get contracted together (only 2 left), whereas the \phi_z 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 \phi^4 theory.
So is it correct to say that the symmetry factor is afterall SF=16?