Can someone check the symmetry factor I've found in the following diagram and verify it?(adsbygoogle = window.adsbygoogle || []).push({});

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 of2since 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 is4x3x2 x4 x3 x2 x 4 = 4! x 4! x 4

The diagram is of order 3 (3 vertices) so there is a factor1/3!and also the1/4!from the coupling constant for [itex]\phi^4[/itex] theory.

So is it correct to say that the symmetry factor is afterall SF=16?

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# SymmetryFactor Calculation

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