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Sorry I am spamming the forum, but I have yet another question on Feynman diagrams -
Please see attached picture.
Apparently the symmetry factor for this FD is 1 - I am trying to understand why.
My notes explain that "the symmetry factor is 1 because:
φ(x1) contracts to φ(y1) in 4 different ways.
φ(x2) contracts to φ(y1) in 3 different ways.
φ(x3) contracts to φ(y1) in 2 different ways.
Similarly for the φ(x4), φ(x5) and φ(x6) with φ(y2).
I don't understand this because if you write out the wick's theorem calculation, we have
[tex]
\frac{k^2}{(4!)^2} \int d^4y_1 d^4y_2 <0|T[\phi (x_1) \phi (x_2) \phi (x_3)\phi (x_4)\phi (x_5)\phi (x_6) \phi^4 (y_1) \phi^4 (y_2)]|0>
[/tex]
Now, the braket is equal to all distinct pairs of contractions, so all of x1, x2, x3 contract with φ(y1) in 4 ways...
I know I can't see something quite trivial, so would be grateful if anyone could enlighten me on this...
Please see attached picture.
Apparently the symmetry factor for this FD is 1 - I am trying to understand why.
My notes explain that "the symmetry factor is 1 because:
φ(x1) contracts to φ(y1) in 4 different ways.
φ(x2) contracts to φ(y1) in 3 different ways.
φ(x3) contracts to φ(y1) in 2 different ways.
Similarly for the φ(x4), φ(x5) and φ(x6) with φ(y2).
I don't understand this because if you write out the wick's theorem calculation, we have
[tex]
\frac{k^2}{(4!)^2} \int d^4y_1 d^4y_2 <0|T[\phi (x_1) \phi (x_2) \phi (x_3)\phi (x_4)\phi (x_5)\phi (x_6) \phi^4 (y_1) \phi^4 (y_2)]|0>
[/tex]
Now, the braket is equal to all distinct pairs of contractions, so all of x1, x2, x3 contract with φ(y1) in 4 ways...
I know I can't see something quite trivial, so would be grateful if anyone could enlighten me on this...
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