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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(x

φ(x

φ(x

Similarly for the φ(x

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

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(x

_{1}) contracts to φ(y_{1}) in 4 different ways.φ(x

_{2}) contracts to φ(y_{1}) in 3 different ways.φ(x

_{3}) contracts to φ(y_{1}) in 2 different ways.Similarly for the φ(x

_{4}), φ(x_{5}) and φ(x_{6}) with φ(y_{2}).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 x_{1}, x_{2}, x_{3}contract with φ(y_{1}) 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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