- #1

CAF123

Gold Member

- 2,948

- 88

Consider the lagrangian of the real scalar field given by $$\mathcal L = \frac{1}{2} (\partial \phi)^2 - \frac{1}{2} m^2 \phi^2 - \frac{\lambda}{4!} \phi^4$$

Disregarding snail contributions, the only diagram contributing to ## \langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle## at one loop order is the so called dinosaur (see attached)

To argue the symmetry factor ##S## of this diagram, I say that there are 4 choice for a ##\phi_y## field to be contracted with one of the final states and then 3 choices for another ##\phi_y## field to be contracted with the remaining final state. Same arguments for the ##\phi_x## fields and their contractions with the initial states. This leaves 2! permutations of the propagators between x and y. Two vertices => have factor (1/4!)

$$S^{-1} = \frac{4 \cdot 3 \cdot 4 \cdot 3 \cdot 2!}{4! \cdot 4! \cdot 2} = \frac{1}{4}$$ I think the answer should be 1/2 so can someone help in seeing where I lost a factor of 2?

I could also evaluate $$\langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + (\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 :| p_1 p_2 \rangle + \dots $$ where dots indicate diagrams generated via this correlator that do not contribute at one loop. (I don't know the latex for the Wick contraction symbol so I just write contract). Is there a way to find out the symmetry factor from computing the term ##(\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2: | p_1 p_2 \rangle?##

Thanks!

Disregarding snail contributions, the only diagram contributing to ## \langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle## at one loop order is the so called dinosaur (see attached)

To argue the symmetry factor ##S## of this diagram, I say that there are 4 choice for a ##\phi_y## field to be contracted with one of the final states and then 3 choices for another ##\phi_y## field to be contracted with the remaining final state. Same arguments for the ##\phi_x## fields and their contractions with the initial states. This leaves 2! permutations of the propagators between x and y. Two vertices => have factor (1/4!)

^{2}and such a diagram would be generated at second order in the Dyson expansion => have factor 1/2. Putting this all together I get$$S^{-1} = \frac{4 \cdot 3 \cdot 4 \cdot 3 \cdot 2!}{4! \cdot 4! \cdot 2} = \frac{1}{4}$$ I think the answer should be 1/2 so can someone help in seeing where I lost a factor of 2?

I could also evaluate $$\langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + (\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 :| p_1 p_2 \rangle + \dots $$ where dots indicate diagrams generated via this correlator that do not contribute at one loop. (I don't know the latex for the Wick contraction symbol so I just write contract). Is there a way to find out the symmetry factor from computing the term ##(\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2: | p_1 p_2 \rangle?##

Thanks!