- #36
Gaussian97
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Yes
transform intoJD_PM said:$$[b (\vec{p}_2), b^{\dagger}(\vec k_3)]=0$$
Gaussian97 said:Well, that depends on how you define the Feynman rule for the vertex, ##\lambda##. If you want to consider all the different ways to connect the external lines to the vertex, as different diagrams, then the computation we have done will be proportional to ##3!\lambda## (because you will have 3! different diagrams). And, therefore, you will obtain that the Feynman rule is $$\lambda = \frac{-i\lambda_3}{3!}$$if otherwise, you want to consider all the possible ways of connecting the lines as the same diagram, then the computation is proportional to ##\lambda## and you have $$\lambda = -i\lambda_3$$personally I have always seen this second one, but there's nothing wrong with the other.
Gaussian97 said:Well, that depends on how you define the Feynman rule for the vertex, ##\lambda##. If you want to consider all the different ways to connect the external lines to the vertex, as different diagrams, then the computation we have done will be proportional to ##3!\lambda## (because you will have 3! different diagrams).
Gaussian97 said:You have 3 external lines (representing the 3 particles at infinity) and 3 legs of the vertex to connect. Therefore 6 possibilities (3 for the first pair, 2 for the second, and the last one is immediately determined).
Gaussian97 said:$$S = \langle \phi\phi | \phi \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle \phi\phi| :\phi(x) \phi(x) \phi (x): |\phi\rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle \phi\phi | :\phi(x) \phi(x) \phi(x) \phi (x) : | \phi \rangle $$
Now show that the first and last terms vanish and then, from all the terms arising from splitting ##\phi = \phi^+ + \phi^-## look which ones survive.
Well, in fact you should studyJD_PM said:$$\langle 0|b(\vec p_2) b(\vec p_3) b(\vec k_2) b(\vec k_3) \phi (x) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$$