# What diagrams to include in Wilson's Approach?

In the Wilson's approach to renormalization we split the field into two parts; a high momentum part and a low momentum part. We then integrate out the high momentum terms. Consider the case of $\phi^3$ theory and the two diagrams below:

The dashed (red) lines represent the high momenta field and the solid (black) the low momenta. It is clear that diagram $A$ contributes to the $C_4\phi^4$ in the new effective Lagrangian by making a change to the $C_4$.

I am confused however about diagram B. Does it contribute to the $C_6\phi^6$ term in the effective Lagrangian or is it a result of the $C_6\phi^6$ term.

I.e. When finding the contributions to the new coupling constants do we only include diagrams of the form A or do we have to also include the form B?

Edit

Colour coded version of B

• Hint: are you sure you should get a closed loop for the slow fields, knowing that you do not average over them ? – Adam Jan 9 '18 at 8:51
• @Adam I would say yes, since in B averaging over the 3 dashed lines means we put them to zero length - does it not? Otherwise I can't possibly see what $B$ must go to? – Quantum spaghettification Jan 9 '18 at 8:55
• As you say, you can shrink the dashed line to zero. But why have you linked two of the black lines together ? – Adam Jan 9 '18 at 8:58
• @Adam I have color coded the diagram. If I take all dashed (red) lines to zero then all four vertexes must meet at the same point. I am not sure which two lines you mean. – Quantum spaghettification Jan 9 '18 at 9:03
• For B, you have a black line linking two vertices. My question is : why have you linked these two vertices with a black line, knowing that you do not average (integrate out) the slow fields. – Adam Jan 9 '18 at 9:05

To implement Wilson's RG for a $\phi^3$ theory, we split the field into a slow $\phi$ and a fast $\hat \phi$ component. There are two kinds of interesting vertices (contributing to the renormalization of the slow field action) : $\phi_1 \hat\phi_1 \hat\phi_1$ and $\phi_1 \phi_1 \hat\phi_1$ (the number corresponding to the space-time/internal indices).
The diagram A of the OP corresponds to (averages over the fast fields only) $$\langle\phi_1 \hat\phi_{1} \hat\phi_{1}\phi_2 \hat\phi_{2} \hat\phi_{2}\phi_3 \hat\phi_{3} \hat\phi_{3}\phi_4 \hat\phi_{4} \hat\phi_{4} \rangle=\phi_{1}\phi_2\phi_3\phi_4 \langle \hat\phi_{1} \hat\phi_{1}\hat\phi_{2} \hat\phi_{2} \hat\phi_{3} \hat\phi_{3}\hat\phi_{4} \hat\phi_{4} \rangle.$$ Here $\langle \hat\phi_{1} \hat\phi_{1}\hat\phi_{2} \hat\phi_{2} \hat\phi_{3} \hat\phi_{3}\hat\phi_{4} \hat\phi_{4} \rangle$ corresponds to the dashed red closed loop of left diagram A of the OP. On the other hand, the (black, full) legs of the diagram correspond to the slow fields $\phi_{1}$, $\phi_2$, $\phi_3$ and $\phi_4$, which are not connected, since they are not integrated over.
In momentum space, thanks to the fact that the incoming momenta are much slower than the flowing mometum in the loop, one can approximate $\langle \hat\phi_{1} \hat\phi_{1}\hat\phi_{2} \hat\phi_{2} \hat\phi_{3} \hat\phi_{3}\hat\phi_{4} \hat\phi_{4} \rangle\propto \delta(1+2)\delta(1+3)\delta(1+4)$ and thus shrink the loop to a single point (corresponding to a $\phi^4$ vertex).
Concerning diagram B, using the same color code, we see that this diagram is made using also the second kind of vertex (that we didn't use for diagram A) : $$\langle\phi_1 \phi_1 \hat\phi_1\phi_2 \hat\phi_2 \hat\phi_2\phi_3 \hat\phi_3 \hat\phi_3\phi_4 \phi_4 \hat\phi_4\rangle=\phi_1 \phi_1\phi_2\phi_3\phi_4 \phi_4\langle \hat\phi_1 \hat\phi_2 \hat\phi_2 \hat\phi_3 \hat\phi_3 \hat\phi_4\rangle.$$
Here $\langle \hat\phi_1 \hat\phi_2 \hat\phi_2 \hat\phi_3 \hat\phi_3 \hat\phi_4\rangle$ corresponds to the three red dashed line of the OP, while the six slow field should correspond to six legs. The OP's problem is that there should not be a full black (or blue) line in diagram B : the slow fields are not integrated over, so the slow fields legs of the vertices should not be connected.
By the same argument than before $\langle \hat\phi_1 \hat\phi_2 \hat\phi_2 \hat\phi_3 \hat\phi_3 \hat\phi_4\rangle\propto\delta(1+2)\delta(1+3)\delta(1+4)$ and thus diargam B corresponds to a $\phi^6$ interaction.