0

I got a follow-up question to my earlier post.

Suppose we have the pseudoscalar Yukawa Lagrangian: $$ L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\bar\psi(i\not\partial-m)\psi-ig\phi\bar\psi\gamma^5\psi. $$ Then we can find its superficial degree of divergence as $D= 4-\frac{3}{2}N_f-N_s$. We can find all divergent amplitudes as follows:

enter image description here

It makes sense to me that the divergent graphs with odd scalar external lines actually vanish due to the symmetry $\phi(t,x)\rightarrow-\phi(t,-x)$, so my question is why do we still have the bottom-left diagram? Isn't in that case we have $N_s = 1$? Is it exist because we have that pseudoscalar vertex? (so odd external scalar is allowed?)

IGY
  • 1,711
  • 2
    Parity acts on all fields. So $\phi(t,x) \to -\phi(t,-x)$ alone is not a symmetry. – Connor Behan Jan 10 '23 at 02:38
  • 1
    Intercalate $γ_5$ between the spinors. – Cosmas Zachos Jan 10 '23 at 03:27
  • 1
  • The divergent part of the bottom-left diagram is of the same type as the interaction term $\phi , \bar{\psi} \gamma_5 \psi$ in your Lagrangian. Therefore, it must be there and renormalizes $g$. 2. A quartic term $\sim \phi^4$ must be included in the Lagrangian you are starting with. What would you do with the divergence in the bottom-right diagram otherwise? 3. We had this discussion already several times.
    • – Hyperon Jan 10 '23 at 09:30