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This is a follow-up question to my earlier post here:

Now suppose we have the pseudoscalar Yukawa Lagrangian: $$ L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\bar\psi(i\not\partial-m)\psi-g\gamma^5\phi\bar\psi\psi. $$ We can find its superficial degree of divergence as $D= 4-\frac{3}{2}N_f-N_s$. From this manual (p.80), we can find all divergent amplitudes as follows: enter image description here We do have other divergent graphs with odd scalar external lines. However, the author ignored them, and claimed they are potentially divergent diagrams that actually vanish. I wonder is there a straightforward way to see they vanish?

And as a consequence, does that imply we will need to add $\phi^4$ term in the Lagrangian and its counterterm $-i\delta_4$ to make the theory normalizable, but don't need to add $\phi^3$ term and its counterterm $-i\delta_3$ to the entire Lagrangian? Does this have anything to do with the fact that this Lagrangian is invariant under the parity transformation?

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IGY
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2 Answers2

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  1. $\gamma_5$ likes to be sandwiched between $\bar{\psi}$ and $\psi$. So, the interaction term should read $-g \phi \bar{\psi} \gamma_5 \psi$.

  2. $\bar{\psi} \gamma_5 \psi$ is a pseudoscalar, consequently also $\phi$ has to be a pseudoscalar.

  3. As a consequence, an interaction term like $\phi^3$ is forbidden by parity invariance.

  4. In order to formulate a consistent renormalizable theory (in 4 space-time dimensions) ALL possible terms up to (operator) dimension 4 invariant under space time symmetries and possibly also other symmetries have to be included. As the interaction term $\phi^4$ is even under parity, it must be included in the Lagrangian you are starting with.

Hyperon
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  • Many thanks for the answer! Are all Lagrangians invariant under parity transformation? Is there an intuition why $\gamma^5$ is sandwiched between two $\psi$'s? – IGY Dec 30 '22 at 13:32
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    Parity is violated by weak interactions. $\gamma_5$ is a $4 \times 4$ matrix, so it must stand between $\bar{\psi}$ (being a $1 \times 4$ matrix) and $\psi$ (being a $4\times 1$ matrix. No intuiton, just correct mathematics. – Hyperon Dec 30 '22 at 13:35
  • Thanks! If the interaction term is $-g\bar\psi\psi\phi$ instead, we still have the transformation $\psi(t,x)\rightarrow\gamma^0\psi(t,-x)$, but $\phi$ is transformed as $\phi(t,x)\rightarrow\phi(t,-x)$, to make the entire term parity-invariant, is that right? So we can have $\phi^3$ in that case. – IGY Dec 31 '22 at 01:54
  • Without the $\gamma_5$, you have a scalar coupling and $\phi^3$ must be included. – Hyperon Dec 31 '22 at 06:56
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Yes, one must for consistency as a minimum include all possible renormalizable terms that are not excluded by symmetry, cf. my related Phys.SE answer here. Pseudoscalar Yukawa theory has a $\mathbb{Z}_2$ parity symmetry that excludes odd $\phi^n$ terms. However the even $\phi^n$ terms with $n=2,4$ must be included.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT. 1995; problem 10.2.
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