Consider pseudoscalar Yukawa theory in 4D:
$$ S =\int d^4x\ \frac{1}{2}(\partial\phi)^2 - \frac{1}{2}m_\phi^2\phi^2 +\bar\psi(i\gamma^\mu\partial_\mu-m_e)\psi - ig\bar\psi\gamma^5\psi\phi -\frac{\lambda}{4!}\phi^4. $$
My question is: Is the 1PI function for 3 bosonic fields $\Gamma_3[\phi(x),\phi(y),\phi(z)]$ zero? How can I see this?
The (original and renormalized) action $S[\phi,\psi]$ is assumed to respect parity symmetry. In the pseudoscalar Yukawa theory, the pseudoscalar $\phi$ transforms $$P^{-1}\phi({\bf x},t)P ~=~ -\phi(-{\bf x},t) $$ under a parity transformation. Therefore the action $S[\phi,\psi]$ cannot contain a $\phi^3$ term, cf. above comment by Cosmas Zachos. (Here it should probably be mentioned that a Lorentz invariant $\phi^3$ derivative term is also not possible, even if the pseudoscalar $\phi$ carries a flavour index. E.g. a $\phi^3$ derivative term with the 4D Levi-Civita tensor $\epsilon^{\mu\nu\lambda\kappa}$ vanishes by antisymmetry because at least 2 derivatives must land on the same $\phi$, cf. comments by Peter Kravchuk.)
The 1PI effective action $\Gamma[\phi_{\rm cl}, \psi_{\rm cl}]$ inherits$^1$ parity-symmetry, so it can also not contain a $\phi_{\rm cl}^3$ term. Hence the 3-point 1PI vertex vanishes $$ \left. \frac{\delta^3\Gamma[\phi_{\rm cl}, \psi_{\rm cl}]}{\delta \phi_{\rm cl}(x)\delta \phi_{\rm cl}(y)\delta \phi_{\rm cl}(z)}\right|_{\phi_{\rm cl}=0=\psi_{\rm cl}}~=~0,$$ cf. OP's question.
References:
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$^1$ More generally, the 1PI effective action $\Gamma$ inherits affine symmetries of the action $S$ and the path integral measure, cf. e.g. Ref. 1.
Here is a more elementary proof which uses only parity of the Yukawa theory and Lorentz-invariance and does not use the concept of the effective action.
The correlation function is $\Gamma(t_1,\vec{r}_1,t_2,\vec{r}_2,t_3,\vec{r}_3) = \langle \mathrm{T} \phi(t_1, \vec{r}_1),\phi(t_1, \vec{r}_1), \phi(t_1, \vec{r}_1)\rangle$. Let me consider the Fourier image of the correlation function $\Gamma(\omega_1, \vec{p}_1, \omega_2, \vec{p}_2, -\omega_1-\omega_2, -\vec{p}_1 - \vec{p}_2)$ and send the three-momentum $\vec{p}_1 + \vec{p}_2$ to zero with a Lorentz transformation. The correlation function now takes form $\Gamma(\omega_1', \vec{p}', \omega_2', -\vec{p}', -\omega_1'-\omega_2', 0)$.
The parity transform changes the sign of $\Gamma$ and the signs of entering 3-momenta. On the other hand, it is possible to perform a rotation which transforms $\vec{p}'$ into $-\vec{p}'$. So, \begin{equation} \begin{cases} \Gamma(\omega_1', \vec{p}', \omega_2', -\vec{p}', -\omega_1'-\omega_2', 0) = -\Gamma(\omega_1', -\vec{p}', \omega_2', \vec{p}', -\omega_1'-\omega_2', 0) & \mathrm{(parity)}\\ \Gamma(\omega_1', \vec{p}', \omega_2', -\vec{p}', -\omega_1'-\omega_2', 0) = \Gamma(\omega_1', -\vec{p}', \omega_2', \vec{p}', -\omega_1'-\omega_2', 0) & \mathrm{(rotation\,\, invariance)} \end{cases} \end{equation} Therefore, $\Gamma$ is zero for all possible values of 4-momenta.
This answer is wrong! (see the comment below.)
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I did not know this statement, so maybe my answer is wrong. Please be careful.
In the following, $n_v$ denotes the number of vertices, $E_F$, $I_F$ denotes the number of fermion’s external/internal line, $E_B$, $I_B$ denotes the number of boson’s external/internal line.
In your theory, a tree level contribution to 1PI vertex for three bosons does not exist, and we can say there is at least one fermion loop.
From a topology of graph, we may notice the following relations: $$n_v=2I_B+E_B=2I_B+3,$$ $$2n_v=2I_F+E_F=2I_F.$$ Therefore, we can say both $n_v$ and $I_F$ are odd integer.
The sub-diagram corresponding to the fermion-loop should consist of an equal number of vertices and fermion’s internal line lines.
However, now that we know that there are only an odd number of vertices and odd number of fermion’s internal lines, at least one of these fermion loops would be composed of an odd number of vertices and an odd number of fermion loops.
We can say this partial diagram must be zero. To show this, remember that each vertex has $\gamma_5$ and each fermion line has one $\gamma_\mu$. Therefore, the total number of gamma matrices along the loop is odd number $$(\mathrm{odd}\ \mathrm{number})\times (\underset{\gamma_5}{\underbrace{4}}+ \underset{\gamma}{\underbrace{1}}),$$ and the trace of the odd-number of gamma matrices will be zero.
I feel like this discussion needs some more detail to be worked out, but roughly from an argument like this, it seems that we can show that some partial diagram is zero if we take the trace of the gamma matrix.
