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?

  • 1
    $\begingroup$ The effective Lagrangian term can only be $m_\phi \phi^3$ on dimensional grounds and Lorentz invariance, but S preserves parity, for $\phi$ pseudoscalar; but then the above cubic term breaks parity, so it cannot arise. $\endgroup$ Commented Apr 19, 2023 at 2:07
  • $\begingroup$ Related: physics.stackexchange.com/q/743151/2451 $\endgroup$
    – Qmechanic
    Commented Apr 19, 2023 at 7:59

3 Answers 3


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.


  1. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4 p. 77 + Section 17.2 p. 84.


$^1$ More generally, the 1PI effective action $\Gamma$ inherits affine symmetries of the action $S$ and the path integral measure, cf. e.g. Ref. 1.

  • $\begingroup$ Where does this argument fail for $7$-point 1PI vertex? (I believe it is generically non-zero) $\endgroup$ Commented Apr 19, 2023 at 9:22
  • $\begingroup$ (Although now I'm not 100% sure about it. It is definitely the case for 5-points in 3d pseduscalar Yukawa.) $\endgroup$ Commented Apr 19, 2023 at 9:24
  • $\begingroup$ Hi @Peter Kravchuk. Thanks for the feedback. Are your claims discussed in a reference? Which page? $\endgroup$
    – Qmechanic
    Commented Apr 19, 2023 at 9:30
  • 1
    $\begingroup$ Perhaps the point is that with enough derivatives I can write down parity-even terms with an odd number of $\phi_\text{cl}$. In that case perhaps your argument needs to be supplemented by the statement that it is not possible for 3-point vertex. $\endgroup$ Commented Apr 19, 2023 at 9:39
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    $\begingroup$ Although I think Siam's answer does show that all odd order 1PI vertices vanish in 4d, but that's confusing to me because it makes it look like there is a global $\mathbb{Z}_2$ symmetry rather than a parity. $\endgroup$ Commented Apr 19, 2023 at 9:40

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.

  • $\begingroup$ Sorry... how does vanishing for $\vec{p}_1 + \vec{p}_2 = 0$ imply vanishing for $\vec{p}_1 + \vec{p}_2 \neq 0$? I would've expected one more step about classifying the possible 3 point structures. $\endgroup$ Commented Apr 20, 2023 at 12:41
  • $\begingroup$ This is because correlation function is Lorentz-invariant. If it is zero in the frame where $\vec{p}_1 + \vec{p}_2 = 0$, it is zero in all other frames including the initial one. $\endgroup$
    – E. Anikin
    Commented Apr 20, 2023 at 15:11
  • $\begingroup$ Ok great. So a configuration that stabilizes parity plus rotation exists for any number of fields, but with 3 fields, it can be reached with just a Lorentz boost. Thanks! $\endgroup$ Commented Apr 20, 2023 at 17:27

This answer is wrong! (see the comment below.)


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.

  • $\begingroup$ I think the caveat predicted by Peter Kravchuk above could arise in the following way. Loop diagrams are most easily regulated by continuing the number of dimensions. And the naive dimensional regularization scheme, where one takes $\gamma_5$ to be anti-commuting like it is in 4 dimensions, is known to only be valid under certain conditions. The number of $\gamma_5$ in a trace probably has to be sufficiently small but I could be misremembering. $\endgroup$ Commented Apr 19, 2023 at 12:00
  • $\begingroup$ In general, I think some of the gamma traces mentioned above will be $O(\epsilon)$ and therefore able to cancel divergent terms. $\endgroup$ Commented Apr 19, 2023 at 12:01
  • $\begingroup$ Thank you for comment. Yes you are true, there can be finite contributions, and from the discussion on parity, they will not mix with the contributions of parity even diagrams (unless you choose a regularization that breaks parity), but will cancel each other out between parity odd diagrams. $\endgroup$
    – Siam
    Commented Apr 19, 2023 at 13:51

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