Given the following Lagrangian

$$\mathscr{L} = c\frac{g}{m}\bar{\psi}_A\Gamma_5\gamma^\mu\psi_B (i\partial_\mu)\phi$$

where $\Gamma_5 \in \{\gamma_5, 1\}$, for two spin one-half particles $A$ and $B$ and a spin $0$ particle $\phi$.

I'm trying to figure out how to determine when $\Gamma_5 = 1$ and $\Gamma_5 = \gamma_5$

I know the Lagrangian should be a scalar. I also know how the bilinear covariants transform.

Let now $A$ and $B$ both have $J^P = \tfrac{1}{2}^+$ and let $\phi$ have $J^P=0^-$.

If $\Gamma_5=1$, then $\bar\psi\gamma^\mu\psi$ transforms like a vector. $\partial_\mu \rightarrow -\partial_\mu$. To my understanding it follows then that $\bar\psi(\partial\!\!/ ~\phi)\psi$ transforms like a pseudoscalar. Likewise, if $\Gamma_5=\gamma_5$, then $\bar\psi\gamma_5(\partial\!\!/~\phi)\psi$ transforms like a scalar.

How can I factor in the parities above?If I just multiply them, I would find that the pion nucleon nucleon interaction $$\mathscr{L}=\frac{g}{m}\bar\psi\gamma_5(\partial\!\!/~\vec\pi)\cdot \vec\tau ~\psi \tag{2}$$ transforms like a pseudoscalar. But the Lagrangian should be scalar. So what am I missing?

Also, taking the $N(1535)$ into account, which has spin parity $J^P=\tfrac{1}{2}^-$, we do not need a $\gamma_5$. How can I see this rigorously?

  • $\begingroup$ Your text does explain how $\psi(x)\to \gamma^0 \psi(-x)$ implies $\bar \psi(x) \to \bar \psi(-x)\gamma^0$, under P, no? $\endgroup$ May 18, 2022 at 13:54
  • $\begingroup$ I can not follow. Perhaps a better question would be, what changes for a $\tfrac{1}{2}^+$ and a $\tfrac{1}{2}^-$ spinor. Likewise a $0^+$ and a $0^-$ spin-zero field. $\endgroup$ May 18, 2022 at 13:57
  • $\begingroup$ ? What is your question? You already conceded that the term in the last sentence of the penultimate paragraph transforms like a scalar. What more do you want? $\endgroup$ May 18, 2022 at 13:58
  • $\begingroup$ There are Lagrangians of the same structure with or without a $\gamma_5$ which clearly depends on the $J^P$ of the involved particles. I just don't know how. That's my question. $\endgroup$ May 18, 2022 at 14:00
  • $\begingroup$ φ and π are both pseudo scalars. You also understand that $\bar \psi_A \psi_B$ is a scalar. What is the problem? Can you state it, clearly, in your question? $\endgroup$ May 18, 2022 at 14:03

1 Answer 1


Perhaps a better question would be, what changes for a ${1\over 2}^+$ and a ${1\over 2}^-$ spinor. Likewise a $0^+$ and a $0^−$ spin-zero field.

I frankly don't see your point: You multiply the intrinsic parities of all fields involved. The γ matrices and derivatives follow the rules you seem to know, and are in your text: so intercalated $\gamma_5$ s produce minus signs under parity. The signs multiply.

So, confirm the following statements:

  1. $\bar \psi_A \psi_B$ and $\bar \psi_A \partial\!\! / ~\psi_B$ are scalars for A and B having the same parity, but pseudoscalars otherwise.

  2. Insertion of a $\gamma_5$ flips the parity of the bilinear.

  3. Insertion of a spin-zero field does not affect the parity if the field is scalar, but flips the parity if it is a pseudoscalar. It does not matter whom the gradient is acting on.

  • You should be able to see how (2) is a scalar, with a product of parities, $-~ -=+$ (π is a pseudoscalar, of course).

But if you had opposite intrinsic parity fermions in the bilinear, that would flip the +, so you'd need to eliminate the $\gamma_5$ to get a scalar term.

  • $\begingroup$ Do I also have to factor in the orbital angular momentum here? $\endgroup$ May 18, 2022 at 15:01
  • $\begingroup$ No, it is hiding in the derivative. If you wish to look at particles, instead of fields, then yes, but there the rules are different, as your text should emphasize. $\endgroup$ May 18, 2022 at 15:02
  • $\begingroup$ Sorry, didn't have time to check this in detail, will accept shortly $\endgroup$ May 21, 2022 at 14:12

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