# Parity in Effective Lagrangians

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?

• 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? Commented May 18, 2022 at 13:54
• 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. Commented May 18, 2022 at 13:57
• ? 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? Commented May 18, 2022 at 13:58
• 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. Commented May 18, 2022 at 14:00
• 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? Commented May 18, 2022 at 14:03

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.

• Do I also have to factor in the orbital angular momentum here? Commented May 18, 2022 at 15:01
• 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. Commented May 18, 2022 at 15:02