At a fixed time (set $t = 0$), we can write the free complex scalar field as \begin{equation*} \begin{split} \phi(x) &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[b_{-p}^\dagger+a_p\right] e^{ipx}\\ &=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[a_pe^{ipx}+b_p^\dagger e^{-ipx}\right]\\ \phi^{\dagger}(x) &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[a^{\dagger}_{-p}+b_p\right] e^{ipx} \end{split} \end{equation*}

The parity reversal operator $P$ flips the sign of the space coordinate $x$. Its transformation rules (on a Dirac field) are \begin{equation*} Pa_p^sP = a_{-p}^s\quad\text{and}\quad Pb_p^sP = -b_{-p}^s\qquad\Rightarrow\qquad P\psi(t,x)P = \gamma^0\psi(t,-x) \end{equation*}

My question is how does the operator affect the $\phi$ field? Is it right we have

\begin{equation*} \begin{split} P\phi(x)P &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[Pb_{-p}^\dagger P+Pa_pP\right] e^{-ipx}\\ &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[-b_{-p}^\dagger +a_{-p}\right] e^{-ipx} \end{split} \end{equation*} I'm not quite sure how to further simplify this.

Now suppose we have the pseudoscalar Yukawa Lagrangian: $$ L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\bar\psi(i\not\partial-m)\psi-g\gamma^5\phi\bar\psi\psi. $$

It is invariant under the parity transformation: $$P\psi(t,x) P = \gamma^0\psi(t,-x), \qquad P\phi(t,x) P = -\phi(t,-x).$$ If we replace the pseudoscalar term with $g\phi\bar\psi\psi$ instead, it seems like the parity transformation on $\phi$ should be $P\phi(t,x) P = \phi(t,-x)$ to make $L$ invariant. Is that correct? If so, how can I understand the difference of $P\phi(t,x) P$ in two cases, based on the expansion of $\phi$ above?

  • 1
    $\begingroup$ The first part of your question has nothing to do with the second. In the first part you have a free complex scalar field, where the possible parity transformations are $\phi(t,\vec{x}) \to e^{i \alpha} \phi(t, -\vec{x})$. (You forgot the possibility of a complex phase $e^{i \alpha}, \, \alpha \in \mathbf{R}$). In the second part, you have real scalar field interacting with a Dirac fermion. As you wish to couple the pseudoscalar $\bar{\psi} \gamma_5 \psi$ (!) to the field $\phi$, also the latter must be a pseudoscalar. $\endgroup$
    – Hyperon
    Dec 31, 2022 at 7:31
  • $\begingroup$ part 2 ctd. On the other hand, if you wish to couple the real scalar $\phi$ to the scalar $\bar{\psi} \psi$, $\phi$ must also transform as a scalar field under parity. $\endgroup$
    – Hyperon
    Dec 31, 2022 at 8:57

1 Answer 1

  1. For a free complex scalar field, the possible parity transformations are $\Phi(x) \to e^{i \alpha} \Phi(\tilde{x})$, where $\tilde{x}=(t,-\vec{x})$.

  2. For a free real scalar field you have the possibilities $\phi(x)\to \pm \, \phi(\tilde{x})$ (scalar/pseudoscalar).

  3. Which of these possibilities survives in an interacting theory, depends on the type of interaction. The pseudoscalar interaction $\bar{\psi} \gamma_5 \psi \, \phi$ requires a pseudoscalar field $\phi$, whereas the the scalar coupling $\bar{\psi} \psi \, \phi$ requires a scalar field $\phi$.


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