The Lagrangian density for the scalar Yukawa theory is given by \begin{equation} L=\frac12(\partial_\mu\phi)(\partial_\nu\phi)\eta^{\mu\nu}-\frac{m^2}{2}\phi^2+\overline\psi(i\gamma_\nu\partial^\mu-M)\psi-g\overline\psi\psi\phi-\frac{\lambda}{4!}\phi^4 \end{equation} where $\eta_{\mu\nu}$ is the Minkowskian metric, $\partial_\mu\equiv\frac{\partial}{\partial x^\mu}$, and $m$ is the mass of the scalar field, while $M$ is the mass of the Dirac field in the Lagrangian above.

Obtain the transformation of the action under Parity ($P$). Is the action invariant under $P$ ?

I was doing this problem and I think the answer is Yes, but I was looking for some easier ways of calculating it. For example, is it possible to use the fact that Hamiltonian is parity invariant to solve it? Or do we have to calculate action $S=\int d^4x L$ under parity explicitly?


1 Answer 1


To determine whether the action of the scalar Yukawa theory is invariant under parity transformation $P$, we should ideally analyze how each term in the Lagrangian density $\mathcal{L}$ transforms under parity. However, before diving into a direct calculation, let's consider the broader context and the nature of parity transformation.

Understanding Parity Transformation ((P))

Parity transformation is a discrete symmetry that involves the inversion of spatial coordinates. In three dimensions, it changes the sign of the spatial part of a four-vector: $\vec{x} \rightarrow -\vec{x}$, but leaves the time component unchanged.

For a field theory, we consider how different fields transform under parity:

  1. Scalar Field $(\phi)$: Scalar fields are invariant under parity: $\phi(x) \rightarrow \phi(-x)$.

  2. Dirac Field $(\psi)$: The Dirac field typically acquires a phase under parity: $\psi(x) \rightarrow \gamma^0 \psi(-x)$, where $\gamma^0$ is the time-like gamma matrix.

  3. Derivatives: The derivative operator transforms as $\partial_\mu \rightarrow (\partial_0, -\vec{\nabla})$ under parity.

Lagrangian Density Terms

The Lagrangian density for scalar Yukawa theory is given by:

$ \mathcal{L} = \frac{1}{2} (\partial_\mu \phi)(\partial^\mu \phi) - \frac{m^2}{2} \phi^2 + \bar{\psi}(i\gamma^\mu \partial_\mu - M)\psi - g\bar{\psi}\psi\phi - \frac{\lambda}{4!}\phi^4 $

To check for parity invariance, we need to see how each term transforms under parity:

  1. Kinetic Term for $\phi$: The kinetic term $(\partial_\mu \phi)(\partial^\mu \phi)$ is invariant under parity, as both the scalar field and its derivative change signs in the spatial components, leaving the term unchanged.

  2. Mass Term for $\phi$: The mass term $\phi^2$ is invariant, as scalar fields do not change under parity.

  3. Kinetic and Mass Terms for $\psi$: The Dirac terms involving $\psi$ and $\partial_\mu \psi$ require careful consideration. The gamma matrices and the derivative must be considered together, but these terms are typically constructed to be parity-invariant.

  4. Yukawa Interaction Term: The Yukawa interaction term $g\bar{\psi}\psi\phi$ should also be invariant under parity if each field transforms correctly.

  5. Quartic Term: The $\phi^4$ term is invariant under parity.

Parity Invariance of the Action

The action $S = \int d^4x \, \mathcal{L}$ is parity-invariant if $\mathcal{L}$ is invariant under parity transformation. Given the transformation properties of scalar and Dirac fields, and their derivatives, it appears that each term in $\mathcal{L}$ is invariant under parity, and hence the action should be invariant as well.

While the Hamiltonian's parity invariance can be a useful indicator, it is generally more straightforward to check parity invariance directly at the level of the Lagrangian.

Based on the transformation properties of the fields and derivatives under parity, the action of the scalar Yukawa theory appears to be invariant under parity.

  • $\begingroup$ Sorry I don’t fully understand, what does your “Scalar fields are invariant under parity: ϕ(x)→ϕ(−x).” mean? Do you mean $P\phi(-x)P^{-1}=\phi(x)$? $\endgroup$
    – Ho-Oh
    Nov 15, 2023 at 8:50
  • $\begingroup$ Also, why do we say that Dirac terms are typically constructed to be parity-invariant? $\endgroup$
    – Ho-Oh
    Nov 15, 2023 at 9:16

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