The spinor inner product in particle physics is given by $\overline{\psi} \psi = \psi^{\dagger} \gamma_0 \psi $, where I take the convention that the zeroth gamma matrix is hermitian while the rest are anti-hermitian. This is invariant under spin group transformations, $\psi \rightarrow e^{\omega_{ab} S^{ab} }\psi$, with $\omega_{ab}$ real parameters and $S^{ab} = \frac{1}{4}[\gamma^a, \gamma^b]$.

However, there is a second invariant inner product given by $\psi^{\dagger} \gamma_0 \gamma_5 \psi$, with $\gamma_5 = i\gamma_0 \gamma_1 \gamma_2 \gamma_3$. My question is, why not the other one? Is it down to the fact that there are observable differences between the two, and one is favoured by experiment?

  • $\begingroup$ What do you mean with "use one inner product and not the other". Where do we use the one and not the other? $\endgroup$ – Hunter Jan 22 '14 at 0:03
  • $\begingroup$ Apologies, I mean in the standard model the spinor inner product is $\psi^{\dagger} \gamma_0 \psi$. My question is why not the other one? $\endgroup$ – Steven Jan 22 '14 at 0:38
  • $\begingroup$ Purely speculative, as I haven't had a proper module on the SM, I guess it has to do with the fact that $\psi^\dagger \gamma_0 \psi$ is used for the Dirac Lagrangian, which yields the famous Dirac equation. This equation is necessary to describe QED (and maybe other processes that I have not yet studied). $\endgroup$ – Hunter Jan 22 '14 at 0:41
  • $\begingroup$ I changed the question according to @Steven : ``why not the other one?'' $\endgroup$ – user28355 Jan 22 '14 at 0:42

Well the $\gamma^5$ term does appear in the standard model.

As you duly note, both $\bar{\psi}\psi$ and $\bar{\psi} \gamma^5 \psi$ are Lorentz invariant, so the question to ask is, how are they different?

It turns out that the difference lies in their transformations under a Parity transformation. Without proof, I claim that under $P:(x,y,z) \to (-x,-y,-z)$, $\bar{\psi}\psi \to \bar{\psi}\psi$, while $\bar{\psi}\gamma^5\psi \to -\bar{\psi}\gamma^5\psi$.

So the first object is a true scalar while the second object is a pseudoscalar (i.e. it picks up a minus sign under $P$). Thus, if we have the second term in the Lagrangian, it breaks Parity symmetry.

It turns out (experimentally) that $P$ is indeed broken in the weak interaction. So we do have a weak interaction vertex (for a electron + neutrino $ \to W^-$ interaction, for example) that looks like

\begin{align} \sim -\frac{-i g_{\mu\nu}}{2\sqrt{2}}\gamma^\mu(1-\gamma^5). \end{align}

This is an axial vector coupling, which is not parity invariant.

However the pseudoscalar $\bar{\psi}\gamma^5\psi$ doesn't appear in the SM because the object $(1-\gamma^5)/2 = \mathbb{P}_L$ is the projector onto the left-handed component of a spinor, so it can only appear if you want a theory in which the mass term of the right-handed and left-handed spinors are different, which we don't observe experimentally.

  • $\begingroup$ As a follow up question: In (3+1)-dimensions, the measure for the action integral transforms with a minus sign under a parity transformation. Wouldn't that mean that the action is only invariant if the Lagrangian is a pseudoscalar under parity? $\endgroup$ – Steven Jan 22 '14 at 17:42
  • $\begingroup$ Hm, why would the measure change? If you change $x\to-x$ the bounds of integration are also flipped: i.e. the integral $\int_{-\infty}^\infty dx \to - \int_{\infty}^{-\infty} dx = \int_{-\infty}^\infty dx$ so it seems to me the measure is invariant? $\endgroup$ – nervxxx Jan 22 '14 at 17:58

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