Is Parity really violated? (Even though neutrinos are massive)

Question

The weak force couples only to left-chiral fields, which is expressed mathematically by a chiral projection operator $P_L = \frac{1-\gamma_5}{2}$ in the corresponding coupling terms in the Lagrangian.

This curious fact of nature is commonly called parity violation and I'm wondering why? Does this name make sense from a modern point of view?

My question is based on the observation:

A Dirac spinor (in the chiral representation) of pure chirality transforms under parity transformations:

$$ \Psi_L = P_L \Psi = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} \rightarrow \Psi_L^P = \begin{pmatrix} 0\\ \chi_L \end{pmatrix} \neq \Psi_R$$

Chirality is a Lorentz invariant quantity and a left-chiral particle is not transformed into a right-chiral particle by parity transformations.(The transformed object lives in a different representation of the Lorentz group, where the lower Weyl spinor denotes the left-chiral part.)

In understand where the name comes from historically (see the last paragraph) but wouldn't from a modern point of view chirality violation make much more sense?

Some background:

Fermions are described by Dirac spinors, transforming according to the $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ representation of the (double cover of the) Lorentz group. Weyl spinors $\chi_L$ transforming according to the $(\frac{1}{2},0) $ representation are called left-chiral and those transforming according to the $(0,\frac{1}{2})$ representation are called right-chiral $\xi_R$. A Dirac spinor is (in the chiral representation)

$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}$$

The effect of a parity transformation is

$$ (\frac{1}{2},0) \underbrace{\leftrightarrow}_P (0,\frac{1}{2}),$$ which means the two irreps of the Lorentz group are exchanged. (This can be seen for example by acting with a parity transformation on the generators of the Lorentz group). That means a parity transformed Dirac spinor, transforms according to the $(0,\frac{1}{2}) \oplus (\frac{1}{2},0) $ representation, which means we have

$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^P = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix} $$

Now we can examine the effect of a parity transformation on a state with pure chirality:

$$ \Psi_L = P_L \Psi = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} \rightarrow \Psi_L^P = \begin{pmatrix} 0\\ \chi_L \end{pmatrix}$$

This means we still have a left-chiral spinor, only written differently, after a parity transformation and not a right-chiral. Chirality is a Lorentz invariant quantity. Nevertheless, the fact that only left-chiral particles interact weakly is commonly called parity violation and I'm wondering if this is still a sensible name or only of historic significance?

Short remark on history

I know that historically neutrinos were assumed to be massless, and for massless particles helicity and chirality are the same. A parity transformation transforms a left-handed particle into a right-handed particle. In the famous Wu experiment, only left-handed neutrinos could be observed, which is were the name parity violation comes from. But does this name make sense today that we know that neutrinos have mass, and therefore chirality $\neq$ helicity.

Answer 1

Yes, parity is really violated, even if neutrinos are massive. You seem to be confusing the relationship between parity, helicity, and chirality in the modern standard model with the physical symmetry operation of spatial inversion.

Wu's experiment did not measure neutrino helicity. Wu and collaborators prepared a thin layer of a beta-emitting nucleus with rather high spin, polarized the nuclei, and observed that the beta particles were more likely to be emitted from the "south pole" of the nucleus than from the "north pole." Why is this a parity violation? The way you define the "north pole" of a rotating object is to grab it with your right hand, so that the fingers of your right hand curl around in the same sense as the rotation; the "north pole" is the one where your thumb goes, and the "south pole" is the other one. Mirror reflection (which is a special case of the parity transformation) turns your right hand into a left hand, and changes which pole you label as "north" for the same sense of rotation.

The following paper in that issue of PRL, by Garwin et al, shows that muons produced in the decay $\pi\to\mu+\nu$ are polarized. Such polarization is a violation of parity symmetry because the pion has zero spin, and a stopped pion can therefore express no preference about the spin directions of its daughters. This experiment was also the first to measure the magnetic moments for the muon and antimuon.

The statement that one can produce polarized $\mu$ from spinless $\pi$ in the decay $\pi\to\mu+\nu$ suggests that in that decay the neutrinos should also be produced polarized. However the first measurement of neutrino helicity is generally taken to be the one by Goldhaber and collaborators, later in the same year. Analysis of the Goldhaber experiment requires you to spend a little time thinking carefully about nuclear spins.

It was actually discovered in 1927 that unpolarized beta sources produce slightly left-handed polarized electrons, though the significance was not understood at the time and the paper was forgotten until unearthed by Grodzins in 1958. Allan Franklin calls it "the non-discovery of parity non-conservation."

Parity violation is real in the visceral sense that if you were to show me a (sufficiently detailed) photograph of a weak interaction experiment with polarized particles, I could in principle tell you whether the photograph had been reflected or not.

Answer 2

Weak interactions with $W$ and $Z$ gauge bosons violate parity simply because the righ-handed and the left-handed fermions coupled differently to $W$ and $Z$. For example, the $W$'s couple only to the left-handed fields. A parity inviariant dynamics would require that both left- and right- fields couple in the same way to the gauge vector since they get exchanged under parity transformation.

Slightly more technically , the gauge bosons must transform as polar vectors under parity (because they form a covariant derivative with $\partial_\mu$ which is polar) and hence in order to preserve parity they should couple to polar vector current $J^\mu$ such as $\bar{\Psi}\gamma^\mu\Psi$. Instead the $W$'s couple to $V-A$ current of the type $\bar{\Psi}\gamma^\mu(1-\gamma^5)\Psi$. While the $\gamma^\mu$ term transforms as polar vector, it's not difficult to see that the $\gamma^\mu\gamma^5$ term transform instead as an axial vector. To restore parity one should add a right-current $\bar{\Psi}\gamma^\mu(1+\gamma^5)\Psi$ that couples with the same strength to $W$ so that the $\gamma^5$ would drop.

Answer 3

Let's check that parity is violated by the weak interaction lagrangian: $$\mathcal{L}(x) = \bar{\psi}(x) \gamma^\mu \frac{(1-\gamma^5)}{2} \psi(x) W_\mu(x)$$ Saying that parity is violated means that the transformed lagrangian $\mathcal{L}'(x)$ is not equal to the old lagrangian resulting from new coordinates $\mathcal{L}(x')$ where $x'^0 = x^0$ and $\vec{x'} = -\vec{x}$. In order to evaluate $\mathcal{L}'(x)$, one has to know the transformations of the fermion field $\psi(x)$ and the boson field $W_\mu(x)$. One can show that $\psi'(x) = \eta \gamma^0 \psi(x')$ with $\eta = \pm 1$ the intrinsic parity of the field $\psi$ and $W'_0(x) = \eta_W W_0(x')$, $W'_i(x) = -\eta_W W_i(x')$ (usual transformation of a vector field) with $\eta_W = \pm 1$ the intrinsic parity of the vector field. Hence: $$\mathcal{L}'(x) = \bar{\psi'}(x) \gamma^0 \frac{(1-\gamma^5)}{2} \psi'(x) W'_0(x)+\bar{\psi'}(x) \gamma^i \frac{(1-\gamma^5)}{2} \psi'(x) W'_i(x)$$ giving: $$\mathcal{L}'(x) = \eta^2 \eta_W \left( \bar{\psi}(x') \gamma^0 \gamma^0 \frac{(1-\gamma^5)}{2} \gamma^0 \psi(x') W_0(x') - \bar{\psi}(x') \gamma^0 \gamma^i \frac{(1-\gamma^5)}{2} \gamma^0 \psi(x') W_i(x')\right)$$ Knowing the commutation rules of the $\gamma$ matrices, $(\gamma^0)^2 = 1$ and $\eta^2 =1$, it becomes: $$\mathcal{L}'(x) = \eta_W \left( \bar{\psi}(x') \gamma^0 \frac{(1+\gamma^5)}{2}\psi(x') W_0(x') + \bar{\psi}(x') \gamma^i \frac{(1+\gamma^5)}{2}\psi(x') W_i(x')\right) = \eta_W \bar{\psi}(x') \gamma^\mu \frac{(1+\gamma^5)}{2} \psi(x') W_\mu(x')$$ As you see, whatever the value of $\eta_W = \pm 1$, the transformed lagrangian cannot be equal to $\mathcal{L}(x')$: $$\mathcal{L}(x') = \bar{\psi}(x') \gamma^\mu \frac{(1-\gamma^5)}{2} \psi(x') W_\mu(x')$$ justifying the violation of the parity by the weak interaction.

Answer 4

Okay, I think I have an idea why the terminology is used, but I think this argument makes little sense:

The Lagrangian term describing weak interactions is of the form

$$ \bar \Psi \gamma_\mu P_L \Psi W^\mu $$

Under parity transformations $ \Psi \rightarrow \gamma_0 \Psi$ and $ \bar \Psi \rightarrow \bar \Psi \gamma_0 $, therefore

$$ \bar \Psi \gamma_\mu P_L \Psi W^\mu \rightarrow \bar \Psi \gamma_0 \gamma_\mu P_L \gamma_0 \Psi W^\mu $$

We can transform the parity transformed Lagrangian, by using the explicit form of $P_L = \frac{1-\gamma_5}{2}$ and $\{\gamma_5, \gamma_\mu \}=0$ :

$$ \bar \Psi \gamma_0 \gamma_\mu P_L \gamma_0 \Psi W^\mu = \bar \Psi \gamma_0 \gamma_\mu \gamma_0 P_R \Psi W^\mu = \bar \Psi \gamma_\mu P_R \Psi W^\mu $$

which shows that the weak interaction term in the Lagrangian is not invariant under parity transformations. I think this argument is wrong!

The above discussion misses that under parity transformations $\gamma_5 \rightarrow - \gamma_5$. If we take this in account the corresponding term in the Lagrangian is invariant und weak interactions are invariant under parity transformations.

We can see this, because for an ordinary Dirac spinor we have the projection operator:

$$P_L \Psi = \begin{pmatrix}1 & 0\\ 0 &0 \end{pmatrix} \begin{pmatrix} \chi_L \\ \xi_R\end{pmatrix} = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} = \Psi_L $$

and for the parity transformed Dirac spinor $\Psi^P$ the left-chiral projection operator is

$$P_L^P \Psi^P = \begin{pmatrix}0 & 0\\ 0 &1\end{pmatrix} \begin{pmatrix} \xi_R \\ \chi_L\end{pmatrix} = \begin{pmatrix} 0\\ \chi_L \end{pmatrix} = \Psi_L^P $$

The parity transformed Dirac spinors live in a different representation and therefore we need different projection operators. The discussion above shows that under parity transformation

$$ P_L \rightarrow P_L^P = \frac{1+\gamma_5}{2} = P_R$$ and therefore the parity transformed in the Lagrangian reads:

$$ \bar \Psi \gamma_0 \gamma_\mu P_R \gamma_0 \Psi W^\mu = \bar \Psi \gamma_0 \gamma_\mu \gamma_0 P_L \Psi W^\mu = \bar \Psi \gamma_\mu P_L \Psi W^\mu $$

which shows the invariance.

Answer 5

I guess it is because you first of all change sign of $\vec x$ to $- \vec x$ in physical space(this is parity transformation in a nutshell). All this peculiar algebra concerning left and right chirality fields comes from $J = 1/2$ Lorenz group representation, so transformation rules are defined as representatives of parity transformation of physical space.