Are there really left-chiral particles?

Question

A chiral eigenstate is always a linear combination of a particle and an antiparticle state and a particle or antiparticle state is always a linear combination of chiral eigenstates. Now, how can we then talk about a left-chiral electron or positron, which are said to take part in weak interactions.

---

Background:

Chiral eigenstates can be identified through the projection operators

$$ P_L = \frac{1 - \gamma_5}{2} \quad P_L = \frac{1 + \gamma_5}{2} $$

The corresponding eigenstates are in the Chiral/Weyl Basis

$$ \Psi_L= \begin{pmatrix} \chi_s \\0 \end{pmatrix} \quad \text{and } \quad \Psi_R = \begin{pmatrix} 0 \\ \xi_s \end{pmatrix} $$

where the index $s$ denotes the different possible spin configuations and with the two component Weyl Spinors $\chi$, $\xi$.

---

Particle states can be identified through the solutions of the Dirac equation. In the Chiral/Weyl Basis the solutions (in the rest frame) are

$$ u_s= \begin{pmatrix} \eta_s \\ \eta_s \end{pmatrix} \quad \text{and } \quad v_s =\begin{pmatrix}\zeta_s \\ - \zeta_s \end{pmatrix} $$

Therefore, four linearly independent solutions of the Dirac equation are

$$ e^-_\uparrow = u_1= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} \quad \quad e^-_\downarrow = u_2 =\begin{pmatrix} 0 \\1 \\0 \\1 \end{pmatrix} \quad e^+_\uparrow = \ v_1= \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} \quad \quad e^+_\downarrow =\begin{pmatrix} 0\\1\\0\\-1 \end{pmatrix} $$

which correspond to, for example electron or positron with spin up or down.

---

The solutions of the Dirac equation are what we use in QFT computations. The chiral structure becomes important for weak interactions, because only left-handed particles interact weakly. This is included through $P_L$ in the vertex factor, for example for an incoming muon, decaying weakly we have a factor $ \propto P_L u_s$.

What is $P_L u_s$? The computation in the Weyl/Chiral basis shows

$$ (u_s)_L = P_L u_s = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \eta_s \\ \eta_s \end{pmatrix} = \begin{pmatrix} \eta_s \\ 0 \end{pmatrix} $$

This is no longer a solution of the Dirac equation, so how do we interpret it? We can see that this is a linear combination of a particle and an antiparticle state. For example

$$ (u_1)_L = \frac{1}{2} ( u_1 + u_2^c) = \frac{1}{2} (u_1 + i\gamma_2 u_2^\star) = \frac{1}{2} (\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 & i \sigma_2 \\ -i \sigma_2 & 0 \end{pmatrix} \begin{pmatrix} 0 \\1 \\0 \\1 \end{pmatrix} ) $$

$$ = \frac{1}{2} (\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} ) = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} $$

with the usual charge conjugation transformation $ i\gamma_2 + $ complex conjugation. Anyway, the moral of the story is that a chiral eigenstate is always a linear combination of a particle and an antiparticle state and particle or antiparticle state is always a linear combination of chiral eigenstates. Now, how can we then talk about a left-chiral electron or positron?

PS: Equivalently we can of course see that the solutions of the Dirac equation are always linear combinations of chiral eigenstates, for example $u_s = (u_s)_L + (u_s)_R $

Answer 1

For anyone with similar problems:

The following observation has helped me immensly: We have in fact four particles directly related to an electron:

• A left-chiral electron $\chi_L$, with isospin $-\frac{1}{2}$ and electric charge $-e$,
• A right-chiral anti-left-chiral-electron $(\chi_L)^c=\chi_R$ with isospin $\frac{1}{2}$, electric charge $+e$
• A right-chiral electron $\xi_R$ with isospin $0$ and electric charge $-e$
• A left-chiral anti-right-chiral-electron $(\xi_R)^c=\xi_L$ with isospin $0$ and electric charge $+e$

These four particles are what we describe by the two Weyl spinors inside a Dirac spinor and its charge conjugate respectively.

The Dirac equation tells us that as time passes on a left-chiral electron transforms into a right-chiral and vice versa. Chirality and therefore weak isospin are not conserved quantities. To be able to talk about electrons evolving in time we therefore need to consider left-chiral and right-chiral electrons at the same time, which is why we use Dirac spinors

$$\Psi_{e^-} = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$ This is what is commonly called a physical electron or simply electron. Its charge conjugate is what we commonly call positron

$$\Psi_e^c = \Psi_{e^+} = \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} $$ Only the two particles from above with isospin interact weakly, which means $\chi_L$ and $\chi_R$.

An electron that was created by weak interactions and is therefore purely left-chiral, is described by the spinor $$(\Psi_{e^-})_L = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} $$

The Dirac equation tells us that as time passes on this transformes into a mixture of $\chi_L$ and $\xi_R$, but not $\chi_R$, which would violate charge conservation anyway.

Answer 2

Roughly sketched, for the quantized Dirac field one has: \begin{equation} \hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg], \end{equation} where $r=\pm1$ denotes helicity.

The ${\hat a^r_\mathbf{p}}^\dagger$ operator creates a helicity-$r$ particle state when it acts on the vacuum, while ${\hat b^r_\mathbf{p}}^\dagger$ creates a helicity $r$ antiparticle state (without the Hermitian conjugation, they destroy/annihilate such states): \begin{align} {\hat a^r_\mathbf{p}}^\dagger |0\rangle &= |\textrm{particle with momentum }\mathbf{p} \textrm{ and helicity }r\rangle \\ {\hat b^r_\mathbf{p}}^\dagger |0\rangle &= |\textrm{antiparticle with momentum }\mathbf{p} \textrm{ and helicity }r\rangle \end{align}

Acting with a left projector $P_L$ on this field results in what is aptly called a left-handed field: \begin{equation} \hat\psi_L(x)\equiv P_L\, \hat\psi(x)\sim \int d\mathbf{p}\, \bigg[ u^{(-1)}(p)\, \hat a^{(-1)}_\mathbf{p}\,e^{-ipx}+v^{(+1)}(p)\, {\hat b^{(+1)}_\mathbf{p}}^\dagger e^{ipx}\bigg], \end{equation} while acting with $P_R$ yields the equally aptly named right-handed field: \begin{equation} \hat\psi_R(x)\equiv P_R\, \hat\psi(x)\sim \int d\mathbf{p}\, \bigg[ u^{(+1)}(p)\, \hat a^{(+1)}_\mathbf{p}\,e^{-ipx}+v^{(-1)}(p)\, {\hat b^{(-1)}_\mathbf{p}}^\dagger e^{ipx}\bigg], \end{equation} i.e. schematically: \begin{align} \hat\psi_L &\sim \hat a_{(-1)} + \hat b_{(+1)}^\dagger \\ \hat\psi_R &\sim \hat a_{(+1)} + \hat b_{(-1)}^\dagger \end{align}

The above expressions result from the fact that $P_{L,R}\, u^r(p) = \frac{1}{2}(1\mp r)u^r(p)=\delta_{r,(\mp 1)}u^{r}(p)$ and $P_{L,R}\, v^r(p) = \frac{1}{2}(1\pm r)v^r(p)=\delta_{r,(\pm 1)}v^{r}(p)$.

Indeed, the left-handed field destroys negative helicity particle states, while it creates positive helicity antiparticle states. On the other hand, the right-handed field destroys positive helicity particle states, while it creates negative helicity antiparticle states.

Key point:

Negative helicity states are (perhaps not so aptly) called left-handed, while positive helicity states are called right-handed. Consider, for example, the neutrino field. One then says that the left-handed field $\hat \psi_L$ destroys left-handed neutrino $\nu_L$ states and creates right-handed antineutrino $\overline{\nu_R}$ states. Similarly, one says that the right-handed field $\hat \psi_R$ destroys right-handed neutrino $\nu_R$ states and creates left-handed antineutrino $\overline{\nu_L}$ states.

The motives behind this terminology seem to be covered by Wolphram jonny's answer.

Answer 3

We then talk about a left-chiral electron we do it in an informal way, you are correct that a massive particle cannot be inherently chiral. To see this, let us remember that he handedness of an elementary particle depends on the correlation between its spin and its momentum (helicity). If the spin and momentum are parallel, the particle can be said to be right-handed (positive helicity). If the spin and momentum are antiparallel, the particle is left-handed (negative helicity).

Photons and electrons differ in that a photon's spin must be exactly aligned with its momentum, whereas an electron's spin points at a velocity-dependent angle from its momentum axis (higher velocities shrink the angle). Therefore, the handedness of an electron depends on the projection of its spin along its momentum. Since electrons travel slower than the speed of light, they (and all massive particles) are not inherently chiral. If we observe a moving electron from two different reference frames – first the lab frame, and then a frame moving faster than the lab velocity of the electron – the electron helicity changes sign. So electron helicity (chirality) is not a Lorentz invariant quantity. These facts notwithstanding, it is still practical to refer to electron chirality.

This is because we normally deal with beams of photons and electrons rather than individual particles, and it is useful to refer to the polarization of the beams to describe average spin-momentum correlations. For instance, if a (coherent) beam of photons has a net positive helicity, the beam is said to be right-circularly polarized. A beam with a net negative helicity is left-circularly polarized. An equal mixture of these right- and left-circular states (when the two are 90º out of phase) yields a linearly polarized beam.

For electrons, a net positive or negative helicity is dubbed longitudinal polarization and is analogous to circular polarization for photons. If the average spin direction is perpendicular to the beam axis, the beam has transverse polarization – in analogy to linear polarization for photons. A transversely polarized electron beam is not chiral.