# Conceptual interpretation of the left- and right-handed spinor representations of the Lorentz group

For massive spinors "right-handed" and "left-handed" chirality isn't tied so much to true rotations, as to the casting of Lorentz transformations as "space-time rotations". In this case, a very popular short answer to the conceptual question is that Lorentz transformations "rotate" $(1/2, 0)$-spinors one way in space-time, and $(0, 1/2)$-spinors the opposite way, while the space inversion corresponding to the parity transformation "rotates" one type of spinor into the other.

But the correct understanding of spinor chirality, and its connection to parity, is closely related to another pair of very familiar concepts, the contravariance and covariance duality of 4-vectors. Recall that in Minkowski space-time a covariant vector $v_\mu$ is the space-inverted of its contravariant counterpart $v^\mu$, and so the parity transformation is the metric itself. The more confusing issue is that, regardless of this connection, each irrep induces its own distinct set of contravariant and covariant spinors, the two constructions being related by a complex conjugation that in a quantum setting compounds the usual role of complex conjugation in defining bra-ket duality.

Now for some details:

**Why the "right-handed" and "left-handed" monikers**:

Say a given space-time Lorentz transformation, for a rotation of angle $\theta$ around axis $\vec{n}_\theta$, $\vec{\theta} = \theta\vec{n}_\theta$, and a boost of rapidity $\zeta$ in direction $\vec{n}_\zeta$, $\vec{\zeta} = \zeta\vec{n}_\zeta$, reads $$ L = e^{\vec{\theta}\cdot\vec{J} + \vec{\zeta}\cdot \vec{K}} $$ where $\vec{J}$, $\vec{K}$ are the usual rotation and boost generators in Minkowski space-time. Then its equivalents on the (1/2,0) and (0,1/2) spinor reps are $$ \Lambda_{(1/2,0)} = e^{\vec{w}\cdot\vec{\sigma}/2} \equiv e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}, \;\;\;\;\;\Lambda_{(0,1/2)} = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = e^{-\vec{w}^*\cdot\vec{\sigma}/2} \equiv e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2},\;\;\;\;\;\vec{w} = -i\vec{\theta} + \vec{\zeta} $$ with generators $\vec{J} \rightarrow i\vec{\sigma}/2$, $\vec{K} \rightarrow -\vec{\sigma}/2$ on $(1/2,0)$, and $\vec{J} \rightarrow -i\vec{\sigma}^*/2$, $\vec{K} \rightarrow -\vec{\sigma}^*/2$ on $(0, 1/2)$.

Now look at the sign of the rapidity in the two transformations. It is exactly opposite: if in one rep the transformation is given by a boost $\vec{\zeta}$, in the other rep it is given by the inverse boost $-\vec{\zeta}$. So although both $\Lambda_{(1/2,0)}$ and $\Lambda_{(0,1/2)}$ correspond to the **same** space-time transform $L$, the change in sign of the boost parameter makes it "as if" they generate opposite formal "rotations": "right-handed" and "left-handed".

**Chirality and parity**:
On the other hand, the switch from the boost $\vec{\zeta}$ to the inverse $-\vec{\zeta}$ is just the usual relationship between contravariant and covariant transformations. Indeed, for
$$
v'^\mu = L^\mu_{\;\;\nu} v^\nu\;\; \leftrightarrow \;\;v'_\mu = v_\nu(L^{-1})^\nu_{\;\;\mu} = [(L^{-1})^T]_\mu^{\;\;\nu} v_\nu
$$
the exponential forms
$$
L = e^{\vec{\theta}\cdot\vec{J} + \vec{\zeta}\cdot\vec{K}} \;\; \leftrightarrow \;\; (L^{-1})^T = e^{\vec{\theta}\cdot\vec{J} - \vec{\zeta}\cdot\vec{K}}
$$
show that if a contravariant 4-vector transforms under a rotation $\vec{\theta}$ and a boost $\vec{\zeta}$, then the parity transformed covariant 4-vector transforms under the same rotation $\vec{\theta}$ and the inverse boost $-\vec{\zeta}$. Likewise, from the spinor transforms
$$
\Lambda_{(1/2,0)} = e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}, \;\;\;\;\;\Lambda_{(0,1/2)} = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2}
$$
we see that, in a completely similar way, if a $(1/2,0)$ spinor transforms under a rotation $\vec{\theta}$ and a boost $\vec{\zeta}$, then its $(0, 1/2)$ counterpart transforms under the same rotation $\vec{\theta}$ and the inverse boost $-\vec{\zeta}$. Hence a $(1/2,0)$ (right-handed) spinor is sometimes referred to as a *contraspinor*, while a $(0,1/2)$ (left-handed) one is then a *cospinor* (see for instance Andrew Steane's intro to spinors; another nice intro is Schulten's Ch.11 in his QM book).

The contravariant to covariant spinor transformation involves a complex conjugation and is given by
$$
{\hat \chi}\;\;\rightarrow \;\;{\hat \eta} = i\sigma_2 {\hat \chi}^* \equiv \left(\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right) {\hat \chi}^*
$$
This is so because if ${\hat\chi} \rightarrow \Lambda_{(1/2,0)} {\hat\chi}$ under a Lorentz transformation, then $\hat \eta$ transforms as
$$
{\hat \eta} = i\sigma_2 {\hat \chi}^* \;\;\rightarrow \;\; i\sigma_2 \left(\Lambda_{(1/2,0)} {\hat\chi}\right)^* = i\sigma_2 \Lambda_{(1/2,0)}^* (i\sigma_2)^\dagger (i\sigma_2){\hat\chi}^* = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger {\hat \eta} \equiv \Lambda_{(0,1/2)} {\hat \eta}
$$
where the latter equality relies on $(i\sigma_2)^\dagger (i\sigma_2) = I$ and the transformation of the spin matrices $\sigma_j$ under $i\sigma_2$. If we define in addition $\sigma_0 = I$, then for the latter we have
$$
(i\sigma_2) \sigma^*_0(i\sigma_2)^\dagger = \sigma_0
$$
$$
(i\sigma_2)\sigma^*_j (i\sigma_2)^\dagger = - \sigma_j, \;\;\;j=1,2,3
$$
which confirms that $i\sigma_2$ is actually a parity transformation, as expected from the transformation of the $\Lambda$-s,
$$
i\sigma_2 \Lambda_{(1/2,0)}^* (i\sigma_2)^\dagger = \left(\Lambda_{(1/2,0)}^{-1}\right)^\dagger = \Lambda_{(0,1/2)} \;\;\; \leftrightarrow \;\;\;i\sigma_2 \left[e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2}\right]^* (i\sigma_2)^\dagger = e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2}
$$
That is, $i\sigma_2$ leaves the rotation in place, but reverses the direction of the boost - *and this is precisely the definition of a space inversion*. Which means that $i\sigma_2$ implements not just a change of rep, but a parity transform, just like the metric in Minkowski space (compare the $\Lambda$ relation above with the Lorentz transform relation $gLg = (L^{-1})^T \Leftrightarrow gLg^T = (L^{-1})^T $). Or turning the statement on its ear, **applying a parity transform changes the spinor rep, hence its chirality**. From this point of view, the right-handed and left-handed terminology can also be understood as labeling spinor transformation properties in right-handed and left-handed 3D reference frames.

But if this is so, why aren't there also 2 chiral irreps for regular 4-vectors? Simply because 4-vectors are real and in this case the contravariant to covariant map is a linear similarity transformation. By Schur's Lemma the reps are then equivalent. Despite the apparent isomorphism with contravariant and covariant representations, respectively, the $(1/2, 0)$ and $(0, 1/2)$ reps are related by an antilinear, not linear, transformation, and formally do not satisfy Schur's Lemma.

**Note added in proof: Another simple way to show the connection between chirality and contravariance/covariance, without going into the full blown formalism**

Let $\hat \chi$ be any normalized spinor. It is always possible to parametrize it as $$ {\hat \chi} = \left(\begin{array}{c}\chi^0 \\ \chi^1\end{array}\right) = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right), \;\;\; {\hat \chi}^\dagger {\hat \chi} = 1 $$ with $\chi^0, \chi^1\in {\mathbb C}$, $u \equiv 2\chi^{0*}\chi^1 = u^1 + iu^2$, $|u|^2 = 4 |\chi^0|^2(1 - |\chi^0|^2)$ (from $|\chi^0|^2 + |\chi^1|^2 = 1$). The upper contravariant indices are foretelling, but otherwise arbitrary at this point.

The first key observation is that the spin projector associated to $\hat \chi$ reads
$$
{\hat \chi}{\hat \chi}^\dagger = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right) \left(\begin{array}{cc}\chi^{0*} & u^*/(2\chi^0)\end{array}\right) = \left(\begin{array}{cc} |\chi^0|^2 & u^*/2 \\ u/2 &1 - |\chi^0|^2) \end{array}\right) = \frac{1}{2} \left(\begin{array}{cc} 1 + \left[\;2|\chi^0|^2-1\;\right] & u^* \\ u &1 - \left[\;2|\chi^0|^2 -1\;\right] \end{array}\right) =
$$
$$
= \frac{1}{2} \left[ I + u^1 \sigma_1 + u^2 \sigma_2 + (2|\chi_0|^2 -1)\sigma_3\right] \equiv \frac{1}{2} \left[ u^0 \sigma_0 + u^1 \sigma_1 + u^2 \sigma_2 + u^3\sigma_3\right]
$$
where $\sigma_\mu$ are Pauli matrices as usual (the lower index has no particular significance), $u^0 = 1$, and $u^3 = 2|\chi_0|^2 -1$, and also
$$
u^\mu = Tr\left(\sigma_\mu {\hat \chi}{\hat \chi}^\dagger\right) = \chi^\dagger \sigma_\mu \chi
$$
A 2nd observation is that we always have
$$(u^1)^2 + (u^2)^2 + (u^3)^2 = |u|^2 + (2|\chi^0|^2 -1 )^2 = 1$$
hence
$$
\det \left({\hat \chi}{\hat \chi}^\dagger \right) = (u^0)^2 - (u^1)^2 - (u^2)^2 - (u^3)^2 = 0
$$
The latter identity obviously suggests that $u^\mu = \chi^\dagger \sigma_\mu \chi$ might behave as a null 4-vector. And indeed, irrespective of the particular rep, under a Lorentz transformation $\Lambda$, $|\det(\Lambda)| = 1$, $\hat \chi$ and its projector transform as
$$
\hat \chi' = \Lambda \hat \chi, \;\;\;{\hat \chi'}({\hat \chi'})^\dagger = \Lambda {\hat \chi}{\hat \chi}^\dagger \Lambda^\dagger
$$
By the same reasoning as above, the transformed projector must again be of the form
$$
{\hat \chi'}({\hat \chi'})^\dagger = \frac{1}{2} \left[ u'^0 I + u'^1 \sigma_1 + u'^2 \sigma_2 + u'^3\sigma_3\right]
$$
$$
u'^\mu = Tr\left[\sigma_\mu {\hat \chi'}({\hat \chi'})^\dagger\right] = (\chi')^\dagger \sigma_\mu \chi'
$$
hence
$$
\det \left[{\hat \chi'}({\hat \chi'})^\dagger \right] = (u'^0)^2 - (u'^1)^2 - (u'^2)^2 - (u'^3)^2 = 0
$$
In other words, $(u^0)^2 - (u^1)^2 - (u^2)^2 - (u^3)^2 = 0$ is a Lorentz invariant, and $u^\mu$ is necessarily a null 4-vector. But it is not clear yet if it is a contravariant or a covariant one. However, from
$$
{\hat \chi'}({\hat \chi'})^\dagger = \Lambda {\hat \chi}{\hat \chi}^\dagger \Lambda^\dagger = \frac{1}{2} \left[ u^0 \Lambda \sigma_0 \Lambda^\dagger + u^1 \Lambda \sigma_1 \Lambda^\dagger + u^2 \Lambda \sigma_2 \Lambda^\dagger + u^3 \Lambda \sigma_3 \Lambda^\dagger \right] \equiv \frac{1}{2} \left[ u'^0 I + u'^1 \sigma_1 + u'^2 \sigma_2 + u'^3\sigma_3\right]
$$
it follows that the transformation law for the $u^\mu$ must be
$$
u'^\mu = \frac{1}{2}\sum_\nu{\left[\text{Tr}\left( \sigma_\mu \Lambda \sigma_\nu \Lambda^\dagger \right) \right] u^\nu}
$$
For the specific forms of $\Lambda_{(1/2,0)}$ and $\Lambda_{(0,1/2)}$ this reads
$$
(1/2, 0):\;\;\;\;\;u'^\mu = \sum_\nu{\left[\text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \right) \right] u^\nu} =
$$
$$
(0, 1/2):\;\;\;\;\;u'^\mu = \sum_\nu{\left[\text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) \right] u^\nu}
$$
Since the $u^\mu$ are 4-vectors, the two transformations above must be Lorentz transforms. But notice that if for $(1/2, 0)$ (with no particular placement of the indices)
$$
L_{\mu\nu}(\vec{\theta},\vec{\zeta}) = \text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} - \vec{\zeta})\cdot\vec{\sigma}/2} \right)
$$
then for $(0, 1/2)$
$$
{\bar L}_{\mu\nu}(\vec{\theta},\vec{\zeta}) = \text{Tr}\left( \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) = \text{Tr}\left( \sigma_\nu e^{(i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \sigma_\mu e^{(-i\vec{\theta} + \vec{\zeta})\cdot\vec{\sigma}/2} \right) = L_{\nu\mu}(-\vec{\theta},-\vec{\zeta}) = \left( \left[ L(\vec{\theta},\vec{\zeta})\right]^{-1}\right)^T_{\mu\nu}
$$
So we always have

$$
(1/2, 0):\;\;\;\;\;u'^\mu = \sum_\nu{L_{\mu\nu}(\vec{\theta},\vec{\zeta}) u^\nu}
$$
$$
(0, 1/2):\;\;\;\;\;u'^\mu = \sum_\nu{ \left( \left[ L(\vec{\theta},\vec{\zeta})\right]^{-1}\right)^T_{\mu\nu} u^\nu}
$$
Compare this with 4-vector contravariant and covariant transforms and it follows that either in $(1/2,0)$ the $u^\mu$ are contravariant and the counterparts in $(0, 1/2)$ are covariant, or conversely. The same applies to any other spinorial bilinears. In other words, we see again that the $(1/2, 0)$ and $(0, 1/2)$ reps are dual to each other. Actually for $(1/2,0)$ it turns out that $(1/2) \text{Tr}\left( \sigma_\mu \Lambda \sigma_\nu \Lambda^\dagger \right) = L^\mu_{\;\;\nu}$, etc, and the corresponding spin 4-vector is indeed contravariant.

**A final note on the nature of parity transformed spinors:**

The parity transformation $i\sigma_2$, usually referred to as the (antisymmetric) spinor metric $\epsilon \equiv i\sigma_2$, takes a $(1/2, 0)$ spinor ${\hat \chi}$ into its $(0, 1/2)$ equivalent ${\hat \eta} = i\sigma_2 {\hat \chi}^*$. Explicitly this amounts to $$ {\hat \chi} = \left(\begin{array}{c}\chi^0 \\ u/(2\chi^{0*})\end{array}\right) \rightarrow {\hat \eta} = \left(\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right)\left(\begin{array}{c}\chi^{0*} \\ u^*/(2\chi^{0})\end{array}\right) = \left(\begin{array}{c}u^*/(2\chi^{0}) \\ -\chi^{0*}\end{array}\right) \equiv \left(\begin{array}{c}{\hat {\bar \chi}}_0\\ -u/2({\hat{\bar\chi}}_0)\end{array}\right) $$ where ${\hat {\bar \chi}}_0 = u^*/(2{\hat\chi}^0)$. Eventually this shows that the chiral dual ${\hat \eta}$ indeed corresponds to the (space inverted) covariant spin 4-vector $u_\mu = g_{\mu\nu}u^\nu$. But equally important, ${\hat \eta}$ turns out to be the (space inverted) orthogonal of the original ${\hat \chi}$: $$ {\hat \chi}^\dagger {\hat \eta} = {\hat \chi}^\dagger \epsilon {\hat\chi}^* = \left( {\hat \chi}^T \epsilon {\hat \chi} \right)^* = {\hat \chi}^\dagger \left(\begin{array}{cc}\chi^{0*} & u^*/(2\chi^0)\end{array}\right) \left(\begin{array}{c}u^*/(2\chi^{0}) \\ -\chi^{0*}\end{array}\right) = 0 $$ Bottom line is, chirally dual spinors are the spin-orthogonal, space inverted of each other.

I'm answering this question very late because I've noticed that a lot of chirality/helicity questions around this site have rather poor answers (though the other answer on this question is excellent!), so this is to help clear the air.

As with most issues surrounding chirality and helicity, the problem is dissolved as long as one remembers that chirality is a property of fields and helicity is a property of particles, and fields are tools to create and annihilate particles.

For example, as stated in great detail here, a right-chiral Weyl field is one which annihilates certain left-helicity massless particles and creates certain right-helicity massless particles, which have opposite charges. The same goes for left-chiral Weyl fields with left and right swapped.

How should I picture the difference between a left-handed and right-handed spinor field?

In terms of *classical fields*, a right-handed field is one which has plane wave solutions where the eigenvalues under rotation about the axis and translation along the axis have the same sign, while a left-handed field is the opposite.

But this picture isn't too useful in quantum field theory. At this level, fields are a tool for bookkeeping the creation and annihilation of particles. A left-handed and right-handed Weyl spinor field correspond to precisely the same particle content, so there isn't a physical difference. For instance, a left-handed Weyl field of positive charge is associated with the same particles as a right-handed Weyl field is negative charge; one creates what the other annihilates.

The question Explaining chirality for spin 1/2 particle does a nice job distinguishing helicity from chirality, but ideally I'd like an explanation of chirality that makes no reference to helicity at all.

Fields transform in representations of the Lorentz group. If one extends this to the full Lorentz group, which includes parity, representations are chiral if they don't map to themselves under parity.

For example,

conceptuallywhy does the parity operator reverse a particle's chirality?

It doesn't, because chirality is a property of fields, not particles. At the level of the fields, parity flips chirality essentially by definition, if you use the definition I gave above.

If one were to be given a particle without being told its chirality, how would one check it? For example, if I were to consider a single Weyl field with a Lagrangian given by Srednicki's eqn. (36.2), $$ \mathcal{L} = i \psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi - \frac{1}{2} m \left( \psi \psi + \psi^\dagger \psi^\dagger \right),$$ how could I determine its chirality experimentally?

Again, this is a meaningless question because chirality is a property of fields, which are used to organize a theory of particles. For example, suppose you detect certain right-helicity particles, and left-helicity particles with the opposite charges, all of which are massless. You can describe them using a Lagrangian containing a left-chiral Weyl field, *or* a Lagrangian containing a right-chiral Weyl field, or even a Lagrangian containing a Dirac field obeying a constraint.

It's kind of like saying, if you detect a photon, can you tell if the associated field is $A_\mu$ or $F_{\mu\nu}$ or something else. The question isn't really meaningful. Either one or both of these fields might appear in the process of describing the very same physical observations of the particle.