I understand mathematically that the Lorentz group's Lie algrebra $\mathfrak{so(3,1)}$ (given by eqns. (33.11)-(33.13) in Srednicki's QFT book) is isomorphic to $\mathfrak{su(2) \times su(2)}$ (given by eqns. (33.18) - (33.20)), and thus that the Lorentz group has two inequivalent irreducible representations with $n + n' = 1/2$. But I don't understand the distinction between these representations at the conceptual level - Srednicki's describes the $N$ operators only as "nonhermitian operators whose physical significance is obscure."
How should I picture the difference between a left-handed and right-handed spinor field? 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. (In my experience, most explanations of chirality start by pretending it's the same thing as helicity, then go on to clarify that they're actually different for massive particles, but never actually go on to properly define chirality except in a formal mathematical way. I feel that it's conceptually clearer to just explain what chirality is rather than explaining what it isn't.)
For example, conceptually why does the parity operator reverse a particle's chirality? (I know that this is usually just treated as part of the parity operator's definition.) 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?
Answer
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.
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