My problem is understanding the transformation behaviour of a Dirac spinor (in the Weyl basis) under parity transformations. The standard textbook answer is

$$\Psi^P = \gamma_0 \Psi = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix}, $$ which I'm trying to understand using the transformation behaviour of the Weyl spinors $\chi_L $ and $\xi_R$. I would understand the above transformation operator if for some reason $\chi \rightarrow \xi$ under parity transformations, but I don't know if and how this can be justified. Is there any interpretation of $\chi $ and $\xi$ that justifies such a behaviour?

Some background:

A Dirac spinor in the Weyl basis is commonly defined as

$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}, $$ where the indices $L$ and $R$ indicate that the two Weyl spinors $\chi_L $ and $\xi_R$, transform according to the $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$ representation of the Lorentz group respectively. A spinor of the form

$$ \Psi = \begin{pmatrix} \chi_L \\ \chi_R \end{pmatrix}, $$ is a special case, called Majorana spinor (which describes particles that are their own anti-particles), but in general $\chi \neq \xi$.

We can easily derive how Weyl spinors behave under Parity transformations. If we act with a parity transformation on a left handed spinor $\chi_L$: $$ \chi_L \rightarrow \chi_L^P$$ we can derive that $\chi_L^P$ transforms under boosts like a right-handed spinor

\begin{equation} \chi_L \rightarrow \chi_L' = {\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L \end{equation}

\begin{equation} \chi_L^P \rightarrow (\chi^P_L)' = ({\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L)^P = {\mathrm{e }}^{ - \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L^P, \end{equation} because we must have under parity transformation $\vec \sigma \rightarrow - \vec \sigma$. We can conclude $ \chi_L^P = \chi_R$ Therefore, a Dirac spinor behaves under parity transformations $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^P= \begin{pmatrix} \chi_R \\ \xi_L \end{pmatrix} , $$ which is wrong. In the textbooks the parity transformation of a Dirac spinor is given by

$$\Psi^P = \gamma_0 \Psi = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix}. $$

This is only equivalent to the transformation described above of $\chi = \xi$, which in my understand is only true for Majorana spinors, or if for some reason under parity transformations $\chi \rightarrow \xi$. I think the latter is true, but I don't know why this should be the case. Maybe this can be understood as soon as one has an interpretation for those two spinors $\chi$ and $ \xi$...

Update: A similar problem appears for charge conjugation: Considering Weyl spinors, one can easily show that $ i \sigma_2 \chi_L^\star$ transforms like a right-handed spinor, i.e. $i \sigma_2 \chi_L^\star = \chi_R $. Again, this can't be fully correct because this would mean that a Dirac spinor transforms under charge conjugation as

$$ \Psi= \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^c = \begin{pmatrix} \chi_R \\ \xi_L \end{pmatrix},$$ which is wrong (and would mean that a parity transformation is the same as charge conjugation). Nevertheless, we could argue, that in order to get the same kind of object, i.e. again a Dirac spinor, we must have

$$ \Psi= \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^c = \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix},$$

because only then $\Psi^c$ transforms like $\Psi$. In other words: We write the right-handed component always below the left-handed component, because only then the spinor transforms like the Dirac spinor we started with.

This is in fact, the standard textbook charge conjugation, which can be written as

$$ \Psi^c = i \gamma_2 \Psi^\star= i \begin{pmatrix} 0 & \sigma_2 \\ -\sigma_2 & 0 \end{pmatrix} \Psi^\star = i \begin{pmatrix} 0 & \sigma_2 \\ -\sigma_2 & 0 \end{pmatrix} \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}^\star= \begin{pmatrix} -i\sigma_2 \xi_R^\star \\ i\sigma_2 \chi_L \end{pmatrix}= \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} .$$ In the last line I used that, $i \sigma_2 \chi_L^\star$ transforms like a right-handed spinor, i.e. $i \sigma_2 \chi_L^\star = \chi_R $. The textbook charge conjugation possible hints us towards an interpretation, like $\chi$ and $\xi$ have opposite charge (as written for example here), because this transformation is basically given by $\chi \rightarrow \xi$.


2 Answers 2


You are looking for a unitary representation of parity on spinors. That it should be unitary can be seen from the fact, that parity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation.

As demonstrated parity transforms a $(\frac{1}{2},0)$ into a $(0,\frac{1}{2})$ representation. Hence it cannot act on any such representation alone in a meaningful way. The Dirac-spinors in the Weyl-basis on the other hand contain a left- and right-handed component $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$

As a linear operator on those spinors - a matrix in a chosen basis - it mixes up the spinor components. After what has been said before, left- and right-handed components should transform into each other. The only matrix one can write down that does this is $\gamma^0$. There could in principle be a phase factor. In a theory with global $U(1)$-symmetry this may be set to one however.

Edit: Statements like $\chi_L \rightarrow P\chi_L=\chi_R $ for a Weyl-Spinor $\chi_L$ are not sensible. The Weyl-spinors are reps. of $\mathrm{Spin(1,3)}$, whereas $P\in \mathrm{Pin(1,3)}$. One cannot expect that some representation is also a representation of a larger group. Dirac-spinors on the other hand are precisely irreps. of $\mathrm{Spin(1,3)}$ including parity, which cannot act in any other sensible way than by exchanging the chiral components.

Think of what representation means. It's a homomorphism from a group to the invertible linear maps on a vector space. $$ \rho: G \rightarrow GL(V)$$ Particularly, for any $g\in G$ and $v\in V$, $\rho(g)v\in V$. Now set $V$ to be the space of say left-handed Weyl-spinors and $g=P\in\mathrm{Pin(1,3)}$ the parity operation. As you have shown above, the image of a potential $\rho(P)$ is not a left-handed Weyl-spinor, hence is not represented.

  • $\begingroup$ Thanks for your answer! I understand that under parity transformations $(\frac{1}{2},0) \leftrightarrow (0,\frac{1}{2})$. Therefore the parity transformed spinor will have a right handed spinor as its top component and a left handed spinor as its lower component. Nevertheless, wouldn't be an equally viable parity transformed Dirac spinor $\Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^P = \begin{pmatrix} \chi_R \\ \xi_L \end{pmatrix}? $ $\endgroup$
    – jak
    Commented Nov 18, 2014 at 7:24
  • $\begingroup$ @JacobH : I've expanded my answer to clarify it in response to your comments. $\endgroup$
    – user7154
    Commented Nov 18, 2014 at 20:34
  • $\begingroup$ @JakobH I've updated the answer. Maybe this'll help. Ask back anytime. $\endgroup$
    – Nephente
    Commented Nov 19, 2014 at 9:40

I find things are clearer using the dotted and undotted spinor notation. The L-spinors $\chi_{L}$ are dotted vectors $\chi^{\dot{A}}$ and the R-spinors $\xi_{R}$ are undotted vectors $\xi^{A}$ with index $A=1,2$. The parity operator has to be a tensor $P^{\dot{A}}_{B}$ and another tensor $P^{A}_{\dot{B}}$ in order to change the way each type of spinor transforms. The action of parity on $\chi^{\dot{A}}$ is to make $P^{A}_{\dot{B}}\chi^{\dot{B}}$ which transforms as an undotted spinor. Similarly, the action of parity on $\xi^{A}$ is to make $P^{\dot{A}}_{B}\xi^{B}$ which transforms as a dotted spinor. It turns out that (presumably in the rest frame of the particles) the parity tensors are $P^{\dot{A}}_{B}=i\delta^{\dot{A}}_{B}$ and $P^{A}_{\dot{B}}=i\delta^{A}_{\dot{B}}$. The action of parity is then, $$ \chi^{\dot{A}}\rightarrow P^{A}_{\dot{B}}\chi^{\dot{B}}=i\delta^{A}_{\dot{B}}\chi^{\dot{B}}=i\chi^{A} $$ $$ \xi^{A}\rightarrow P^{\dot{A}}_{B}\xi^{B}=i\delta^{\dot{A}}_{B}\xi^{B}=i\xi^{\dot{A}} $$ and this means that the components of the spinors get a phase and the way the components transform is changed. The dots are a reminder of how each component transforms. The action of parity on a Dirac spinor is obtained from the above transformations by stacking the Weyl spinors.

\begin{equation} \left[ \begin{array}{c} \chi^{\dot{1}}\\ \chi^{\dot{2}}\\ \xi^{1}\\ \xi^{2} \end{array} \right] \rightarrow i\left[ \begin{array}{c} \xi^{\dot{1}}\\ \xi^{\dot{2}}\\ \chi^{1}\\ \chi^{2} \end{array} \right] \end{equation}

Edit : Clarification. The Dirac spinor has four components. Components one and two transform as the two components of a dotted Weyl spinor and components three and four transform as the components of an undotted Weyl spinor. If we remember that is how the Weyl spinors stack into a Dirac spinor then we can remove the dots and the L and R labels and then the last equation on the action of parity on a Dirac spinor is, \begin{equation} \left[ \begin{array}{c} \chi^{1}\\ \chi^{2}\\ \xi^{1}\\ \xi^{2} \end{array} \right] \rightarrow i\left[ \begin{array}{c} \xi^{1}\\ \xi^{2}\\ \chi^{1}\\ \chi^{2} \end{array} \right] \end{equation} In matrix notation this is, \begin{equation} \left[ \begin{array}{cc} i & 0\\ 0 & i \end{array} \right] \left[ \begin{array}{c} \chi\\ \xi \end{array} \right]=i\left[ \begin{array}{c} \xi\\ \chi \end{array} \right] \end{equation} Modulo the phase factor $i$ (because I think that the parity operator on spinors has to be PP=-1), this agrees with the action of parity given by the gamma matrix $\gamma_{0}$ as in the standard texts such as equation (1.4.42) on page 19 of Ramond's "Field Theory", Second Edition.


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