Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let $P$ be the parity operator of the Lorentz group,

$$P=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$$ the commutation relations of $so(3,1)$ be :

$$[M_i,M_j] = \epsilon_{ijk}M_k$$ $$[M_i,N_j] = \epsilon_{ijk}N_k$$ $$ [N_i,N_j] = -\epsilon_{ijk}M_k$$

and define : $L_i=1/2(M_i+iN_i)$ and $\overline{L_i}=1/2(M_i - iN_i)$.

From these, we have : $[P,M_i]=0$, $PN_i=-N_iP$. Hence : $PL_i=\overline{L_i}P$

We have the more general representation of $so(3,1)$ being labelled by two indices $(j,j')$ because of the following isomorphism : $so(3,1)=so(3)\oplus \overline{so(3)}$. We can by exponentiation get the representations of $L_{+}^{\uparrow}$ (the proper Lorentz transformation preserving direction of time). I try to get from these representations those of $L^{\uparrow}$ (generated by just adding $P$ in $L_{+}^{\uparrow}$). Assuming we have an irreducible representation of $L^{\uparrow}$, it is also a representation of $L_{+}^{\uparrow}$ and we can generally write it as $T=\oplus_{j,j'} (j,j')$.

Until here I think it is clear. But then the textbook says that if we take $v$ a vector in a invariant subspace associated to the representation $(j,j')$ then $T(P)v$ transforms with $(j',j)$.

I imagine that we can start with :


But I don't understand how from it we can show that it is a transformation with $(j',j)$.

Edit : I think we can also see it this way (not as complete as the answer of @Stephen Blake, but it gives the idea) :


and because of the equivalence between the complex conjugate of these representations. We have : $$\overline{(j,j')}\sim (j',j)$$ where it is the complex conjugate representation of $M_i$ and $N_i$ which was taken (we have a real vector space for $so(3,1)$) and hence : $$\overline{T}_{(j,j')}(L_i)= T_{(j,j')}(\overline{L_i})\ast$$ (where $*$ denotes the complex conjugate matrix).

Eventually :

$$T_{(j,j')}(P)\left(\overline{T}_{(j,j')}(L_i)*\right)v = T_{(j,j')}(P)\left(T_{(j',j)}(L_i)*\right)v$$

Here, we see that it is $v$ that transforms by $(j',j)$, not $T(P)v$ but it seems that Stephen Blake comes to the same result. Can I make it better in this way ?

share|improve this question

1 Answer 1

Parity flips the sign of a Lorentz boost $\eta$, $$ T(P)T(\eta)T(P^{-1})=T(-\eta) $$ and commutes with a spatial rotation R, $$ T(P)T(R)T(P^{-1})=T(R) \ . $$ Consider a Lorentz spinor $\psi^{A}\in V_{2}$ with indices A=1,2 and try to set up a 2x2 matrix $T(P)$ for parity. A Lorentz boost along the z-axis is, $$ T(\eta)=\begin{pmatrix}\ e^{-\eta/2}&0\\0&\ e^{\eta/2}\end{pmatrix} $$ and a rotation about z by angle $\phi$ is the unitary matrix, $$ T(R)=\begin{pmatrix}\ e^{-i\phi/2}&0\\0&\ e^{i\phi/2}\end{pmatrix} $$ It's easy to see that the conditions imposed upon $T(P)$ by the boost and rotation are impossible to satisfy so the matrix $[T(P)]^{A}_{B}$ does not exist. However, there are also dotted Lorentz spinors $\phi^{\dot{A}}\in \tilde{V}^{*}_{2}$ and so it's natural to try the parity matrix as $[T(P)]^{\dot{A}}_{B}$. A dotted spinor transforms as, $$ \phi'^{\dot{A}}=[T(g^{-\dagger})]^{\dot{A}}_{\dot{B}}\phi^{\dot{B}} $$ and taking parity as the diagonal matrix, $$ [T(P)]^{\dot{A}}_{B}=i\delta^{\dot{A}}_{B} $$ and, $$ [T(P)]^{A}_{\dot{B}}=i\delta^{A}_{\dot{B}} $$ so that $P^{2}=-1$ for Lorentz spinors makes the boosts and rotations work nicely.

A general (m,n) spinor is a symmetric tensor, $$ F^{A_{1}\ldots A_{m}\dot{B}_{1}\ldots \dot{B}_{n}} $$ and if this is transformed under parity by tensoring $[T(P)]^{A}_{\dot{B}}$ and $[T(P)]^{\dot{A}}_{B}$ the (m,n) spinor is changed into a (n,m) spinor.

share|improve this answer
Why do dotted spinor transform with the hermitian conjugate of $T(g)$ ? Doesn't it have to be the complexe conjugate of it ? (and to be sure, did you use 2 times the metric $\eta_{\mu\nu}$ to write your second form of the representation of $P$ ?) –  faero Dec 24 '13 at 18:24
@faero reply to "Why do dotted spinors transform with the conjugate of $T(g)?" : Upstairs and downstairs dotted spinors are equivalent SL(2,C) irreps. In the notation I use (see physics.stackexchange.com/questions/55350/… ) the downstairs dotted indices transform with the complex conjugate of T(g). –  Stephen Blake Dec 28 '13 at 11:02
@faero reply to "Did you use 2 times the metric?" : The rep of P was a bit of a mystery to me (see physics.stackexchange.com/questions/78935/… ). I did not get it using the metric. Instead I used the first two equations in m answer to show P must be diagonal. Then, I restricted it further by looking at how four vectors and antisymmetric tensors with spacetime indices transformed under P, these are $X^{\dot{A}}_{B}$ and symmetric $Y^{AB}$ Lorentz spinors. –  Stephen Blake Dec 28 '13 at 11:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.