Exploring Left & Right Spinor Fields in Srednicki

[LAHLH]

Hi,

I'm just looking at the stuff on left and right handed spinor fields in Srednicki. Srednciki distinguishes fields in the left rep from those in the right rep by putting a dot over them. Since hermitian conjugation swaps the two SU(2) algebras, the hermitian conj of a left spinor is a right spinor etc. e.g. $$\left[\psi_a(x)\right]^{\dag}=\psi^{\dag}_{\dot{a}}(x)$$

Under a Lorentz transformation we have, $$U(\Lambda)^{-1}\psi^{\dag}_{\dot{a}}(x)U(\Lambda)=R^{\dot{b}}_{\dot{a}}(\Lambda)\psi^{\dag}_{\dot{b}}(\Lambda^{-1}x)$$.

Under infinitesimal transformations we have, $$R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}$$. $$R^{\dot{b}}_{\dot{a}}(1+\delta\omega)=\delta^{\dot{b}}_{\dot{a}}+\frac{i}{2}\delta\omega_{\mu\nu}\left(S^{\mu\nu}_{R}\right)^{\dot{b}}_{\dot{a}}$$.
From this one finds,

$$\left[\psi^{\dag}_{\dot{a}}(0),M^{\mu\nu}\right]=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)$$Now I want to take the Hermitian conjugate of this to get Srednicki's 34.16. I presume this conjugate is w.r.t. the a,b, internal indices not the spacetime indices,

$$\psi^{\dag}_{\dot{a}}(0)M^{\mu\nu}-M^{\mu\nu}\psi^{\dag}_{\dot{a}}(0)=(S^{\mu\nu}_{R})^{\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)$$

OK, so taking the h.c.:

$$(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\psi^{\dag}_{\dot{b}}(0)\right]^{\dag}$$

$$(M^{\mu\nu})^{\dag}\psi_{a}(0)-\psi_{a}(0)(M^{\mu\nu})^{\dag}=\psi_{b}(0)\left[(S^{\mu\nu}_{R})^{..\dot{b}}_{\dot{a}}\right]^{\dag}$$
But given that the M's have no a,b type indices I don't really see what this would mean.

Could anyone point me in the right direction on how to derive this 34.16, thanks

 

[schieghoven]

Hi,

"the hermitian conj of a left spinor is a right spinor"

To my understanding this is not correct. If $$\chi$$ is a left-spinor, then let

$$\varepsilon$$ =
( 0 -1 )
( 1 0 )
Then $$\varepsilon \chi^c$$ transforms as a right-spinor, where the superscript-c denotes complex conjugation (without transpose). Left- and right-handed spinors transform as $$R$$ and $$(R^{-1})^\dagger$$ respectively, where $$R \in SL(2)$$ and the dagger denotes hermitian conjugate. The reason it works is because
$$\varepsilon R = (R^{-1})^T \varepsilon$$
which you can show to be true by writing it out longhand and using det R = 1.

I'm sorry I can't post in more detail because I have to run off urgently... I hope I got all the details right but at least in broad sweep I think what I've said is correct. Hope it helps get you part way towards the rest of the question which I haven't had a chance to get to,

Cheers
Dave

 

[LAHLH]

Hi,

thanks for the reply schieghoven. I see, it's strange then that Srednicki should say on p210 "the Hermitian conjugate of a field in the (2,1) representation should be a field in the (1,2) representation"

Is this an error then, or is he just being sloppy? How does one get from his 34.15 to 34.16 with either definition of a RH spinor?

cheers

 

[LAHLH]

After reading my nearest group theory book (Jones). I wonder if what you are describing here is the transformations of upper and lower spinors under SU(2), where $$C_{ab}=\epsilon_{ab}$$ is the (invariant analogue of the GR metric) and takes upper spinors that transform under $$U*$$ of SU(2) to lower spinors that transform under U of SU(2).

But I am under the impression that these spinors are not the same as a Weyl spinor? since Weyl LH/RH spinors (2,1) or (1,2) transform under the Poincare group which can be thought of as a product of SU(2)XSU(2). So it's kind of like, one "component" (probably not the correct word here) of the Weyl spinor transforms under U and the other under U* (i.e. we have two SU(2) algebras that are complex conjugates of each other)

hmm, somewhat confused.

 

[schieghoven]

Hi,

I disagree with books that say SU(2) x SU(2) is a rep of the Lorentz group. It's transparently not the case, because SU(2) x SU(2) is compact; the Lorentz group SO(1,3) is not. I kind of understand why authors try to do it; it is because if J_j, K_k are generators for rotations and boosts respectively, these authors construct the quantities $J_j \pm iK_j$ and show that they obey SU(2) commutation relations. Doing so belies the fact that J_j and K_j are generators of the real (i.e. not complex) Lie group SO(1,3). $J_j \pm iK_j$ is not a real linear combination of the J_j, K_j, so it's not correct to change basis this way.

(Note that you referred to the Poincare group, which generally speaking refers to something different again:
Lorentz = rotations and boosts = 6 generators
Poincare = rotations, boosts, and translations = 10 generators)

I think of spinors (= Weyl spinors I think??) as transforming under the group SL(2,C) = 2-by-2 complex matrices with unit determinant. This group has six real dimensions and the generators obey the commutation relations for the Lorentz group, given in e.g. Weinberg's text or Foldy (PR 102, 568, 1956). There are two distinct representations which you can switch between using $U \mapsto U^{-1}^*$. I don't think you can use U* because it wouldn't preserve the order of operation: $U^*V^* \neq (UV)^*$, so it's not a representation.

I don't have a copy of Srednicki, so I can't comment on the other statements. I'm afraid we seem to get a step further away from your original question with each reply :-) .

Cheers

Dave

 

[LAHLH]

Hi,

OK admittedley you kind of lost me, as all I know is what I've read in Srednicki/Jones, who both do exactly the process you described to argue we can split the Lorentz algebra into two SU(2) algebras etc etc. I'm aware the Poincare and Lorentz group are different in the way you stated, however I think if I remember correctly from Jones, using the Poincare group he still drew the same conclusions about this splitting into su(2)xsu(2), his algebra including not just the J,K generators but also P and H.

I've heard people saying SL(2,C) is a rep of Lorentz, but don't really understand this idea, could you perhaps explain it at all?

As for my original question a free copy of Srednicki is available here: http://www.physics.ucsb.edu/~mark/qft.html if anyone is interested

thanks

 

[haushofer]

Perhaps the first chapter of Bilal's SUY intro helps answering your question about SL(2,C):

http://arxiv.org/abs/hep-th/0101055

 

[schieghoven]

Cool, I hadn't seen either of these papers before.

I like Bilal's construction more than Srednicki's... Still I don't know why Bilal didn't simply state

$$J_j = \sigma_j, \qquad K_j = i\sigma_j, \qquad (j=1,2,3)$$

from which it can be verified directly that they satisfy the Lie bracket (Bilal eq 2.1) for the Lorentz group. To construct the Lie group from the Lie algebra, take the exponential
$$\exp( a_j J_j + b_j K_j )$$
(Bilal eq 2.8) for arbitrary $a_j, b_j \in \mathbb{R}$. The exponential of a traceless matrix is a matrix with unit determinant --- which is why we get to $SL(2,\mathbb{C})$.

My take on the spin-half representation of the Lorentz group is that you need to find a group whose elements can be mapped into the Lorentz group, such that the group operation is preserved. This group is called a covering group. In other words, if U_1, U_2 are elements of the covering group, then you have a map P into the Lorentz group such that
P(U_1) P(U_2) = P(U_1 U_2)
where on the RHS the product is the group operation of the Lorentz group, on the LHS the product is the group operation of the covering group. So while U_1, U_2 represent transformations on some other space (the space of spinors), they can be mapped to a unique Lorentz transformation, and the effect of multiple transformations is reflected consistently under that mapping. The map is not one-to-one, so I think that Srednicki's (33.8) (a kind of "inverse" to the above) is misleading at best.

The group and the covering group are locally isomorphic, so they have isomorphic Lie algebras. So you first satisfy the Lie bracket, then exponentiate to construct the group from the Lie algebra.

I'm afraid I can't remember a good reference off the top of my head, which is why I wrote out this fairly hashed-up overview. Perhaps you can find an topology text with a chapter on topological groups... but depends how much time you have to throw at it.


Hope that helps
Dave