I'm trying to understand the argument that extended supersymmetry cannot produce the chiral structure of the Standard Model, as explained on page 25 of these notes. My impression of the argument goes like this:

  • All particles in a supermultiplet must transform the same way under internal symmetries.
  • All massless particles with helicity $|\lambda| = 1$ must be created by gauge fields, because gauge symmetry is the only way to get rid of the extra degrees of freedom.
  • Gauge fields must transform under the adjoint representation of any gauge group, and for all matrix Lie groups the adjoint representation is real.
  • In extended SUSY, with one exception, all multiplets containing $|\lambda| = 1/2$ also contain $|\lambda| = 1$. Thus we can only get fermions in real representations of any gauge group.
  • This is incompatible with the chiral structure of the Standard Model, which requires that fermions live in complex representations, such as left-handed quarks, which are in the doublet representation of $SU(2)_L$.

Granted that I paraphrased the argument correctly, I'm fine with every step except for the last step, which just doesn't seem right to me.

Here are the preliminary thoughts I've had about this:

  • It's ambiguous whether we're talking about the representations of fields or representations of particles. These are different but I haven't been able to make the argument work either way.
  • It's ambiguous whether 'complex' means $\text{complex}_1$, i.e. the base field is the complex numbers, or $\text{complex}_2$, i.e. the representation is $\text{complex}_1$ and not equivalent to its conjugate. I think the only sensible option is $\text{complex}_2$.
  • The doublet representation of $SU(2)_L$ is not $\text{complex}_2$! There is only one two-dimensional representation of $SU(2)_L$ so it must necessarily be equivalent to its conjugate.
  • In this answer it is argued that if $q_L$ transforms in a representation $R$, then $q_R$ transforms in the conjugate representation $\bar{R}$. Therefore if $R$ is real, $q_L$ and $q_R$ transform the same way, which contradicts observation. But this seems clearly wrong, it's not true for $q_L$ and $q_R$ in the Standard Model!
  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    May 4, 2018 at 14:19
  • $\begingroup$ Might specify by "internal symmetry" you mean one commuting with the extended SUSY algebra, whose global bosonic symmetries you are ignoring here... $\endgroup$ May 4, 2018 at 14:25
  • 1
    $\begingroup$ Note that the $SU(2)$ doublet is pseudoreal (quaternionic), not real. $\endgroup$ May 4, 2018 at 14:33
  • 1
    $\begingroup$ @RyanThorngren Yes, but I think the common usage of "complex" is "not real or pseudoreal". $\endgroup$
    – knzhou
    May 4, 2018 at 14:36
  • $\begingroup$ @CosmasZachos Yes, that's what I mean here! $\endgroup$
    – knzhou
    May 4, 2018 at 14:36

1 Answer 1

  1. You must be talking about the representations of the fields, since the particles are quantum states that always live in infinite-dimensional complex Hilbert space with corresponding unitary representations on it.

  2. Let us make clear what we really mean by "chirality" of the Standard Model. Given a Dirac fermion $\psi$ with its chiral parts/Weyl fermions $\psi_L,\psi_R$, chirality means that $\psi_L$ transforms under a different representation of at least one gauge group (the weak one in the case of the SM) than $\psi_R$. For how exactly this translates to our ordinary natural language of left/right-handed electrons and positron, see this answer of mine to a previous question of yours.

  3. The meaning of "complex" here is "not real", not "complex" as in "not real or pseudoreal". For the difference between real, pseudoreal and complex, see this answer of mine to another previous question of yours.

  4. When we say that "a Weyl spinor" $\psi_{R/L}$ transforms in the representation $\rho$ of a gauge group, what we really formally would have to say that the object transforms in the combined representation $(1/2,0)\otimes \rho$ where $(1/2,0)$ is the usual notation for the Weyl spinor representation of the Lorentz group.

  5. Finally, we are in a position to observe the following: Given a massless multiplet with a Weyl fermion of any handedness in it and a real representation of the gauge group, we can split the result of the tensor product since we can split the real representation $\rho$ into two representations on real vector spaces as $\rho_\mathbb{R} + J\rho_\mathbb{R}$, where $J$ is basically a complex conjugation map: $$ (1/2,0) \otimes \rho = \left( (1/2,0) \otimes \rho_\mathbb{R} \right) \oplus \left( (1/2,0) \otimes J\rho_\mathbb{R}\right)$$

All that is left to do is to observe that $(1/2,0)\otimes J\rho_\mathbb{R}$ is equivalent to $J(1/2,0)\otimes \rho_\mathbb{R}$, and that $J(1/2,0) = (0,1/2)$ - the conjugate representation of a left-handed Weyl fermion is a right-handed Weyl fermion. Therefore, choosing a real representation $\rho$ of the gauge group for a Weyl fermion actually results in the Weyl fermion transforming in $\rho_\mathbb{R}$ and its opposite-chirality version also transforming in $\rho_\mathbb{R}$, explaining why real representations can never lead to chirally symmetric theories.

Observe also that this argument fails for a pseudoreal representation, which also resolves the apparent contradiction with the doublet of $\mathrm{SU}(2)_L$ being "not complex".

  • $\begingroup$ I agree with everything you said in the first few points. Over the years I've asked several questions about charge conjugation, chirality, and helicity, and I have to thank you for patiently teaching me the essential points. (I think I finally have a full understanding, written up here.) $\endgroup$
    – knzhou
    May 5, 2018 at 13:10
  • $\begingroup$ However, I think we're working from different definitions of what a "chiral gauge theory" is. I agree with your reasoning given your definition, but consider the particles in a theory with just a left-chiral Weyl field transforming in an $SU(2)$ doublet. We have four, corresponding to helicity $\pm 1/2$, and "$I_3$" value $\pm 1/2$. Everything is perfectly symmetric under the charge conjugation operator $\hat{C}$, as well as $\hat{P}$. I would expect a "chiral gauge theory" to not be $\hat{C}$ and $\hat{P}$ symmetric. $\endgroup$
    – knzhou
    May 5, 2018 at 13:12
  • $\begingroup$ To summarize, I think you define a chiral gauge theory so that a non-chiral gauge theory is one where you can pair up left-chiral and right-chiral Weyl spinors in real representations. But I instead define a chiral gauge theory to be one that violates $\hat{C}$ and $\hat{P}$; then a pseudoreal representation can't be chiral. Papers on this seem to just say a chiral gauge theory has a complex representation, but we already know this word is ambiguous. Do you have a definitive reference here? $\endgroup$
    – knzhou
    May 5, 2018 at 13:13
  • $\begingroup$ @knzhou: FWIW, Wikipedia's definition agrees with mine. I'll see if I can find better references, but it might take a while. $\endgroup$
    – ACuriousMind
    May 5, 2018 at 13:15
  • $\begingroup$ No worries, it's just a semantics issue! In any case the full SM fermion representation is not pseudoreal, so either definition works fine there. $\endgroup$
    – knzhou
    May 5, 2018 at 13:20

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