5
$\begingroup$

Consider $SU(2)$ supergauge theory with $A$, a doublet of two chiral superfields in the fundamental representation.

$$A= \begin{pmatrix} \Phi_1\\ \Phi_2 \end{pmatrix}$$ where $\Phi_1$ and $\Phi_2$ are chiral superfields. Since we have said it is in the fundamental representation it transforms

$$A\to{}A'=e^{i\sigma_j\phi_j}A$$

where $\sigma_j$ are Pauli matrices. If $e^{i\sigma_j\phi_j}$ were to be a $SU(2)$ matrix then the $\phi_j$ ought to be real. Nonetheless, this would make $A'$ no longer chiral unless $\phi_j$ are themselves chiral superfields, that is complex functions making $e^{i\sigma_j\phi_j}$ not a $SU(2)$ matrix.

Therefore this is no longer a proper $SU(2)$ theory. What is going on?

$\endgroup$
5
  • $\begingroup$ Could you present your precise definition of "chiral superfield"? $\endgroup$
    – ACuriousMind
    Nov 29, 2015 at 18:04
  • $\begingroup$ @ACuriousMind Following section 4.4 of arxiv.org/abs/hep-ph/9709356 a chiral superfield is a superfield $\Phi$ satisfying $\bar{D}_{\dot{\alpha}}\Phi=0$ $\endgroup$
    – Yossarian
    Nov 29, 2015 at 20:01
  • $\begingroup$ I'm not sure what your question really is, then. As far as I can see, the linear combination of chiral superfields is a chiral superfield (the constraint they fulfill is linear), so for any matrix $M$, the components of $MA$ are chiral superfields if the components of $A$ were, no? $\endgroup$
    – ACuriousMind
    Nov 30, 2015 at 10:11
  • $\begingroup$ @ACuriousMind notice that the elements of the matrix $M$ need not be constants so $\bar{D}_{\dot{\alpha}}=-\frac{\partial}{\partial\bar{\theta}^{\dot{\alpha}}}+i(\theta\sigma^{\mu})_{\dot{\alpha}}\partial_{\mu}$ applies to $M$ as well $\endgroup$
    – Yossarian
    Nov 30, 2015 at 11:57
  • $\begingroup$ Ah, yes, indeed. $\endgroup$
    – ACuriousMind
    Nov 30, 2015 at 12:01

1 Answer 1

6
+50
$\begingroup$

You are correct in stating that the doublet $A'$ ceases to be chiral if your gauge parameters $\phi_j$ are real. In fact, in an $\mathcal{N} = 1$ (global) supersymmetric gauge theory, a gauge transformation is not given by what you wrote above. Working as you are in the superfield formalism, a supersymmetric gauge transformation of a chiral superfield $\Phi$ in a representation $R$ of the gauge group $G$ is given by $$ \Phi \rightarrow \Phi' = e^{i \Lambda} \Phi, $$ where $\Lambda \equiv \Lambda_a T_R^a$, with $T_R^a$ the generators in the appropriate representation (the Pauli matrices, in your example).

These gauge parameters $\Lambda_a$ are, as you suspected, chiral superfields, whose 'lowest components' $\lambda_a$ (see the equation below) are complex fields.

The $\Lambda_a$ are indeed complex functions of superspace coordinates, namely spacetime $x^\mu$ and the Grassmann numbers $\theta_\alpha, \overline\theta_{\dot \alpha}$. Explicitly (different conventions may be at play): $$ \Lambda_a \equiv \Lambda_a(x,\theta,\overline\theta) = \lambda_a(x) + \sqrt{2} \theta \psi^\lambda_a(x) + i \theta \sigma^\mu \overline \theta \partial_\mu \lambda_a(x)- \theta \theta F^\lambda_a(x)\\ \qquad- \frac{i}{\sqrt{2}}\theta\theta \partial_\mu \psi^\lambda_a(x) \sigma^\mu \overline \theta - \frac{1}{4} \theta\theta\overline\theta\overline\theta \square \lambda_a(x). $$ Similarly, $\Phi(x,\theta,\overline\theta) = \phi(x)+ \sqrt{2} \theta \psi(x) - \theta \theta F(x) + (\ldots) $.

If my gauge group is, say, $G = \textrm{SU}(N)$, you claim that the theory is no longer a proper $\textrm{SU}(N)$ theory, since $e^{i\Lambda}$ is not (taking $R$ to be the fundamental representation) an $\textrm{SU}(N)$ matrix.

However, the fact is that after integrating over Grassmann coordinates -- thus hiding the superfield construction -- you end up with the actual Lagrangian of the theory where gauge invariance is manifest in the usual way.

In particular, the matter kinetic part of the Lagrangian of a gauge-matter SUSY theory is given, using superfields, by: $$ \mathcal{L}_\textrm{kin} = \int d\theta^2 d\overline\theta^2 \, \overline\Phi e^{V} \Phi, $$ where $V \equiv V_a T_R^a$, with $V_a$ real/vector superfields (general superfields constrained by $\overline V_ a = V_a$).

The gauge transformation acts on the vector superfields as follows: $$ e^V \rightarrow e^{i\overline\Lambda} e^V e^{-i\Lambda}, $$ ensuring that in superspace our theory is SUSY-gauge invariant.

Integrating now over the Grassmann coordinates $\theta, \overline\theta$, one finds (omitting gaugino interactions and terms with auxiliary fields): $$ \mathcal{L}_\textrm{kin} \supset |D_\mu \phi(x)|^2 -i\overline\psi(x) \overline\sigma^\mu D_\mu \psi(x), $$ where (up to a factor of $2g$, where $g$ is the gauge coupling) the covariant derivative $D_\mu$ is the usual one. This is a perfectly normal gauge-invariant Lagrangian, i.e. invariant under usual gauge transformations. Here, it is $\phi(x)$ (the $\Phi$ superfield's lowest component) which transforms as you wrote, in the same representation $R$.


N.B.: I have systematically omitted gauge indices, $\Phi$ is $\Phi^i$ (and thus $\phi$ is $\phi^i$, $i=1,\ldots,\textrm{dim }R$), corresponding to $A$ in your example ($i=1,2$).

Recommended reading: chapter 4.3.1 of R. Argurio's ``Introduction to Supersymmetry'', available online.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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