Many textbooks present expressions for superfields in $4$ dimensions. For my current project, I have to find out how things work in $2$ dimensions.

Let me summarise in short what we know about $4$d (the equations below are taken from the book by Muller-Kirsten and Wiedemann).

The most general form of a superfield is: \begin{equation}\begin{gathered} \Phi(x, \theta, \bar{\theta}) = f(x) + \theta \varphi(x) + \bar{\theta} \bar{\chi} (x) \\ + (\theta \sigma^\mu \bar{\theta}) V_\mu(x) + \theta \theta m(x) + \bar{\theta} \bar{\theta} n(x) \\ + (\theta \theta) \bar{\theta} \bar{\lambda}(x) + (\bar{\theta} \bar{\theta}) \theta \psi(x) + (\theta \theta) (\bar{\theta} \bar{\theta}) d(x) \quad. \end{gathered}\end{equation} Here and hereafter, I am using the two-component formalism (i.e. Weyl in $4$d and Majorana in $2$d). If we set a superfield to be left-chiral by requiring \begin{equation} \bar{D} \Phi = 0 \quad, \end{equation} with \begin{equation} \bar{D}_{\dot{a}} = - \bar{\partial}_{\dot{a}} - i \theta_b \sigma^\mu_{b \dot{a}} \partial_\mu \quad, \end{equation} then the superfield will take the form of \begin{equation}\begin{gathered} \Phi = A(y) + \sqrt{2} \theta \psi(y) + \theta \theta F(y) \\ = A(x) + i (\theta \sigma^\mu \bar{\theta}) \partial_\mu A(x) - \dfrac{1}{4} (\theta \theta) (\bar{\theta}\bar{\theta}) \Box A(x) + \sqrt{2} \theta \psi(x) \\+ \dfrac{i}{\sqrt{2}} (\theta \theta) \bar{\theta}_{\dot{a}}\partial_\mu \psi^a(x) \sigma^\mu_{a \dot{b}} \epsilon^{\dot{b}\dot{a}} + (\theta \theta) F(x) \quad. \end{gathered}\end{equation} As ever, we here use the notation \begin{equation} y^\mu = x^\mu + i \theta \sigma^\mu \bar{\theta} \quad. \end{equation}

Imposition of the reality condition, \begin{equation} V (x, \theta, \bar{\theta}) = V^\dagger (x, \theta, \bar{\theta}) \quad, \end{equation} will render us the vector superfield \begin{equation}\begin{gathered} V (x, \theta, \bar{\theta}) = C(x) + \theta \varphi(x) + \bar{\theta} \bar{\varphi} (x) \\ + (\theta \sigma^\mu \bar{\theta}) V_\mu(x) + \theta \theta M(x) + \bar{\theta} \bar{\theta} M^*(x) \\ + (\theta \theta) \bar{\theta} \bar{\psi}(x) + (\bar{\theta} \bar{\theta}) \theta \psi(x) + (\theta \theta) (\bar{\theta} \bar{\theta}) D(x) \quad, \end{gathered}\end{equation} where the conditions of reality for the component fields are imposed as well: \begin{equation} C(x) = C^*(x), \quad,\qquad V^*_\mu(x) = V_\mu(x) \quad,\qquad D^*(x) = D(x) \quad. \end{equation}

I want to know how all these equations look in $2$d. So, let me state the questions once again:

  1. What is the most general form of the superfield? (assuming it is a Lorentz scalar)

  2. What is the most general form of the left-chiral superfield?

  3. What is the most general form of the vector superfield?

I know that there emerge some differences between the $4$d and $2$d cases, as one starts constructing Lorentz scalars. Indeed, in the former case the charge conjugation does not preserve handedness, while in the latter it does. How exactly does this difference affect the number of ways in which we can write a Lorentz-invariant expression? (Basically, this should guide us on how to choose the component fields, right?..)

Most papers on $2$d SUSY begin with the statement that the most general superfield has the shape of \begin{equation} \Phi(x, \theta, \bar{\theta}) = \phi(x) + \sqrt{2} \bar{\theta}\psi(x) + \theta \bar{\theta} F(x) \quad. \end{equation} Is this the most general form of an unconstrained superfield? What precludes us from adding to it a term of the form $\sqrt{2}\theta \bar{\chi}(x)$? Or is it left-chiral?... I am confused.

Any comments or references will be greatly appreciated.

  • $\begingroup$ I am really sorry that you didn't get an answer until now. Are you here yet? $\endgroup$ – user259412 Jul 26 '16 at 6:09
  • $\begingroup$ I think I got some understanding. It will be more efficient if I try to answer the question myself and then get your feedback. Thanks! $\endgroup$ – mavzolej Jul 26 '16 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.