% File: susy.tex
% Section: QFT
% Title: Supersymmetry
% Last modified: 14.11.2002
%
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\begin{document}

\begin{center}
\large\textbf{SUPERSYMMETRY}
\end{center}

\vspace{0cm}

\begin{center}
\large\textbf{Formulas with $\mathbf{SL(2,C)}$ matrices}
\end{center}

\vspace{.1cm}

In its fundamental (spinor) representation, the $SL(2,C)$ group acts
by $2\times2$ unimodular matrices with complex elements,
\begin{equation}
\label{sl2c}
M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , \ \ \ \
\det M = ad-bc = 1 \ \ \ \ \ \ \Rightarrow \ \ \ \ \
M^{-1} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \,.
\end{equation}
If $M\in SL(2,C)$ then $M^{-1}, M^T, M^*, M^{\dag}$ also $\in SL(2,C)$\,.
With the help of the antisymmetric matrix
\begin{equation}
\label{eps}
\varepsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \ \ \ \ \ \
\varepsilon^{-1} = \varepsilon^T = \varepsilon^\dag  = -\varepsilon\,,
\end{equation}
the definition (\ref{sl2c}) for $M\in SL(2,C)$ can be equivalently
rewritten as
\begin{equation}
\label{epsM}
\varepsilon M^{-1} = M^T \varepsilon \ \ \ \ \ \ \text{or} \ \ \ \ \ \
M^T \varepsilon M = \varepsilon\,.
\end{equation}
We also introduce
\begin{equation}
\label{Pauli}
\sigma_\mu = (\mathbf{1},\vec{\sigma})\,, \ \ \ \
\bar{\sigma}_\mu = (\mathbf{1},-\vec{\sigma})\,, \ \ \ \
\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} , \ \ \
\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} , \ \ \
\sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}  \ \ \
\end{equation}
(note that $\sigma_\mu=\sigma^\mu$\,, etc., so as we do not distinguish
between upper and lower Lorentz 4-indices)\,, which obey
\begin{gather}
\label{sig1}
\text{tr}(\sigma_\mu \bar{\sigma}_\nu) = 2g_{\mu\nu}\,, \ \ \ \ \ \
\sigma_\mu \bar{\sigma}_\nu + \sigma_\nu \bar{\sigma}_\mu
 = \bar{\sigma}_\mu \sigma_\nu + \bar{\sigma}_\nu \sigma_\mu
 = 2g_{\mu\nu}\,\mathbf{1}\,, \\
\label{sigeps}
\varepsilon \bar{\sigma}_\mu = \sigma_{\mu}^T \varepsilon \ \ \ \ \ \
\Rightarrow \ \ \ \ \
 \bar{\sigma}_\mu = \varepsilon\sigma_{\mu}^T\varepsilon^T, \ \ \ \ \
 \sigma_\mu = \varepsilon^T\bar{\sigma}_{\mu}^T\varepsilon\,.
\end{gather}
In what follows, we shall need the component versions of the above
(and some extra) relations. Don't worry for the moment about dotted
vs.\ undotted indices, as well as about their upper and lower positions:
this will be explained in the next section.
\begin{gather}
\label{eps1}
\varepsilon_{\alpha\beta} = -\varepsilon_{\beta\alpha}
 = \varepsilon^{\alpha\beta} = -\varepsilon^{\beta\alpha}\,, \ \ \ \ \
\varepsilon_{12} = \varepsilon^{12} =1\,, \ \ \ \ \
\varepsilon^{\alpha\gamma}\varepsilon_{\gamma\beta}
 = -\delta^{\alpha}_{\beta}\,, \\
\label{eps2}
\varepsilon^{\alpha\gamma}\varepsilon^{\beta\rho}\varepsilon_{\gamma\rho}
 = \varepsilon^{\alpha\beta}\,, \ \ \ \ \
\varepsilon_{\gamma\alpha}\varepsilon_{\rho\beta}\varepsilon^{\gamma\rho}
 = \varepsilon_{\alpha\beta}\,, \ \ \ \ \
\varepsilon_{\alpha\gamma}\varepsilon^{\beta\rho}
 = \delta_{\alpha}^{\beta}\delta_{\gamma}^{\rho}
 - \delta_{\alpha}^{\rho}\delta_{\gamma}^{\beta}\,, \\
\label{epsM1}
\varepsilon^{\alpha\gamma}\,(M^{-1})_\gamma^\beta
 = M^\alpha_\gamma\,\varepsilon^{\gamma\beta}\,, \ \ \ \ \
\varepsilon_{\alpha\gamma}\,(M^{-1})^\gamma_\beta
 = M_\alpha^\gamma\,\varepsilon_{\gamma\beta}\,, \ \ \ \ \
M^\gamma_\alpha\,\varepsilon_{\gamma\rho}M^\rho_\beta
 = \varepsilon_{\alpha\beta}\,, \\
\label{orthog}
\bar{\sigma}_\mu^{\dot{\alpha}\alpha}\sigma^\mu_{\beta\dot{\beta}}
 = 2\delta^{\alpha}_{\beta}\delta^{\dot{\alpha}}_{\dot{\beta}}\,, \ \ \ \
\sigma^\mu_{\alpha\dot{\alpha}}\sigma^\mu_{\beta\dot{\beta}}
 = 2\varepsilon_{\alpha\beta}\varepsilon_{\dot{\alpha}\dot{\beta}}\,, \ \ \ \
\bar{\sigma}_\mu^{\dot{\alpha}\alpha}\bar{\sigma}_\mu^{\dot{\beta}\beta}
 = 2\varepsilon^{\alpha\beta}\varepsilon^{\dot{\alpha}\dot{\beta}}\,, \\
\label{trace}
\sigma^\mu_{\alpha\dot{\alpha}}\bar{\sigma}_\nu^{\dot{\alpha}\alpha}
 = 2g_{\mu\nu}\,, \ \ \ \ \
\varepsilon^{\alpha\gamma}\sigma^\mu_{\gamma\dot{\alpha}}
 = \varepsilon_{\dot{\gamma}\dot{\alpha}}
  \bar{\sigma}_\mu^{\dot{\gamma}\alpha}\,, \ \ \ \
\varepsilon^{\dot{\alpha}\dot{\gamma}}\sigma^\mu_{\alpha\dot{\gamma}}
 = \varepsilon_{\gamma\alpha}
  \bar{\sigma}_\mu^{\dot{\alpha}\gamma}\,, \\
\label{sig2}
\bar{\sigma}_\mu^{\dot{\alpha}\alpha}
 = \varepsilon^{\alpha\gamma}\varepsilon^{\dot{\alpha}\dot{\gamma}}
  \sigma^\mu_{\gamma\dot{\gamma}}\,, \ \ \ \ \ \
\sigma^\mu_{\alpha\dot{\alpha}}
 = \varepsilon_{\gamma\alpha}\varepsilon_{\dot{\gamma}\dot{\alpha}}
  \bar{\sigma}_\mu^{\dot{\gamma}\gamma}\,.
\end{gather}

Let us associate with a Lorentz 4-vector $x$ a matrix
\begin{equation}
\label{xchi}
\chi = x_\mu\sigma_\mu = x_0\sigma_0 - \vec{x}\vec{\sigma}
= \begin{pmatrix} x_0-x_3 & -x_1+ix_2 \\ -x_1-ix_2 & x_0+x_3 \end{pmatrix}\,.
\end{equation}
Obviously,
\begin{equation}
\label{}
\chi^{\dag}=\chi\,, \ \ \ \ \
\det \chi = x_0x_0 - \vec{x}\vec{x} = x_\mu x_\mu
 = x^2\,, \ \ \ \ \
x_\mu = \frac{1}{2}\,\text{tr}(\bar{\sigma}_\mu\chi)\,.
\end{equation}
If now $\chi$ is transformed with the use of the $SL(2,C)$ matrix $M$
as follows,
\begin{equation}
\label{transchi}
\chi \rightarrow \chi' = M\chi M^{\dag} \ \ \ \ \ \
\Rightarrow \ \ \ \ \ (\chi')^{\dag} = \chi'\,, \ \ \ \
\det\chi' = \det\chi = x^2\,,
\end{equation}
then the associated $x$ is also linearly transformed,
\begin{equation}
\label{transx}
x_\mu \rightarrow x'_\mu = \frac{1}{2}\,\text{tr}(\bar{\sigma}_\mu\chi')
 = \Lambda_{\mu\nu}x_\nu\,, \ \ \ \ \ (x')^2 = \det\chi' = x^2\,,
\end{equation}
with $\Lambda_{\mu\nu}$ being the Lorentz transformation matrix:
\begin{equation}
\label{Lambda}
\Lambda_{\mu\nu}
 = \frac{1}{2}\,\text{tr}(\bar{\sigma}_\mu M\sigma_\nu M^{\dag})\,, \ \ \ \ \
\Lambda_{\mu\lambda}\Lambda_{\nu\lambda} = g_{\mu\nu}\,.
\end{equation}
The last identity follows from the invariance of $x^2$. It can
also be verified explicitly, using (\ref{epsM}), (\ref{sigeps}) and
(\ref{orthog})\,:
\begin{multline}
\label{}
\Lambda_{\mu\lambda}\Lambda_{\nu\lambda}
= \frac{1}{4}\,\text{tr}(\bar{\sigma}_\mu M\sigma_\lambda M^{\dag})
 \,\text{tr}(\bar{\sigma}_\nu M\sigma_\lambda M^{\dag})
= \frac{1}{4}\,\text{tr}(\bar{\sigma}_\mu M
  \varepsilon^T\bar\sigma_\lambda^T\varepsilon M^{\dag})
 \,\text{tr}(\bar{\sigma}_\nu M\sigma_\lambda M^{\dag}) \\
= \frac{1}{4}(\bar\sigma_\lambda)^{\dot\alpha\alpha}
   (\varepsilon M^\dag\bar\sigma_\mu M \varepsilon^T)_{\dot{\alpha}\alpha}
    (\sigma_\lambda)_{\beta\dot{\beta}}
   (M^\dag\bar\sigma_\nu M)^{\dot{\beta}\beta}
= \frac{1}{2}\,\text{tr}(\varepsilon M^{\dag}\bar{\sigma}_\mu M
   \varepsilon^T M^T\bar{\sigma}_{\nu}^T (M^{\dag})^T) \\
= \frac{1}{2}\,\text{tr}(\varepsilon \bar{\sigma}_\mu
   \varepsilon^T \bar{\sigma}_{\nu}^T)
= \frac{1}{2}\,\text{tr}(\sigma_\mu^T \bar{\sigma}_{\nu}^T)
= \frac{1}{2}\,\text{tr}(\bar{\sigma}_{\nu}\sigma_\mu)
= g_{\mu\nu}\,.
\end{multline}

%\newpage
\vspace{.3cm}

\begin{center}
\large\textbf{Anticommuting spinors}
\end{center}

\vspace{.1cm}

At the most elementary level, supersymmetry deals with two-component (Weyl)
anticommuting spinors: any spinorial component anticommutes with any other
(including itself)\,. These spinors belong to irreducible two-dimensional
representations of $SL(2,C)$\,. Contrary to the Lorentz 4-indices, we do
distinguish between upper and lower spinorial 2-indices:
\begin{equation}
\label{raise}
\varepsilon^{\alpha\beta}\theta_\beta = \theta^\alpha\,, \ \ \ \ \ \
\varepsilon_{\beta\alpha}\theta^\beta = \theta_\alpha \ \ \ \ \ \
\Rightarrow \ \ \ \ \ \theta^1 = \theta_2\,, \ \ \ \ \theta^2 = -\theta_1\,,
\end{equation}
which is consistent with (\ref{eps1})\,. Eq.\,(\ref{eps2}) shows that
$\varepsilon$-tensor can also be used to raise, or lower, its own indices
(as well as spinorial indices of any other object)\,. By the way, we see
from (\ref{sig2}) that $\bar\sigma_\mu$ is nothing but $\sigma_\mu^T$ with
both indices raised, and vice versa.

Mutually complex conjugated representations of $SL(2,C)$ are non-equivalent,
so it is general practice to use dotted indices (and a bar) for one of these,
\begin{equation}
\label{dot}
(\theta^*)_\alpha \doteq \bar{\theta}_{\dot{\alpha}}\,, \ \ \ \ \
(\theta^*)^\alpha \doteq \bar{\theta}^{\dot{\alpha}}\,, \ \ \ \ \
\varepsilon_{\dot{\alpha}\dot{\beta}} = \varepsilon^*_{\alpha\beta}
 = \varepsilon_{\alpha\beta}\,, \ \ \ \ \
\varepsilon^{\dot{\alpha}\dot{\beta}} = \varepsilon^{\alpha\beta}\,,
\end{equation}
so that the dotted raising (lowering) formulas look precisely like
(\ref{raise}) (the same is true for (\ref{eps1}) and (\ref{eps2}))\,.
In the definitions (\ref{dot})\,, $\alpha$ and $\dot{\alpha}$ are assumed
to be numerically equal, but in formally covariant relations like
(\ref{trace}) they (being summed or not) are, of course, treated as
independent.

Another widely accepted convention about dotted and undotted indices is
to number rows by upper dotted but lower undotted indices (resp., columns
by lower dotted and upper undotted ones)\,. This agrees with the notation
chosen in (\ref{dot}) for mutually conjugate spinors
(Hermitean conjugation is to send a column to a row, and vice versa)\,,
\begin{equation}
\label{Herm}
\begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix}^{\dag}
 = \begin{pmatrix} \theta_1^* \\ \theta_2^* \end{pmatrix}^T
 = \begin{pmatrix} \bar{\theta}_{\dot{1}} & \bar{\theta}_{\dot{2}}
\end{pmatrix}\,, \ \ \ \ \ \
\begin{pmatrix} \theta^1 & \theta^2 \end{pmatrix}^{\dag}
 = \begin{pmatrix} \theta^{*1} & \theta^{*2} \end{pmatrix}^T
 = \begin{pmatrix} \bar{\theta}^{\dot{1}} \\ \bar{\theta}^{\dot{2}}
\end{pmatrix}\,,
\end{equation}
and in (\ref{orthog}) -- (\ref{sig2}) for $\sigma$ and $\bar{\sigma}$\,.
Adopting this convention, we, at last, write down how two-component
spinors transform under the action of $SL(2,C)$\,:
\begin{align}
\theta'_\alpha &= M_\alpha^\beta \theta_\beta     &\text{or} \qquad
\theta' &= M\theta  &(\text{column}) \\
{\theta'}^\alpha &= \theta^\beta (M^{-1})^\alpha_\beta  &\text{or} \qquad
\theta' &= \theta M^{-1} &(\text{row}) \\
\bar\theta'_{\dot{\alpha}} &= \bar\theta_{\dot{\beta}}
 (M^{\dag})_{\dot{\alpha}}^{\dot{\beta}}  &\text{or} \qquad
\bar\theta' &= \bar\theta M^{\dag} &(\text{row}) \\
\bar\theta'{}^{\dot\alpha} &= (M^\dag{}^{-1})^{\dot{\alpha}}_{\dot{\beta}}
 \bar\theta^{\dot\beta}    &\text{or} \qquad
\bar\theta' &= (M^\dag{}^{-1})\bar\theta &(\text{column})
\end{align}
These transformation laws ensure that for any two spinors their contraction
\begin{equation}
\label{contr1}
\theta\varphi \doteq \theta^\alpha\varphi_\alpha = \varphi^\alpha\theta_\alpha
 = \varphi\theta = -\theta_\alpha\varphi^\alpha = -\varphi_\alpha\theta^\alpha
 = \theta^2\varphi^1 - \theta^1\varphi^2
 = \theta_2\varphi_1 - \theta_1\varphi_2
\end{equation}
and its Hermitean conjugate
\begin{equation}
\label{contr2}
\bar{\theta}\bar{\varphi}
 \doteq \bar{\theta}_{\dot{\alpha}}\bar{\varphi}^{\dot{\alpha}}
 = \bar{\varphi}\bar{\theta} = (\theta\varphi)^{\dag}
 = \bar{\theta}_{\dot{1}}\bar{\varphi}_{\dot{2}}
  - \bar{\theta}_{\dot{2}}\bar{\varphi}_{\dot{1}}
\end{equation}
are both scalars.

The higher tensors transform accordingly. For instance, the last
equality in (\ref{epsM1}) implies that $\varepsilon_{\alpha\beta}$ itself
is an invariant tensor. It should be also noted that an object
\begin{equation}
\label{vector}
\theta\sigma_\mu\bar{\varphi} \doteq
 \theta^\alpha\sigma^\mu_{\alpha\dot{\alpha}}\bar{\varphi}^{\dot{\alpha}}
\end{equation}
transforms as a Lorentz 4-vector. Really, by definition
\begin{equation}
\label{}
(\theta\sigma_\mu\bar{\varphi})' = \theta'\sigma_\mu\bar{\varphi}'
= \theta M^{-1}\sigma_\mu(M^\dag{}^{-1})\bar\varphi\,,
\end{equation}
which coincides with the result of the following computation:
\begin{multline}
\label{}
\Lambda_{\mu\nu}\theta\sigma_\nu\bar{\varphi}
= \frac{1}{2}\,\text{tr}(\bar{\sigma}_\mu M\sigma_\nu M^{\dag})\,
   \theta\sigma_\nu\bar{\varphi}
= \frac{1}{2}\,\text{tr}(\bar{\sigma}_\mu M\varepsilon^T\bar{\sigma}_\nu^T
   \varepsilon M^{\dag})\,\theta\sigma_\nu\bar{\varphi}
= \frac{1}{2}\,(\ldots)_{\dot{\alpha}\alpha}
   (\bar{\sigma}_\nu)^{\dot{\alpha}\alpha}\theta\sigma_\nu\bar{\varphi} \\
= \theta(\ldots)^T\bar{\varphi}
= \theta\varepsilon M^T\bar{\sigma}_\mu^T {M^{\dag}}^T
   \varepsilon^T\bar{\varphi}
= \theta M^{-1}\varepsilon\bar{\sigma}_\mu^T\varepsilon^T
   (M^\dag{}^{-1})\bar\varphi
= \theta M^{-1}\sigma_\mu(M^\dag{}^{-1})\bar\varphi\,.
\end{multline}

The following identities with anticommuting spinors are
straightforwardly derived:
\begin{gather}
\theta\theta \doteq \theta^\alpha\theta_\alpha
 = -2\theta^1\theta^2 = -2\theta_1\theta_2\,, \ \ \ \ \ \ \
\bar{\theta}\bar{\theta}
 \doteq \bar{\theta}_{\dot{\alpha}}\bar{\theta}^{\dot{\alpha}}
 = 2\bar{\theta}_{\dot{1}}\bar{\theta}_{\dot{2}}
 = 2\bar{\theta}^{\dot{1}}\bar{\theta}^{\dot{2}}\,, \\
\theta^\alpha\theta^\beta
 = -\frac{1}{2}\varepsilon^{\alpha\beta}\,\theta\theta\,, \ \ \ \
\theta_\alpha\theta_\beta
 = -\frac{1}{2}\varepsilon_{\alpha\beta}\,\theta\theta\,, \ \ \ \
\bar{\theta}^{\dot{\alpha}}\bar{\theta}^{\dot{\beta}}
 = \frac{1}{2}\varepsilon^{\dot{\alpha}\dot{\beta}}\,
    \bar{\theta}\bar{\theta}\,, \ \ \ \
\bar{\theta}_{\dot{\alpha}}\bar{\theta}_{\dot{\beta}}
 = \frac{1}{2}\varepsilon_{\dot{\alpha}\dot{\beta}}\,
    \bar{\theta}\bar{\theta}\,, \\
(\theta\xi)(\theta\varphi) = -\frac{1}{2}(\theta\theta)(\xi\varphi)\,, \ \ \ \
(\bar{\theta}\bar{\xi})(\bar{\theta}\bar{\varphi})
 = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\xi\bar\varphi)\,, \\
(\theta\xi)(\theta\sigma_\mu\bar\varphi)
 = -\frac{1}{2}(\theta\theta)(\xi\sigma_\mu\bar\varphi)\,, \ \ \ \
(\bar{\theta}\bar{\xi})(\bar{\theta}\bar\sigma_\mu\varphi)
 = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\xi\bar\sigma_\mu\varphi)\,, \\
\theta\sigma_\mu\bar\varphi = (\varphi\sigma_\mu\bar\theta)^\dag
 = -\bar\varphi\bar\sigma_\mu\theta\,, \ \ \ \
\theta\sigma_\mu\bar\sigma_\nu\varphi
 = (\bar\varphi\bar\sigma_\nu\sigma_\mu\bar\theta)^\dag
 = \varphi\sigma_\nu\bar\sigma_\mu\theta\,, \\
(\theta\sigma_\mu\bar\varphi)(\xi\sigma_\mu\bar\chi)
 = -(\bar\varphi\bar\sigma_\mu\theta)(\xi\sigma_\mu\bar\chi)
 = (\bar\varphi\bar\sigma_\mu\theta)(\bar\chi\bar\sigma_\mu\xi)
 = 2(\theta\xi)(\bar\varphi\bar\chi)\,, \\
(\theta\sigma_\mu\bar\theta)(\theta\sigma_\nu\bar\theta)
 = -(\bar\theta\bar\sigma_\mu\theta)(\theta\sigma_\mu\bar\theta)
 = (\bar\theta\bar\sigma_\mu\theta)(\bar\theta\bar\sigma_\nu\theta)
 = \frac{1}{2}g_{\mu\nu}(\theta\theta)(\bar\theta\bar\theta)\,, \\
\theta^\alpha\bar\varphi^{\dot\alpha}
 = \frac{1}{2}\bar\sigma_\mu^{\dot\alpha\alpha}
  (\theta\sigma_\mu\bar\varphi)\,, \ \ \ \ \
\theta_\alpha\bar\varphi_{\dot\alpha}
 = \frac{1}{2}\sigma^\mu_{\alpha\dot\alpha}(\theta\sigma_\mu\bar\varphi)\,.
\end{gather}

Dirac $\gamma$-matrices and four-component spinors can be constructed
from the above objects in the following way:
\begin{equation}
\label{Dirac}
\gamma_\mu =
 \begin{pmatrix} 0 & \sigma_\mu \\ \bar\sigma_\mu & 0 \end{pmatrix}\,, \ \ \ \
\psi =
 \begin{pmatrix} \theta_\alpha \\ \bar\xi^{\dot\alpha} \end{pmatrix}\,, \ \ \ \
\psi^\dag =
 \begin{pmatrix} \bar\theta_{\dot\alpha} & \xi^\alpha \end{pmatrix}\,, \ \ \ \
\bar\psi = \psi^\dag \gamma_0 =
 \begin{pmatrix} \xi^\alpha & \bar\theta_{\dot\alpha} \end{pmatrix}\,, \ \ \ \
\end{equation}
so that
\begin{equation}
\label{}
\bar\psi\psi = \xi^\alpha\theta_\alpha +
  \bar\theta_{\dot\alpha}\bar\xi^{\dot\alpha}
 = \theta\xi + \bar\theta\bar\xi = \theta\xi + (\theta\xi)^\dag\,.
\end{equation}
A Majorana spinor, specified by the requirement of charge-conjugation
invariance,
\begin{equation}
\label{C}
\psi^c \doteq C\bar\psi^T = \psi\,, \ \ \ \ \ \ C^T = -C\,, \ \ \ \ \
 C^{-1}\gamma_\mu C =-\gamma_\mu^T\,,
\end{equation}
is, owing to (\ref{raise}), given by
\begin{equation}
\label{Maj}
\psi_M = \begin{pmatrix} \theta_\alpha \\
 \bar\theta^{\dot\alpha} \end{pmatrix}
 = \begin{pmatrix} \theta_1 \\ \theta_2 \\ \theta^{*1} \\ \theta^{*2}
  \end{pmatrix}\,, \ \ \ \ \ \ \ \
 C = i\gamma_0\gamma_2 =
  \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\
   0 & 0 & -1 & 0 \end{pmatrix}\,.
\end{equation}
It actually have only two independent (complex) components.

%\newpage
\vspace{.3cm}

\begin{center}
\large\textbf{Superspace}
\end{center}

\vspace{.1cm}

The simplest ($N=1$) superspace is obtained by adding four odd
(anticommuting, or Grassmann) degrees of freedom
$\theta_\alpha,\bar\theta_{\dot\alpha}$
to the (even) space-time coordinates $x_\mu$\,. Standard operations in
superspace are differentiation and integration. Let's begin with the latter.
Basic definitions of Grassmann integration
(over one odd variable $\eta$) are very simple:
\begin{equation}
\label{int1}
\int\!d\eta = 0\,, \ \ \ \ \ \int\!d\eta\,\eta = 1\,, \ \ \ \ \
\delta(\eta) = \eta\,.
\end{equation}
Differential $d\eta$ is also assumed to be odd.
For the two-component spinors we introduce
\begin{equation}
\label{int2}
d^2\theta \doteq -\frac{1}{4}d\theta^\alpha d\theta_\alpha
 = \frac{1}{2}d\theta_1 d\theta_2\,, \ \ \ \ \
d^2\bar\theta = (d^2\theta)^\dag
 = -\frac{1}{4}d\bar\theta_{\dot\alpha}d\bar\theta^{\dot\alpha}
 = -\frac{1}{2}d\bar\theta_{\dot 1}d\bar\theta_{\dot 2}\,,
\end{equation}
and obtain
\begin{equation}
\label{int3}
\int\!d^2\theta\ \theta\theta
 = \int\!d^2\bar\theta\ \bar\theta\bar\theta = 1\,, \ \ \ \ \
\delta(\theta) = \theta\theta\,, \ \ \ \
\delta(\bar\theta) = \bar\theta\bar\theta\,.
\end{equation}

We define Grassmann derivatives via
\begin{equation}
\label{diff1}
f(\theta+\xi) = f(\theta) + \xi^\alpha\partial_\alpha f(\theta)
 + o(\xi)\,, \ \ \ \ \ \
f(\bar\theta+\bar\xi) = f(\bar\theta)
 + \bar\xi_{\dot\alpha}\bar\partial^{\dot\alpha}\!f(\bar\theta)
 + o(\bar\xi)\,,
\end{equation}
and
\begin{equation}
\label{}
\partial^\alpha = \varepsilon^{\alpha\beta}\partial_\beta\,, \ \ \ \ \ \
\partial_\alpha = \varepsilon_{\beta\alpha}\partial^\beta\,, \ \ \ \ \ \
\bar\partial^{\dot\alpha}
 = \varepsilon^{\dot\alpha\dot\beta}\bar\partial_{\dot\beta}\,, \ \ \ \ \ \
\bar\partial_{\dot\alpha}
 = \varepsilon_{\dot\beta\dot\alpha}\bar\partial^{\dot\beta}\,.
\end{equation}
This entails
\begin{gather}
\label{}
\partial_\alpha\theta^\beta = -\partial^\beta\theta_\alpha
 = \delta_\alpha^\beta\,, \ \ \ \ \
\bar\partial^{\dot\beta}\bar\theta_{\dot\alpha}
 = -\bar\partial_{\dot\alpha}\bar\theta^{\dot\beta}
 = \delta_{\dot\alpha}^{\dot\beta}\,, \ \ \ \ \
\partial_\alpha\theta_\beta = \varepsilon_{\alpha\beta}\,, \ \ \ \ \
\bar\partial_{\dot\alpha}\bar\theta_{\dot\beta}
 = \varepsilon_{\dot\beta\dot\alpha}\,, \\
\partial_\alpha(\theta\xi) = \xi_\alpha\,, \ \ \ \ \
\bar\partial_{\dot\alpha}(\bar\theta\bar\xi)
 = \bar\xi_{\dot\alpha}\,, \ \ \ \ \
\partial_\alpha(\theta\theta) = 2\theta_\alpha\,, \ \ \ \ \
\bar\partial_{\dot\alpha}(\bar\theta\bar\theta)
 = 2\bar\theta_{\dot\alpha}
\end{gather}
(plus, of course, the corresponding relations with upper indices). It also
appears that Grassmann differentiation and integration operations are,
in a sense, interchangeable:
\begin{equation}
\label{intdiff}
\int\!d^2\theta = \frac{1}{4}\varepsilon^{\alpha\beta}
 \partial_\alpha\partial_\beta\,, \ \ \ \ \ \
\int\!d^2\bar\theta\ = -\frac{1}{4}\varepsilon^{\dot\alpha\dot\beta}
 \bar\partial_{\dot\alpha}\bar\partial_{\dot\beta}\,.
\end{equation}

The following explicit realization of Grassmann derivatives is often used.
If one, quite naturally, assumes
\begin{equation}
\label{}
\frac{\partial}{\partial\theta^\alpha}\theta^\beta
 = \frac{\partial}{\partial\theta_\beta}\theta_\alpha
 = \delta_\alpha^\beta\,, \ \ \ \ \ \
\frac{\partial}{\partial\bar\theta^{\dot\alpha}}\bar\theta^{\dot\beta}
 = \frac{\partial}{\partial\bar\theta_{\dot\beta}}\bar\theta_{\dot\alpha}
 = \delta_{\dot\alpha}^{\dot\beta}\,,
\end{equation}
then
\begin{equation}
\label{}
\partial_\alpha = \frac{\partial}{\partial\theta^\alpha}\,, \ \ \ \ \
\partial^\alpha = -\frac{\partial}{\partial\theta_\alpha}\,, \ \ \ \ \
\bar\partial_{\dot\alpha}
 = -\frac{\partial}{\partial\bar\theta^{\dot\alpha}}\,, \ \ \ \ \
\bar\partial^{\dot\alpha}
 = \frac{\partial}{\partial\bar\theta_{\dot\alpha}}\ .
\end{equation}

\vspace{0.4cm}

Supersymmetry transformations in superspace look like
\begin{align}
x_\mu\rightarrow \tilde{x}_\mu
 &= \,x_\mu + i\theta\sigma_\mu\bar\xi - i\xi\sigma_\mu\bar\theta
 = x_\mu - i\bar\xi\bar\sigma_\mu\theta - i\xi\sigma_\mu\bar\theta
\notag \\
\label{shift}
\theta\rightarrow\tilde\theta \ \,&= \ \theta + \xi \\
\bar\theta\rightarrow\tilde{\bar\theta} \ \,&= \ \bar\theta + \bar\xi\,,
\notag
\end{align}
or $\tilde{x}_\mu=x_\mu-i\bar\eta\gamma_\mu\psi$ in terms of Majorana
spinors $\eta=\binom{\xi_\alpha}{\bar\xi^{\dot\alpha}},
\psi=\binom{\theta_\alpha}{\bar\theta^{\dot\alpha}}$\,.

Expanding in powers of $\xi,\bar\xi$ as follows\,,
\begin{equation}
\label{}
f(\tilde{x},\tilde{\theta},\tilde{\bar\theta}) = f(x,\theta,\bar\theta)
 + (\xi^\alpha Q_\alpha + \bar\xi_{\dot\alpha}\bar{Q}^{\dot\alpha})f
 +\ldots \ ,
\end{equation}
we obtain the supersymmetry generators
\begin{align}
Q_\alpha &= \partial_\alpha
 - i\sigma^\mu_{\alpha\dot\alpha}\bar\theta^{\dot\alpha}\partial_\mu &
Q^\alpha &= \partial^\alpha
 + i\bar\theta_{\dot\alpha}\bar\sigma_\mu^{\dot\alpha\alpha}\partial_\mu \\
\bar Q_{\dot\alpha} &= \bar\partial_{\dot\alpha}
 + i\theta^\alpha\sigma^\mu_{\alpha\dot\alpha}\partial_\mu &
\bar Q^{\dot\alpha} &= \bar\partial^{\dot\alpha}
 - i\bar\sigma_\mu^{\dot\alpha\alpha}\theta_\alpha\partial_\mu
\end{align}
obeying the famous relation
\begin{equation}
\label{}
[Q_\alpha,\,\bar Q_{\dot\alpha}]_+
 = 2i\sigma^\mu_{\alpha\dot\alpha}\partial_\mu
 \equiv 2\sigma^\mu_{\alpha\dot\alpha}P_\mu\,.
\end{equation}

However, Grassmann derivatives $\partial_\alpha,\,\bar\partial_{\dot\alpha}$
are not super-invariant (do not anticommute with the generators $Q$)\,.
Instead, by the chain rule
\begin{equation}
\label{chain}
\tilde{\partial} = (\tilde{\partial}x_\mu)\,\partial_\mu
 + (\tilde{\partial}\theta^\alpha)\,\partial_\alpha
 + (\tilde{\partial}\bar\theta_{\dot\alpha})\,\bar\partial^{\dot\alpha}\,,
\end{equation}
they transform as
\begin{align}
\tilde\partial_\mu &= \partial_\mu \\
\tilde\partial_\alpha &= \partial_\alpha
 - i\sigma^\mu_{\alpha\dot\alpha}\bar\xi^{\dot\alpha}\partial_\mu \\
\tilde{\bar\partial}_{\dot\alpha} &= \bar\partial_{\dot\alpha}
 + i\xi^\alpha\sigma^\mu_{\alpha\dot\alpha}\partial_\mu
\end{align}
To remedy the non-invariance, one introduces super-covariant derivatives
\begin{align}
D_\alpha &= \partial_\alpha
 + i\sigma^\mu_{\alpha\dot\alpha}\bar\theta^{\dot\alpha}\partial_\mu &
D^\alpha &= \partial^\alpha
 - i\bar\theta_{\dot\alpha}\bar\sigma_\mu^{\dot\alpha\alpha}\partial_\mu \\
\bar D_{\dot\alpha} &= \bar\partial_{\dot\alpha}
 - i\theta^\alpha\sigma^\mu_{\alpha\dot\alpha}\partial_\mu &
\bar{D}^{\dot\alpha} &= \bar\partial^{\dot\alpha}
 + i\bar\sigma_\mu^{\dot\alpha\alpha}\theta_\alpha\partial_\mu
\end{align}
which satisfy
\begin{equation}
\label{}
[D_\alpha,\,\bar{D}_{\dot\alpha}]_+
 = -2i\sigma^\mu_{\alpha\dot\alpha}\partial_\mu\,, \ \ \ \ \
[D^\alpha,\,\bar{D}^{\dot\alpha}]_+
 = -2i\bar\sigma_\mu^{\dot\alpha\alpha}\partial_\mu\,, \ \ \ \ \
[D,Q]_+ = 0\,.
\end{equation}

Functions on superspace are called superfields.
Extremely useful class of them is provided by chiral superfield $\Phi$
and antichiral $\Phi^\dag$ specified by the conditions
\begin{equation}
\label{chiral}
\bar D_{\dot\alpha}\Phi = D_\alpha\Phi^\dag = 0\,.
\end{equation}
Because of
\begin{equation}
\label{}
\bar D_{\dot\alpha}\theta
 = \bar D_{\dot\alpha}(x_\mu + i\theta\sigma_\mu\bar\theta) = 0\,, \ \ \ \ \ \
D_\alpha\bar\theta
 = D_\alpha(x_\mu - i\theta\sigma_\mu\bar\theta) = 0
\end{equation}
we actually have
\begin{equation}
\label{}
\Phi = \Phi\,(x + i\theta\sigma\bar\theta,\,\theta)\,, \ \ \ \ \ \
\Phi^\dag = \Phi^\dag(x - i\theta\sigma\bar\theta,\,\bar\theta)\,.
\end{equation}
Really, a change of coordinates
$(x,\theta,\bar\theta)\rightarrow
(x'=x + i\theta\sigma\bar\theta,\theta'=\theta,\bar\theta'=\bar\theta)$
causes $\partial\rightarrow\partial'$\,:
\begin{align}
\partial'_\mu &= \partial_\mu \\
\partial'_\alpha &= \partial_\alpha
 - i(\sigma_\mu\bar\theta)_\alpha\,\partial_\mu \\
{\bar\partial}'_{\dot\alpha} &= \bar\partial_{\dot\alpha}
 - i(\theta\sigma_\mu)_{\dot\alpha}\partial_\mu
\end{align}
so that, in the new variables $(x',\theta',\bar\theta')$\,,
\begin{equation}
\label{}
D'_\alpha = \partial'_\alpha
 + 2i\sigma^\mu_{\alpha\dot\alpha}\bar\theta'{}^{\dot\alpha}\partial'_\mu\,,
\ \ \ \ \ \ \bar D'_{\dot\alpha} = \bar\partial'_{\dot\alpha}\,.
\end{equation}
Hence, $\Phi$ does not depend on $\bar\theta$ (except via $x'$)\,.
Now, expanding in $\theta$\,, we express chiral superfield $\Phi$ in terms
of scalar ($\varphi, F$) and fermionic ($\psi$) component fields:
\begin{multline}
\label{comp}
\Phi\,(x + i\theta\sigma\bar\theta,\,\theta)
 = \varphi(x') + \sqrt{2}\,\theta\psi(x') + \theta\theta F(x') \\
= \varphi(x) + \sqrt{2}\,\theta\psi(x) + \theta\theta F(x)
 + i\theta\sigma_\mu \bar{\theta}\,\partial_\mu\varphi(x)
 - \frac{1}{4}\,\theta\theta\,\bar{\theta}\bar{\theta}\,\partial^2\varphi(x)
 - \frac{i}{\sqrt{2}}\,\theta\theta\,
  \partial_\mu\psi(x)\sigma_\mu\bar{\theta}
\end{multline}
Analogously for $\Phi^\dag$\,:
\begin{equation}
\label{comp'}
\Phi^\dag = \varphi^\dag + \sqrt{2}\,\bar{\theta}\bar{\psi}
 + \bar{\theta}\bar{\theta} F^\dag
 - i\theta\sigma_\mu\bar{\theta}\,\partial_\mu \varphi^\dag
 - \frac{1}{4}\,\theta\theta\,\bar{\theta}\bar{\theta}\,\partial^2\varphi^\dag
 + \frac{i}{\sqrt{2}}\,\bar{\theta}\bar{\theta}\,\theta\sigma_\mu\partial_\mu
\bar{\psi}\ .
\end{equation}

Let us now return to the superspace shift (\ref{shift})\,. Taking the chain
rule for differentials in the form
\begin{equation}
\label{chain2}
dz = dx_\mu(\partial_\mu z) + d\theta^\alpha\partial_\alpha z
 + d\bar\theta_{\dot\alpha}\bar\partial^{\dot\alpha}z\,,
\end{equation}
one comes to
\begin{equation}
\label{}
\begin{pmatrix} d\tilde x_\mu & d\tilde\theta^\alpha
 & d\tilde{\bar\theta}_{\dot\alpha} \end{pmatrix}
 = \begin{pmatrix} d x_\nu & d\theta^\beta & d\bar\theta_{\dot\beta}
 \end{pmatrix}
\begin{pmatrix} g_{\mu\nu} & 0 & 0 \\
 i(\sigma_\mu\bar\xi)_\beta & \delta^\alpha_\beta & 0 \\
 i(\bar\sigma_\mu\xi)^{\dot\beta} & 0 & \delta_{\dot\alpha}^{\dot\beta}
  \end{pmatrix} \ .
\end{equation}
Clearly, the transformation matrix here is equal to
\begin{equation}
\label{}
\exp \begin{pmatrix} 0 & 0 & 0 \\
 i(\sigma_\mu\bar\xi)_\beta & 0 & 0 \\
 i(\bar\sigma_\mu\xi)^{\dot\beta} & 0 & 0 \end{pmatrix} \ ,
\end{equation}
so its Jacobian is unity.

\end{document}