# Crystal structure informed mesoscale deformation model for HCP Cu6Sn5
intermetallic compound

Anil Kunwar1,∗, Haoran Ma2,∗∗, Johan Hektor3
1Faculty of Mechanical Engineering, Silesian University of Technology,
Konarskiego 18A, 44-100 Gliwice, Poland
2 School of Microelectronics, Dalian University of Technology, 116024 Dalian,
China
3Department of Materials Science and Applied Mathematics, Faculty of
Technology and Society,
Malmö University, 21119 Malmö, Sweden
∗Email (Corresponding Author<EMAIL_ADDRESS>; ∗∗Email (Corresponding
Author<EMAIL_ADDRESS>
###### Abstract

In the electronic packaging and energy storage sectors, the study of Cu6Sn5
intermetallic compound (IMC) is getting more attention. At temperatures above
186 ∘C, this IMC exists in a hexagonal closed packed (HCP) crystalline
structure. Crystal plasticity finite element simulations are performed on
Cu6Sn5 IMC by taking the information about its lattice parameters and
direction dependent elastic properties. Three types of models corresponding to
deformations in basal, prismatic and pyramidal modes are developed. With the
same type of loading in the elastic regime and boundary conditions, the
results of the computations reveal the differences in displacement magnitudes
among the three model types.

###### Index Terms:

Hexagonal closed packed ; Anisotropy; Deformation; Crystal plasticity finite
element; Intermetallic compound; Elastic regime

##  I Introduction

The use of Pb-free solders joints is getting promotion and significant
research interest because of the immense benefit to environment and human
health [1, 2]. For Sn-based solder joints using Cu substrates, the formation
of joint is guaranteed by the growth of Cu6Sn5 intermetallic compound (IMC) at
the interface. For miniaturized (Pb-free) Sn-based solder joints, the volume
fraction of the Cu6Sn5 IMC in the overall composite joint can be
proportionately significant. For an example, in context of 2.5D IC packaging,
the diameters of solder balls can be considered in the values well below 10
$\mu$m, and so during the reflow joining procedure with Cu substrate, the
major portion of the liquid solder is consumed for the interfacial reaction
and formation of the IMC. This suggests that the knowledge of the structure
and properties of Cu6Sn5 IMC can eventually help in the design of strong and
reliable solder joints. Besides the electronic packaging sector, the Cu6Sn5
IMC is known as a promising candidate materials for the electrode of the
Lithium-ion battery [3].

Figure 1: Perspective view of a unit cell of $\eta$-Cu6Sn5 compound is
presented in (i). The images in (ii), (iii) and (iv) respectively show the
view along the planes normal to respectively <a>, <b> and <c> directions.

Cu6Sn5 IMC can exist in two forms of crystalline structures [4]. At
temperatures below 186 ∘C [3], it has a monoclinic crystalline structure, and
is referred to as $\eta^{\prime}$-Cu6Sn5. However, for temperatures higher
than 186 ∘C, it undergoes a structural transformation to have a hexagonal
closed packed structure [5, 6]. The IMC with hexagonal closed packed (HCP)
crystalline structure is also termed as $\eta$-Cu6Sn5. During reflow soldering
procedure, the Sn-based solder has to undergo melting and as the melting point
of pure Sn is 523.15 K, $\eta$-Cu6Sn5 is formed during the interfacial
reaction between the solder and substrate. In this study, $\eta$-Cu6Sn5 is
considered as the material for computational analysis. The currently
widespread applications such as high-powered electronic equipment and third
generation semiconducctor power devices have high service temperature ( T >
523.15 K) requirements [7], and this makes the study of $\eta$-Cu6Sn5 phase
more relevant.

##  II Computational study of the IMC

###  II-A Crystal structure

The location of the the Cu atoms and Sn atoms as viewed on the unit cell of
HCP Cu6Sn5 compound is presented in Fig. 1 and that for the 2 $\times$ 2
$\times$ 4 supercell is provided in Fig. 2. The visualization is done using
the Visualisation for Electronic Structural Analysis (VESTA) software [8]. The
lattice parameters for the HCP crystal are a = b = 4.190 Å, c=5.086 Å,
$\alpha$ = $\beta$ = 90 ∘, and $\gamma$ = 120 ∘ [4]. The lengths a, b and c
can are the dimensions of the unit cell corresponding to <a>,<b> and <c>
directions respectively. In Fig. 1(i),the height of the unit cell is 5.086
Å.The views provided in Fig. 1(ii)-(iv) illustrate the angles $\alpha$,
$\beta$ and $\gamma$.

Figure 2:  The 2 $\times$ 2 $\times$ 4 supercell of the unit cell presented in
Fig. 1 (i).

###  II-B Elastic properties of $\eta$-Cu6Sn5

The $\eta$-Cu6Sn5, with HCP structure, is characterized by anisotropy in
elastic properties. The elasticity (stiffness) tensor (Cijkl) represents the
elastic behavior of a crystalline matrix. For simplifying the models,
different materials symmetry options are utilized. In this work, orthotropic
symmetry is assumed for the IMC crystal. The values for the 9 components of
the elasticity tensor of the IMC crystal are obtained from the work of Lee et
al [9]. From the stiffness tensor, it is possible to compute and visualize
direction dependent Young’s modulii, Poisson’s ratios and shear modulii. The
images corresponding to the visualization of Young’s modulii, Poisson’s ratios
and shear modulii are presented in Figs. 3 and 4.

The visualization of Young’s modulii (Fig. 3) is done with the help of Multi-
phase Elastic Aggregates (MELASA) software [10]. From the figure, it can be
understood that the Young’s modulus of Cu6Sn5 IMC crystal is direction
dependent, and its magnitude fluctuates in the range 18.15-131.75 GPa. The
visualization of Poisson’s ratios and shear modulii (Fig. 4) is performed
using ELATE software [11]. The 3D image for variation of Poisson’s ratio along
the different directions is shown in Fig. 4(a). The magnitudes of the
Poisson’s ratio are in the range from 0.03 to 0.82. The image showing the
variation of shear modulus along different direction is presented as 3D figure
in Fig. 4(b) and the corresponding projections on 2D planes in Fig. 4(c). The
minimum value of shear modulus is 4.8 GPa and the maximum magnitude is 51.4
GPa.

Figure 3: Visualization of directional dependence of Young’s modulus of Cu6Sn5
intermetallic compound. Figure 4: Assuming the IMC as an orthotropic material,
the directional dependence of poisson’s ratio is illustrated in (a) whereas
the spatial variation of shear modulus is presented in (b)-(c). The image in
(c) corresponds to the 2D visualization of the 3D image provided in (b). Th
unit corresponding to the numerical values represented in (b) and (c) is GPa.

##  III Crystal plasticity finite element simulation

In context of finite strain inelastic mechanics of crystal plasticity, the
deformation gradient (F) in IMC is assumed to be composed of elastic
deformation gradient (Fe) and plastic deformation gradient (Fp), as following:

$\textbf{F}=\textbf{F${}^{e}$}\textbf{F${}^{p}$}$ (1)

where, det(Fe) > 0 and det(Fp) = 1. With Cauchy stress denoted as $\sigma$,
the following equation can be utilized to represent the Second Piola-Kirchhoff
Stress:

$\textbf{S}=det(\textbf{F})\textbf{F${}^{-1}$}\sigma\textbf{(F${}^{-1}$)${}^{T}$}$
(2)

The HCP crystal of $\eta$-Cu6Sn5 consists of three slip modes - (i) basal
(slip system {0001} <11$\overline{2}$0>), (ii) prismatic (slip system
{10$\overline{1}$0} <11$\overline{2}$0> ) ; and <a> pyramidal (slip system
{10$\overline{1}$1} <11$\overline{2}$0>) , and these modes provide deformation
only to <a> direction [12]. In addition to this, the deformation can can occur
via combined <c+a> pyramidal slips (slip systems {10$\overline{1}$1}
<11$\overline{2}$3> and {10$\overline{2}$2} <11$\overline{2}$3> ) and
deformation twinning.

In this work, the deformation behavior along basal, prismatic and pyramidal
modes is inspected for an IMC block. The finite element method based numerical
simulations were performed in Multiphysics Object Oriented Simulation
Environment (MOOSE) software [13, 14]. A cube of dimension (1 mm $\times$ 1 mm
$\times$ 1 mm) is chosen as the geometry and the mesh is constructed on this
geometry. A total number of 216 HEX8 mesh elements were assigned in the cube
mesh. Appropriate boundaries were assigned to establish the three modes. A
strain rate of 5.0E-4 s-1 was applied to one face of the cube for all of the
three simulations. The elasticity tensor was supplied in the Materials block,
and symmetric9 fill method is utilized to consider orthotropic symmetry of the
IMC material. The information of slip systems and lattice constants (a, b and
c) of the HCP crystal for Cu6Sn5 IMC also was provided in the Materials block.
For the current simulation, the grain size of 15 $\mu$m is considered for the
IMC grains in the cube. The temperature of the computational domain is
considered to be equal to 470.0 K.

The yield strength of Cu6Sn5 IMC is 2009 MPa [15]. Thus it should be noted
that the loading described above will not be sufficient to initiate the
plastic slip based deformation in the HCP crystal. Although this work will
utilize the methodology of crystal plasticity finite element simulations, the
following results and discussions sections will be confined within the elastic
regime. That is, the stress discussed in this study will be so small that it
will not induce any significant plastic deformations. The choice of elastic
regime is favorable from the viewpoint of numerical convergence and
computational efficiency.

##  IV Results and discussions

The results for displacements at t=0.01 s for simulations corresponding to
basal and prismatic modes of deformation are presented in Fig. 5(a)-(b). It
has to be noted that for a given strain rate, the deformation profiles for the
two modes is different. The entire frontal part of the IMC cube for basal mode
undergoes displacement with magnitudes larger than 6.0e-6 mm. Some points in
the cube reach a displacement as high as 7.6e-6 mm. However, in case of
prismatic mode, the maximum displacement reached at the top face of the cube
is only 5.24E-6 mm. Moreover, instead of entire frontal region, the
deformation is more prevalent in the top face of the cube. The displacement
magnitude at t=0.01 s in context of deformation corresponding to <c+a>
pyramidal mode, is presented in Fig. 6. As anticipated, for the same magnitude
of applied strain rate, the cube is characterized by larger deformation at the
top frontal region (as compared to two other modes). The maximum displacement
magnitude at the top regimes is around 1.3E-05 mm .

Figure 5: Computed displacement at t=0.01 s for deformation slip along <a>
direction for the (a) basal and (b) prismatic modes.

For the loading in the [001] direction, it is important to compare the
magnitudes of effective Second Piola-Kirchoff stress between basal and
pyramidal modes, and such comparison is provided in Fig. 7. As shown in the
figure, when the strain along this direction is 5.0E-5, the average magnitude
of stress for Basal mode is 2.5 MPa whereas the effective stress for Pyramidal
mode is above 4 MPa.

Figure 6: Simulated displacement at t=0.01 s for deformation along <c+a>
direction for the pyramidal deformation mode.

These facts reveal that the information obtained from crystal structure when
used together with mesoscale simulations, can provide vast range of data about
the materials behavior under external loading.

As illustrated by the magnitudes of stress above,it can be said that the
current models are confined within the elastic regime. In context of
applications of HCP Cu6Sn5 IMC undergoing elasto-plastic loadings; these
existing models have the capability to incorporate loadings that are large
enough to induce stresses beyond the yield strength of the material. In
future, the numerical simulations will be performed in the plastic regime.

Figure 7: Effective Second Piola-Kirchoff Stress versus strain along
Z-direction for basal and pyramidal deformation modes.

##  V Conclusions

The following conclusions have been derived from this study :

  1. 1.

At temperature above 186 ∘C, Cu6Sn5 IMC has a HCP structure. In this study,
this structural information is utilized in the finite element method based
deformation model at mesoscale.

  2. 2.

The lattice constants and the elasticity tensors of Cu6Sn5 IMC have been
supplied to the mesoscale model as input materials parameters.

  3. 3.

Three different modes of deformation - basal, prismatic and pyramidal, are
considered in the mesoscale model. The displacement magnitudes in the three
modes are studied for a given strain rate of 5.0E-04 s-1 in the Z-axis
direction. It is found that the maximum displacement of the computational
domain (Cu6Sn5 IMC cube) under basal mode is larger as compared to that of
prismatic mode. While all regions of the cube undergo a displacement in
prismatic mode for the given loading condition, it is possible for some local
regions in basal mode to not undergo deformation. In pyramidal deformation
mode, all locations of the cube undergo displacement; and the magnitudes are
larger compared to the other two modes.

  4. 4.

For a given strain magnitude, the average value of Second Piola-Kirchoff
stress in pyramidal deformation mode is larger as compared to the basal mode.
This facts illustrate the importance of using atomistic informations in
mesoscale simulations.

## Acknowledgment

This work was supported by the National Science Centre, Poland (Grant Number:
2021/42/E/ST5/00339) and the National Natural Science Foundation of China
(Grant Number: 52101035).

## References

  * [1] Chong Dong, Jinye Yao, Min Shang, Haoran Ma, Jun Chen, Haitao Ma, and Yunpeng Wang.  New phenomenon: Cellular boundary during the isothermal stage in the Sn-3Ag/(0 0 1)Cu and Sn-3.5Ag/(0 0 1)Cu joint.  Materials Letters, 317(March):132110, 2022.
  * [2] Anil Kunwar, Johan Hektor, Sukeharu Nomoto, Yuri Amorim Coutinho, and Nele Moelans.  Combining multi-phase field simulation with neural network analysis to unravel thermomigration accelerated growth behavior of Cu6Sn5 IMC at cold side Cu–Sn interface.  International Journal of Mechanical Sciences, 184(May):105843, 2020\.
  * [3] Xin F Tan, Shiwei Tao, Lingbing Ran, Ruth Knibbe, and Kazuhiro Nogita.  Cobalt-doped Cu 6 Sn 5 lithium-ion battery anodes with enhanced electrochemical properties.  Nano Select, (April):1–13, 2022.
  * [4] D K Mu, S D McDonald, J Read, H Huang, and K Nogita.  Critical properties of Cu 6 Sn 5 in electronic devices: Recent progress and a review.  Curr. Opin. Solid State Mater. Sci., 20(2):55–76, 2016.
  * [5] Shuibao Liang, Anil Kunwar, Changqing Liu, Han Jiang, and Zhaoxia Zhou.  Preferential growth of intermetallics under temperature gradient at Cu–Sn interface during transient liquid phase bonding: insights from phase field simulation.  Journal of Materials Research and Technology, 19:345–353, 2022\.
  * [6] Johan Hektor, Matti Ristinmaa, Håkan Hallberg, Stephen A Hall, and Srinivasan Iyengar.  Coupled diffusion-deformation multiphase field model for elastoplastic materials applied to the growth of Cu6Sn5.  Acta Materialia, 108:98–109, apr 2016.
  * [7] Chao Ding, Jian Wang, Tianhan Liu, Hongbo Qin, Daoguo Yang, and Guoqi Zhang.  The Mechanical Properties and Elastic Anisotropy of $\eta$’-Cu6Sn5 and Cu3Sn Intermetallic Compounds.  Crystals, 11:1562, 2021.
  * [8] Koichi Momma and Fujio Izumi.  VESTA 3 for three-dimensional visualization of crystal, volumetric and morphology data.  Journal of Applied Crystallography, 44(6):1272–1276, 2011.
  * [9] N. T.S. Lee, V. B.C. Tan, and K. M. Lim.  First-principles calculations of structural and mechanical properties of Cu6Sn5.  Applied Physics Letters, 88(3):1–3, 2006.
  * [10] M. Friák, D. Lago, N. Koutná, D. Holec, T. Rebok, and M. Šob.  Multi-phase ELAStic Aggregates (MELASA) software tool for modeling anisotropic elastic properties of lamellar composites.  Computer Physics Communications, 247:106863, 2020.
  * [11] Romain Gaillac, Pluton Pullumbi, and François Xavier Coudert.  ELATE: An open-source online application for analysis and visualization of elastic tensors.  Journal of Physics Condensed Matter, 28(27), 2016.
  * [12] Soud Farhan Choudhury and Leila Ladani.  Single Crystal Plasticity Finite Element Analysis of Cu6Sn5 Intermetallic.  Metallurgical and Materials Transactions A, 46(3):1108–1118, mar 2015.
  * [13] Cody J. Permann, Derek R. Gaston, David Andrš, Robert W. Carlsen, Fande Kong, Alexander D. Lindsay, Jason M. Miller, John W. Peterson, Andrew E. Slaughter, Roy H. Stogner, and Richard C. Martineau.  MOOSE: Enabling massively parallel multiphysics simulation.  SoftwareX, 11:100430, 2020.
  * [14] Stephanie A. Pitts, Wen Jiang, Davide Pizzocri, Erin I. Barker, and Hussein M. Zbib.  A Continuum Dislocation Dynamics Crystal Plasticity Approach to Irradiated Body-Centered Cubic $\alpha$-Iron.  Journal of Engineering Materials and Technology, Transactions of the ASME, 144(1):1–9, 2022.
  * [15] X. Deng, N. Chawla, K. K. Chawla, and M. Koopman.  Deformation behavior of (Cu, Ag)-Sn intermetallics by nanoindentation.  Acta Materialia, 52(14):4291–4303, 2004.


# PLAYING WITH THE ENVELOPING ALGEBRA OF SUPERSYMMETRY

E. Cattaruzza  INFN, Sezione di Trieste, Trieste, Italy
<EMAIL_ADDRESS>E. Gozzi  Theoretical Section Dept. of Physics,
University of Trieste, Trieste, Italy
INFN, Sezione di Trieste, Trieste, Italy
<EMAIL_ADDRESS>
###### Abstract

In this paper we show how to obtain from a scalar superfield its first
component via a similarity transformation. We prove that in $D=4$ the
generators of this similarity transformation live in the enveloping algebra of
supersymmetry while for $D=1$ they belong to the basic algebra.

###### keywords:

Supersymmetry; Superfield.

PACS numbers: 11.30.Pb, 31.15.xK

##  1 INTRODUCTION

Supersymmetry [1, 2, 3] is, in our opinion, one of the most interesting idea
which has appeared in physics over the last 50 years. It is the only other
space-time symmetry, beside the Poincaré invariance, which has been proposed
for flat space-time. Moreover it has a unique role in taming the ultraviolet
divergences of quantum field theories [4, 5].

Unfortunately so far there is no experimental indication of it. It may be that
the phenomenological models that have been tested are not the ones which
nature uses. But it may also be that there are further theoretical aspects of
supersymmetry (SUSY) which have not been investigated in enough details. This
paper is a modest attempt in this direction.

SUSY is a symmetry which put in the same multiplet fields of different
statistics, bosons and fermions. A short manner to write up all these fields
in a single multiplet is via the concept [6] of superfield. This concept was
crucial also in order to bring to light the geometry hiding behind SUSY. This
geometry is related to the concept of superspace which is an extension via
grassmannian coordinates of ordinary space-time. The superfields usually have
a first component which is a known physical field like the electron field or
the quark or the Higgs, while the other components are partner fields which
should somehow appear in nature.

The manner to obtain the first component from the superfield is to bring to
zero the grassmannian partners of space-time. This is a very rude manner to
proceed which does not throw any light on the interplay between the superfield
and its first component which is usually also the only physical component
discovered so far. In this paper we shall study another manner to go from the
superfield to its first component. It is based on a similarity transformation
on the superfield. The generators of this transformation belong to the SUSY
algebra in the case of 1-dimensional system and to the enveloping algebra in
the 4-dim case.

For the 1-dim case the similarity transformation is basically a sort of change
of picture from the Heisenberg to the Schrödinger one with respect to the
grassmannian partners of time. In the 4-dim case we have not succeeded in
giving such a clear-cut interpretation.

The paper is organised as follows: in Sect.2 we analyse a particular 1-dim.
supersymmetric model which is basically a modern way to rewrite classical
mechanics. In Sect.3 we analyse the 4-dim Wess-Zumino[1] model and its
associated[6] scalar superfield. In this case first we perform, like for the
1-dim case, a similarity transformation which is a field theory implementation
of translation in the grassmannian part of superspace. We prove that this
transformation does not bring the superfield to its first component but to the
set of its scalar components. In Sect.4 we apply to this first transformation
a second one but built with elements of the enveloping algebra of SUSY. This
set of combined transformations brings the superfield to its first component.

Some long and detailed calculations are confined to few appendices.

##  2 1-DIMENSIONAL CASE

The 1-dim model we will analyse in this section made its appearance in a path-
integral approach to classical mechanics reviewed in ref.7. Its lagrangian is
the following one:

$\widetilde{\mathcal{L}}=\lambda_{a}\,\dot{\varphi}^{a}+i\,\bar{c}_{a}\,\dot{c}^{a}-\lambda_{a}\omega^{ab}\frac{\partial
H}{\partial\varphi^{b}}-i\bar{c}_{a}\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}c^{b}.$
(1)

The ”fields” $\varphi^{a}(t)$ are the $2n$-phase space coordinates of a
classical point particle

$\varphi^{a}=\left(q^{1},\cdots,q^{n},p^{1},\cdots,p^{n}\right),\,a=1,\cdots,2n.$

while the $\lambda_{a}(t),\,c^{a}(t),\,\bar{c}_{a}(t)$ are auxiliary
coordinates depending on the 1-dim. time variable $t$. We point out that the
variables $c^{a},\bar{c}_{a}$ have odd grassmannian parity[11] . Also on these
auxiliary variables the index ”$a$” runs from 1 to $2\,n$ . The symbol
$\omega^{ab}$ indicates the symplectic matrix of the original
$\varphi^{a}$-space:

$\omega^{ab}=\left(\begin{matrix}0&\mathbb{I}\\\
-\mathbb{I}&0\end{matrix}\right)$

and $H(\varphi)$ is a function of $\varphi$ which plays the role of a
superpotential in supersymmetry but originally was the Hamiltonian of the
classical system living in the $\varphi^{a}$-space. The readers interested in
the relation between $\tilde{\mathcal{L}}$ and the original classical system
moving in the $\varphi$-space can consult ref.7. Here let us only mention that
the $c^{a}$ turn out to be the basis of the differential forms on $\varphi$
-space, $\bar{c}_{a}$ the basis of the totally anti-symmetric tensors and that
$\tilde{\mathcal{L}}$ is related to the Lie-derivative of the Hamiltonian
flow[10] . As $\tilde{\mathcal{L}}$ makes its appearance in a path-integral so
we can calculate [8, 9] the graded commutators [11] among the $8\,n$ variables
$(\varphi^{a},\,\lambda_{a},\,c^{a},\,\bar{c}_{a})$. The only non-zero graded
commutators are the following ones:

$\begin{matrix}\left[\varphi^{a},\lambda_{b}\right]&=&i\,\delta_{b}^{a}\\\
\left[c^{a},\bar{c}_{b}\right]&=&\delta_{b}^{a}.\\\ \end{matrix}$ (2)

The Hamiltonian associated to the Lagrangian (1) is:

$\mathcal{\tilde{H}}=\lambda_{a}\omega^{ab}\frac{\partial
H}{\partial\varphi^{b}}+i\bar{c}_{a}\omega^{ac}\frac{\partial^{2}H}{\partial\varphi^{c}\partial\varphi^{b}}c^{b}$
(3)

and it is easy to prove that it commutes with the following charges:

$\begin{matrix}Q&\equiv&i\,c^{a}\,\lambda_{a}\\\
\widebar{Q}&\equiv&i\,\bar{c}_{a}\,\omega^{ab}\,\lambda_{b}\\\
Q_{H}&=&Q-c^{a}\,\partial_{a}H\\\
\widebar{Q}_{H}&=&\widebar{Q}+\bar{c}_{a}\,\omega^{ab}\,\partial_{b}H.\end{matrix}$
(4)

Besides commuting with $\tilde{\mathcal{H}}$ it is easy to check that

$\left[Q_{H},\widebar{Q}_{H}\right]=2\,i\,\widetilde{\mathcal{H}}.$ (5)

So the $Q_{H},\,\widebar{Q}_{H}$ are generators of an $N=2$ supersymmetry[12]
. There are other charges which commute [12] with $\widetilde{\mathcal{H}}$
but we are not interested in them in this paper.

An important tool introduced in ref.6 for supersymmetry is the one of
superfield. This is basically a manner to put in the same multiplet all the
$8\,n$ variables $(\varphi^{a},\,\lambda_{a},\,c^{a},\,\bar{c}_{a})$. First we
extend the space $t$ via two grassmannian partners $(\theta,\bar{\theta})$ to
what is called a superspace $(t,\,\theta,\,\bar{\theta})$. Next we introduce a
function $\Phi^{a}(t,\theta,\bar{\theta})$ in this superspace defined as
follows:

$\Phi^{a}(t,\theta,\bar{\theta})\equiv\varphi^{a}(t)+\theta\,c^{a}(t)+\bar{\theta}\,\omega^{ab}\,\bar{c}_{b}(t)+i\,\theta\,\bar{\theta}\,\omega^{ab}\,\lambda_{b}.$
(6)

In this superspace it is easy to give to the charges a differential-operator
representation. This is done in the following manner. Let us call the super-
space $(t,\,\theta,\,\bar{\theta})$ as base-space and
$(\varphi,\,\lambda,\,c,\,\bar{c})$ as target-space. It is then natural to
think that the transformations on the target-space generated by the charges of
eq.(4) are induced by transformations on the base-space generated by some
differential operator. The simplest request we can make on the superfield
$\Phi$ is that it is a scalar under the transformations generated by the
operators in eq.(4) which means :

$\Phi^{\prime
a}(t^{\prime},\theta^{\prime},\bar{\theta}^{\prime})=\Phi^{a}(t,\theta,\bar{\theta}),$
(7)

where the $\Phi^{\prime}$ indicates the superfield built out of the
transformed target-space
$(\varphi^{\prime},\,\lambda^{\prime},\,c^{\prime},\,\bar{c}^{\prime})$. If we
indicate with $\mathcal{O}$ any of the charges in eq.(4) and with
$\widehat{\mathcal{O}}$ the associated differential operator in the base-
space, the relation (7) is satisfied if and only if the following equation
holds:

$\left[\mathcal{O},\Phi^{a}\right]=-\widehat{\mathcal{O}}\,\Phi^{a}.$ (8)

The $\mathcal{O}$ on the L.H.S. of (8) acts on the target space while the
$\widehat{\mathcal{O}}$ on the R.H.S. acts on the base-space. As we have the
explicit form of the various charges $Q$ , $\widebar{Q}$, etc. of eq.(4) it is
easy to derive from eq. (8) the expressions of the various $\widehat{Q}$,
$\widehat{\widebar{Q}}$ etc. associated to those charges. They are:

$\begin{matrix}\widehat{Q}&=&-\cfrac{\partial}{\partial\theta}\equiv-\partial_{\theta}\\\
\widehat{\widebar{Q}}&=&+\cfrac{\partial}{\partial\bar{\theta}}\equiv\partial_{\bar{\theta}}\quad\\\
\,\,\widehat{Q}_{H}&=&-\partial_{\theta}-\bar{\theta}\,\partial_{t}\,\,\\\
\,\,\widehat{\widebar{Q}}_{H}&=&\partial_{\bar{\theta}}+\theta\,\partial_{t}.\end{matrix}$
(9)

We would like to stress that the $\mathcal{O}$ and $\widehat{\mathcal{O}}$
have the same algebra but are reppresented in two different spaces: the
$\mathcal{O}$ in the target-space and the $\widehat{\mathcal{O}}$ in the base-
space. In A we will prove the first of eq.(9). The others are easy to derive.

Now let us turn to the goal of this paper, i.e. to build a similarity
transformation which project out of the superfield $\Phi^{a}$ in (6) its first
component $\varphi^{a}$. Trivially it could be obtained by freezing to zero
$\theta,\bar{\theta}$:

$\left.\Phi^{a}\right|_{\theta,\bar{\theta}=0}=\varphi^{a}.$

We would like to do it in a different manner to which a physical
interpretation could be associated. What we will do here is already contained
in ref.7. but we present it again for completeness and in order to help the
reader not familiar with that reference.
It is easy to prove that the similarity transformation we are looking for is
the following one

$e^{-\theta\,Q-\bar{Q}\,\bar{\theta}}\,\Phi^{a}(t,\theta,\bar{\theta})\,e^{\theta\,Q+\bar{Q}\,\bar{\theta}}=\varphi^{a}(t).$
(10)

The proof of the formula above is contained in ref.7. and for completeness we
have reproduced it in B. Using (10) it is not difficult to prove a
generalisation of that formula which is the following one :

$e^{-\theta\,Q-\bar{Q}\,\bar{\theta}}\,F\left(\Phi^{a}(t,\theta,\bar{\theta})\right)\,e^{\theta\,Q+\bar{Q}\,\bar{\theta}}=F\left(\varphi^{a}(t)\right)$
(11)

where $F$ is a generic function. The proof of this formula has been done in
ref.7 and we shall not do it again here. It is also easy to prove the inverse
of (10) and (11), i.e.

$\Phi^{a}(t,\theta,\bar{\theta})=e^{\theta\,Q+\bar{Q}\,\bar{\theta}}\varphi^{a}(t)\,e^{-\theta\,Q-\bar{Q}\,\bar{\theta}}$
(12)

and

$F\left(\Phi^{a}(t,\theta,\bar{\theta})\right)=e^{\theta\,Q+\bar{Q}\,\bar{\theta}}\,F\left(\varphi^{a}(t)\right)\,e^{-\theta\,Q-\bar{Q}\,\bar{\theta}}.$
(13)

Note that this inverse relation cannot be obtained if we replace the operators
$Q$ and $\widebar{Q}$ with their ”hat” partners given in eq.(9). This seems to
imply that the $Q$, ${\widebar{Q}}$ and $\widehat{Q}$, $\widehat{\widebar{Q}}$
are not completely equivalent111 We wish to thank one referee for having
brought this point to our attention.. Anyhow in this paper we will be using
only relation (10) and for these $Q$, $\widebar{Q}$ and $\widehat{Q}$,
$\widehat{\widebar{Q}}$ are equivalent. We will mostly use the hat operators
in the last steps of the calculations (like in B) because they are easier to
use than the non-hat (or target-space ) operators.

It is now amusing to give a physical interpretation of the formulas above. We
said that $\theta,\bar{\theta}$ are grassmannian partners of $t$ and more
details can be found in ref.7. With respect to $t$ we can have the usual
Heisenberg and Scroedinger picture of the system and the same should be
possiible for its partners $\theta$ and $\bar{\theta}$. The observables in the
Heisenberg picture depend on $t$ while they don’t in the Schroedinger picture.
The same is possible for the grassmannian partners $\theta,\bar{\theta}$ of
$t$. We can have a picture in which the observables depend on
$\theta,\bar{\theta}$ (which we shall call Heisenberg picture for
$\theta,\bar{\theta}$) and one in which the observables do not depend on
$\theta,\bar{\theta}$ (the Schroedinger picture in $\theta,\bar{\theta}$).
Looking now at (12) and (13) we can say that the superfield
$\Phi^{a}(t,\theta,\bar{\theta})$ is the Heisenberg picture in
$\theta,\bar{\theta}$ of the variables $\varphi^{a}(t)$. The same for the
general observable $F\left(\Phi(t,\theta,\bar{\theta}\right))$ as proved by
formula (13). This is the physical interpretation of the similarity
transformation we mentioned at the beginning and we are amused with it !!. It
may throw further light on the interplay between the superfield and its first
component. For sure this throws more light that the naive procedure of getting
one from the other by naively putting $\theta,\bar{\theta}$ to zero.

##  3 4-DIMENSIONAL CASE

In this section we will try to do the same as in Section 2 but for a
relativistic field theory in $4$-dim.

We will use the Wess-Zumino model [1] in the notation of ref.13. The
superspace in this case is made of the usual $4$-dim. space-time $x^{\mu}$ and
of four grassmannian variables $\left\\{\theta_{A}\right\\}|_{A=1,2}$,
$\left\\{\bar{\theta}_{\dot{A}}\right\\}|_{\dot{A}=1,2}$, where we have used
the Weyl two components notation. The $\theta_{A}$ carries the self-
representation of SL($2;\mathbb{C})$, while $\bar{\theta}_{\dot{A}}$ carries
the conjugate self-representation. They are all anticommuting variables:

$\left[\theta_{A},\theta_{B}\right]=\left[\bar{\theta}_{\dot{A}},\bar{\theta}_{\dot{B}}\right]=\left[\theta_{A},\bar{\theta}_{\dot{B}}\right]=0,$
(14)

where $\left[(\cdot),(\cdot)\right]$ indicates the graded commutators. The
indices are raised by the symbols $\epsilon^{AB}$ and
$\epsilon^{\dot{A}\dot{B}}$, defined as:

$\begin{matrix}\left\\{\epsilon^{AB}\right\\}_{A,B=1,2}=\left(\begin{matrix}0&1\\\
-1&0\end{matrix}\right)=\left\\{\epsilon^{\dot{A}\dot{B}}\right\\}_{\dot{A},\dot{B}=1,2}\\\
\left\\{\epsilon_{AB}\right\\}_{A,B=1,2}=\left(\begin{matrix}0&-1\\\
1&0\end{matrix}\right)=\left\\{\epsilon_{\dot{A}\dot{B}}\right\\}_{\dot{A},\dot{B}=1,2}.\\\
\end{matrix}$ (15)

So the objects with the indices up are defined as:

$\psi^{A}\equiv\epsilon^{AB}\,\psi_{B}=-\psi_{B}\,\epsilon^{BA},\quad\bar{\psi}^{\bar{A}}\equiv\epsilon^{\dot{A}\dot{B}}\bar{\psi}_{\dot{B}}.$

It is then easy to obtain the following relations:

$\begin{matrix}\psi_{A}&=&\epsilon_{AB}\,\psi^{B}&=&-\psi^{B}\,\epsilon_{BA}\\\
\bar{\psi}_{\dot{A}}&=&\epsilon_{\dot{A}\dot{B}}\,\bar{\psi}^{\dot{B}}&=&-\bar{\psi}^{\dot{B}}\,\epsilon_{\dot{B}\dot{A}}.\end{matrix}$
(16)

In performing the scalar products among Weyl spinors we will use the
convention of summing upper left indices with lower right indices and lower
left dotted indices with upper right dotted indices:

$\left(\psi\,\chi\right)\equiv\psi^{A}\,\chi_{A},\quad\left(\bar{\psi}\,\bar{\chi}\right)\equiv\bar{\psi}_{\dot{A}}\,\bar{\chi}^{\dot{A}}.$
(17)

In the superspace $\\{x^{\mu};\theta^{\alpha},\bar{\theta}_{\dot{\alpha}}\\}$
we can define a scalar superfiel as follows

$\begin{matrix}\Phi(x;\theta,\bar{\theta})&=&f(x)+\theta\,\phi(x)+\bar{\theta}\,\bar{\chi}(x)+(\theta\theta)\,m(x)\\\
&+&(\bar{\theta}\bar{\theta})\,n(x)+\left(\theta\sigma^{\mu}\bar{\theta}\right)\,V_{\mu}(x)+(\theta\theta)\bar{\theta}\,\bar{\lambda}\\\
&+&(\bar{\theta}\bar{\theta})\theta\,\psi+(\theta\theta)(\bar{\theta}\bar{\theta})\,d(x),\quad\quad\quad\quad\quad\,\end{matrix}$
(18)

where $f(x),m(x),n(x),d(x)$ are scalar and pseudoscalar complex fields,
$\psi^{\alpha}(x),\,\phi^{\alpha}(x)$ are spinor fields which transform
according to the self-representation of SL$(2;\mathbb{C})$ (left-handed Weyl
spinors), $\bar{\chi}(x)^{\dot{\alpha}},\,\bar{\lambda}(x)^{\dot{\alpha}}$ are
spinor fields which transform according to the conjugate self-representation
of SL$(2;\mathbb{C})$ (right-handed Weyl spinors) and $V_{\mu}$ is a Lorentz
vector field. In what follows we will use the notation
$\sigma^{\mu}=(\mathbb{I},\sigma^{i})$, where $\sigma^{i}$ are the Pauli
matrices. Notice that the $\sigma^{\mu}$ carry mixed indices
$(\sigma^{\mu})_{A\dot{A}}$ while its adjoint is defined as

$\left(\bar{\sigma}^{\mu}\right)^{\dot{A}A}\equiv\epsilon^{AB}\epsilon^{\dot{A}\dot{B}}\sigma_{B\dot{B}}^{\mu}.$

The various expressions without explicit index notation appearing in (18) are
defined as follows:

$\begin{matrix}\theta\,\phi=\theta^{\alpha}\,\phi_{\alpha}\quad\quad\quad\quad\quad\quad\\\
\bar{\theta}\,\bar{\chi}=\bar{\theta}_{\dot{\alpha}}\,\bar{\chi}^{\dot{\alpha}}\quad\quad\quad\quad\quad\quad\\\
(\theta\sigma^{\mu}\bar{\theta})\,V_{\mu}=(\theta^{\alpha}\,\sigma^{\mu}_{\alpha\dot{\alpha}}\,\bar{\theta}^{\dot{\alpha}})\,V_{\mu}\\\
\bar{\theta}\bar{\lambda}=\bar{\theta}_{\dot{\alpha}}\bar{\lambda}^{\dot{\alpha}}\quad\quad\quad\quad\quad\quad\,\,\\\
\theta\psi=\theta^{\alpha}\psi_{\alpha}\quad\quad\quad\quad\quad\quad\,\,\end{matrix}.$
(19)

Note that differently than the scalar superfiend in 1-dim of eq.(6), the
scalar superfield in 4-dim of eq.(18) contains up to 4 powers in
$\theta,\,\bar{\theta}$, because these $\theta$ and $\bar{\theta}$ carry two
indices. This point will be crucial in calculations we will perform later on.

In the Weyl-formulation the supersymmetry charges $Q_{H},\,\bar{Q}_{H}$ are
two and each carries two indices:

$\begin{matrix}(Q_{H})_{A},&\,&A=1,2\\\
(\bar{Q}_{H})_{\dot{A}},&\,&\dot{A}=1,2.\end{matrix}$

The supersymmetry (or Super-Poincaré algebra) is the following one

$\begin{matrix}\left[(Q_{H})_{A},(Q_{H})_{B}\right]&=&0\\\
\left[(\bar{Q}_{H})_{\dot{A}},(\bar{Q}_{H})_{\dot{B}}\right]&=&0\\\
\left[(Q_{H})_{A},(\bar{Q}_{H})_{\dot{B}}\right]&=&2\,\left(\sigma^{\mu}\right)_{A\dot{B}}\,P_{\mu}\end{matrix},$
(20)

where $P_{\mu}$ is the 4-momentum operator. The representation as differential
operators in super-space of the supersymmetry operators (20) is the following
[13] one:

$\begin{matrix}\left(\widehat{Q}_{H}\right)_{A}&=&-i\left(\partial_{A}-\sigma^{\mu}_{A\dot{B}}\,\bar{\theta}^{\dot{B}}\,\partial_{\mu}\right)\\\
\left(\widehat{\widebar{Q}}_{H}\right)^{\dot{A}}&=&-i\left(\bar{\partial}^{\dot{A}}-i\,\left(\bar{\sigma}^{\mu}\right)^{\dot{A}B}\,\theta_{\dot{B}}\,\partial_{\mu}\right)\end{matrix}$
(21)

where

$\partial_{A}\equiv\frac{\partial}{\partial\theta^{A}},\quad\bar{\partial}^{\dot{A}}\equiv\frac{\bar{\partial}}{\partial\bar{\theta}_{\dot{A}}},\quad\partial_{\mu}\equiv\frac{\partial}{\partial
x^{\mu}}.$

From now on , in analogy with the 1-dim.case, let us indicate with
$\widehat{Q}$ and $\widehat{\widebar{Q}}$ the generators:

$\begin{matrix}\widehat{Q}_{A}&\equiv&+i\,\partial_{A}\\\
\widehat{\widebar{Q}}^{\dot{A}}&\equiv&+i\,\bar{\partial}^{\dot{A}}.\end{matrix}$
(22)

Note that differently than $\widehat{Q}_{H}$ and $\widehat{\widebar{Q}}_{H}$,
the $\widehat{Q}$ and $\widehat{\widebar{Q}}$ anticommute

$\left[\widehat{Q}_{A},\widehat{\widebar{Q}}^{\dot{B}}\right]=0$ (23)

and the same happens with the $\widehat{Q}_{A}$ among themselves and the
$\widehat{\widebar{Q}}^{\dot{B}}$ as well. Via the analog of eq.(8) we can
build the target-space counterpart of
$\widehat{Q}_{A},\,\widehat{\widebar{Q}}^{\dot{B}}$ which we shall indicate
without ” $\widehat{\,\,\,}$ ” as $Q_{A}$ and $\widebar{Q}^{\dot{B}}$.
Following what we did in the 1-dim case, the next step consists in performing
the following similarity transformation:

$e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}\,\Phi\,e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}$
(24)

on the superfield (18) in 4-dim. Let us stress again that
${Q},\,{\widebar{Q}}$ are not the supersymmetry charges but the generators of
translations in $\theta$ and $\bar{\theta}$ in analogy to the 1-dim case. In
this manner, if (24) would give the first component of the superfield, we
could say that $\Phi$ is the Heisenberg picture representation of its first
component.

In order to perform the calculation (24) we need first to find out the
expression for the exponentials surrounding $\Phi$. It is easy to prove that :

$\begin{matrix}e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}&=&\left(1-\theta\,{Q}+\cfrac{1}{2}\,\left(\theta\,Q\right)^{2}\right)\,\left(1-\bar{\theta}\,{\widebar{Q}}+\cfrac{1}{2}\,\left(\bar{\theta}\,\widebar{Q}\right)^{2}\right)\quad\quad\quad\quad\,\,\\\
&=&\left(1-\theta\,{Q}-\bar{\theta}\,{\widebar{Q}}-\theta\bar{\theta}\,{Q}{\widebar{Q}}-\cfrac{1}{4}\,(\theta\theta)\left({Q}{Q}\right)-\cfrac{1}{4}\,(\bar{\theta}\bar{\theta})\left({\widebar{Q}}{\widebar{Q}}\right)\right.\\\
&+&\left.\cfrac{1}{4}\,(\theta\theta)\bar{\theta}\left({\widebar{Q}}{\widebar{Q}}\right){\widebar{Q}}+\cfrac{1}{4}\,\theta\,(\bar{\theta}\bar{\theta}){Q}\left({\widebar{Q}}{\widebar{Q}}\right)\right.\\\
&+&\left.\cfrac{1}{16}\,(\theta\theta)\,(\bar{\theta}\bar{\theta})\left({Q}{Q}\right)\left({\widebar{Q}}{\widebar{Q}}\right)\right).\end{matrix}.$
(25)

The other exponential $e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}$ is the
same as the R.H.S. of (25) with the replacement $\theta\to-\theta$ and
$\bar{\theta}\to-\bar{\theta}$. We note that, because $\theta_{\alpha}$
carries two indices, in the R.H.S. of (25) we have up to quartic terms like
$(\theta\theta)(\bar{\theta}\bar{\theta})$. This is the crucial difference
with respect to the 1-dim case. In (25) the convention we have used is to put
round brackets $(\,\,)$ around quantities which have their indices saturated
with each other like it is done in (17). Moreover in (25) we use the
convention of putting in front all the strings of $\theta$ and $\bar{\theta}$
and behind the operators $Q$ and ${\widebar{Q}}$.

Let us now go back to (25): in order to understand how the indices are
saturated among quantities which are not incapsulated inside round brackets,
let us explicitly write few terms present in (25) :

$\begin{matrix}\theta\bar{\theta}\,{Q}{\widebar{Q}}&\equiv&\theta^{\alpha}\bar{\theta}_{\dot{\beta}}\,{Q}_{\alpha}{\widebar{Q}}^{\dot{\beta}}\\\
(\theta\theta)\bar{\theta}\,\left({Q}{Q}\right){\widebar{Q}}&\equiv&(\theta\theta)\bar{\theta}_{\dot{\alpha}}\,\left({Q}{Q}\right){\widebar{Q}}^{\dot{\alpha}}\\\
\theta(\bar{\theta}\bar{\theta})\,Q\left({\widebar{Q}}{\widebar{Q}}\right)&\equiv&\theta^{\alpha}\left(\bar{\theta}\bar{\theta}\right)\,{Q}_{\alpha}\left({\widebar{Q}}{\widebar{Q}}\right).\end{matrix}$
(26)

Let us now insert (25) and its inverse into eq. (24) and let us perform the
calculations. These are rather long and some samples of them are reproduced in
C. Looking at this appendix we want to point out for the reader that, once we
get to (44), its R.H.S. can be calculated in the following manner. First we
use recoursevly the analog of eq.(8) and turn all commutators in actions of
the derivative operators on the superfield, second we perform these simple
derivations like we did in the last steps of B. The final result is the
following:

$e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}\,\Phi\,e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}=f(x)-2\,(\theta\theta)\,m(x)-2\,(\bar{\theta}\bar{\theta})\,n(x)+4\,(\theta\theta)(\bar{\theta}\bar{\theta})\,d(x),$
(27)

where $f(x),\,m(x),\,n(x),\,d(x)$ are the scalar and pseudocalar fields
component of the superfield (18).

From eq.(27) we notice unfortunatly that our similarity transformation does
not give us the first component $f(x)$ of the superfield like in the 1-dim.
case (see (10)). So this is not the transformation we are looking for. Anyhow
the result (27) is consistent with (10). In fact if the $\theta$ and
$\bar{\theta}$ in (27) had no index, then
$\theta\theta=0=\bar{\theta}\bar{\theta}$ and the last three terms on the
R.H.S. of (27) would be zero leaving only the first term like in (10).

At this point we were tempted by the idea that if we had started with a
superfield containing less components than the one in (18), we may have ended
up in the R.H.S. of (27) with less fields or even just one. The superfield
with less components that we know are the chiral and antichiral
superfields.They are also called lefthanded (L) or righthanded (R) chiral
superfields. We will use the notation of ref.13 and they are given by:

$\begin{matrix}\Phi_{L}&=&A(x)+\sqrt{2}\,\theta\,\psi(x)+i\,\left(\theta\sigma^{\mu}\bar{\theta}\right)\partial_{\mu}A(x)+(\theta\theta)\,F(x)\\\
&+&\sqrt{2}\,i\,\left(\theta\sigma^{\mu}\bar{\theta}\right)\partial_{\mu}\left(\theta\,\psi\right)-\cfrac{1}{4}\,(\theta\theta)(\bar{\theta}\bar{\theta})\,\Box\,A(x)\\\
&&\\\
\Phi_{R}&=&A^{*}(x)+\sqrt{2}\,\bar{\theta}\,\bar{\psi}(x)-i\,\left(\theta\sigma^{\mu}\bar{\theta}\right)\partial_{\mu}A^{*}(x)+(\bar{\theta}\bar{\theta})\,F^{*}(x)\\\
&-&\sqrt{2}\,i\,\left(\theta\sigma^{\mu}\bar{\theta}\right)\partial_{\mu}\left(\bar{\theta}\,\bar{\psi}\right)-\cfrac{1}{4}\,(\theta\theta)(\bar{\theta}\bar{\theta})\,\Box\,A^{*}(x),\\\
\end{matrix}$ (28)

where $A$ and $F$ are complex scalar fields of no definite parity and $\psi$
is a Weyl fermion field. Doing the similarity transformation analog of (24)
for $\Phi_{L}$ and $\Phi_{R}$ we get:

$\begin{matrix}e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}\,\Phi_{R}\,e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}&=&A^{*}(x)-2\,(\theta\theta)\,F^{*}(x)-(\theta\theta)(\bar{\theta}\bar{\theta})\,\Box\,A^{*}(x),\\\
e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}\,\Phi_{L}\,e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}&=&A(x)-2\,(\theta\theta)\,F(x)-(\theta\theta)(\bar{\theta}\bar{\theta})\,\Box\,A(x).\end{matrix}$
(29)

So we conclude that also for the chiral superfields our similarity
transformation does not lead us to the first component but to a combination of
the scalar fields present in the original superfield. The calculations that
lead to the results (29) are rather easy. Basically we started identifying the
fields present on the R.H.S. of (18) with the analog ones present on the
R.H.S. of (LABEL:eq:24-1). Some of them turned out to be zero. Once this
identification was done we plugged it in the R.H.S. of (27) and immediately we
got eqs.(29).

The reader may be tempted to explore another route that is the one of using
linear combinations of $Q,\widebar{Q},Q_{H},\widebar{Q}_{H}$ instead of just
$Q,\,\widebar{Q}$. Also this route does not work because these combinations
can only get rid of terms linear in $\theta$ or in $\bar{\theta}$ and not
quadratic terms like those which survive on the R.H.S. of eq.(27). In the next
section we will attack this problem of the quadratic terms.

##  4 PLAYING WITH THE ENVELOPING ALGEBRA

Let us go back to (27) and let us define with
$\widetilde{\Phi}(x,\theta,\bar{\theta})$ the R.H.S. of (27):

$\widetilde{\Phi}\equiv
f(x)-2\,(\theta\theta)\,m(x)-2\,(\bar{\theta}\bar{\theta})\,n(x)+4\,(\theta\theta)(\bar{\theta}\bar{\theta})\,d(x).$
(30)

So (27) can be written as:

$e^{-\theta\,{Q}-{\widebar{Q}}\,\bar{\theta}}\,\Phi\,e^{\theta\,{Q}+{\widebar{Q}}\,\bar{\theta}}=\widetilde{\Phi}.$
(31)

As we said in the previous section the reason that the R.H.S. of (30) is
different from $f(x)$ is because the grassmannian variables $\theta$ are two:
$\theta^{A},\,A=1,2$, and the same for the $\bar{\theta}$ and so terms
quadratic in $\theta$ and $\bar{\theta}$ are not identically zero like it
happened instead in $D=1$. We should then look for an operator whose base-
space differential counterpart kills terms like $(\theta\theta)$ and
$(\bar{\theta}\bar{\theta})$ present on the R.H.S. of eq.(30). The base-space
differential counterpart of $Q$ is the $\widehat{Q}$ of eq.(22) and it
annihilates only one power of $\theta$. We need something like a double
derivative, we can try on $\widetilde{\Phi}$ the following similarity
transformation

$e^{-\alpha\,(\theta\theta)\,({Q}{Q})-\beta\,(\bar{\theta}\bar{\theta})\,({\widebar{Q}}{\widebar{Q}})}\,\widetilde{\Phi}\,e^{\alpha\,(\theta\theta)\,({Q}{Q})+\beta\,(\bar{\theta}\bar{\theta})\,({\widebar{Q}}{\widebar{Q}})},$
(32)

where $(QQ)$ is the target-space version of a double derivative. The problem
here is that $\widetilde{\Phi}$ is not a scalar field and we cannot use eq.(8)
in order to turn the target-field operators $Q,\widebar{Q}$ into the
derivative operators $\widehat{Q},\widehat{\widebar{Q}}$. To solve this
problem let us indicate with $\widetilde{Q},\widetilde{\widebar{Q}}$ some
abstract operators which would act as translation operators in $\theta$ and
$\bar{\theta}$. They would be the analog of the abstract translation operators
$\hat{P}$ which on a function $F(x)$ of $x$ acts as follows:

$e^{-i\hat{P}a}\,F(x)\,e^{i\hat{P}a}=F(x+a).$

So eq.(32) will be replaced by the following one:

$e^{-\alpha\,(\theta\theta)\,(\widetilde{Q}\widetilde{Q})-\beta\,(\bar{\theta}\bar{\theta})\,(\widetilde{\widebar{Q}}\widetilde{\widebar{Q}})}\,\widetilde{\Phi}\,e^{\alpha\,(\theta\theta)\,(\widetilde{Q}\widetilde{Q})+\beta\,(\bar{\theta}\bar{\theta})\,(\widetilde{\widebar{Q}}\widetilde{\widebar{Q}})},$
(33)

where $\alpha$ and $\beta$ are parameters to be adjusted in order for the
result of (33) to be the $f(x)$ of (30). If this were possible then the
combination of the similarity transformation (31) and (32) would manage to
turn the superfield into its first component which was our original goal. A
long set of calculations (of which some samples are reported in D) shows that
a choice of $\alpha,\,\beta$ exists, which leads to the result we want and it
is $\alpha=\beta=1/4$. Inserting these values we can say that we have
basically proved the following:

$e^{\frac{1}{4}\,\left[-\,(\theta\theta)\,(\widehat{Q}\widehat{Q})-\,(\bar{\theta}\bar{\theta})\,(\widehat{\widebar{Q}}\widehat{\widebar{Q}})\right]}\,\widetilde{\Phi}\,e^{\frac{1}{4}\,\left[\,(\theta\theta)\,(\widehat{Q}\widehat{Q})+\,(\bar{\theta}\bar{\theta})\,(\widehat{\widebar{Q}}\widehat{\widebar{Q}})\right]}=f(x).$
(34)

In the formula above we have replaced the $\widetilde{Q}$ with $\widehat{Q}$
because their action is the same. Note that the operators which we used are
$\widehat{Q}\widehat{Q}$ and $\widehat{\widebar{Q}}\widehat{\widebar{Q}}$
which do not belong to the algebra of (21) and (22) but to their enveloping
algebra. In general ”enveloping algebras” are defined as an extension of Lie
algebras which contain, beside the operation of commutation, also the one of
multiplication of operators. So enveloping algebras contain the basic
operators of the Lie algebras plus all of its higher powers and linear
combinations of them.

Let us now summarize what we did and combine into a single operation the
transformation (27) which brings us from $\Phi$ to $\widetilde{\Phi}$ with the
(34) which carries us from $\widetilde{\Phi}$ to $f(x)$ :

$e^{\left[-\frac{1}{4}\,(\theta\theta)\,(\widehat{Q}\widehat{Q})-\frac{1}{4}\,(\bar{\theta}\bar{\theta})\,(\widehat{\widebar{Q}}\widehat{\widebar{Q}})-\theta\widehat{Q}-\bar{\theta}\widehat{\widebar{Q}}\right]}\,\Phi\,e^{\left[\frac{1}{4}\,(\theta\theta)\,(\widehat{Q}\widehat{Q})+\frac{1}{4}\,(\bar{\theta}\bar{\theta})\,(\widehat{\widebar{Q}}\widehat{\widebar{Q}})+\theta\widehat{Q}+\bar{\theta}\widehat{\widebar{Q}}\right]}=f(x).$
(35)

We have written above all over the ”hat” operators because by now it should be
clear that its action is the same as the ”non-hat” operators if one uses the
relation (8). The combined transformation above does not seem to be the
Heisenberg picture transformation of the form (10) . One last attempt we can
do is to modify the $\widehat{Q},\widehat{\widebar{Q}},\theta,\bar{\theta}$ in
order to reduce (35) to the form (10). Let us look at the exponential on the
left of $\Phi$ in (35) and expand it in $\theta,\bar{\theta}$:

$\begin{matrix}e^{\left[-\frac{1}{4}\,(\theta\theta)\,(\widehat{Q}\widehat{Q})-\frac{1}{4}\,(\bar{\theta}\bar{\theta})\,(\widehat{\widebar{Q}}\widehat{\widebar{Q}})-\theta\widehat{Q}-\bar{\theta}\widehat{\widebar{Q}}\right]}&=&\\\
&=&1-\theta\widehat{Q}-\left(\bar{\theta}\widehat{\widebar{Q}}\right)-\frac{1}{2}\,(\theta\theta)\left(\widehat{Q}\widehat{Q}\right)-(\theta\bar{\theta})\,\left(\widehat{Q}\widehat{\widebar{Q}}\right)\\\
&-&\frac{1}{2}\,(\bar{\theta}\bar{\theta})\left(\widehat{\widebar{Q}}\widehat{\widebar{Q}}\right)+\frac{1}{2}\,(\theta\theta)\bar{\theta}\left(\widehat{Q}\widehat{Q}\right)\widehat{\widebar{Q}}\\\
&+&\frac{1}{2}\,(\bar{\theta}\bar{\theta})\left(\widehat{\widebar{Q}}\widehat{\widebar{Q}}\right)\left(\theta\widehat{Q}\right)+\frac{1}{4}\,(\theta\theta)(\bar{\theta}\bar{\theta})\left(\widehat{Q}\widehat{Q}\right)\left(\widehat{\widebar{Q}}\widehat{\widebar{Q}}\right).\end{matrix}$
(36)

This should be compared with the first exponential in (10) (but with $\theta$
and $\bar{\theta}$ carrying indices like in (25)). Note that in (36) the
coefficients of the fourth, the sixth, seventh and eight terms are different
than those in (25). We may try to fix things by rescaling $\theta$ and
$\bar{\theta}$ but, as the second and third terms are the same in (36) and
(25), we would have to rescale also the $\widehat{Q}$ and
$\widehat{\widebar{Q}}$ in an inverse manner. This rescaling unfortunately
leaves invariant also the fourth term in (36) which instead has to be changed
to match the analogous one in (25). So a rescaling does not work. We tried
other kind of transformations like:

$\widehat{Q}_{\alpha}\to\widehat{Q}_{\alpha}+\theta_{\alpha}\left(\widehat{Q}^{\beta}\widehat{Q}_{\beta}\right).$
(37)

This new charge generates still a translation in $\theta_{\alpha}$ like the
$\widehat{Q}_{\alpha}$ but it does not act as a translation on quadratic
function of $\theta$, so its geometric meaning is lost.

##  5 CONCLUSIONS

In this paper we have built the similarity transformation which reduces a
superfield to its first component. While in the 1-dim. this transformation can
be built from the generators of the supersymmetry algebra (or better from a
supertranslation), in 4-dim this was not possible and we had to use the
generators of the enveloping algebra. This took away the possibility of
interpreting the similarity transformation as an Heisenberg picture
transformation in the grassmannian time like in 1-dim.

Some extra work could be done along the directions indicated at the end of
Section 4. In particular we should try to see if there are changes on the
generators and the space $\theta,\bar{\theta}$ which would turn our similarity
transformation in some translation in a modified space. The few attempts we
tried do not seem to work but there may be other more interesting ones to
explore. For sure they will be transformations with generators on the
enveloping algebra analogous to those in (37). In general we feel that the
enveloping algebra is a space which deserves to be explored more.

## Appendix A

In this appendix we will check relation (8) with the choice of operators given
in the first of eqs.(4) and (9), i.e.:

$\left[Q,\Phi\right]=-\widehat{Q}\,\Phi$ (38)

Using eq.(9), the R.H.S. of (38) turns out to be

$\begin{matrix}-\widehat{Q}\,\Phi&=&\partial_{\theta}\left[\varphi^{a}+\theta\,c^{a}+\bar{\theta}\,\omega^{ab}\,\bar{c}_{b}+i\,\bar{\theta}\theta\,\omega^{ab}\lambda_{b}\right]\\\
&=&c^{a}-i\,\bar{\theta}\,\omega^{ab}\lambda_{b}.\end{matrix}$

while the L.H.S. is

$\begin{matrix}\left[i\,\lambda_{d}\,c^{d},\varphi^{a}+\theta\,c^{a}+\bar{\theta}\,\omega^{ab}\,\bar{c}_{b}+i\,\bar{\theta}\theta\,\omega^{ab}\lambda_{b}\right]&=&\\\
&=&\left[i\,\lambda_{d}\,c^{d},\varphi^{a}+i,\bar{\theta},\omega^{ab}\bar{c}_{b}\right]\\\
&=&i\,\left[\lambda_{d},\varphi^{a}\right]\,c^{d}-i\,\lambda_{d}\left[c^{d},\bar{c}_{b}\right]\bar{\theta}\,\omega^{ab}\\\
&=&i\,(-i)\delta^{a}_{d}\,c^{d}-i\lambda_{d}\,\delta^{d}_{b}\,\bar{\theta}\,\omega^{ab}\\\
&=&c^{a}-i\,\bar{\theta}\,\omega^{ab}\lambda_{b},\end{matrix}$

where we have used the commutation relations (2). So the R.H.S.. and L.H.S. of
(38) are equal and this concludes our proof.

## Appendix B

In this appendix we will reproduce formula (10):

$e^{-\theta\,Q-\bar{Q}\,\bar{\theta}}\,\Phi^{a}(t,\theta,\bar{\theta})\,e^{\theta\,Q+\bar{Q}\,\bar{\theta}}=\varphi^{a}(t).$
(39)

The L.H.S. of the previous equation, because of the anticommuting character of
$\theta$ and $\bar{\theta}$, is equal to:

$\left(1-\theta
Q\right)\left(1-\widebar{Q}\bar{\theta}\right)\Phi^{A}\left(1+\theta
Q\right)\left(1+\widebar{Q}\bar{\theta}\right)$ (40)

and performing the products contained in (40) we get that it is equal to:

$\Phi^{a}-\theta\left[Q,\Phi^{a}\right]+\bar{\theta}\left[\widebar{Q},\Phi^{a}\right]+\bar{\theta}\theta\left[\left[Q,\Phi^{a}\right],\widebar{Q}\right],$
(41)

where the commutators are intended as graded commutators. We can perform the
calculations of the various commutators in (41) using the expressions of $Q$
and $\bar{Q}$ from eq.(4) and the commutators of the basic variables given in
eq.(2) but we will obtain the same result if we use eq.(8) and the
differential operator version of $Q$ and $\widebar{Q}$. These last
calculations are easier to do. The relation (41) can then be written as:

$\Phi^{a}-\theta\,\partial_{\theta}\Phi^{a}-\bar{\theta}\,\partial_{\bar{\theta}}\Phi^{a}-\bar{\theta}\theta\,\partial_{\bar{\theta}}\partial_{\theta}\Phi^{a}.$
(42)

It is easy to perform in (42) the derivative with respect to $\theta$ and
$\bar{\theta}$ of the superfields $\Phi^{a}$ using its expression (6). The
result turns out to be:

$\Phi^{a}-\theta\,\partial_{\theta}\Phi^{a}-\bar{\theta}\,\partial_{\bar{\theta}}\Phi^{a}-\bar{\theta}\theta\,\partial_{\bar{\theta}}\partial_{\theta}\Phi^{a}=\varphi^{a}.$
(43)

This proves relation (39).

## Appendix C

In this appendix we will report a sample of the calculations which lead to
eq.(27). It is hard to report all of them because they are extremely long.
Those which lead to (25) are relatively easy and short and we will not present
them. Inserting the (25) and its inverse into (24) we get:

$\begin{matrix}e^{-\theta\,{Q}-\bar{\theta}\,{\widebar{Q}}}\,\Phi\,e^{\theta\,{Q}+\bar{\theta}{\widebar{Q}}}&=&\Phi-\theta\,\left[{Q},\Phi\right]-\bar{\theta}\,\left[{\widebar{Q}},\Phi\right]\\\
&+&\cfrac{(\theta\theta)}{4}\,\left[\left[{Q},\Phi\right],{Q}\right]+\cfrac{(\bar{\theta}\bar{\theta})}{4}\,\left[\left[{\widebar{Q}},\Phi\right],{\widebar{Q}}\right]\\\
&+&\theta\bar{\theta}\,\left[\left[{Q},\Phi\right],{\widebar{Q}}\right]+\cfrac{(\theta\theta)\bar{\theta}}{4}\left[\left[\left[{Q},\Phi\right],\widehat{Q}\right],{\widebar{Q}}\right]\\\
&+&\cfrac{\theta(\bar{\theta}\bar{\theta})}{4}\left[\left[\left[{Q},\Phi\right],{\widebar{Q}}\right],{\widebar{Q}}\right]-\cfrac{(\theta\theta)(\bar{\theta}\bar{\theta})}{16}\left[\left[\left[\left[{Q},\Phi\right],{Q}\right],{\widebar{Q}}\right],{\widebar{Q}}\right].\end{matrix}$
(44)

These commutators can be easily performed by using the four-dimensional analog
of eq.(8) which turns the commutators into actions of the derivative operators
$\widehat{Q},\widehat{\widebar{Q}}$ on $\Phi$ and the result is eq.(27),

Going now back to eq.(44) let us show how we got, for example, the seventh
term on the R.H.S. i.e.:

$\cfrac{(\theta\theta)\bar{\theta}}{4}\left[\left[\left[{Q},\Phi\right],{Q}\right],{\widebar{Q}}\right].$

Consider that in (24), inserting (25) and its inverse and performing the
products, the terms which are proportional to $\theta\theta\bar{\theta}$ are
given by

$\begin{matrix}-\cfrac{1}{4}\,(\theta\theta\bar{\theta})\Phi\left({Q}{Q}\right){\widebar{Q}}+\left(\theta{Q}\right)\Phi\left(\theta\bar{\theta}\,{Q}{\widebar{Q}}\right)+\cfrac{1}{4}\,\left(\bar{\theta}{\widebar{Q}}\right)\Phi(\theta\theta)\left({Q}{Q}\right)\\\
-\cfrac{1}{4}\,(\theta\theta)\left({Q}{Q}\right)\Phi\left(\bar{\theta}{\widebar{Q}}\right)+\cfrac{1}{4}\,(\theta\theta)\bar{\theta}\left({Q}{Q}\right){\widebar{Q}}\Phi-\theta\bar{\theta}\,{Q}{\widebar{Q}}\,\Phi\left(\theta{Q}\right).\quad\end{matrix}$
(45)

The first term in (45) can be written as

$-\cfrac{1}{4}\left(\theta^{A}\theta_{A}\,\bar{\theta}_{\dot{A}}\,\Phi\,{Q}^{B}{Q}_{B}{\widebar{Q}}^{\dot{A}}\right)=-\cfrac{1}{4}\left((\theta\theta)\left(\bar{\theta}\,\Phi\left({Q}{Q}\right){\widebar{Q}}\right)\right).$
(46)

The second term, using the identity

$\theta^{A}\theta^{B}=-\frac{1}{2}\epsilon^{AB}(\theta\theta),$

can be turned into

$\begin{matrix}\theta^{A}\,{Q}_{A}\Phi\,\theta^{B}\bar{\theta}_{\dot{A}}\,{Q}_{B}{\widebar{Q}}^{\dot{A}}&=&\theta^{A}\theta^{B}\bar{\theta}_{\dot{A}}\,{Q}_{A}\Phi\,{Q}_{B}{\widebar{Q}}^{\dot{A}}\\\
&=&-\cfrac{1}{2}\,\epsilon^{AB}(\theta\theta)\bar{\theta}_{\dot{A}}\,{Q}_{A}\Phi\,{Q}_{B}{\widebar{Q}}^{\dot{A}}\\\
&=&\cfrac{1}{2}\,(\theta\theta)\bar{\theta}\left({Q}\,\Phi\,{Q}\right){\widebar{Q}},\quad\end{matrix}$
(47)

where

$\left({Q}\,\Phi\,{Q}\right)={Q}^{A}\Phi\,{Q}_{A}.$

In the same manner the third term in (45) can be re-written as:

$\frac{1}{4}\,\left(\bar{\theta}{\widebar{Q}}\right)\Phi(\theta\theta)\left({Q}{Q}\right)=\frac{1}{4}\,(\theta\theta)\,\left(\bar{\theta}\,{\widebar{Q}}\Phi\left({Q}{Q}\right)\right)$
(48)

and the same for the fourth term:

$-\cfrac{1}{4}\,(\theta\theta)\left({Q}{Q}\right)\Phi\left(\bar{\theta}{\widebar{Q}}\right)=-\frac{1}{4}\,(\theta\theta)\left(\bar{\theta}\left({Q}{Q}\right)\Phi\,{\widebar{Q}}\right).$
(49)

The sixth term of (45) gives the following expression:

$\begin{matrix}-\theta\bar{\theta}\,{Q}{\widebar{Q}}\,\Phi\left(\theta{Q}\right)&=&-\theta^{A}\bar{\theta}_{\dot{A}}\,{Q}_{A}{\widebar{Q}}^{\dot{A}}\Phi\left(\theta^{B}{Q}_{B}\right)\\\
&=&\theta^{A}\theta^{B}\bar{\theta}_{\dot{A}}\,{Q}_{A}{\widebar{Q}}^{\dot{A}}\Phi\,Q_{B}\\\
&=&\frac{1}{2}\,(\theta\theta)\left(\bar{\theta}\,{Q}{\widebar{Q}}\Phi\,{Q}\right),\end{matrix}$
(50)

where again we have used the identity
$\theta^{A}\theta^{B}=-\frac{1}{2}\epsilon^{AB}(\theta\theta)$.

The fifth term is :

$\frac{1}{4}\,(\theta\theta)\bar{\theta}\left({Q}{Q}\right){\widebar{Q}}\Phi.$
(51)

Summing up (46),(47),(48),(49),(50) we get :

$\begin{matrix}\cfrac{1}{4}\,(\theta\theta)\,\bar{\theta}_{\dot{A}}\left[-\Phi\left({Q}{Q}\right){\widebar{Q}}^{\dot{A}}+2\,\left({Q}\Phi{Q}\right){\widebar{Q}}^{\dot{A}}+{\widebar{Q}}^{\dot{A}}\Phi\left({Q}{Q}\right)\right.\\\
&&\\\
\left.-\left({Q}{Q}\right)\Phi{\widebar{Q}}^{\dot{A}}+2\,{Q^{A}}{\widebar{Q}}^{\dot{A}}\Phi\,{Q}_{A}+\left({Q}{Q}\right){\widebar{Q}}^{\dot{A}}\Phi+{\widebar{Q}}^{\dot{A}}\left({Q}{Q}\right)\Phi\right].\end{matrix}$
(52)

We have to prove that this set of terms is equal to the seventh term in the
R.H.S. of (44). This term can be written as

$\left[\left[\left[{Q}^{A},\Phi\right],{Q}_{A}\right],{\widebar{Q}}^{\dot{A}}\right]=\left[\left[{Q}^{A},\Phi\right],Q_{A}\right]{\widebar{Q}}^{\dot{A}}-{\widebar{Q}}^{\dot{A}}\left[\left[{Q}^{A},\Phi\right],Q_{A}\right],$
(53)

where the $[\,,\,]$ are always intended as graded commutators. The first term
on the R.H.S. of (53) can be re-written as

$\begin{matrix}\left(\left[Q^{A},\Phi\right]{Q}_{A}+{Q}_{A}\left[{Q}^{A},\Phi\right]\right){\widebar{Q}}^{\dot{A}}&=&\\\
&=&\left({Q}^{A}\Phi{Q}_{A}-\Phi\,{Q}^{A}{Q}_{A}+{Q}_{A}{Q}^{A}\Phi-{Q}_{A}\Phi{Q}^{A}\right){\widebar{Q}}^{\dot{A}}\\\
&=&2\left({Q}\Phi{Q}\right){\widebar{Q}}^{\dot{A}}-\Phi\left({Q}{Q}\right){\widebar{Q}}^{\dot{A}}-\left({Q}{Q}\right)\Phi{\widebar{Q}}^{\dot{A}}.\end{matrix}$
(54)

The second term on the R.H.S. of (53) can be re-written as

$\begin{matrix}-{\widebar{Q}}^{\dot{A}}\left(\left[{Q}^{A},\Phi\right]{Q}_{A}+{Q}_{A}\left[{Q}^{A},\Phi\right]\right)&=&\\\
&=&-{\widebar{Q}}^{\dot{A}}\left({Q}^{A}\Phi{Q}_{A}-\Phi{Q}^{A}{Q}_{A}\right.\\\
&+&\left.{Q}_{A}{Q^{A}}\Phi-{Q}_{A}\Phi{Q}^{A}\right)\quad\quad\\\
&=&-{\widebar{Q}}^{\dot{A}}\left(\left({Q}\Phi{Q}\right)-\Phi\left({Q}{Q}\right)-\right.\\\
&-&\left.\left({Q}{Q}\right)\Phi+\left({Q}\Phi{Q}\right)\right)\quad\quad\\\
&=&-2\,{\widebar{Q}}^{\dot{A}}\left({Q}\Phi{Q}\right)+{\widebar{Q}}^{\dot{A}}\Phi\left({Q}{Q}\right)+\\\
&+&{\widebar{Q}}^{\dot{A}}\left({Q}{Q}\right)\Phi.\quad\quad\quad\quad\quad\quad\end{matrix}$
(55)

Inserting (55) and (54) into (53) we get that it is exactly equal to (52).

## Appendix D

In this appendix we will give a small sample of the calculations which are
needed in order to obtain formula (34). It would be too long to give all of
them. It is easy to prove that the RHS of eq. (34) (with $\alpha$ and $\beta$
not yet specified) can be written as

$\begin{matrix}&&e^{-\alpha\,(\theta\theta)\left(\widehat{Q}\widehat{Q}\right)-\beta\,(\bar{\theta}\bar{\theta})\left({\widehat{\widebar{Q}}}{\widehat{\widebar{Q}}}\right)}\widetilde{\Phi}\,e^{\alpha\,(\theta\theta)\left(\widehat{Q}\widehat{Q}\right)+\beta\,(\bar{\theta}\bar{\theta})\left({\widehat{\widebar{Q}}}{\widehat{\widebar{Q}}}\right)}=\quad\quad\quad\quad\quad\quad\quad\quad\quad\\\
&&=\widetilde{\Phi}-\alpha\,(\theta\theta)\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]-\beta\,(\bar{\theta}\bar{\theta})\left[\widehat{\widebar{Q}}\widehat{\widebar{Q}},\widetilde{\Phi}\right]+\alpha\beta\,(\theta\theta)(\bar{\theta}\bar{\theta})\left[\widehat{\widebar{Q}}\widehat{\widebar{Q}},\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]\right].\\\
\end{matrix}$ (56)

Next we should evaluate the various terms on the R.H.S. of (56). Let us for
example calculate the second term
$(\theta\theta)\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]$. As it
involves a differential operator we should check what the above quantity does
when applied to a funcion $F$, i.e.

$\begin{matrix}(\theta\theta)\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]F&=&(\theta\theta)\left(\widehat{Q}^{A}\widehat{Q}_{A}\,\Phi-\widetilde{\Phi}\,\widehat{Q}^{A}\widehat{Q}_{A}\right)F\quad\quad\quad\quad\\\
&=&(\theta\theta)\left(\widehat{Q}^{A}\left(\widehat{Q}_{A}(\widetilde{\Phi})F\right)+\widehat{Q}^{A}\left(\widetilde{\Phi}\,\widehat{Q}_{A}(F)\right)\right.\\\
&-&\left.\widetilde{\Phi}\,\widehat{Q}^{A}\left(\widehat{Q}_{A}(F)\right)\right)\quad\quad\quad\quad\quad\quad\quad\quad\quad\\\
&=&(\theta\theta)\left(\widehat{Q}^{A}\left(\widehat{Q}_{A}(\widetilde{\Phi})\right)F-\widehat{Q}_{A}(\widetilde{\Phi})\,\widehat{Q}^{A}(F)+\right.\\\
&+&\left.\left(\widehat{Q}^{A}\widetilde{\Phi}\right)\widehat{Q}_{A}(F)+\widetilde{\Phi}\,\widehat{Q}^{A}\left(\widehat{Q}_{A}(F)\right)-\quad\quad\quad\right.\\\
&-&\left.\widetilde{\Phi}\,\widehat{Q}^{A}\left(\widehat{Q}_{A}(F)\right)\right)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\\
&=&(\theta\theta)\left(\widehat{Q}^{A}\left(\widehat{Q}_{A}(\widetilde{\Phi})\right)F+2\,\widehat{Q}^{A}(\widetilde{\Phi})\,\widehat{Q}_{A}(F)\right).\end{matrix}$
(57)

So we get

$\begin{matrix}\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]=\widehat{Q}\widehat{Q}\,\widetilde{\Phi}+2\,\widehat{Q}^{A}(\widetilde{\Phi})\,\widehat{Q}_{A},\end{matrix}$
(58)

and remembering the form of $\widetilde{\Phi}$ in (30) we have

$\widehat{Q}^{A}(\widetilde{\Phi})=4\,\theta^{A}\,m-8\,\theta^{A}\,\bar{\theta}\bar{\theta}\,d.$
(59)

Inserting this in (58) and multiplying by $(\theta\theta)$ like in the L.H.S.
of (57) we get that the last term on the R.H.S. of (58) gives zero, so we
conclude that :

$(\theta\theta)\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]=(\theta\theta)\,\widehat{Q}\widehat{Q}\,\widetilde{\Phi}.$

Analogously, for the third term in (56) , we can prove that:

$(\bar{\theta}\bar{\theta})\,\left[\widehat{\widebar{Q}}\widehat{\widebar{Q}},\widetilde{\Phi}\right]=(\bar{\theta}\bar{\theta})\widehat{\widebar{Q}}\widehat{\widebar{Q}}\widetilde{\Phi}.$

A little more involved is the last term on the R.H.S. of (56), but the result
is:

$(\theta\theta)(\bar{\theta}\bar{\theta})\left[\widehat{\widebar{Q}}\widehat{\widebar{Q}},\left[\widehat{Q}\widehat{Q},\widetilde{\Phi}\right]\right]=(\bar{\theta}\bar{\theta})\,(\theta\theta)\left(\widehat{\widebar{Q}}\widehat{\widebar{Q}}\right)\left(\widehat{Q}\widehat{Q}\right)\,\widetilde{\Phi}.$

Combining all these terms into the R.H.S. of (56) and using the expression
(30) of $\widetilde{\Phi}$, the R.H.S. of (56) turns out to be:

$\begin{matrix}f(x)-2\,(\theta\theta)\,n(x)\,[1-4\,\alpha]-2\,(\bar{\theta}\bar{\theta})\,m(x)\,[1-4\,\beta]+\\\
+4\,(\theta\theta)(\bar{\theta}\bar{\theta})\,d(x)\,[1-4\,\alpha-4\,\beta+16\,\alpha\,\beta].\quad\quad\quad\quad\quad\end{matrix}$
(60)

It is easy to see that with the choice $\alpha=\beta=\frac{1}{4}$ the
expression (60) is reduced to just $f(x)$, which is the first term of the
original superfield $\Phi$ of (30).

## Acknowledgments

The work contained here has been supported by a FRA grant of the University of
Trieste and by the grant GEOSYMQFT of INFN. This paper is dedicated to the
loving memory of Pino Furlan.

## References

  * [1] J. Wess and B. Zumino, Nucl. Phys. B 70, 39 (1974)
  * [2] Yu. A. Golfand and E. P. Likhtman, JETP. Lett.  13, 323 (1971)
  * [3] D. Volkov and V. Akulov, JETP. Lett.  16, 621 (1972)
  * [4] J. Wess and B. Zumino, Phys.Lett. B 49, 52 (1974)
  * [5] J. Iliopoulos and B. Zumino, Nucl. Phys. B 76, 310 (1974)
  * [6] A. Salam and J. Strathdee, Nucl. Phys. B 76, 477 (1974)
  * [7] A. A. Abrikosov Jr., E. Gozzi, and D. Mauro, Ann. Phys. 317, 24 (2005)
  * [8] E. Gozzi, M. Reuter, and W. D. Thacker, Phys. Rev. D 40, 3363 (1989)
  * [9] E. Gozzi, Phys.Lett. B 201, 525 (1988)
  * [10] R. Abraham and J. Marsden, Foundations of Mechanics, (Benjamin Press, New York, 1978)
  * [11] B. De Witt, Supermanifolds, (Cambridge University Press, Cambridge, 1987)
  * [12] E. Gozzi, M. Reuter Phys. Lett. B 233, 383 (1984)
  * [13] H. J. W. Müller-Kirsten and A. Widemann, Supersymmetry: an introduction with conceptual and calculational details, (World Scientific, Singapore, 1987)