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\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{SUSY breaking}
\begin{center}\begin{huge}\begin{center}\Purple{\bf  and the MSSM}\end{center}\end{huge}\end{center}

\setcounter{equation}{0}
\setcounter{footnote}{0}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\myhead{Spontaneous SUSY breaking at tree-level}
\label{Chap5:break:sec:tree}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[htb]
\centerline{   \includegraphics[width=0.7\hsize]{slideimages/ORaighFayetIlliopoul}}
\centerline{ O'Raifeartaigh, Fayet, Iliopoulos }
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myhead{Spontaneous SUSY Breaking}
\beq
\langle0| H |0\rangle   > 0
\eeq
implies that SUSY is broken. 

\beq
V=  {\mathcal F}^{i *}  {\mathcal F}_i+\frac{g^2}{2} D^a D^a~,
\label{Chap5:break:scalarpot}
\eeq
 find  models where 
${\mathcal F}_i=0$ or $D^a=0$ cannot be simultaneously solved 

\vspace{36pt}
\noindent then use this SUSY breaking
sector to generate the soft SUSY breaking 


\myhead{O'Raifeartaigh model}
have nonzero ${\mathcal F}$-terms
\beq
W_{O'R} = -k^2 \Phi_1 + m \Phi_2 \Phi_3 + {y\over 2} \Phi_1 \Phi_3^2 .
\eeq
 scalar potential is
\beq
V &=& |{\mathcal F}_1|^2 + |{\mathcal F}_2|^2 + |{\mathcal F}_3|^2 \\
&=&
|k^2 - {y\over 2} \phi_3^{*2} |^2+
| m \phi^*_3 |^2
+|m \phi^*_2 + y \phi^*_1 \phi^*_3|^2 .
\label{Chap5:ORV}
\eeq
no solution where both ${\mathcal F}_1=0$ and ${\mathcal F}_2=0$

\noindent
For large  $m$,  minimum is
at $\phi_2=\phi_3=0$ with $\phi_1$ undetermined

\vspace{24pt}
\noindent
vacuum energy density is
\beq
V=|{\mathcal F}_1|^2=k^4~.
\label{Chap5:ORvac}
\eeq 

\myhead{O'Raifeartaigh model}
Around $\phi_1=0$, the mass
spectrum
of scalars is
\beq
0,\>\> 0,\>\> m^2,\>\> m^2,\>\> m^2 - yk^2,\>\> m^2 + yk^2 .
\label{Chap5:ORscalars}
\eeq
There are also three fermions with masses
\beq
0,\>\> m,\>\> m.
\label{Chap5:ORfermions}
\eeq
Note that these masses satisfy a sum rule for
tree-level breaking
\beq
{\rm Tr}[ M^2_{\rm scalars}] = 2 {\rm Tr}[ M^2_{\rm fermions}]
\label{Chap5:sumrule1}
\eeq

\myhead{O'Raifeartaigh model: One Loop}
For $k^2\ne0$, loop corrections will give a mass
to  $\phi_1$ 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\hsize]{slideimages/Oraiff}}
\caption[]{ Crosses mark an insertion of $y k^2$.}
\label{Chap5:fig:masscorr}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$y k^2$ insertions must appear with an even number in order to preserve the orientation of the arrows flowing into the 
vertices
\newline correction to the $\phi_1$ mass from the top three graphs 
 vanishes by SUSY

\myhead{O'Raifeartaigh model: One Loop}
bottom two graphs give
\beq
-i m_1^2=\int \frac{d^4 p}{2 \pi^4} (-iy^2) \frac{i  y^2 k^4}{(p^2-m^2)^3} +(i y m)^2\frac{i}{p^2-m^2}
\frac{i y^2 k^4}{(p^2-m^2)^3}~,
\eeq
yields a
 finite, positive, result
\beq
m_1^2= \frac{y^4 k^4 }{48  \pi^2 m^2} = \frac{y^4}{48  \pi^2 } \frac{|{\mathcal F}_1|^2 }{m^2}~.
\eeq
 the classical flat direction is lifted by quantum corrections, the potential is stable around $\phi_1=0$

the massless fermion $\psi_1$ stays massless since it is the 
Nambu--Goldstone particle for the broken SUSY generator,  a
{\em goldstino}

\vspace{12pt}
\noindent
 $\psi_1$ is the fermion in the multiplet with
the nonzero ${\mathcal F}$ component.

\myhead{Fayet--Iliopoulos mechanism}

uses a nonzero $D$-term for a $U(1)$ gauge group

\noindent
add a term
linear in the auxiliary field to the theory:
\beq
\lagr_{\rm FI}
= \kappa^2 D~,
\eeq
where $\kappa$ is a constant parameter with dimensions of mass
\newline
scalar potential is
\beq
V = {1\over 2} D^2 - \kappa^2 D + g D \sum_i q_i \phi^{i*} \phi_i~,
\eeq
and the $D$ equation of motion gives
\beq
D = \kappa^2 - g \sum_i q_i \phi^{i*} \phi_i~.
\eeq
If the $\phi_i$s  have large positive mass squared terms, 
$\langle \phi \rangle =0$ and  $D=\kappa^2$

\vspace{12pt}
\noindent in the MSSM, however,
squarks and sleptons cannot 
have superpotential mass terms

\myhead{Problems}

 Fayet--Iliopoulos and O'Rai\-f\-ear\-taigh models set the scale of
SUSY breaking  by hand. To get a SUSY
breaking scale that is naturally small compared to the Planck scale, $M_{Pl}$, we 
 need
an  asymptotically free gauge theory that gets
strong through RG evolution at some much smaller 
scale
\beq
\Lambda \sim
e^{-8\pi^2/(b g_0^2)} M_{Pl}~,
\eeq
and breaks SUSY nonperturbatively
\newline
can't use
renormalizable tree-level couplings to  transmit
SUSY breaking, since 
SUSY does not allow
scalar--gaugino--gaugino couplings


we expect that SUSY breaking occurs dynamically in a ``hidden sector'' and is
communicated by non-renormalizable interactions or through loop effects.  If the
interactions that communicate SUSY breaking to the MSSM (``visible") sector are flavor-blind it is possible to suppress FCNCs

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{Gauge-Mediated Scenario}
\label{Chap5:break:sec:scenarios}




add ``messenger" chiral supermultiplets where the fermions and bosons are split and 
which
couple to the SM gauge groups

\noindent
 MSSM superpartners get masses
through loops:
\beq
m_{\rm soft} \sim \frac{\alpha_i}{ 4\pi} \frac{\langle {\mathcal F} \rangle}
{M_{\rm mess}}
\eeq

If $M_{\rm mess} \sim \sqrt{\langle {\mathcal F}\rangle}$,
then
the SUSY breaking scale can be as low as ${\sqrt{\langle {\mathcal F}\rangle}} \sim 10^4$--$10^5$ GeV. 

\myhead{Gravity-Mediated Scenario}
interactions with the SUSY breaking sector are suppressed by
powers of $M_{Pl}$
\noindent
 hidden sector field $X$ with a nonzero  $\langle
{\mathcal F}_X\rangle$, then MSSM soft terms 
of the order
\beq
m_{\rm soft} \sim {{\langle {\mathcal F}_X \rangle}\over {M_{Pl}}}~.
\eeq
To get $m_{\rm soft} $ to come out around the weak scale we need
${\sqrt{\langle {\mathcal F}_X\rangle}} \sim 10^{10}$--$10^{11}$ GeV.
Alternatively, if SUSY is broken by a gaugino  condensate $\langle 0| \lambda^a \lambda^b
|0 \rangle = \delta^{ab} \Lambda^3\not= 0$,
then
\beq
m_{\rm soft} \sim
{\Lambda^3 \over M_{Pl}^2} ,
\eeq
which requires $\Lambda \sim 10^{13}$~GeV. This can, of course, be rewritten as: 
$ \langle {\mathcal F}_X \rangle=\Lambda^3/M_{Pl}$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\myhead{Gravity-mediated SUSY Breaking}
%\label{Chap5:break:sec:gravity}

\myhead{Effective  Lagrangian }
Below the $M_{Pl}$ 
is:
\beq
\lagr_{\rm eff} 
&=&- \int d^4 \theta \,  \frac{X^* }{ M_{Pl}}  \hat b^{\prime  ij}  \psi_i   \psi_j +
 \frac{X X^*}{M_{Pl}^2}   \, \left( \hat m^i_j \psi_i  \psi^{j*}+
 \hat b^{ ij}  \psi_i   \psi_j \right)+ h.c.
\nonumber
\\
 && 
-\int d^2 \theta \, \frac{X}{2 M_{Pl}}  \left(\hat  M_3 G^\alpha  G_\alpha+\hat M_2 W^\alpha  W_\alpha+ 
\hat M_1 B^\alpha B_\alpha \right)+h.c.
\nonumber
\\
&&-\int d^2 \theta  \, \frac{X }{ M_{Pl}}  \hat a^{ijk}  \psi_i  \psi_j  \psi_k 
+ h.c.
\eeq
where $G_\alpha$, $W_\alpha$,  $B_\alpha$, and $\psi_i$ are the chiral superfields
of the MSSM, and the hatted symbols are dimensionless

\vspace{12pt}
\noindent
If $\langle X \rangle=\langle {\mathcal F}\rangle$ then
\beq
\lagr_{\rm eff} &=& 
-\frac{ \langle {\mathcal F}_X \rangle}{ 2 M_{Pl}} \left(\hat M_3 {\wtilde G}  {\wtilde G}+ 
\hat M_2{\wtilde W}  {\wtilde W}+ 
\hat M_1 {\wtilde B}  {\wtilde B} \right)+ h.c.
\nonumber
\\
&&- \frac{\langle {\mathcal F}_X \rangle \langle{\mathcal F}_X^* \rangle}{M_{Pl}^2}  \,
\left(\hat m^i_j {\wtilde \psi_i} {\wtilde \psi^{j*}} 
+ \hat b^{ ij} {\wtilde \psi_i}  {\wtilde \psi_j} \right)
+h.c. \nonumber
\\
&&-\frac{\langle{\mathcal F}_X \rangle}{ M_{Pl}} 
\hat a^{ ijk} {\wtilde \psi_i}  {\wtilde \psi_j} 
{\wtilde \psi_k}  
- \frac{\langle {\mathcal F}_X^* \rangle }{ M_{Pl}}  \int d^2 \theta \,  \hat b^{\prime ij}   \psi_i   \psi_j 
+ h.c.
\eeq

\myhead{Assumptions}
assume $\hat M_i=\hat M$, 
$\hat m^i_j =\hat m \delta^i_j$

\noindent we have generated a $\mu$-term 
with $\mu^{ij}=\hat b^{\prime}\delta^i_{H_u}\delta^j_{H_d} \langle {\mathcal F}_X^* \rangle / M_{Pl}$
assuming
$\hat a^{ ijk} =\hat a Y^{ijk}$ and $\hat b^{ij} = \hat b \delta^i_{H_u}\delta^j_{H_d}$, then 
soft parameters
have a  universal form  (when renormalized at $ M_{Pl}$)  \newline gaugino masses are  equal
\beq
&&\!\!\!\! M_i  = m_{1/2}=\hat M {\langle {\mathcal F}_X\rangle\over M_{Pl}}~,
\eeq
the scalar masses are universal 
\beq
&&\!\!\!\! {\bf m^2_{f}} =
 m_{H_u}^2 = m^2_{H_d} = m_0^2= \hat m{|\langle {\mathcal F}_X\rangle|^2\over M_{Pl}^2}~,\eeq
$A$ and $b$ terms are given by
\beq
 {\bf A_f} &=& A \,{\bf Y_f}=\hat a{\langle {\mathcal F}_X\rangle\over M_{Pl}} \,{\bf Y_f}~,\,\,\,
b = B \mu= \frac{\hat b}{\hat b^\prime} {\langle {\mathcal F}_X\rangle\over M_{Pl}} \mu~.
\eeq
$\mu^2$  and $b$ are naturally of the same order of magnitude if $\hat b$ and $\hat b^\prime$ are of the same order of magnitude 

\myhead{Justified Assumptions?}
the assumptions  avoid problems  with FCNCs.
Since gravity is flavor-blind, it might seem that this  a natural result of gravity mediation.
However, the equivalence principle 
does not guarantee these universal terms, since nothing forbids a \Kahler function of the form
\beq
K_{\rm bad}=f(X^\dagger,X)^i_j \, \psi^{\dagger j}\psi_i~,
\eeq
which leads directly to off-diagonal terms in the matrix $\hat m^i_j$

\vspace{12pt}
\noindent
Taking $\mu$ and the four SUSY breaking parameters  and running
them down from the unification scale (rather than the Planck scale
as one would expect) is referred to as the {\em minimal supergravity}
scenario

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{MSugra}

\vspace{-12pt}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.75\hsize]{slideimages/msugra}}
\end{figure}

\begin{center}scalar mass $m^2$, gaugino mass $M$, $A=0$\end{center}
\begin{center}\Blue{Giudice, Rattazzi, hep-ph/0606105}\end{center}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{The goldstino }
\label{Chap5:break:sec:goldstino}

Consider the fermions in a general SUSY gauge theory. Take a basis  $\Psi =(\lambda^a,\psi_i)$.  
The mass matrix is
\beq
{\bf M}_{\rm fermion} =
\pmatrix{
0   &
\sqrt{2} g_a  (\langle \phi^{*}\rangle T^a)^i    \cr
\sqrt{2} g_a  (\langle \phi^{*}\rangle T^a)^j  &
\langle W^{ij} \rangle
}
\eeq
eigenvector with eigenvalue zero:
\beq
\pmatrix{{\langle D^a \rangle /\sqrt{2}} \cr \langle {\mathcal F}_i
\rangle
}
\eeq
eigenvector is only nontrivial if  SUSY is broken. The corresponding
canonically normalized massless fermion field is the goldstino:
\beq
\Pi= \frac{1}{F_\Pi}  
\left(\frac{\langle D^a \rangle}{ \sqrt{2}}\lambda^a  + \langle {\mathcal F}_i \rangle \psi_i  \right)
\eeq
where
\beq
F_\Pi^2=\sum_a \frac{\langle D^a \rangle^2}{2} + \sum_i \langle {\mathcal F}_i \rangle^2
\eeq
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{The Goldstino }
masslessness of the goldstino follows from two facts. First the superpotential is gauge invariant, 
\beq
 ( \phi^{*} T^a)^i  W^*_i 
 = - ( \phi^{*} T^a)^i  {\mathcal F}_i=0
\eeq
second, the first derivative of the scalar potential
\beq
\frac{\partial V}{\partial \phi_i}  =  - 
W^*_i   \frac{\partial W^i}{\partial \phi_i}  -g_a ( \phi^{*} T^a)^j D^a~
\eeq
vanishes at its minimum
\beq
\langle \frac{\partial V}{\partial \phi_i} \rangle = 
\langle {\mathcal F}_i  \rangle  \langle W^{ij}   \rangle -g_a (\langle \phi^{*}\rangle T^a)^j 
\langle  D^a \rangle=0
\eeq
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{The Supercurrent  }

\beq
J_\alpha^\mu &=& i F_{\Pi} (\sigma^\mu {\bar  \Pi})_\alpha\, 
+(\sigma^\nu\sigmabar^\mu \psi_i)_\alpha\, D_\nu \phi^{*i}
- {1\over 2 \sqrt{2}}
(\sigma^\nu \sigmabar^\rho \sigma^\mu
{\bar \lambda}^{a})_\alpha\, F^a_{\nu\rho}~,\nonumber \\
&\equiv& i F_{\Pi} (\sigma^\mu {\bar  \Pi})_\alpha\, + j_\alpha^\mu~.
\eeq
 terms included in $j_\alpha^\mu$ contain two  or more fields.
\newline supercurrent conservation:
\beq
 \partial_\mu J^\mu_\alpha =
i F_\Pi (\sigma^\mu \partial_\mu{\bar \Pi})_\alpha +
\partial_\mu
j^\mu_\alpha=0~~~~~~(*)
\label{Chap5:supercurrentcons}
\eeq
effective Lagrangian for the goldstino
\beq
\lagr_{\rm goldstino}
= i {\bar \Pi} \sigmabar^\mu \partial_\mu  \Pi +
\frac{1}{F_\Pi} (\Pi \, \partial_\mu j^\mu
+ h.c.).
\label{Chap5:goldstinoeff}
\eeq
The EQOM  for  $\Pi$ is just eqn (*)

\vspace{12pt}
\noindent goldstino--scalar--fermion 
and goldstino--gaugino--gauge boson interactions allow the heavier superpartner  to decay
interaction terms have two derivatives, coupling is proportional to the difference of mass squared 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{Eat the Goldstino}
Nambu--Goldstone boson can be eaten by a gauge boson 
\newline
for gravity, Poincar\'e symmetry, and hence SUSY, must
be a local
\newline SUSY spinor $\epsilon^\alpha\rightarrow \epsilon^\alpha(x)$: {\it supergravity}
\newline spin-2 graviton has spin-3/2 fermionic superpartner,
 {\it gravitino}, $\wtilde \Psi_\mu^\alpha$,
which transforms inhomogeneously under local
SUSY transformations:
\beq
\delta \wtilde\Psi_\mu^\alpha = -\partial_\mu\epsilon^\alpha
+\ldots~.
\eeq
gravitino is  the ``gauge" particle
of local SUSY transformations

\vspace{12pt}
\noindent when SUSY is spontaneously broken, the gravitino acquires a mass
by ``eating" the goldstino: the other {\it super Higgs} mechanism
\newline gravitino  mass:
\beq
m_{3/2} \sim \frac{\langle {\mathcal F}_X \rangle}{ M_{Pl}}
\eeq


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{Gravitino Mass}
In gravity-mediated SUSY breaking, the gravitino
mass  $\sim m_{\rm soft}$
\newline
In gauge-mediated SUSY  breaking the
gravitino
is much lighter than the MSSM sparticles if $M_{\rm mess} \ll
M_{Pl}$, so
the gravitino is the LSP. 
%The longitudinal components of the gravitino
%(the goldstino) have non-gravitational interactions.
%The decay rate of any sparticle $\wtilde X$ into
%its Standard Model partner $X$ plus an approximately massless goldstino $\Pi$
%is given by
%\beq
%\Gamma(\wtilde X \rightarrow X \Pi) =
%{m^5_{\wtilde X} \over 16 \pi \langle {\mathcal F} \rangle^2}
%\left ( 1 - {m_X^2\over m_{\wtilde X}^2} \right )^4 .
%\qquad\>\>{}
%\label{Chap5:generalgravdecay}
%\eeq

\vspace{12pt}
\noindent
For a superpartner of mass $m_{\wtilde \psi} \approx 100$~GeV, and $\sqrt{
\langle {\mathcal F}_X\rangle }
< 10^6$~GeV
\beq m_{3/2} < 1~{\rm keV}\eeq

 the decay $\wtilde \psi \rightarrow
\psi \,\Pi$ can be observed inside a collider detector


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\myhead{The goldstino theorem}
\label{Chap5:break:sec:goldstinotheorem}


no matter how SUSY is spontaneously broken, even if it is dynamical, there is a goldstino.
Using the SUSY algebra it follows
\beq
 \langle 0| \{ Q_\alpha , J^{\mu\dagger}_{\dot{\alpha}} (y) \}  |0\rangle
= \sqrt{2} \sigma^\nu_{\alpha \dot{\alpha}}  \langle 0| T^\mu_\nu(y)  |0\rangle
=\sqrt{2} \sigma^\nu_{\alpha \dot{\alpha}}\,  E \,\eta^\mu_\nu~,
\label{Chap5:SUSYanticom}
\eeq
where $E$ is the vacuum energy density.  When $E\ne 0$, SUSY is spontaneously broken.
Taking the location of the current to be at the origin, and writing out $Q_\alpha$ as an integral over
a dummy spatial variable
\beq
\label{Chap5:SUSYanticom2}
\sqrt{2} \sigma^\mu_{\alpha \dot{\alpha}}\,  E
&=& \langle 0| \{ \int d^3x J^0_\alpha(x) , J^{\mu\dagger}_{\dot{\alpha}}(0) \}  |0\rangle  \\
&=& \sum_n  \int d^3x \left(\langle 0| J^0_\alpha (x) | n \rangle \langle n| J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle +
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) | n \rangle \langle n|  J^0_\alpha(x)  |0\rangle\right),
 \nonumber
\eeq
where we have inserted a sum over a complete set of states. Choosing $x^0=0$,  use the generator of translations ($P^\mu$) to show that
\beq
\langle 0| J^0_\alpha (x) | n \rangle &=& \langle 0| e^{ i P.x}J^0_\alpha (0) e^{ - i P.x} | n \rangle \nonumber \\
&=& \langle 0| J^0_\alpha (0) e^{ - i {\vec p_n}.{\vec x}} | n \rangle
\label{Chap5:translation}
\eeq

\myhead{The goldstino theorem}
So we have
\beq
\sqrt{2} \sigma^\mu_{\alpha \dot{\alpha}}\,  E
&=& \sum_n  (2 \pi)^3 \delta({\vec p_n})  \left(\begin{array}{l}\langle 0| J^0_\alpha (0) | n \rangle \langle n| J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle \\+
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) | n \rangle \langle n|  J^0_\alpha(0)  |0\rangle
 \end{array}\right)
  \label{Chap5:SUSYanticom3D}
\eeq
write the term in parenthesis as $f_n(E_n,{\vec p_n,})$
We can also write our anticommutator as
\beq
\sqrt{2} \sigma^\mu_{\alpha \dot{\alpha}}\,  E
&=&\!\!   \int d^4x \left(\langle 0| J^0_\alpha (x)J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle +
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) J^0_\alpha(x)  |0\rangle\right) \delta(x^0)
\\
&=&\!\!  \int d^4x \,\partial_\rho \left(\langle 0| J^\rho_\alpha (x)J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle \Theta(x^0)-
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) J^\rho_\alpha(x)  |0\rangle \Theta(-x^0) \right),
  \nonumber
\eeq
where $\Theta(x^0)$ is the step function \newline  $E$ is related to the integral of a total divergence. 
Nonvanishing if there is  a  massless particle contributing to the two-point function.

\myhead{The goldstino theorem}
Inserting a sum over a complete set of states  we have
\beq
 \label{Chap5:SUSYanticom4D}
\sqrt{2} \sigma^\mu_{\alpha \dot{\alpha}}\,  E \!\! 
&=&\!\! \sum_n \int d^4x \partial_\rho \left(\begin{array}{l}
\langle 0| J^\rho_\alpha (0) e^{ - i {\vec p_n}.{\vec x}} | n \rangle \langle n| J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle \Theta(x^0)\\-
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) | n \rangle \langle n|  e^{  i {\vec p_n}.{\vec x}} J^\rho_\alpha(0)  |0\rangle \Theta(-x^0) \end{array}  \right)\\
&=&\!\!  \sum_n \int d^4x\left[\begin{array}{l}  -i p_{n\rho} \left(\begin{array}{l} e^{ - i {\vec p_n}.{\vec x}}\langle 0| J^\rho_\alpha (0) | n \rangle \langle n| J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle \Theta(x^0)\\+
e^{  i {\vec p_n}.{\vec x}} 
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) | n \rangle \langle n|  J^\rho_\alpha(0)  |0\rangle \Theta(-x^0)  \end{array}\right)
 \\+
 \delta(x^0) \left(\begin{array}{l} e^{ - i {\vec p_n}.{\vec x}}\langle 0| J^\rho_\alpha (0) | n \rangle \langle n| J^{\mu\dagger}_{\dot{\alpha}}(0)   |0\rangle \\+ e^{  i {\vec p_n}.{\vec x}} 
 \langle 0|  J^{\mu\dagger}_{\dot{\alpha}}(0) | n \rangle \langle n|  J^\rho_\alpha(0)  |0\rangle \end{array}   \right) \end{array}  \right]
 \nonumber\\
&=&\!\!  \sum_n (2 \pi)^3 \delta({\vec p_n}) \, \left(  f_n(E_n,{\vec p_n,}) -i \int_0^\infty dx^0\, 
e^{  i {\vec E_n}.{\vec x}} E_n  f_n(E_n,{\vec p_n,})
   \right)
 \nonumber
\eeq

\myhead{The goldstino theorem}
Comparing the two  eqns 
we see that  
\beq 
\int_0^\infty dx^0\, 
e^{  i E_n  x^0} E_n  f_n(E_n,{\vec 0,})=0
\eeq	
and if SUSY is spontaneously broken 
\beq 
  f_n(E_n,{\vec 0,})\ne 0
\eeq	
The only possibility is that
\beq
  f_n(E_n,{\vec 0,})\propto \delta(E_n)
  \eeq
so a state contributes to our two-point function with the quantum numbers of $J^0_\alpha$ (i.e. a fermion) with ${\vec p}=0$ and $E=0$.
In other words there must be a goldstino!



 
 \end{document}
