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  How to count flat directions after supersymmetry breaking

+ 3 like - 0 dislike
732 views

In short my question is how do I show that a superpotential of the form,
W=NYi=1Yifi(X1,...,XNX) 
(generic O'Raifertaigh models) lead to NYNX flat directions after spontaneous SUSY breaking? I describe the context and the details below.

In Weinberg Vol III (pg. 84) he introduces two fields, 

Y=(Y1,...,YNY),X=(X1,...,XNX)  

where Y has an R charge of 2 and X has an R charge of 0. Then superpotential has to take the form,
W=iYifi(X1,...,XNX) 
which gives the SUSY conserving conditions,
fi(X)=0 for i=1,...,NYiYif(X)Xn=0 for n=1,...,NX 
The first set of equations is made up of NX unknowns and NY equations, thus if NY>NX it can't be generically solved. He goes on to define 
vn,ifixn|x=x0 
which gives the potential,
V(x,y)=i|fi|2+n|iyivn,i|2 

Up to this point I have no problems. However, then he goes on to say that the second term vanishes if y is orthogonal to vn. Since ``vn cannot span the space of yis and there will be at least NYNX flat directions''. Where does this conclusion come from? How can I see these ``flat directions''?  

asked May 13, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]

1 Answer

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V(x,y) is a function on the space of fields x and y. This space of fields is the product of the space of x, of dimension NX by the space of y, of dimension NY. A minimum point of V is of the form (x0,0) (with x0 minimizing the first term in V and y=0 giving the vanishing of the second term). We have spontaneous symmetry breaking at this point if V(x0,0) is non-zero. We want to show that there exists at least NYNX flat directions i.e. that there exists at least NYNX directions in the space of fields around (x0,0) where V remains at the value V(x0,0 (the graph of the function V is flat in restriction to these directions). We will find these directions in the space of y : we fix x at the value x0 and we consider V(x0,y) as a function on the space of y, which is of dimension NY. In the formula defining V, the first term does not depend on y so we can forget about it. The second vanishes at y=0. So we are looking at directions in the space of y such that the second term still vanishes. This means we are looking for y such that the scalar products y.vn are 0 for every n, i.e to the space inside the space of y orthogonal to the space generated by the vn. As the space of y is of dimension NY and the space generated by the vn is of dimension at most NX, the orthogonal of the space generated by the vn is at least of dimension NYNX.

answered May 14, 2014 by 40227 (5,140 points) [ revision history ]

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