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Ques. F Term SuSy Breaking Bailin & Love 
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#1
May914, 10:11 PM

P: 754

I have a problem understanding the attached part of the book... Specifically how they derived this result for the Fs...
For example I'll try to write: [itex] F_{1}^{\dagger}=  \frac{\partial W(Φ_{1},Φ_{2},Φ_{3})}{\partial Φ_{1}}_{Φ_{1}=φ_{1}}[/itex] If I do that for the O'Raifeartaigh superpotential I'll get: [itex] F_{1}^{\dagger}= λ_{1} (Φ_{3}^{2}Μ^{2})[/itex] However they choose [itex]Φ_{3}=φ_{3}[/itex] instead, and I don't know why they do that... 


#2
May1014, 11:56 AM

P: 1,020

Write the most general form of chiral superfield with it's expansion in terms of the scalar field, fermionic and auxiliary field. Constructing a superpotential consists of some expression like Φ^{n} with different values of n and a number of superfields and writing a general expression.
You need to put the superpotential term along with superlagrangian (which contains terms of auxiliary fields like F*F) and integrate with something like ∫d^{2}θ[...], to get the general lagrangian. In doing so you will encounter terms like Nφ^{N1}F and φ^{N2}ψψ, you just concentrate on the term Nφ^{N1}F which can be written as ##\frac{∂W}{∂φ}##F. You just also add the complex conjugate of it to satisfy the hermtiicity condition. Along with F*F term you have the general form of potential in which you have to get rid of auxiliary field by EOM. Doing that gives you, ##F^*=\frac{∂W}{∂φ}##. So it is already assumed that when you differentiate with respect to φ, you get a term which is dependent on the superfield but then you have to make a replacement of Φ by φ. 


#3
May1014, 12:04 PM

P: 90




#4
May1014, 12:28 PM

P: 754

Ques. F Term SuSy Breaking Bailin & Love
Yes, that what it does... Also F could not be proportional to a chiral sfield, because F is a scalar one...
However in such a definition of F I gave in my post, there is no way for me to plug in the scalar field φ_{3} instead of the chiral sfield Φ_{3}. Andrien, I would agree with that if the chiral field was one... In that case, you indeed get [itex] Nφ^{N1} F[/itex] terms... (the FF* comes from the Kahler part of the Lagrangian). However in the case you have different chiral fields (as you do here), you get different terms... for example the: [itex] Φ_{1}Φ_{2} _{θθwanted}= φ_{1}F_{2} + F_{1} φ_{2} [/itex] and [itex] Φ_{1}Φ_{2} Φ_{3} _{θθwanted}= φ_{1}φ_{2}F_{3} + φ_{1}F_{2}φ_{3}+ F_{1}φ_{2}φ_{3}[/itex] But enough of that, let me write the spotential of the current O'Raifeartaigh model... [itex] W= λ(Φ_{1} Φ_{3}^{2}  Φ_{1} Μ^{2} )+ g Φ_{2}Φ_{3}[/itex] By the above, the terms containing F's after a ∫d^{2}θ are: [itex] W= λ[(2φ_{1}φ_{3}F_{3}+ F_{1} φ_{3}^{2})  F_{1} M^{2}] + g (φ_{2}F_{3}+φ_{3}F_{2})[/itex] (but question, how could you drop the ψ fields appearing?) From that I see for example that the coefficient of [itex]F_{2}[/itex] is: [itex] g φ_{3}= \frac{∂W}{∂Φ_{2}}[/itex] but not for [itex]Φ_{2}=φ_{2}[/itex], but for all the rest fields also reduced down to their bosonic component.... 


#5
May1014, 12:29 PM

P: 754

I guess I had the derivative definition, I used in my 1st post, wrong in my lecture notes :/



#6
May1014, 01:10 PM

P: 1,020

The differentiation is performed with respect to scalar part as opposed to superfield as you have written in OP.



#7
May1014, 01:15 PM

P: 754

then [itex] dW/dφ_{2}= g F_{3}[/itex]?



#8
May1014, 01:48 PM

P: 1,020

If you write it in terms of F's like, 


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