Ques. F Term SuSy Breaking -Bailin & Love

[ChrisVer]

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:
$F_{1}^{\dagger}= - \frac{\partial W(Φ_{1},Φ_{2},Φ_{3})}{\partial Φ_{1}}|_{Φ_{1}=φ_{1}}$
If I do that for the O'Raifeartaigh superpotential I'll get:
$F_{1}^{\dagger}= -λ_{1} (Φ_{3}^{2}-Μ^{2})$
However they choose $Φ_{3}=φ_{3}$ instead, and I don't know why they do that...

 

[andrien]

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 Φⁿ 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 ∫d²θ[...], to get the general lagrangian.

In doing so you will encounter terms like Nφ^N-1F and φ^N-2ψψ, you just concentrate on the term Nφ^N-1F 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.

 

[ofirg]

However they choose Φ3=φ3 instead, and I don't know why they do that...

I think $\Phi_{3}$ is the superfield where $\varphi_{3}$ is the scalar field component of the superfield. This formula enables to compute the scalar potential of the theory (which is $\sum{F_{i}^{\dagger}F_{i}}$) from the superpotential.

 

[ChrisVer]

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 φ₃ instead of the chiral sfield Φ₃.

Andrien, I would agree with that if the chiral field was one... In that case, you indeed get $Nφ^{N-1} F$ 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:
$Φ_{1}Φ_{2} |_{θθ-wanted}= φ_{1}F_{2} + F_{1} φ_{2}$
and
$Φ_{1}Φ_{2} Φ_{3} |_{θθ-wanted}= φ_{1}φ_{2}F_{3} + φ_{1}F_{2}φ_{3}+ F_{1}φ_{2}φ_{3}$

But enough of that, let me write the spotential of the current O'Raifeartaigh model...
$W= λ(Φ_{1} Φ_{3}^{2} - Φ_{1} Μ^{2} )+ g Φ_{2}Φ_{3}$
By the above, the terms containing F's after a ∫d²θ are:
$W= λ[(2φ_{1}φ_{3}F_{3}+ F_{1} φ_{3}^{2}) - F_{1} M^{2}] + g (φ_{2}F_{3}+φ_{3}F_{2})$

(but question, how could you drop the ψ fields appearing?)

From that I see for example that the coefficient of $F_{2}$ is:
$g φ_{3}= \frac{∂W}{∂Φ_{2}}$
but not for $Φ_{2}=φ_{2}$, but for all the rest fields also reduced down to their bosonic component...

 

[ChrisVer]

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

 

[andrien]

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

how could you drop the ψ fields appearing?

For the same reason you don't take VEV of spinor fields because it's zero.

 

[ChrisVer]

then $dW/dφ_{2}= g F_{3}$?

 

[andrien]

then $dW/dφ_{2}= g F_{3}$?

You use this on original superpotential treating Φ as φ and you get right result.
If you write it in terms of F's like,

$W= λ[(2φ_{1}φ_{3}F_{3}+ F_{1} φ_{3}^{2}) - F_{1} M^{2}] + g (φ_{2}F_{3}+φ_{3}F_{2})$

then you need to use eqn. of motion for F's to get the same thing ( with F*F term included), you get by directly differentiating superpotential with respect to scalar part. Both give equivalent result but don't confuse one by another.