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Ques. F Term SuSy Breaking -Bailin & Love

by ChrisVer
Tags: bailin, love, ques, susy, term
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ChrisVer
#1
May9-14, 10:11 PM
P: 855
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...
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andrien
#2
May10-14, 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 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 NφN-1F and φN-2ψψ, you just concentrate on the term NφN-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
#3
May10-14, 12:04 PM
P: 93
However they choose Φ3=φ3 instead, and I don't know why they do that...
I think [itex]\Phi_{3}[/itex] is the superfield where [itex]\varphi_{3}[/itex] is the scalar field component of the superfield. This formula enables to compute the scalar potential of the theory (which is [itex]\sum{F_{i}^{\dagger}F_{i}}[/itex]) from the superpotential.

ChrisVer
#4
May10-14, 12:28 PM
P: 855
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 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 [itex] Nφ^{N-1} 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} |_{θθ-wanted}= φ_{1}F_{2} + F_{1} φ_{2} [/itex]
and
[itex] 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}[/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 ∫d2θ 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....
ChrisVer
#5
May10-14, 12:29 PM
P: 855
I guess I had the derivative definition, I used in my 1st post, wrong in my lecture notes :/
andrien
#6
May10-14, 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.
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
#7
May10-14, 01:15 PM
P: 855
then [itex] dW/dφ_{2}= g F_{3}[/itex]?
andrien
#8
May10-14, 01:48 PM
P: 1,020
Quote Quote by ChrisVer View Post
then [itex] dW/dφ_{2}= g F_{3}[/itex]?
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.


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