Register to reply

Ques. F Term SuSy Breaking -Bailin & Love

by ChrisVer
Tags: bailin, love, ques, susy, term
Share this thread:
ChrisVer
#1
May9-14, 10:11 PM
P: 919
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...
Attached Thumbnails
LatEx.jpg  
Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
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: 919
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: 919
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: 919
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.


Register to reply

Related Discussions
Spin-half mass term with symmetry breaking Quantum Physics 3
F term breaking potential Beyond the Standard Model 1
If SUSY explains E-w scale breaking, does it require a little Higgs at Tevatron energ Beyond the Standard Model 1
If spinfoam incorporated unbroken SUSY formalism, would SUSY Beyond the Standard Model 2
GUT-SUSY SU(5) is falsified, does SO(10) SUSY predict superpartner masses & proton de High Energy, Nuclear, Particle Physics 4