Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

How to count flat directions after supersymmetry breaking

+ 3 like - 0 dislike
98 views

In short my question is how do I show that a superpotential of the form,
\begin{equation} 
W = \sum _{i = 1 } ^{ N _Y } Y _i f _i  ( X _1 , ... , X _{ N _X } ) 
\end{equation} 
(generic O'Raifertaigh models) lead to $ N _Y - N _X $ flat directions after spontaneous SUSY breaking? I describe the context and the details below.

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

\begin{equation} Y = \left( Y _1, ... , Y _{N _Y } \right) , \quad X = \left( X _1 , ... , X _{ N _X } \right) \end{equation}  

where $ Y $ has an $R$ charge of $2 $ and $ X $ has an $R$ charge of $0$. Then superpotential has to take the form,
\begin{equation} 
W = \sum _i Y _i f _i  ( X _1 , ... , X _{ N _X } ) 
\end{equation} 
which gives the SUSY conserving conditions,
\begin{align} 
 f _i  ( X ) & = 0 \mbox{ for $ i = 1 , ... , N _Y $}\\ 
 \sum _i Y _i \frac{ \partial f ( X ) }{ \partial X _n } & = 0 \mbox{ for $n = 1 , ... , N _X $}
\end{align} 
The first set of equations is made up of $ N _X $ unknowns and $ N _Y $ equations, thus if $ N _Y > N _X $ it can't be generically solved. He goes on to define 
\begin{equation} 
 v _{n,i} \equiv \frac{ \partial f _i }{ \partial x _n } \bigg|_{ x = x _0 }
\end{equation} 
which gives the potential,
\begin{equation} 
V ( x, y ) = \sum _i \left| f _i \right| ^2 + \sum _n \left| \sum _i y _i v _{n,i}  \right| ^2 
\end{equation} 

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 $ v _{ n} $. Since ``$ v _n $ cannot span the space of $ y _i $s and there will be at least $ N _Y - N _X $ 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

+ 3 like - 0 dislike

$V(x,y)$ 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 $N_X$ by the space of $y$, of dimension $N_Y$. A minimum point of $V$ is of the form $(x_0,0)$ (with $x_0$ 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 $V(x_0,0)$ is non-zero. We want to show that there exists at least $N_Y-N_X$ flat directions i.e. that there exists at least $N_Y-N_X$ directions in the space of fields around $(x_0,0)$ where $V$ remains at the value $V(x_0,0$ (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 $x_0$ and we consider $V(x_0,y)$ as a function on the space of $y$, which is of dimension $N_Y$. 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 $y.v_n$ are 0 for every n, i.e to the space inside the space of $y$ orthogonal to the space generated by the $v_n$. As the space of y is of dimension $N_Y$ and the space generated by the $v_n$ is of dimension at most $N_X$, the orthogonal of the space generated by the $v_n$ is at least of dimension $N_Y-N_X$.

answered May 14, 2014 by 40227 (4,660 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...