Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Couplings give the strength of interactions. They appear in a Lagrangian as a coefficient to the two (or more) interacting fields. Unfortunately, I'm struggling to find a source that describes what it is that the Fayet-Iliopoulos coupling actually couples.

As far as I know, it is simply the coefficient (given by the greek letter 'xi') in the Fayet-Iliopoulos D-term. The D-term is just one of the terms that appears in the expansion of a superfield and doesn't describe an interaction.

My apologies if the answer is obvious.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Let's focus on global supersymmetry; Fayet-Iliopoulos terms in supergravity are a lot more subtle.

You are right that 'the $D$-term' is just a particular piece in the expansion of a superfield. When people talk about a "Fayet-Iliopoulos $D$-term", it refers to adding some multiple of the $D$-term of a $U(1)$ gauge field to the Lagrangian: $$ \mathcal{L}_{\mathrm{FI}} = \xi\, D ~, $$ where $\xi$ is a constant with dimensions of mass squared. This is gauge-invariant for a $U(1)$ gauge field, and under supersymmetry, it transforms by a total derivative, so the action is invariant.

Such a term may look strange, but remember that $D$ is an auxiliary field, which means it doesn't have a kinetic term. This means that its equation of motion is just algebraic, and we can solve it and substitute the solution back into the Lagrangian. Let the scalar fields in the theory be denoted $\{\phi_i\}_{i=1,\ldots,N}$, where $\phi_i$ has charge $q_i$ under the $U(1)$. With the FI term included, the solution for $D$ is $$ D = -\left( \sum_i q_i |\phi_i|^2 +\xi \right) ~. $$ The scalar potential includes a term $\frac{1}{2}D^2$, and when you expand this out, you get a bunch of quartic interactions which are independent of $\xi$, a bunch of mass terms proportional to $\xi$, and a constant $\xi^2$ term.

To see why this is an interesting thing to do, remember that supersymmetry is unbroken if and only if $D = 0$, and also $F_i = 0$ for each $i$, where $F_i$ is the auxiliary component of the chiral superfield $\Phi_i$. You can see that when $\xi \neq 0$, $D=0$ is impossible unless $\phi_i \neq 0$ for at least one value of $i$ (with $q_i \neq 0$). But $\phi_i \neq 0$ means that the gauge symmetry is broken. So either gauge symmetry or supersymmetry must be broken!

Actually, gauge symmetry or supersymmetry or both may be broken in a model like this; I recommend playing around with a simple example -- two chiral fields, $q_1 = e, q_2 = -e$, with mass term $\int d^2\theta\, m \Phi_1 \Phi_2$ -- and work out for which values of $m, \xi$ supersymmetry and/or gauge symmetry is broken.

share|improve this answer
    
Thank you for the very insightful answer, it's helped me get a much better grip on the FI term/coupling. However just to make sure I've really got it; is it fair then to say that the FI coupling "xi" isn't a coupling between two or more fields that gives a strength of some interaction. Rather it is just called a coupling because it appears in front of the D-term auxiliary field in the Lagrangian (in the same way a coupling would appear in front of field terms in the Lagrangian)? –  Siraj R Khan Nov 19 '12 at 9:58
1  
Yes, I think what you're saying is fair enough: the terms proportional to $\xi$ are mass terms and a constant, and neither of these correspond to an 'interaction'. –  Rhys Nov 19 '12 at 11:13
    
Fantastic, thanks once again! –  Siraj R Khan Nov 19 '12 at 12:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.