Supersymmetric gauge actions and D terms
Normalize a vector multiplet and its matter couplings, eliminate the real auxiliary D, and distinguish gauge fixing from physical polarization counting.
On this page
The question for today
What you’ll understand
- Identify the propagating and auxiliary components of a vector multiplet.
- Fix a canonical gauge and D-term normalization.
- Compute a charge-weighted D potential including a declared Abelian FI term.
A supersymmetric gauge field belongs to a real vector superfield with a Weyl gaugino and a real auxiliary field D. Gauge redundancy removes some superfield components before the familiar physical photon polarizations are counted. Coupling charged chiral fields then generates an additional positive scalar potential after D is eliminated. We use canonically normalized Abelian gauge fields with the coupling in the matter interactions, and define the sign of a possible Fayet-Iliopoulos term directly through its component coupling.
A real superfield contains a gauge multiplet
In Wess-Zumino gauge the surviving components are , , and D. The vector has four off-shell real components, one ordinary gauge redundancy, and the real D supplies one additional bosonic component, matching the four real Grassmann gaugino components. On shell a massless vector has two transverse polarizations, matched by its gaugino states. Wess-Zumino gauge removes extra superfield components but does not itself fix the remaining ordinary gauge freedom or impose the vector field equation.
A supersymmetry transformation generally leaves this superfield gauge slice. A compensating gauge transformation restores it, so the component algebra closes up to gauge transformations. This feature is a bookkeeping consequence of gauge fixing rather than a failure of supersymmetry. For an Abelian vector choose the chiral field strength . Its gauge-invariant action supplies the Maxwell term, gaugino kinetic term, and positive auxiliary square. We distinguish this spinor field strength from the scalar superpotential W by its calligraphic symbol and index.
Complete the auxiliary square before reading the potential
For scalar backgrounds the D-dependent matter term is g D times the charge-weighted sum of scalar moduli squared. Add xi D and define . The auxiliary Lagrangian is one half D squared plus D J. Its equation gives D equal to minus J; substitution gives minus one half J squared. Thus the static potential is positive one half J squared. The sign of D is convention dependent, but its potential and the explicitly defined flatness equation are unambiguous together.
Unlike an electric field propagating through space, D has no derivative kinetic term and no independent particle polarization. It enforces a local relation between scalar backgrounds and a charge moment map. Oppositely charged fields contribute with opposite signs even though each modulus square is nonnegative. Their contributions can cancel. In a non-Abelian canonical theory the same construction uses Hermitian generators in each matter representation, one real auxiliary per generator, and a sum of auxiliary squares. An arbitrary FI term is not available for every simple non-Abelian generator.
Gauge symmetry and supersymmetry are different tests
A charged scalar expectation can Higgs the gauge field while preserving supersymmetry if both F and D vanish. Conversely, a nonzero D expectation in a genuine stable vacuum breaks rigid supersymmetry through the gaugino transformation. A nonzero value at an arbitrary background proves neither a stable vacuum nor a broken ground state. Finding the vacuum requires minimizing the full potential and respecting gauge equivalence. This distinction prevents a convenient charge assignment or one incomplete square from being mistaken for a full mechanism of symmetry breaking.
For a vectorlike pair with charges plus one and minus one, the Abelian cubic and mixed gravitational gauge anomalies cancel between the fermions. Such a pair is a useful consistent toy matter sector, though a realistic model may need additional fields and interactions. A positive FI parameter can be cancelled by the negatively charged scalar in our sign convention; reversing the FI sign reverses that role. The next lesson uses the zero-F and zero-D equations to identify inequivalent vacuum configurations rather than treating gauge-related field values as distinct vacua.
Work it through
A vectorlike pair away from D flatness
Take charges +1 and −1, , xi zero, and scalar magnitudes 3 GeV and 1 GeV respectively. Compute J, D, and the D-term potential, assuming canonical normalization.
The charge-weighted modulus sum is . The minus sign comes from the charge rather than a negative modulus square.
Multiplication by g gives and the algebraic equation gives . There is no separate dynamical initial condition for D.
The potential is . The eliminated auxiliary contributes minus that amount to the Lagrangian, consistently with the static-potential sign.
The background has J = 4 GeV², D = −4 GeV², and VD = 8 GeV⁴; it is not D flat.
The living notebook
Explore the model
Interactive lab · function
A charge changes the sign of a quadratic contribution
Reset and set BOTH Coefficient a = 1 and Position x = 1; predict y = +1. For a fresh comparison, Reset and set a = −1 and x = 1; predict y = −1. Inspect the value before summing any hypothetical contributions.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe quadratic models a signed charge times a real scalar magnitude squared. It does not compute a gauge multiplet, sum D sources, eliminate auxiliaries, or provide a SUSY field solver.
Pause & reason
Check your understanding
Your turn
Try it yourself
Use g = 1, charges +1 and −1, and . Set the positively charged scalar to zero and the negatively charged scalar magnitude to 2 GeV. Compute J, D, and VD. Does D flatness alone determine F flatness?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the declared positive xi D convention, then remember that the superpotential supplies separate equations.
Compare with the solution
The charge-weighted contribution is minus four GeV squared, cancelled by the positive FI term. Thus , , and . This is a D-flat background and may Higgs the gauge field through the charged scalar expectation. F flatness still depends on the superpotential derivatives. With no superpotential these derivatives vanish, but adding allowed interactions can change that conclusion. D flatness by itself is only one part of the supersymmetric-vacuum condition.
Keep the idea
Take it with you
- Wess-Zumino gauge and on-shell polarization counting remove different redundancies or constraints.
- Canonical D elimination gives VD equal to one half the squared charge moment map.
- Higgsing a gauge field can coexist with unbroken supersymmetry when all F and D terms vanish.
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