Build a simple supersymmetric Abelian gauge theory

We now have all the ingredients for a supersymmetric gauge theory: charged chiral multiplets, a vector multiplet, and a way to assemble their interactions. The model will remain fixed through the next quantum lessons:

$$\boxed{\text{3d }\mathcal N=2,\quad G=U(1),\quad q_+=+1,\quad q_-=-1,\quad k=0,\quad W=0.}$$

There is one chiral multiplet of each charge. Both real masses and the Fayet–Iliopoulos parameter are initially zero. The Yang–Mills coupling $g_{\rm YM}^2>0$ has mass dimension one. The gauge group is compact $U(1)$ with unit electric charge; a gauge parameter has period $2\pi$. The theory has four real Lorentzian supercharges and gauge rank one. We calculate using their flat Euclidean complexification, with no fixed supergravity background fields.

This model is called one-flavor supersymmetric quantum electrodynamics, or SQED. “One flavor” counts a charge-opposite pair, not a single chiral field. We use $\Phi_+$ and $\Phi_-$ for the two chiral superfields. Their antichiral partners are $\widetilde\Phi_+$ and $\widetilde\Phi_-$. In particular, $\Phi_-$ is not the antichiral partner of $\Phi_+$. This avoids confusing the conventional names $Q,\widetilde Q$ for the two chiral matter multiplets with a Euclidean conjugation condition.

The immediate prerequisites are superspace and the vector multiplet. We will first find the potential, then derive the complete action that produces it. This is a classical construction; quantum corrections and the global description of its vacuum space are separate questions.

First calculation: eliminate the auxiliaries

For each sign $s=+,-$, write $\Phi_s=(\phi_s,\psi_s,F_s)$ and $\widetilde\Phi_s=(\widetilde\phi_s,\widetilde\psi_s,\widetilde F_s)$. The vector fields are $(a_\mu,\sigma,\lambda,\widetilde\lambda,D)$. Here $a_\mu$ is dynamical; it is not the background R-symmetry connection used in curved-space lessons. The auxiliary $D$ is a scalar, whereas $D_\mu^{(q)}$ below denotes a covariant derivative.

The supersymmetric coupling between $D$ and matter is weighted by the electric charge. Define the algebraic combination

$$\mu_{\rm alg} =\widetilde\phi_+\phi_+ -\widetilde\phi_-\phi_-.$$

The subscript reminds us that the Euclidean partners are still independent. The auxiliary part of the action is

$$\mathcal L_{\rm aux} =-\frac{D^2}{2g_{\rm YM}^2} +D\mu_{\rm alg} -\widetilde F_+F_+-\widetilde F_-F_-.$$

The negative Euclidean quadratic coefficients are the ones already fixed in the free chiral and vector lessons. Varying each auxiliary independently gives

$$D_*=g_{\rm YM}^2\mu_{\rm alg}, \qquad F_+=F_-=\widetilde F_+=\widetilde F_-=0.$$

Completing the square makes the resulting sign explicit:

$$-\frac{D^2}{2g_{\rm YM}^2}+D\mu_{\rm alg} =-\frac{(D-g_{\rm YM}^2\mu_{\rm alg})^2}{2g_{\rm YM}^2} +\frac{g_{\rm YM}^2}{2}\mu_{\rm alg}^2.$$

To discuss the physical bosonic potential, choose $\widetilde\phi_s=\phi_s^*$ and real $\sigma,a_\mu$. Then $\mu_{\rm alg}$ becomes the real charge-weighted density

$$\mu=|\phi_+|^2-|\phi_-|^2.$$

The scalar coupling to $\sigma$ contributes $\sigma^2(|\phi_+|^2+|\phi_-|^2)$. Therefore the reduced scalar potential is

$$\boxed{ U=\sigma^2\left(|\phi_+|^2+|\phi_-|^2\right) +\frac{g_{\rm YM}^2}{2} \left(|\phi_+|^2-|\phi_-|^2\right)^2.}$$

Both terms are nonnegative on this contour. The homogeneous zero-energy equations are

$$\boxed{\sigma\phi_+=\sigma\phi_-=0, \qquad |\phi_+|^2=|\phi_-|^2.}$$

These are equations for representatives of classical vacua. Gauge-equivalent representatives must still be identified. We will not infer a full quantum vacuum space merely from these equations.

Algebraic elimination is compatible with a convergent auxiliary integral. For the vector auxiliary, a shifted steepest-descent contour is

$$D=g_{\rm YM}^2\mu+i h,\qquad h\in\mathbb R.$$

On it, $\mathcal L_D=h^2/(2g_{\rm YM}^2)+g_{\rm YM}^2\mu^2/2$. Equivalently, begin with the decaying contour $D=ih$ and perform the entire Gaussian contour shift. For the matter auxiliaries at $W=0$, $\widetilde F_s=-F_s^*$ gives a positive quadratic form. These contour statements do not impose a conjugation condition on the independent Euclidean fermions.

Specify the gauge and symmetry assignments

For any field $X_q$ of charge $q$, our local gauge convention is

$$X_q\longmapsto e^{iq\alpha(x)}X_q, \qquad a_\mu\longmapsto a_\mu+\partial_\mu\alpha, \qquad D_\mu^{(q)}X_q=(\partial_\mu-iq a_\mu)X_q.$$

Every component of $\Phi_s$ has charge $q_s$; every component of $\widetilde\Phi_s$ has charge $-q_s$. The Abelian vector multiplet is gauge neutral, apart from the connection’s inhomogeneous shift. Its field strength is $f_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu$.

There is also an axial flavor symmetry $U(1)_A$ that rotates the two chiral multiplets by the same phase. It is distinct from the gauge rotation, which rotates them oppositely. With a common trial scalar R-charge $r$, the assignments are

Component	Gauge charge	Axial charge	R-charge
$\phi_s$	$q_s$	$1$	$r$
$\psi_s$	$q_s$	$1$	$r-1$
$F_s$	$q_s$	$1$	$r-2$
$a_\mu,\sigma,D$	$0$ for homogeneous transformations	$0$	$0$
$\lambda,\widetilde\lambda$	$0$	$0$	$1,-1$

Antichiral matter has the opposite charges to its chiral partner. In flat space with $W=0$, the classical action allows this common R-charge to vary by mixing with the axial symmetry. This freedom does not determine an infrared scaling dimension.

Aharony and collaborators discuss the gauge multiplet and real masses in AHISS v1, §§2.2–2.3, and introduce this charged-pair SQED theory in §3. Their adjoint scalar called $\phi$ corresponds to our $\sigma$, not to either matter scalar.

Recover the complete action

Let $\mathcal V$ be the vector superfield in the Wess–Zumino gauge of the superspace lesson, and $\Sigma$ its gauge-invariant linear field-strength superfield. With the normalized Berezin measures fixed there, take

$$S=\int d^3x\,d^4\theta\left[ -\frac1{g_{\rm YM}^2}\Sigma^2 +\widetilde\Phi_+e^{-2\mathcal V}\Phi_+ +\widetilde\Phi_-e^{2\mathcal V}\Phi_- \right].$$

The vector term is the flat specialization of P v3, (6.30). Each matter term is gauge invariant by the exponent cancellation in that lesson. There are no half-superspace terms because $W=0$, no FI term, and no gauge Chern–Simons term.

We can recover the component coefficients without leaving the gauge coupling implicit. In this gauge $\mathcal V$ begins at odd-coordinate degree two, so $\mathcal V^3=0$ and

$$e^{-2q\mathcal V} =1-2q\mathcal V+2q^2\mathcal V^2.$$

The degree-two part is $\mathcal V_{(2)}=b^\mu a_\mu-i\theta\widetilde\theta\,\sigma$, where $b^\mu=\theta\gamma^\mu\widetilde\theta$. Its square is

$$\mathcal V^2 =\frac12\theta^2\widetilde\theta^2 (a_\mu a^\mu+\sigma^2).$$

Indeed, $b^\mu b^\nu=\delta^{\mu\nu}\theta^2\widetilde\theta^2/2$, $(\theta\widetilde\theta)^2=-\theta^2\widetilde\theta^2/2$, and the mixed product vanishes. The factor $(-i)^2$ converts the last negative sign into the positive $\sigma^2$ contribution.

For a single charge-$q$ chiral multiplet, the three powers in the exponential contribute as follows. The free term includes the surface term already displayed in the superspace lesson.

Superspace contribution	Component terms added to the free Lagrangian
$-2q\,\widetilde\Phi\mathcal V\Phi$: scalar derivative terms	$iq a^\mu(\widetilde\phi\,\partial_\mu\phi-\partial_\mu\widetilde\phi\,\phi)$
The same term: fermion and auxiliary couplings	$-q a_\mu\widetilde\psi\gamma^\mu\psi-iq\sigma\widetilde\psi\psi+qD\widetilde\phi\phi$
The same term: gaugino couplings	$\sqrt2\,iq(\widetilde\phi\lambda\psi+\phi\widetilde\lambda\widetilde\psi)$
$2q^2\,\widetilde\Phi\mathcal V^2\Phi$	$q^2(a_\mu a^\mu+\sigma^2)\widetilde\phi\phi$

For example, the $D$ term follows from $(-2q)(-\theta^2\widetilde\theta^2D/2)\widetilde\phi\phi$. The quadratic connection term joins the first row and the ordinary scalar kinetic term to form $(D_\mu^{(-q)}\widetilde\phi)(D^{(q)\mu}\phi)$. The fermion connection term similarly joins its ordinary derivative. Thus the complete matter Lagrangian is

$$\begin{aligned} \mathcal L_q={}& (D_\mu^{(-q)}\widetilde\phi)(D^{(q)\mu}\phi) -i\widetilde\psi\gamma^\mu D_\mu^{(q)}\psi -\widetilde F F\\ &+qD\widetilde\phi\phi +q^2\sigma^2\widetilde\phi\phi -iq\sigma\widetilde\psi\psi\\ &+\sqrt2\,iq \left(\widetilde\phi\lambda\psi +\phi\widetilde\lambda\widetilde\psi\right). \end{aligned}$$

All spinor products retain the order shown, with $uv=u^TCv$ and the matrices of the three-dimensional spinor lesson. This agrees term by term with the zero-background, zero-central-charge specialization of P v3, (6.31)–(6.32). The gaugino couplings are required by supersymmetry; replacing ordinary derivatives with gauge-covariant ones alone would miss them, the $\sigma$ terms, and the $D$ coupling.

For completeness, the vector contribution is

$$\mathcal L_{\rm YM} =\frac1{g_{\rm YM}^2}\left[ \frac14 f_{\mu\nu}f^{\mu\nu} +\frac12\partial_\mu\sigma\,\partial^\mu\sigma -i\widetilde\lambda\gamma^\mu\partial_\mu\lambda -\frac12D^2\right].$$

The full off-shell action is $S=\int d^3x\,(\mathcal L_{\rm YM}+\mathcal L_{+1}+\mathcal L_{-1})$, with the field labels restored separately in each matter term. The superspace expansion and this component expression differ by an explicitly known divergence:

$$\begin{aligned} B^\mu_{\rm total}={}& \frac1{g_{\rm YM}^2} \left(-\frac12\sigma\partial^\mu\sigma +\frac i2\widetilde\lambda\gamma^\mu\lambda\right)\\ &+\sum_{s=\pm}\left[ \frac i2\widetilde\psi_s\gamma^\mu\psi_s -\frac14\partial^\mu(\widetilde\phi_s\phi_s)\right]. \end{aligned}$$

Specifically, the superspace density equals the displayed component density plus $\partial_\mu B^\mu_{\rm total}$. We assume periodic fields or sufficient falloff to discard its integral. A boundary theory requires its own boundary terms and conditions.

Several checks are immediate. Every term is gauge invariant. The first gaugino term, for example, has charge $-q+0+q=0$, axial charge $-1+0+1=0$, and R-charge $-r+1+(r-1)=0$. With $[\phi]=1/2$, $[\psi]=1$, $[\sigma]=[a]=1$, $[\lambda]=3/2$, and $[D]=2$, every matter term has dimension three; the vector terms have dimension four before division by $g_{\rm YM}^2$. Setting $q=0$ in the matter formula recovers the free chiral action. The compact superspace construction guarantees the flat supersymmetry of the assembled action, and the explicit expansion verifies which component action that statement concerns.

Test a classical vacuum

For homogeneous bosonic configurations, the potential vanishes when the boxed equations hold. The same conclusion follows from the fermion transformations. With constant scalars and a gauge choice $a_\mu=0$, the charged chiral rule reduces to

$$\delta\psi_s =\sqrt2\,\zeta F_s +\sqrt2\,i q_s\sigma\widetilde\zeta\,\phi_s.$$

The two independent commuting spinor labels require $F_s=0$ and $\sigma\phi_s=0$ if all supersymmetries are to be preserved. For a constant vector multiplet with $f_{\mu\nu}=0$, its gaugino variations are $\delta\lambda=i\zeta D$ and $\delta\widetilde\lambda=-i\widetilde\zeta D$, requiring $D=0$. Combining this with its algebraic equation gives $\mu=0$. This uses the flat specialization of P v3, (6.20) and (6.23).

As an explicit smooth local example, choose a real constant $v>0$ and any smooth real function $\alpha(x)$:

$$\phi_+=v e^{i\alpha(x)},\qquad \phi_-=v e^{-i\alpha(x)},\qquad a_\mu=\partial_\mu\alpha,\qquad \sigma=D=F_s=\widetilde F_s=0,$$

with all fermions zero and antichiral scalars conjugated on the stated contour. Then

$$D_\mu^{(\pm1)}\phi_\pm =\pm i(\partial_\mu\alpha)\phi_\pm \mp i(\partial_\mu\alpha)\phi_\pm=0, \qquad f_{\mu\nu}=0,\qquad\mu=0.$$

The scalar gradients in this representative do not cost energy: the connection cancels them. Locally it is gauge equivalent to $\phi_+=\phi_-=v$, $a_\mu=0$. No statement about nontrivial global holonomy follows from this local example.

Now perturb only the positive-charge magnitude to $\phi_+=v+\varepsilon$, with $\phi_-=v$, $\sigma=a_\mu=0$, and real small $\varepsilon$. The potential becomes

$$U=\frac{g_{\rm YM}^2}{2}(2v\varepsilon+\varepsilon^2)^2 =2g_{\rm YM}^2v^2\varepsilon^2 +2g_{\rm YM}^2v\varepsilon^3 +\frac{g_{\rm YM}^2}{2}\varepsilon^4.$$

A relative magnitude is a physical deformation, not a gauge phase. Its positive quadratic cost is a check of the auxiliary sign. In contrast, setting both matter scalars to zero leaves any constant $\sigma$ at zero classical potential. Describing the gauge quotient and the additional low-energy gauge variable belongs to the next vacuum-space calculation.

Fix what the level statement means

The classical phrase “$k=0$” needs a quantum convention. A Chern–Simons level is the coefficient of the local gauge term normalized here as $ik(4\pi)^{-1}\int a\wedge da$, with its supersymmetric partners. We work on spin manifolds, as appropriate to the fermions, with the compact unit-charge normalization already stated.

For charged complex two-component fermions, gauge-invariant regularization requires

$$k+\frac12\sum_f q_f^2\in\mathbb Z.$$

For our two matter fermions the sum is $1^2+(-1)^2=2$, so $k=0$ is allowed. A single unit-charge chiral would instead require a half-integer offset. This is the relevant gauge parity-anomaly condition, not a count of scalar fields. It is the sourced quantum input of AHISS v1, §2.4, (2.12).

Our level convention uses the massive-fermion shift

$$k_{\rm eff}=k+\frac12\sum_f q_f^2\,\operatorname{sign}(M_f),$$

as in AHISS (2.11). The physical Dirac mass in our component convention is $M_s=q_s\sigma-z_s$; the central-charge mass parameters $z_s$ are zero here. To see the sign, introduce the independent row $\chi_s=-i\widetilde\psi_s^TC$. Its quadratic action is $\chi_s(\gamma^\mu D_\mu+q_s\sigma-z_s)\psi_s$. At nonzero constant $\sigma$, the pair has masses $+\sigma,-\sigma$, so its two contributions to the gauge level cancel.

This formula concerns integrating out massive fermions. It is not applied at $\sigma=0$, where the charged fields are massless and remain in the theory. The bare gauge-level choice is $k=0$ in a gauge-invariant paired prescription with zero net gauge-level offset. An opposite-sign regulator pair makes this possible.

For future source calculations, take the integer-normalized reference symmetry $R_0(\phi_\pm)=0$ and axial charges $A(\Phi_\pm)=1$, as in AHISS (3.1); a trial $R$ is $R_0+rA$. We add no optional integer pure-background Chern–Simons counterterms in this reference convention. No background sources are turned on in the action above. This statement does not set regulator-induced fractional contact terms to zero, and it does not fix them by a classical calculation. When a background gauge or R field is used, its mixed gauge/background counterterms and determinant phases must be stated explicitly before comparing partition functions. The gauge cancellation just checked is not a claim that every background contact term cancels.

Exercises

1. Check a Higgs mass normalization

Expand near $\phi_+=\phi_-=v>0$, $\sigma=a_\mu=0$:

$$\phi_\pm=v+\frac{h_\pm+i\pi_\pm}{\sqrt2}.$$

Use canonical vector fields $a_{\mu,c}=a_\mu/g_{\rm YM}$ and $\sigma_c=\sigma/g_{\rm YM}$. Find the vector mass squared, the $\sigma_c$ mass squared, and the mass squared of $h_D=(h_+-h_-)/\sqrt2$. Identify which phase combination is removed by the local gauge symmetry.

Hint 1

At quadratic order the two scalar kinetic terms contribute $2v^2 a_\mu a^\mu$. Compare a canonically normalized vector mass term with $\frac12m_a^2a_{\mu,c}a_c^\mu$.

Hint 2

The density difference begins as $\mu=\sqrt2v(h_+-h_-)=2v h_D$. The $\sigma$ potential begins as $2v^2\sigma^2$.

Solution

The vector kinetic term is $f_c^2/4$. Its quadratic mass coefficient is $2g_{\rm YM}^2v^2a_c^2$, giving

$$m_a^2=4g_{\rm YM}^2v^2.$$

The $\sigma_c$ kinetic term is $(\partial\sigma_c)^2/2$. Since $2v^2\sigma^2=2g_{\rm YM}^2v^2\sigma_c^2$, we obtain the same mass squared. Finally,

$$U_D^{(2)} =\frac{g_{\rm YM}^2}{2}(2v h_D)^2 =\frac12(4g_{\rm YM}^2v^2)h_D^2.$$

The real fluctuations have canonical kinetic coefficients $1/2$, so no further rescaling of $h_D$ is needed. All three masses agree.

To linear order, $\delta\pi_\pm=\pm\sqrt2v\alpha$. Hence $\pi_D=(\pi_+-\pi_-)/\sqrt2$ shifts by $2v\alpha$ and is the phase removed in unitary gauge. The sums $(h_++h_-)/\sqrt2$ and $(\pi_++\pi_-)/\sqrt2$ have no quadratic potential here. This is a local classical spectrum check, not a global description of the moduli space. The dimensions also agree: $[g_{\rm YM}^2v^2]=2$.

2. Turn on a specified FI deformation

Change the initial theory by adding $\mathcal L_{\rm FI}=+\xi D$, with real $\xi$ of dimension one. This sign is the flat limit of P v3, (6.27). Derive the new potential and homogeneous supersymmetric vacuum equations. For $\xi>0$, give one vacuum with only one chiral scalar nonzero. Does the configuration $\phi_+=\phi_-=0$ still have zero energy?

Hint 1

Replace the density multiplying $D$ by $\mu+\xi$. The sign convention for $\xi$ has been specified rather than inferred from a remembered potential.

Hint 2

The equation is $\mu=-\xi$. Decide which charge has the appropriate contribution to $\mu$.

Solution

The saddle becomes $D_*=g_{\rm YM}^2(\mu+\xi)$ and the potential is

$$U_\xi=\sigma^2(|\phi_+|^2+|\phi_-|^2) +\frac{g_{\rm YM}^2}{2} (|\phi_+|^2-|\phi_-|^2+\xi)^2.$$

The vacuum equations are $\sigma\phi_\pm=0$ and $|\phi_+|^2-|\phi_-|^2=-\xi$. For $\xi>0$, choose $\phi_+=0$, $|\phi_-|=\sqrt{\xi}$, and $\sigma=0$. The arbitrary phase is gauge. The matter auxiliaries still vanish because $W=0$.

With both matter scalars zero the energy density is $g_{\rm YM}^2\xi^2/2>0$, independently of a constant $\sigma$. Thus that previously flat family no longer consists of supersymmetric vacua. Reversing the sign chosen for $\mathcal L_{\rm FI}$ would exchange the roles of the two charges; the displayed answer follows the convention actually stated.

3. Remove one charged multiplet

Suppose the negative-charge chiral and its antichiral partner are deleted while the gauge field remains compact with unit charge. First write the remaining classical potential at zero FI parameter. Then test whether the statement “the same theory has $k=0$” remains compatible with the displayed gauge parity-anomaly condition. Explain why a consistent classical action does not answer this quantum question.

Hint 1

The classical potential follows by setting the deleted scalar to zero. The anomaly sum runs over the remaining charged complex fermions.

Hint 2

The condition is $k+q^2/2\in\mathbb Z$, not $k+q/2\in\mathbb Z$.

Solution

The remaining classical potential is

$$U_{\rm one}=\sigma^2|\phi_+|^2 +\frac{g_{\rm YM}^2}{2}|\phi_+|^4.$$

It is nonnegative and comes from a locally supersymmetric classical action. But the quantum gauge consistency condition becomes $k+1/2\in\mathbb Z$. The original $k=0$ prescription is therefore unavailable for this single-chiral theory. A half-integer level offset, together with the corresponding fermion determinant prescription and supersymmetric Chern–Simons completion, is needed.

The classical action only tests local field transformations and their polynomial couplings. The determinant phase under gauge transformations is additional quantum data. Deleting a multiplet changes that data even though the remaining classical formulas still make sense.

The result is a specified classical theory, not just a field list: its gauge group and charge lattice, complete off-shell action, auxiliary contours, initial parameter values, and vacuum equations are fixed. Continue with Higgs and Coulomb branches to solve those equations modulo gauge transformations and determine what the massless photon contributes to the low-energy vacuum space.

Chern–Simons and FI couplings provide a separate continuation using this same flat-space preparation. Derive their supersymmetric partners and check the mass generated when a Maxwell field acquires a Chern–Simons term.