Superspace, interactions, and supersymmetric vacua

The free multiplet lesson checked supersymmetry by canceling component variations. Can we add a Yukawa interaction without guessing which scalar interaction must accompany it? Superspace answers this by putting the fields into a finite polynomial in odd coordinates. An integral selects the component combination whose variation is a boundary term.

This lesson requires that free multiplet calculation and matrix-valued matter, representations, and traces. We will expand the compact expressions, derive their scalar potentials, and solve two classical vacuum problems. Superspace and vacuum geometry are introduced here; three-dimensional conventions are not needed.

We work in flat four-dimensional Lorentzian signature $+---$, with manifest $\mathcal N=1$ supersymmetry and the sigma matrices of the preceding lessons. The worked theories are a neutral Wess–Zumino model and a $U(1)$ theory with two oppositely charged chirals. All kinetic metrics are canonical and positive. Statements about minima and particle masses below are classical; no quantum effective potential is being assumed.

Superspace packages a multiplet

Add odd coordinates $\theta^\alpha,\bar\theta^{\dot\alpha}$ to spacetime. A superfield is an ordinary finite Grassmann polynomial in these coordinates, whose coefficients are spacetime fields. Define $\theta^2=\theta^\alpha\theta_\alpha$ and $\bar\theta^2=\bar\theta_{\dot\alpha}\bar\theta^{\dot\alpha}$, and normalize the Berezin measures by

$$\int d^2\theta\,\theta^2=1,\qquad \int d^2\bar\theta\,\bar\theta^2=1,\qquad d^4\theta=d^2\theta\,d^2\bar\theta.$$

A superspace derivative removes one odd coordinate. Using left derivatives, the operators

$$\mathbb D_\alpha= \frac{\partial}{\partial\theta^\alpha} +i\sigma^\mu_{\alpha\dot\beta}\bar\theta^{\dot\beta}\partial_\mu, \qquad \bar{\mathbb D}_{\dot\alpha}= \frac{\partial}{\partial\bar\theta^{\dot\alpha}} +i\theta^\beta\sigma^\mu_{\beta\dot\alpha}\partial_\mu$$

anticommute with the supercharges. Hence the constraint $\bar{\mathbb D}_{\dot\alpha}\Phi=0$ is preserved by supersymmetry. Its solution is a chiral superfield depending on $y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$ and $\theta$:

$$\Phi(y,\theta)=\phi(y)+\sqrt2\,\theta\psi(y)+\theta^2F(y).$$

To check the coordinate choice, a left derivative with respect to $\bar\theta$ crosses $\theta$ with a minus sign. Its action on $i\theta\sigma^\mu\bar\theta$ cancels the second term of $\bar{\mathbb D}y^\mu$. Chirality therefore removes the independent $\bar\theta$ coefficients without requiring an equation of motion. The component auxiliary is the preceding lesson’s $F=-f_{\rm Bilal}$. Bilal, (4.14)–(4.24)

Full superspace integration extracts a $\theta^2\bar\theta^2$ coefficient, called a D-term. A chiral integral extracts a $\theta^2$ coefficient, called an F-term. These names describe integration operations; a D-term need not be the vector auxiliary $D$ alone.

Why do these operations produce invariant actions? Supersymmetry acts by odd-coordinate derivatives plus spacetime derivatives. The Berezin integral of an odd-coordinate derivative vanishes, while the spacetime derivative integrates to a boundary term. For a chiral integral the same reasoning works in $(y,\theta)$ coordinates. With suitable boundary conditions,

$$S=\int d^4x\left[ \int d^4\theta\,K +\left(\int d^2\theta\,W(\Phi)+\text{h.c.}\right)\right]$$

is supersymmetric when $K$ is a real superfield and $W$ is chiral. “h.c.” means the Hermitian conjugate term. Bilal, (4.26)–(4.28); Tong, section 3.2.1.

For canonical matter, $K=\sum_i\Phi_i^\dagger\Phi_i$. This is a particular Kähler potential, the function specifying the scalar kinetic metric. We keep this canonical choice throughout. The dimensions $[\theta]=-1/2$, $[d^4\theta]=2$, and $[d^2\theta]=1$ require $[K]=2$ and $[W]=3$.

Expand an interaction before using it

Let $W$ be a holomorphic function: it depends on $\Phi_i$ but not their conjugates. At fixed $y$, Taylor expansion terminates:

$$W(\Phi)=W(\phi) +W_i\left(\sqrt2\,\theta\psi_i+\theta^2F_i\right) +\frac12W_{ij}(\sqrt2\,\theta\psi_i)(\sqrt2\,\theta\psi_j),$$

where $W_i=\partial W/\partial\phi_i$ and $W_{ij}=\partial_i\partial_jW$. The Grassmann identity

$$(\theta\psi_i)(\theta\psi_j) =-\frac12\theta^2\,\psi_i\psi_j$$

then gives the full chiral integral:

$$\int d^2\theta\,W(\Phi) =W_iF_i-\frac12W_{ij}\psi_i\psi_j.$$

The same function fixes both the auxiliary coupling and the Yukawa coupling. This is Bilal’s equation (4.25), translated by $F=-f$.

For one neutral chiral choose

$$W(\Phi)=\frac m2\Phi^2+\frac h3\Phi^3,\qquad [m]=1,\quad[h]=0.$$

Its interaction density is

$$\mathcal L_W=(m\phi+h\phi^2)F -\frac12(m+2h\phi)\psi\psi+\text{h.c.}$$

The term $-h\phi\psi\psi$ is the desired Yukawa interaction. Its scalar interaction emerges when we combine the first term with $\bar F F$ from $K$:

$$\bar F F+W'F+\overline{W'}\,\bar F =(\bar F+W')(F+\overline{W'})-|W'|^2.$$

Varying $\bar F$ gives $F=-\overline{W'}$. Since a Lorentzian Lagrangian contains minus the potential,

$$V_F=|W'|^2=|m\phi+h\phi^2|^2.$$

For nonzero $m,h$, it vanishes at $\phi=0$ and $\phi=-m/h$. The quartic term and its interference with the mass term are fixed by supersymmetry. A chosen supersymmetric interaction does not automatically preserve a continuous R-symmetry: fixed nonzero quadratic and cubic coefficients cannot both accommodate one common R-charge for $\Phi$ without additional transforming source data.

Gauge interactions from a finite exponential

Keep the vector fields canonically normalized as in the preceding lesson and take $g>0$. With Hermitian generators $t_R^a$ in the matter representation $R$, write

$$\mathscr D_\mu\phi =\left(\partial_\mu-igA_\mu^at_R^a\right)\phi.$$

Thus the connection $a_\mu$ of the matrix lesson is $a_\mu=gA_\mu$. This is an explicit change of field normalization, not a change of its gauge action.

In Wess–Zumino gauge the real vector superfield is

$$V=\theta\sigma^\mu\bar\theta\,A_\mu +i\theta^2\bar\theta\bar\lambda -i\bar\theta^2\theta\lambda +\frac12\theta^2\bar\theta^2D.$$

For a $U(1)$ chiral of charge $q$, the matter D-term is $\Phi^\dagger e^{2gqV}\Phi$. There are only four odd coordinates and every term of $V$ has degree at least two, so $V^3=0$ and

$$e^{2gqV}=1+2gqV+2g^2q^2V^2.$$

The exponential is finite here, even though it packages arbitrarily many powers of an ordinary matrix before the odd-coordinate expansion. In a non-Abelian representation the corresponding expression is $\Phi^\dagger e^{2gV_R}\Phi$, with $V_R=V^at_R^a$.

Its gauge invariance can be checked directly. For a chiral matrix parameter $\Lambda$, take

$$\Phi'=e^{i\Lambda}\Phi,\qquad e^{2gV_R'}=e^{i\Lambda^\dagger}e^{2gV_R}e^{-i\Lambda}.$$

The group factors cancel in $\Phi'^\dagger e^{2gV_R'}\Phi'$. A superpotential must separately be gauge invariant; covariantizing derivatives cannot repair a charged monomial in $W$. Bilal, (5.1), (5.19)–(5.21)

To multiply the factors at a common spacetime point, Taylor-expand the chiral coordinates:

$$\begin{aligned} \Phi(x,\theta,\bar\theta) ={}&\phi+\sqrt2\,\theta\psi+\theta^2F +i\theta\sigma^\mu\bar\theta\,\partial_\mu\phi\\ &-\frac{i}{\sqrt2}\theta^2 \partial_\mu\psi\,\sigma^\mu\bar\theta -\frac14\theta^2\bar\theta^2\Box\phi. \end{aligned}$$

Use its Hermitian conjugate for $\Phi^\dagger$. Multiplication now gives the following additional components beyond $\Phi^\dagger\Phi$. In the table, $|\phi|^2=\bar\phi\phi$ and all spinor contractions use the preceding lesson’s convention.

Contribution	Additional component density
$2gq\,\Phi^\dagger V\Phi$	$igqA^\mu(\bar\phi\,\partial_\mu\phi-\partial_\mu\bar\phi\,\phi)+gqA_\mu\psi\sigma^\mu\bar\psi$
The gaugino and auxiliary pieces of the same term	$i\sqrt2gq(\bar\phi\lambda\psi-\bar\psi\bar\lambda\phi)+gqD\lvert\phi\rvert^2$
$2g^2q^2\,\Phi^\dagger V^2\Phi$	$g^2q^2A_\mu A^\mu\lvert\phi\rvert^2$

For example, the top component $D/2$ in $V$ multiplied by $2gq$ gives $gqD|\phi|^2$. The quadratic term uses $(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) =\theta^2\bar\theta^2\eta^{\mu\nu}/2$. The terms containing one matter fermion come from multiplying $\sqrt2\theta\psi$ by the gaugino pieces of $V$; the same Grassmann rearrangement that produced the superpotential Hessian fixes their relative minus sign.

Adding the free D-term reorganizes the derivatives and gives, up to a spacetime boundary term,

$$\begin{aligned} \int d^4\theta\,\Phi^\dagger e^{2gqV}\Phi ={}&\mathscr D_\mu\bar\phi\,\mathscr D^\mu\phi -i\psi\sigma^\mu\mathscr D_\mu\bar\psi+\bar F F\\ &+i\sqrt2gq(\bar\phi\lambda\psi-\bar\psi\bar\lambda\phi) +gqD|\phi|^2 . \end{aligned}$$

The barred fields use the opposite charge. In particular $-i\psi\sigma^\mu(\partial_\mu+igqA_\mu)\bar\psi$ has the positive $gqA_\mu\psi\sigma^\mu\bar\psi$ term shown in the table. This is a useful sign check. Bilal, (5.22)–(5.25)

The gauge action and the trace convention

The chiral gauge field-strength superfield, denoted $\mathcal W_\alpha$ to distinguish it from the superpotential, can be defined by

$$\mathcal W_\alpha=-\frac1{8g}\bar{\mathbb D}^{\,2} \left(e^{-2gV}\mathbb D_\alpha e^{2gV}\right).$$

Here $\bar{\mathbb D}^{\,2}=\bar{\mathbb D}_{\dot\alpha}\bar{\mathbb D}^{\dot\alpha}$. The field strength transforms by conjugation, and its lowest component is $-i\lambda_\alpha$. The factor $1/(2g)$ relative to Bilal’s rescaled field-strength superfield keeps our vector components canonical. Its Abelian limit is $-\bar{\mathbb D}^{\,2}\mathbb D_\alpha V/4$. Bilal, (5.3)–(5.5) and (5.14)

With a canonical basis of gauge components and no theta angle, the action contains

$$\frac14\int d^2\theta\, \sum_a\mathcal W^{a\alpha}\mathcal W^a_\alpha+\text{h.c.}$$

For $SU(N)$ we use the ordinary fundamental trace $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$. The same term is therefore $\frac12\int d^2\theta\,\operatorname{Tr}_N(\mathcal W^\alpha\mathcal W_\alpha)+\text{h.c.}$. In components its bosonic terms are

$$-\frac12\operatorname{Tr}_N(F_{\mu\nu}F^{\mu\nu}) +\operatorname{Tr}_N(D^2) =-\frac14\sum_aF^a_{\mu\nu}F^{a\mu\nu} +\frac12\sum_a(D^a)^2,$$

where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu]$. A trace convention must be carried through this coefficient; replacing $\operatorname{Tr}_N$ by $\operatorname{Tr}_N/N$ would change the action unless its prefactor changed too.

The potential and the moment map

For canonical matter in any unitary representation, define the real quantities

$$\mu^a(\phi,\bar\phi)=\sum_i\phi_i^\dagger t_{R_i}^a\phi_i.$$

They are called the moment map for the gauge action on the scalar fields. Here this name has a concrete use: $\mu^a$ is the coefficient of $gD^a$ in the component matter action. For $U(1)$, it is simply $\sum_iq_i|\phi_i|^2$.

The complete auxiliary density is

$$\mathcal L_{\rm auxiliary} =\sum_i\left(\bar F_iF_i+W_iF_i+\bar W_i\bar F_i\right) +\frac12\sum_a(D^a)^2+g\sum_aD^a\mu^a.$$

Completing its squares gives

$$F_i=-\bar W_i,\qquad D^a=-g\mu^a,\qquad V(\phi,\bar\phi)=\sum_i|W_i|^2+\frac{g^2}{2}\sum_a(\mu^a)^2.$$

This is a positive sum on the Lorentzian real field space with canonical kinetic terms. For a vacuum on Minkowski space, choose a gauge with $A_\mu=0$, spacetime-constant scalars, and vanishing fermions. The preceding component transformations then preserve all four real supercharges precisely when

$$W_i=0,\qquad \mu^a=0.$$

These are F-flatness and D-flatness. A solution has zero classical potential and is a classical minimum. Configurations related by a gauge transformation describe the same vacuum. Failure to solve these equations is not by itself a stability analysis of a proposed nonsupersymmetric state. Bilal, (5.28)–(5.31) and section 6.1

For a matrix example, a canonical adjoint $\Phi=\Phi^at^a$ has $K=2\operatorname{Tr}_N(\Phi^\dagger\Phi)$. Its gauged form is $2\operatorname{Tr}_N(\Phi^\dagger e^{2gV}\Phi e^{-2gV})$. Expanding the two exponentials to first order in $D$ gives

$$2g\operatorname{Tr}_N\big(\phi^\dagger[D,\phi]\big) =2g\operatorname{Tr}_N\big(D[\phi,\phi^\dagger]\big).$$

Thus

$$\mu^a=2\operatorname{Tr}_N(t^a[\phi,\phi^\dagger]),\qquad V_D=g^2\operatorname{Tr}_N([\phi,\phi^\dagger]^2).$$

The commutator is Hermitian, so its squared trace is nonnegative. The factor two in the Kähler potential is necessary to make the components $\phi^a$ canonical.

Worked charged example: one flavor of SQED

Take two independent chirals $\Phi_+,\Phi_-$ with charges $+1,-1$ and $W=m\Phi_+\Phi_-$. Opposite charges make the superpotential invariant and the fermion matter content vectorlike. A single charged field’s quadratic mass monomial would fail this gauge test.

The chiral expansion is

$$\int d^2\theta\,m\Phi_+\Phi_- =m(\phi_+F_-+\phi_-F_+-\psi_+\psi_-).$$

Combining it with the two gauge-matter D-terms gives

$$V=|m|^2(|\phi_+|^2+|\phi_-|^2) +\frac{g^2}{2}(|\phi_+|^2-|\phi_-|^2)^2.$$

For $m\ne0$, the only supersymmetric classical vacuum is the origin. For $m=0$, the D-flat configurations have $\phi_+=v e^{i\alpha}$, $\phi_-=v e^{i\beta}$ with $v\ge0$. A gauge transformation shifts $\alpha$ and $\beta$ oppositely. The surviving complex coordinate is

$$M=\phi_+\phi_-=v^2e^{i(\alpha+\beta)}.$$

Every complex $M$ has a representative, and equal $M$ identifies the same orbit, including $M=0$. Equivalently, one real constraint and one gauge identification reduce four real scalar components to two vacuum parameters. This identifies the classical set of vacua; it does not assert a flat metric on that set. Tong, (4.15) and section 4.3.

Exercises

1. Check the masses at both Wess–Zumino vacua

For nonzero $m,h$, expand around each zero of $W'=m\phi+h\phi^2$. Determine the scalar and fermion masses and explain which interaction coefficient controls them.

Hint 1

Write $\phi=\phi_0+\eta$ and expand $W'$ to first order in $\eta$.

Hint 2

The quadratic scalar potential comes from $|W''(\phi_0)\eta|^2$. The fermion bilinear is already visible in the superpotential component expansion.

Solution

At $\phi_0=0$ and $\phi_0=-m/h$,

$$W''(\phi_0)=m+2h\phi_0=m,\ -m,$$

respectively. Since $W'(\phi_0)=0$, $V=|m|^2|\eta|^2+O(\eta^3)$. Writing $\eta=(\eta_1+i\eta_2)/\sqrt2$ shows that both real scalars have mass squared $|m|^2$. The Weyl mass term is $-\tfrac12W''(\phi_0)\psi\psi+\text{h.c.}$, giving mass $|m|$ at either vacuum. The phase or sign of the complex fermion mass does not change its positive mass magnitude. Both spectra are paired as expected in a supersymmetric vacuum. At a point where $W'\ne0$, the omitted cross terms can contribute to the scalar Hessian, so this argument cannot be reused without checking its premise.

2. Shift the Abelian auxiliary equation

Add a Fayet–Iliopoulos term $2\xi\int d^4\theta\,V=\xi D$, with real $\xi$ of dimension two, to the two-chiral example. Derive the new potential. Can the F-flatness and D-flatness equations be solved for $m=0$? For $m\ne0$ and $\xi\ne0$?

Hint 1

The terms containing $D$ become $D^2/2+(g\mu+\xi)D$.

Hint 2

For $m=0$, a difference of two nonnegative squared magnitudes can take either sign. For $m\ne0$, first solve the two F equations.

Solution

The auxiliary solution is $D=-g\mu-\xi$, so

$$V=|m|^2(|\phi_+|^2+|\phi_-|^2) +\frac12\left[g(|\phi_+|^2-|\phi_-|^2)+\xi\right]^2.$$

When $m=0$, choose $|\phi_+|^2-|\phi_-|^2=-\xi/g$. For $\xi>0$, for example, $\phi_+=0$ and $|\phi_-|^2=\xi/g$ solve it; the opposite choice works for $\xi<0$. There are also solutions with both fields nonzero.

For $m\ne0$, the F equations force $\phi_+=\phi_-=0$. Then the D equation is unsatisfied if $\xi\ne0$. There is no supersymmetric classical vacuum under these assumptions. Determining the actual minimum and its stability is a further problem. The FI operator is available for an Abelian factor; a fixed nonzero coefficient of an individual $SU(N)$ adjoint component $D^a$ would not be gauge invariant. Bilal, (5.26)

3. Test the adjoint normalization

Use $SU(2)$ generators $t^a=\tau_a/2$, canonical $K=2\operatorname{Tr}_2(\Phi^\dagger\Phi)$, and $W=m\operatorname{Tr}_2(\Phi^2)$ with $m\ne0$. On the scalar configuration $\phi=u t^1+iw t^2$, with $u,w$ real, calculate the three moment-map components and the full potential. Which configurations in this family are D-flat, and which are supersymmetric vacua?

Hint 1

First calculate $[\phi,\phi^\dagger]$. Keep the factor two in the moment-map definition.

Hint 2

The component superpotential is $m\sum_a(\phi^a)^2/2$, so its F-potential is $|m|^2\sum_a|\phi^a|^2$.

Solution

Using $[t^1,t^2]=it^3$ gives

$$[\phi,\phi^\dagger]=2uw\,t^3,\qquad (\mu^1,\mu^2,\mu^3)=(0,0,2uw).$$

Consequently

$$V_D=\frac{g^2}{2}(2uw)^2=2g^2u^2w^2,\qquad V_F=|m|^2(u^2+w^2).$$

The matrix formula independently gives $g^2\operatorname{Tr}_2[(2uwt^3)^2]=2g^2u^2w^2$. D-flatness requires $uw=0$, but with nonzero $m$ the F equations additionally require $u=w=0$. Thus many D-flat configurations fail the full vacuum test. Omitting the factor two in the canonical adjoint Kähler potential while retaining canonical component formulas would give an inconsistent normalization.

From a potential to a vacuum question

Superspace has produced the Yukawa couplings, gauge couplings, and auxiliary terms together. Eliminating the auxiliaries then gave a potential whose zeros can be found by solving F and D equations and identifying gauge-related configurations.

These positivity and vacuum statements use the Lorentzian adjoint and positive kinetic metric. Euclidean continuation complexifies the variables and requires integration-contour choices, as explained in the multiplet lesson. The next lesson asks whether a candidate is a supersymmetric vacuum when those equations have no simultaneous solution. An extremum, a stable vacuum, and a runaway require different tests.

For a quantum-dynamics continuation, begin with holomorphy and the effective action. Remove a heavy chiral multiplet, derive its induced interaction, and compare what the superpotential determines with the information in the kinetic terms.

To place this kind of theory on a fixed curved geometry, continue with the four-dimensional new minimal rigid background. It starts from a conserved R-current and solves for the preserved supersymmetry parameters on a sphere times a circle. Fixed background auxiliaries and dynamical matter auxiliaries play different roles.