Four-dimensional chiral and vector multiplets

How can a scalar, a spinor, and a gauge field belong to supersymmetric theories when their component counts differ? The answer requires two distinctions: components before imposing equations are not particle states, and an auxiliary field can complete a transformation law without adding a particle.

We will construct two free theories and check their actions directly. The required preparation is four-dimensional Weyl spinors and the supersymmetry algebra. Keep the multiplet comparison nearby for the larger picture. This lesson supplies the transformations behind its four-dimensional $\mathcal N=1$ rows.

Our setting is flat four-dimensional Lorentzian spacetime, with all four real $\mathcal N=1$ Poincaré supercharges manifest. One theory is a gauge-neutral chiral multiplet; the other is a free $U(1)$ vector multiplet. They are independent examples, with canonical kinetic normalization, no interactions, and no background sources.

Components and conventions

Use $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and

$$\sigma^\mu=(I,-\tau_1,-\tau_2,-\tau_3),\qquad \bar\sigma^\mu=(I,+\tau_1,+\tau_2,+\tau_3), \qquad \sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu =2\eta^{\mu\nu}I.$$

Here $\tau_i$ are the Pauli matrices. Spinor indices are raised with $\epsilon^{12}=+1$ and lowered with $\epsilon_{12}=-1$, also for dotted indices. In particular,

$$\epsilon\psi=\epsilon^\alpha\psi_\alpha,\qquad \bar\epsilon\bar\psi=\bar\epsilon_{\dot\alpha}\bar\psi^{\dot\alpha}, \qquad \epsilon\sigma^\mu\bar\psi =\epsilon^\alpha\sigma^\mu_{\alpha\dot\alpha}\bar\psi^{\dot\alpha}.$$

The parameter $\epsilon_\alpha$ and the fermion fields are Grassmann odd. The transformation $\delta_\epsilon$, which includes its odd parameter, is even and obeys the ordinary product rule. Moving the parameter past a fermion still introduces a minus sign. Parameters are constant in spacetime.

Multiplet	Bosonic fields	Fermionic field	Auxiliary field
Chiral	Complex scalar $\phi$, dimension $1$	Weyl spinor $\psi_\alpha$, dimension $3/2$	Complex scalar $F$, dimension $2$
Vector	Real potential $A_\mu$, dimension $1$	Weyl gaugino $\lambda_\alpha$, dimension $3/2$	Real scalar $D$, dimension $2$

The parameters have dimension $-1/2$. A bar denotes Lorentzian conjugation: $\bar\phi=\phi^*$, $\bar F=F^*$, and $\bar\psi_{\dot\alpha}=(\psi_\alpha)^\dagger$. Conjugation reverses the order of odd factors. Thus $\bar\psi$ belongs to the same Weyl field, rather than being a second independent Lorentzian field.

The chiral transformation law

A scalar can transform into a spinor through $\delta\phi=\sqrt2\,\epsilon\psi$. To transform the spinor back into a derivative of the scalar requires $\sigma^\mu\bar\epsilon\,\partial_\mu\phi$. Dimension and Lorentz type also permit $\epsilon F$. The complete rules are

$$\begin{aligned} \delta\phi&=\sqrt2\,\epsilon\psi,\\ \delta\psi_\alpha &=\sqrt2\,\epsilon_\alpha F +i\sqrt2\,\sigma^\mu_{\alpha\dot\alpha} \bar\epsilon^{\dot\alpha}\partial_\mu\phi,\\ \delta F &=i\sqrt2\,\bar\epsilon_{\dot\alpha} \bar\sigma^{\mu\dot\alpha\alpha}\partial_\mu\psi_\alpha, \end{aligned}$$

Their conjugates are

$$\begin{aligned} \delta\bar\phi&=\sqrt2\,\bar\epsilon\bar\psi,\\ \delta\bar\psi^{\dot\alpha} &=\sqrt2\,\bar\epsilon^{\dot\alpha}\bar F +i\sqrt2\,\bar\sigma^{\mu\dot\alpha\alpha} \epsilon_\alpha\partial_\mu\bar\phi,\\ \delta\bar F&=i\sqrt2\,\epsilon\sigma^\mu\partial_\mu\bar\psi. \end{aligned}$$

These are Bilal’s chiral rules with $\phi=z$ and $F=-f$. For example, $\bar\epsilon\bar\sigma^\mu\partial_\mu\psi=-(\partial_\mu\psi)\sigma^\mu\bar\epsilon$, so the last unbarred rule has exactly the source’s sign after this translation. Bilal, equation (4.24)

One way to determine the derivative coefficient is to require two operations to give a translation. With $[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$, direct substitution gives

$$\delta_1\delta_2\phi =2\epsilon_2\epsilon_1 F +2i\epsilon_2\sigma^\mu\bar\epsilon_1\,\partial_\mu\phi.$$

The scalar contraction of two odd undotted spinors is symmetric. Consequently the $F$ terms cancel in the commutator, leaving

$$[\delta_1,\delta_2]\phi=a^\mu\partial_\mu\phi,\qquad a^\mu=2i\left( \epsilon_2\sigma^\mu\bar\epsilon_1- \epsilon_1\sigma^\mu\bar\epsilon_2\right).$$

For the spinor, inspect the mixed sector $\bar\epsilon_1=\epsilon_2=0$. Substituting both $\delta\phi$ and $\delta F$ gives

$$[\delta_1,\delta_2]\psi =2i\left[ \sigma^\mu\bar\epsilon_2(\epsilon_1\partial_\mu\psi) -\epsilon_1(\bar\epsilon_2\bar\sigma^\mu\partial_\mu\psi) \right] =-2i(\epsilon_1\sigma^\mu\bar\epsilon_2)\partial_\mu\psi.$$

The last equality is the two-component completeness identity, with the odd factors kept in their displayed order. The second term inside the brackets came from varying $F$; dropping it would spoil this rearrangement for an arbitrary fermion. Including the other mixed sector gives $a^\mu\partial_\mu\psi$. Substitution in $\delta F$ gives the same translation on $F$; terms with two identical chiralities cancel by the symmetric second derivatives and antisymmetric spinor tensors.

No field equation is used: this is off-shell closure. The auxiliary transformation is part of the representation, even though its action will have no kinetic term. Exercise 1 tests what fails if we prematurely set it to zero.

A complete free-action cancellation

Consider the density

$$\mathcal L_{\mathrm{ch},0} =\partial_\mu\bar\phi\,\partial^\mu\phi -i\psi\sigma^\mu\partial_\mu\bar\psi+\bar F F.$$

It agrees with the free part of Bilal’s equation (4.31). The explicitly Hermitian representative is

$$\mathcal L_{\mathrm{ch}} =\partial_\mu\bar\phi\,\partial^\mu\phi +\frac i2\left( \partial_\mu\psi\,\sigma^\mu\bar\psi -\psi\sigma^\mu\partial_\mu\bar\psi\right)+\bar F F =\mathcal L_{\mathrm{ch},0} +\frac i2\partial_\mu(\psi\sigma^\mu\bar\psi).$$

First keep the $\epsilon$ part of the variation, formally putting $\bar\epsilon=0$. This is a way to separate coefficients in a polynomial, not a Lorentzian reality condition. The three contributions are

$$\begin{aligned} \delta_\epsilon\mathcal L_{\rm scalar} &=\sqrt2\,\epsilon\,\partial_\mu\psi\,\partial^\mu\bar\phi,\\ \delta_\epsilon\mathcal L_{\rm fermion} &=-i\sqrt2 F\,\epsilon\sigma^\mu\partial_\mu\bar\psi +\sqrt2\,\epsilon\psi\,\Box\bar\phi,\\ \delta_\epsilon\mathcal L_{\rm auxiliary} &=+i\sqrt2 F\,\epsilon\sigma^\mu\partial_\mu\bar\psi. \end{aligned}$$

For the second term on the middle line, insert $\delta_\epsilon\bar\psi^{\dot\alpha} =i\sqrt2\bar\sigma^{\nu\dot\alpha\alpha}\epsilon_\alpha\partial_\nu\bar\phi$. The symmetric second derivatives select the symmetric part of $\sigma^\mu\bar\sigma^\nu$, namely $\eta^{\mu\nu}I$. The odd scalar contraction satisfies $\psi\epsilon=\epsilon\psi$. These two steps produce $+\sqrt2\epsilon\psi\Box\bar\phi$.

The two $F$ terms cancel, while the remaining terms combine into

$$\delta_\epsilon\mathcal L_{\mathrm{ch},0} =\sqrt2\,\partial_\mu \left(\epsilon\psi\,\partial^\mu\bar\phi\right).$$

Nothing has required $\Box\phi=0$, a fermion equation, or $F=0$. For completeness, define

$$\sigma^{\mu\nu}=\frac14 (\sigma^\mu\bar\sigma^\nu-\sigma^\nu\bar\sigma^\mu).$$

Including the boundary improvement that makes the density Hermitian gives

$$\begin{aligned} \delta\mathcal L_{\mathrm{ch}} &=\partial_\mu(j_{\mathrm{ch},\epsilon}^\mu+ (j_{\mathrm{ch},\epsilon}^\mu)^\dagger),\\ j_{\mathrm{ch},\epsilon}^\mu &=\frac{\sqrt2}{2}\epsilon\psi\,\partial^\mu\bar\phi +\sqrt2\,\epsilon\sigma^{\mu\nu}\psi\,\partial_\nu\bar\phi +\frac{i\sqrt2}{2}F\,\epsilon\sigma^\mu\bar\psi . \end{aligned}$$

Conjugation supplies the entire $\bar\epsilon$ sector because the density and the paired transformation rules are Hermitian. These are boundary terms in the variation, not a definition of the conserved supercurrent. With boundary conditions for which their flux vanishes, the action is invariant.

The vector multiplet and its gauge freedom

For the independent free $U(1)$ theory define

$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu,\qquad A_\mu\longmapsto A_\mu+\partial_\mu\alpha.$$

The two-index $F_{\mu\nu}$ is a field strength; the index-free $F$ above is a chiral auxiliary scalar. The gaugino and $D$ are gauge neutral in this Abelian theory. Define also $\bar\sigma^{\mu\nu}=(\bar\sigma^\mu\sigma^\nu-\bar\sigma^\nu\sigma^\mu)/4$. The rules are

$$\begin{aligned} \delta A_\mu &=i\epsilon\sigma_\mu\bar\lambda -i\lambda\sigma_\mu\bar\epsilon,\\ \delta\lambda_\alpha &=i\epsilon_\alpha D-(\sigma^{\mu\nu}\epsilon)_\alpha F_{\mu\nu},\\ \delta\bar\lambda^{\dot\alpha} &=-i\bar\epsilon^{\dot\alpha}D -(\bar\sigma^{\mu\nu}\bar\epsilon)^{\dot\alpha}F_{\mu\nu},\\ \delta D &=-\epsilon\sigma^\mu\partial_\mu\bar\lambda -(\partial_\mu\lambda)\sigma^\mu\bar\epsilon. \end{aligned}$$

Both orders of the antisymmetric pair $\mu,\nu$ are summed; there is no extra factor of one-half in $\delta\lambda$. The tensor $\sigma^{\mu\nu}$ converts the antisymmetric field strength into the spinor representation, while $D$ supplies the scalar part.

These component rules implement the vector multiplet encoded in Bilal’s equations (4.39) and (4.46). The source’s field-strength superfield has lowest component $-i\lambda_\alpha$, and its terms linear in the odd coordinate are $\theta_\alpha D+i(\sigma^{\mu\nu}\theta)_\alpha F_{\mu\nu}$. Replacing that coordinate by the variation parameter yields $\delta(-i\lambda)=\epsilon D+i\sigma^{\mu\nu}\epsilon F_{\mu\nu}$, fixing the displayed gaugino signs. Superspace will later package this calculation systematically; it is not needed for the component checks below.

The displayed field content uses Wess–Zumino gauge, which removes additional gauge components of the superfield. Ordinary gauge freedom $A\mapsto A+d\alpha$ remains. Supersymmetry includes the compensating transformation that keeps this gauge choice, so closure on the potential takes the form

$$\begin{aligned} [\delta_1,\delta_2]A_\mu &=a^\nu F_{\nu\mu}\\ &=a^\nu\partial_\nu A_\mu +\partial_\mu(-a^\nu A_\nu). \end{aligned}$$

Thus the commutator is a translation plus a gauge transformation with parameter $-a^\nu A_\nu$. Substituting the rules gives a plain translation on $\lambda$, $D$, and the gauge-invariant field strength. This closure is again off shell.

Checking the vector action

The canonical free density, with no theta term, is

$$\mathcal L_{\mathrm v,0} =-\frac14F_{\mu\nu}F^{\mu\nu} -i\lambda\sigma^\mu\partial_\mu\bar\lambda+\frac12D^2.$$

The relative coefficients agree with the real part of Bilal’s equation (4.50). As above, add $(i/2)\partial_\mu(\lambda\sigma^\mu\bar\lambda)$ to obtain the Hermitian density $\mathcal L_{\mathrm v}$.

We can check every term without a gauge condition on $A_\mu$. Define $\widetilde F^{\mu\nu}=\frac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$, with $\epsilon^{0123}=+1$. The needed matrix identity is

$$\sigma^{\rho\sigma}\sigma^\mu =\frac12\left( \eta^{\sigma\mu}\sigma^\rho-\eta^{\rho\mu}\sigma^\sigma -i\epsilon^{\rho\sigma\mu\kappa}\sigma_\kappa\right).$$

It follows by multiplying the displayed Pauli matrices. Combining it with the raising tensors, the $\epsilon$ variation separates as

$$\begin{aligned} \delta_\epsilon\mathcal L_{\rm Maxwell} &=-iF^{\mu\nu}\epsilon\sigma_\nu\partial_\mu\bar\lambda,\\ \delta_\epsilon\mathcal L_{\rm gaugino} &=+iF^{\mu\nu}\epsilon\sigma_\nu\partial_\mu\bar\lambda -\widetilde F^{\mu\nu}\epsilon\sigma_\nu\partial_\mu\bar\lambda +D\epsilon\sigma^\mu\partial_\mu\bar\lambda,\\ \delta_\epsilon\mathcal L_D &=-D\epsilon\sigma^\mu\partial_\mu\bar\lambda. \end{aligned}$$

The Maxwell and $D$ terms cancel their partners. Since $F=dA$, commuting partial derivatives gives the Bianchi identity $\partial_\mu\widetilde F^{\mu\nu}=0$. It is an identity, not Maxwell’s equation. Therefore

$$\delta_\epsilon\mathcal L_{\mathrm v,0} =-\partial_\mu (\widetilde F^{\mu\nu}\epsilon\sigma_\nu\bar\lambda).$$

For the Hermitian density the full result is

$$\begin{aligned} \delta\mathcal L_{\mathrm v} &=\partial_\mu(j_{\mathrm v,\epsilon}^\mu+ (j_{\mathrm v,\epsilon}^\mu)^\dagger),\\ j_{\mathrm v,\epsilon}^\mu &=-\frac12\left[ (\widetilde F^{\mu\nu}+iF^{\mu\nu}) \epsilon\sigma_\nu\bar\lambda +D\epsilon\sigma^\mu\bar\lambda\right]. \end{aligned}$$

This establishes the free action’s invariance with the same boundary qualification as for the chiral theory.

What actually propagates?

The auxiliary equations are algebraic: varying $\bar F$ gives $F=0$, and varying $D$ gives $D=0$. These fields have no wave equation or independent initial data. Eliminating them is legitimate when studying the free solutions; keeping them is what made the preceding symmetry calculations independent of equations of motion.

Off shell, a complex $\phi$ and complex $F$ have $2+2=4$ real components, matching a Weyl field’s four real Grassmann components. For the vector, four potential components modulo one gauge function, together with one real $D$, give $4-1+1=4$, again matching the gaugino. This is local component bookkeeping before dynamical constraints, not a polarization count.

For a concrete physical count take a massless wave with $p^\mu=(E,0,0,E)$, $E>0$. The Weyl equation gives

$$\bar\sigma^\mu p_\mu\psi =E(I-\tau_3)\psi=0, \qquad \psi=\begin{pmatrix}c\\0\end{pmatrix}.$$

One spinor amplitude remains for this frequency sector; the conjugate field supplies the antiparticle sector, not another independent Weyl field. Together there are two fermionic one-particle states. A complex scalar likewise has particle and antiparticle states, giving two bosonic states.

For a vector polarization $a^\mu$, the massless Maxwell equation requires $p_\mu a^\mu=E(a^0-a^3)=0$. The gauge shift $a^\mu\mapsto a^\mu+c\,p^\mu$ removes the common longitudinal/time component. Only $a^1$ and $a^2$ remain: two photon polarizations. The gaugino has two fermionic states as well. Both multiplets therefore have $2+2$ propagating states, although each had $4+4$ off-shell components. Bilal, section 3.3

Continuing to Euclidean signature

The explicit spinor continuation uses $x^0=-ix^4$ and treats dotted and undotted Euclidean spinors independently. Apply that continuation to the complete Lorentzian formulas. Do not impose $\bar\psi=\psi^\dagger$ as a relation preserved by all Euclidean rotations, or import a three-dimensional transformation law.

Reality of an action and the choice of an integration contour are additional questions. For example, $S_E=-iS_L$ after the time rotation sends the Lorentzian auxiliary terms to

$$\mathcal L_{E,\rm auxiliary}=-\bar F F-\frac12D^2.$$

The analytic variables $F,\bar F,D$ can be complexified. For a convergent free auxiliary Gaussian one may choose $\bar F=-F^*$ and $D=iD_E$, with $D_E$ real. This gives $|F|^2+D_E^2/2$. This auxiliary contour is separate from the physical scalar contour $\bar\phi=\phi^*$, which gives a positive scalar kinetic term. Exercise 3 checks the Gaussian explicitly. A chosen Euclidean contour should never be used to redefine the Lorentzian adjoint or double its physical-state count.

Exercises

1. Can the auxiliary field be discarded before varying?

Set $F=\bar F=0$, but leave the fermion arbitrary. Is this restricted field space preserved by every supersymmetry transformation? Test $\psi_1(t)=\chi t$, $\psi_2(t)=0$, where $\chi$ is a constant odd number, in a neighborhood of a point. What extra condition restores preservation of $F=0$?

Hint 1

An algebraic equation such as $F=0$ defines an invariant restriction only if its variation also vanishes there.

Hint 2

Use $\bar\sigma^0=I$ in $\delta F$ and keep $\bar\epsilon_{\dot1}$ arbitrary.

Solution

For this test field,

$$\delta F=i\sqrt2\,\bar\epsilon_{\dot1}\chi,$$

which need not vanish. The restriction is therefore not invariant on arbitrary configurations. For all $\bar\epsilon$, preservation requires $\bar\sigma^\mu\partial_\mu\psi=0$, precisely the free Weyl equation; conjugation gives the partner condition. One may eliminate the auxiliaries and use the resulting on-shell representation, but then equations of motion enter closure. The example explains why an auxiliary can be necessary for an off-shell calculation while carrying no physical state.

2. A magnetic field and a false vacuum conclusion

At a point take $\lambda=0$, $F_{12}=B$ with all other independent components zero, and real constant $D=d$. Find when a nonzero Lorentzian supersymmetry parameter obeys $\delta\lambda=\delta\bar\lambda=0$. Does this by itself establish a supersymmetric vacuum of the free theory?

Hint 1

Calculate $\sigma^{12}$ from the Pauli matrices. Both $F_{12}$ and $F_{21}$ contribute to the summed transformation.

Hint 2

Diagonalize the resulting two-by-two matrix, then separately impose the auxiliary field equation of the free action.

Solution

The matrices give $\sigma^{12}=-i\tau_3/2$, hence

$$\delta\lambda=i(dI+B\tau_3)\epsilon =i\begin{pmatrix}d+B&0\\0&d-B\end{pmatrix}\epsilon.$$

For $B\ne0$, a nonzero kernel exists when $d=-B$ or $d=B$. In the first case $\epsilon_2=0$; in the second $\epsilon_1=0$. Each leaves one complex parameter, or two real Lorentzian supersymmetry parameters. The lower dotted equation is its conjugate and imposes the conjugate restriction, not a second independent choice. With constant field strength and $D$, and an identically vanishing gaugino profile, $\delta A$ and $\delta D$ vanish too.

This is an off-shell algebraic preservation test. In the free theory the auxiliary equation requires $d=0$, incompatible with either nonzero-$B$ kernel. For $d=0$ and $B\ne0$ the gaugino-variation matrix is invertible. Moreover a spatially uniform magnetic field on infinite flat space is not the zero-energy vacuum. A transformation annihilating a chosen off-shell configuration does not establish a physical supersymmetric vacuum. For $B=d=0$, all four real parameters survive, as expected.

3. A convergent auxiliary Gaussian

For a single regulated Euclidean cell, let $t>0$ multiply $-\bar F F-D^2/2$. Compare the naive contours $\bar F=F^*$, $D\in\mathbb R$ with $F=x+iy$, $\bar F=-x+iy$, $D=id$, where $x,y,d$ are real. Evaluate the two Gaussian integrals on the latter contours using the real-coordinate measures $dx\,dy$ and $dd$.

Hint 1

The weight is $e^{-S_E}$, so a negative quadratic Euclidean action grows along a real contour.

Hint 2

On the second contours the quadratic forms are $x^2+y^2$ and $d^2/2$. Use the ordinary real Gaussian separately in each variable.

Solution

The naive contours give weights $e^{+t|F|^2}$ and $e^{+tD^2/2}$ and hence divergent integrals. On the specified complex contours,

$$-\bar F F=x^2+y^2,\qquad -D^2/2=d^2/2,$$

so

$$\int_{\mathbb R^2}dx\,dy\,e^{-t(x^2+y^2)} =\frac{\pi}{t},\qquad \int_{\mathbb R}dd\,e^{-td^2/2} =\sqrt{\frac{2\pi}{t}}.$$

These formulas use the real-coordinate measures stated in the question. A measure written in the original complex variables can carry a constant contour-orientation phase, which must be fixed along with its normalization. Convergence fixes neither that phase nor a conjugation law for Euclidean spinors. The exercise is a free Gaussian check, not a claim that every supersymmetry transformation preserves this contour as a real subspace.

What the two examples establish

We now have explicit realizations of the four-dimensional algebra on matter and gauge fields. Their free actions are invariant before using equations of motion, while their physical spectra contain equal bosonic and fermionic states after eliminating auxiliaries and imposing dynamics. The multiplet comparison uses exactly this distinction when organizing extended supersymmetry.

Continue with Superspace, interactions, and supersymmetric vacua to couple these multiplets, recover the component interactions, and solve the resulting vacuum equations. Bilal’s sections 4.2–5.2 supply the source formulas; the translation to keep in mind is $\phi=z$, $F=-f$, and $A_\mu=v_\mu$ in the Abelian formulas used here.