Supersymmetric QCD and its classical vacua

Quark expectation values carry gauge indices. They are useful for solving equations, but they cannot by themselves label distinct physical vacua: a gauge transformation changes their components without changing the state. Can mesons and baryons give a complete set of coordinates, and can every proposed coordinate value actually come from a zero-energy field configuration?

We will answer both questions for two-color SQCD with two flavor pairs. The preparation is chiral operators and vacuum relations and anomalies. We use their four-dimensional $\mathcal N=1$ setting: flat Lorentzian spacetime, four real supercharges, all manifest in $\mathcal N=1$ superspace. The vacuum calculation is classical throughout.

Define the theory and its indices

Supersymmetric QCD, abbreviated SQCD, has gauge group $SU(N_c)$, one vector multiplet, and $N_f$ pairs of chiral multiplets

$$Q_i^a,\qquad \widetilde Q^{\,j}_a, \qquad a=1,\ldots,N_c,\quad i,j=1,\ldots,N_f.$$

$N_c$ counts colors; the gauge rank is $N_c-1$, and there are $N_c^2-1$ gauge generators. Each flavor pair contains a fundamental and an antifundamental. Their two Weyl fermions can supply the two chiralities of a Dirac quark, but the theory also contains their scalar partners and an adjoint Weyl gaugino. There is no adjoint chiral multiplet and no extra supersymmetry being assumed.

Take massless matter, $W_{\rm tree}=0$, and canonical positive matter kinetic terms. We use $Q$ for the $N_c$-by-$N_f$ matrix of fundamental scalar components and $\widetilde Q$ for the $N_f$-by-$N_c$ matrix of antifundamental scalar components. The tilde means a separate chiral field, not complex conjugation. A gauge transformation acts as

$$Q\longmapsto UQ,\qquad \widetilde Q\longmapsto\widetilde Q U^{-1}.$$

With Hermitian generators $\operatorname{Tr}_{N_c}(t^At^B)=\delta^{AB}/2$, keep the canonical vector convention $D_\mu Q=\partial_\mu Q-igA_\mu^At^AQ$. The auxiliary terms and their elimination give

$$\mathcal L_D=\frac12D^AD^A+gD^A\mu^A,\qquad \mu^A=\operatorname{Tr}(t^AH),\qquad H=QQ^\dagger-\widetilde Q^\dagger\widetilde Q,$$ $$D^A=-g\mu^A,\qquad V_D=\frac{g^2}{2}\sum_A(\mu^A)^2 =\frac{g^2}{4}\operatorname{Tr} \left(H-\frac{\operatorname{Tr}H}{N_c}\mathbf1\right)^2.$$

The last equality follows by expanding a traceless Hermitian matrix in the normalized generators. All $F$ auxiliaries vanish because $W_{\rm tree}=0$. Thus homogeneous classical supersymmetric vacua satisfy

$$H=\frac{\operatorname{Tr}H}{N_c}\mathbf1, \qquad\text{modulo }SU(N_c).$$

Do not set $\operatorname{Tr}H=0$: there is no gauged $U(1)$ generator to impose that additional condition. This distinction will allow a vacuum with a baryon but no meson. Tong, (4.36)

Which global symmetries survive?

For general $N_c\geq3$, a convenient continuous symmetry algebra is $SU(N_f)_L\times SU(N_f)_R\times U(1)_B\times U(1)_R$. Choose the right flavor convention in which $\widetilde Q$ is an antifundamental of $SU(N_f)_R$. The scalar charges are

Multiplet	Gauge representation	Left flavor	Right flavor	Baryon charge	R-charge
$Q$	$\mathbf{N_c}$	$\mathbf{N_f}$	$\mathbf1$	$+1$	$r$
$\widetilde Q$	$\overline{\mathbf{N_c}}$	$\mathbf1$	$\overline{\mathbf{N_f}}$	$-1$	$r$
Vector	Adjoint	$\mathbf1$	$\mathbf1$	$0$	Gaugino: $+1$

Here $N_f>0$. Why choose $r=1-N_c/N_f$? The mixed R–gauge–gauge anomaly counts left-handed fermions, whose matter R-charges are $r-1$. With $T(\mathbf{N_c})=1/2$ and $T(\mathrm{adj})=N_c$,

$$k_{SU(N_c)^2R} =N_c+N_f\frac12(r-1)+N_f\frac12(r-1) =N_c+N_f(r-1).$$

Setting this coefficient to zero gives the stated $r$. The ordinary baryon current also has zero mixed gauge anomaly, since its two matter contributions cancel. An axial rotation giving both chirals charge $+1$ instead has coefficient $N_f$ and is anomalously broken as a continuous symmetry of the quantum theory. The fundamental and antifundamental contributions to the local cubic gauge anomaly cancel.

This R assignment removes a dynamical-gauge obstruction; it need not remove global background anomalies. It can mix with baryon number, giving scalar charges $r+s,r-s$, though we use the symmetric choice $s=0$. These are symmetry charges, not a claim about scaling dimensions at an infrared fixed point. We discuss the continuous algebra here; finite identifications with the gauge center and discrete symmetries require additional global data. Intriligator–Seiberg, (4.1)–(4.2)

For $N_c=2$ the fundamental is pseudoreal: its conjugate representation can be converted to the fundamental with an invariant antisymmetric matrix. All $2N_f$ chirals therefore admit an enhanced $SU(2N_f)$ flavor rotation. In our $N_f=2$ example this is $SU(4)$, not merely the displayed left/right subgroup. Baryon number embeds into it. We will exhibit that conversion explicitly. Tong, §6.2.1

Mesons and baryons are composite coordinates

A meson contracts a fundamental with an antifundamental:

$$M^{\,j}{}_i=\widetilde Q^{\,j}_a Q_i^a, \qquad M=\widetilde Q Q.$$

A baryon contracts $N_c$ fundamentals with an alternating color tensor,

$$B_{i_1\cdots i_{N_c}} =\varepsilon_{a_1\cdots a_{N_c}} Q_{i_1}^{a_1}\cdots Q_{i_{N_c}}^{a_{N_c}},$$

and $\widetilde B$ is built from $N_c$ antifundamentals. These are antisymmetric in flavor indices, so baryons vanish identically if $N_f<N_c$. We choose the color orientation so that, when $N_f=N_c$, $B=\det Q$ and $\widetilde B=\det\widetilde Q$.

Their ultraviolet engineering dimensions are $[M]=2$ and $[B]=[\widetilde B]=N_c$; their baryon charges are $0,N_c,-N_c$, and their scalar R-charges are $2r,N_cr,N_cr$. Naming a composite “meson” does not assert that it is an elementary particle with a canonical kinetic term.

Now set $N_c=N_f=2$. There are four entries of $M$ and two baryons. Matrix multiplication gives

$$\det M=\det(\widetilde Q Q)=B\widetilde B.$$

The determinant fixes the sign in our orientation. This is a classical polynomial identity, not an equation obtained by varying a superpotential. It is not yet enough: we must show that it describes the D-flat quotient.

Turn four doublets into six minors

Introduce the color matrix

$$J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad u_1=q_1,\quad u_2=q_2,\quad u_3=-J\widetilde q^{\,1T},\quad u_4=-J\widetilde q^{\,2T},$$

where $q_i$ is a column of $Q$ and $\widetilde q^{\,j}$ a row of $\widetilde Q$. This $J$ concerns color and is unrelated to the Lorentz spinor epsilon. For any $g\in SL(2,\mathbb C)$, $g^TJg=J$, so all four columns transform as $u_I\mapsto gu_I$. For unitary $g$ their norms are also preserved. Hence the canonical action permits unitary flavor rotations of the four columns; the nonanomalous non-Abelian part is the announced $SU(4)$.

Put the columns into a $2$-by-$4$ matrix $\mathcal U$ and form its six minors

$$p_{IJ}=u_I^TJu_J=-p_{JI}.$$

Direct contraction gives

$$p_{12}=B,\quad p_{34}=\widetilde B,\quad p_{13}=M^1{}_1,\quad p_{23}=M^1{}_2,\quad p_{14}=M^2{}_1,\quad p_{24}=M^2{}_2.$$

Holomorphic $SU(2)$ singlets contract doublet color indices with $J$, so their polynomial contractions are products of these minors. Only two color directions exist. Expanding the two-by-two determinants gives the Plücker relation

$$p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0,$$

which is exactly $B\widetilde B-\det M=0$.

To see that there is no extra relation away from the origin, choose a patch $p_{12}\ne0$. With any fixed nonzero reference $v$ of mass dimension one, take

$$\mathcal U_0= \begin{pmatrix} v&0&-vp_{23}/p_{12}&-vp_{24}/p_{12}\\ 0&p_{12}/v&p_{13}/v&p_{14}/v \end{pmatrix}.$$

Five minors have the specified values; the sixth is $p_{34}=(p_{13}p_{24}-p_{14}p_{23})/p_{12}$. An invertible pair of columns also fixes the complex gauge transformation uniquely. Changing $v$ is a determinant-one gauge transformation, not another modulus. Every nonzero solution has some nonzero minor and an analogous patch. The all-zero solution is represented by $\mathcal U=0$.

Solve the real D equation

A complex gauge transformation is a tool for finding representatives; it is not an extra physical gauge redundancy on arbitrary scalar configurations. We now find where its orbit meets the real D-flat equations.

Set $S=\mathcal U\mathcal U^\dagger$. The identity

$$(-J\widetilde q^T)(-J\widetilde q^T)^\dagger =|\widetilde q|^2\mathbf1-\widetilde q^\dagger\widetilde q$$

shows that $S$ and $H$ have the same traceless part. D-flatness is therefore $S=c\mathbf1$, $c\geq0$.

For rank-two $\mathcal U$, $S$ is positive definite. Define

$$g=(\det S)^{1/4}S^{-1/2},\qquad \mathcal U_D=g\mathcal U.$$

$S^{-1/2}$ is the positive inverse square root, found by diagonalizing the Hermitian matrix $S$. Its determinant makes $\det g=1$, while

$$\mathcal U_D\mathcal U_D^\dagger =\sqrt{\det S}\,\mathbf1.$$

Thus every reconstructed nonzero point has a D-flat representative, with unchanged minors. The dimensions also work: $S$ has dimension two, so $g$ is dimensionless.

Uniqueness is equally concrete. If two D-flat representatives with the same minors are related by $h\in SL(2,\mathbb C)$, their constants $c$ agree because $\det S$ is unchanged. Then $h(c\mathbf1)h^\dagger=c\mathbf1$, so $h$ is unitary and lies in $SU(2)$. There is precisely one physical gauge orbit.

For example, take real $v>0$ and

$$\mathcal U=v\begin{pmatrix}2&0&1&1\\0&1&1&-1\end{pmatrix}.$$

Then $S=v^2\operatorname{diag}(6,3)$ and $g=\operatorname{diag}(2^{-1/4},2^{1/4})$. The resulting $S_D=3\sqrt2\,v^2\mathbf1$ is D-flat, with

$$B=2v^2,\quad \widetilde B=-2v^2,\qquad M=v^2\begin{pmatrix}2&-1\\-2&-1\end{pmatrix}.$$

Both sides of $\det M=B\widetilde B$ equal $-4v^4$. This checks a configuration with nonzero mesons and both baryons.

There is one essential exception to “divide by the complex gauge group.” A nonzero rank-one $\mathcal U$ has all minors zero but cannot be D-flat: a nonzero rank-one $S$ cannot equal $c\mathbf1$. After aligning its columns along the first color direction, $\operatorname{diag}(t,t^{-1})$ sends it to zero as $t\to0$. Its complex orbit is not closed; zero lies in its closure. The invariant quotient identifies such orbits with that closed zero orbit. It does not create an extra vacuum for each rank-one configuration. This is the meaning of the double slash in

$$\mathcal M_{\rm cl} =\{\mu^A=0\}/SU(2) \simeq\mathbb C^8/\!/SL(2,\mathbb C).$$

The general correspondence is explained in Intriligator–Seiberg, §3; here the balancing construction and rank-one limit display it explicitly.

Count dimensions and inspect the origin

At a rank-two point the gauge stabilizer is trivial. Eight complex scalar components lose three real D constraints and three real gauge directions, leaving ten real, or five complex, moduli. The six invariant coordinates with one relation give the same answer:

$$\mathbb C[\mathcal M_{\rm cl}] =\frac{\mathbb C[M^1{}_1,M^1{}_2,M^2{}_1,M^2{}_2,B,\widetilde B]} {(\det M-B\widetilde B)}.$$

This is the classical matter invariant ring, not a claim to list all gauge-field chiral operators. We have neither discarded an F-term nilpotent nor guessed a quantum relation: $W_{\rm tree}=0$ supplies no F equation, and the displayed relation comes from color contractions.

The gradient of $\det M-B\widetilde B$ vanishes only when all six coordinates vanish. Thus the variety is smooth of complex dimension five away from the origin and singular there. Its linearized tangent space at the origin is six-dimensional because the constraint starts quadratically.

Physically, every nonzero D-flat point here completely Higgses $SU(2)$. At the origin the gauge group is unbroken and the vector multiplet is classically massless. A description containing only five local moduli coordinates cannot provide a smooth low-energy description through that point. Flavor rotations can relate different vacua; unlike gauge transformations, they are not divided out.

These statements concern classical equations and classical masses. A weakly coupled interpretation of the Higgsed spectrum requires a sufficiently large Higgs scale compared with the strong scale. In particular, the classical singularity does not establish the quantum behavior at the origin. The relation we have derived is the starting point for a later quantum calculation, not its answer.

Practice

1. A baryonic vacuum and the missing trace equation

Take $Q=v\mathbf1_2$, $\widetilde Q=0$, with $v\ne0$. Determine its invariants and D terms. Compare with a $U(2)$ theory at zero FI parameter, normalizing its additional generator as $t^0=\mathbf1_2/2$. Why is the distinction physical?

Hint 1

$H$ is proportional to the identity. Only traceless generators occur in $SU(2)$.

Hint 2

For $U(2)$ evaluate $\operatorname{Tr}(t^0H)$ separately. Also ask how $\det Q$ transforms under a gauge matrix with determinant different from one.

Solution

$H=|v|^2\mathbf1_2$ gives $\mu^A=0$ for all three $SU(2)$ generators, so $V_D=0$. The invariants are

$$M=0,\qquad B=v^2,\qquad \widetilde B=0.$$

They obey the relation even though the meson vanishes. Since $Q$ is invertible, $UQ=Q$ forces $U=\mathbf1$: no gauge generator remains unbroken.

In $U(2)$, $\mu^0=|v|^2$ is nonzero. With the same canonical coupling for that generator its contribution is $g^2|v|^4/2$, so this configuration is not D-flat. Moreover $B$ transforms as $B\mapsto(\det U)B$, and is not a $U(2)$ gauge invariant. Adding one gauged generator changes both the equations and the allowed coordinates; it is not a harmless change of notation.

2. A mesonic patch without a baryon

Consider $M=\operatorname{diag}(v^2,0)$, $B=\widetilde B=0$, with real $v>0$. Find a D-flat representative and its gauge stabilizer. Show that the surrounding patch $p_{13}=M^1{}_1\ne0$ still has five complex coordinates.

Hint 1

Try $Q=\widetilde Q=\operatorname{diag}(v,0)$. Convert its nonzero rows and columns into $\mathcal U$.

Hint 2

The two nonzero columns of $\mathcal U$ are independent. In the Plücker relation solve for $p_{24}$, not for $p_{34}$.

Solution

For the suggested representative, $H=0$ and $\mathcal U=(ve_1,0,ve_2,0)$, where $e_1,e_2$ are the two unit color columns. Hence $S=v^2\mathbf1$ and the configuration is D-flat. A gauge matrix preserving both independent columns is the identity. The rank-one meson does not mean rank-one $\mathcal U$ or an unbroken $U(1)$ gauge group.

On this patch the relation becomes

$$p_{24}=\frac{p_{12}p_{34}+p_{14}p_{23}}{p_{13}}.$$

Thus $p_{13}\ne0$ and the four unrestricted coordinates $p_{12},p_{14},p_{23},p_{34}$ describe a five-complex-dimensional neighborhood. The baryons can vary away from zero in this neighborhood. This point is smooth even though the particular $p_{12}\ne0$ chart used earlier does not cover it.

3. R-charge zero does not mean dimension zero

For $SU(2)$ with two flavor pairs, compute the scalar and matter-fermion charges under the symmetric gauge-anomaly-free R symmetry, and check its mixed gauge anomaly. What are the engineering dimensions and R-charges of the six invariants? If a fixed mass term $W=mM^1{}_1$ is added, does it preserve that same R symmetry?

Hint 1

There are four doublet Weyl fermions and an adjoint gaugino. Use $T(\mathbf2)=1/2$ and $T(\mathrm{adj})=2$.

Hint 2

A superpotential must have R-charge two, while the fermion R-charge is one less than its scalar’s. A formally charged mass parameter is different from a fixed neutral coupling.

Solution

$r=1-2/2=0$ for all four chiral scalars, so their Weyl fermions have charge $-1$. The gaugino has charge $+1$, giving

$$k_{SU(2)^2R}=2(1)+4\left(\frac12\right)(-1)=0.$$

All six quadratic invariants have engineering dimension two and scalar R-charge zero. There is no contradiction: the relation between chiral scaling dimension and R-charge requires a superconformal R generator at a fixed point. We have only selected a gauge-anomaly-free symmetry of this theory.

With a fixed nonzero neutral $m$, the operator $mM^1{}_1$ has R-charge zero, so its superpotential integral breaks the chosen R symmetry. Assigning a spurion charge $R(m)=2$ supplies a formal selection rule across different parameter values; it does not make that fixed neutral mass invariant. The supersymmetric mass interaction itself remains an allowed $\mathcal N=1$ interaction.

What the classical calculation leaves open

You should now be able to start from any nonzero six-tuple satisfying $\det M=B\widetilde B$, reconstruct a quark configuration, and balance it into a unique D-flat gauge orbit. You should also be able to explain why a nonzero configuration with all six invariants zero is not a new vacuum.

The next question is dynamical: when a flavor becomes heavy, how does the holomorphic strong scale change, and can quantum effects alter the classical relation? The classical quotient supplies the coordinates and symmetry charges needed to ask that question precisely.