Chiral operators, Ward identities, and vacuum relations

For one chiral field with superpotential $W=gX^3/3$, the classical supersymmetric vacuum equation is $x^2=0$. As an equation for a complex number, this means $x=0$. As a relation among operators, it need not mean that the class of $X$ vanishes. Understanding that difference is our first task.

The preparation is Holomorphy and the quantum effective action. We use flat four-dimensional Lorentzian $\mathcal N=1$ supersymmetry, with four real supercharges and no gravitational background. The worked theories have gauge-neutral chiral multiplets, canonical positive Kähler metrics, and the component convention $\mathcal L_W=W_iF_i-W_{ij}\psi_i\psi_j/2+\text{h.c.}$ All explicit polynomial quotients and vacuum potentials below are classical. The operator argument explains what their quantum counterparts must respect; it does not assert that classical relations always survive quantization.

Closed operators and exact changes

For operators of definite fermion parity, define the graded bracket

$$[A,B\}=AB-(-1)^{|A||B|}BA.$$

It is a commutator unless both entries are odd, in which case it is an anticommutator. The supercharges are odd. A local operator $\mathcal O$ is chiral when

$$[\bar Q_{\dot\alpha},\mathcal O\}=0,\qquad \dot\alpha=\dot1,\dot2.$$

The lowest component of a chiral superfield has this property. We will concentrate on bosonic Lorentz-scalar operators; in a gauge theory they must also be gauge invariant.

An exact change is a sum of barred-supercharge variations,

$$\mathcal O\ \sim\ \mathcal O+ \sum_{\dot\alpha}[\bar Q_{\dot\alpha},Y^{\dot\alpha}\},$$

where both representatives are chiral. For bosonic $\mathcal O$, the ancestors $Y^{\dot\alpha}$ are odd. The equivalence classes form the chiral ring. “Ring” means that we can add classes and multiply them, with an identity class $[1]$; it does not refer to a geometric circle.

Why does multiplication respect the equivalence? Write $\bar\delta A=[\bar Q,A\}$ for either barred charge. Its graded product rule is

$$\bar\delta(AB)=(\bar\delta A)B+(-1)^{|A|}A(\bar\delta B).$$

If $B$ is closed and $A=Y$ is odd, then $(\bar\delta Y)B=\bar\delta(YB)$. Thus multiplying an exact change by a closed operator gives another exact change. The scalar examples below have a commutative ring. Their products are understood as appropriately defined local chiral composites, not as an instruction to multiply arbitrary singular distributions at the same point. Cachazo–Douglas–Seiberg–Witten, §2.1

A translation is exact in this sector

Recall the site’s normalization,

$$\{Q_\alpha,\bar Q_{\dot\beta}\} =2\sigma^\mu_{\alpha\dot\beta}P_\mu,\qquad [P_\mu,\mathcal O]=-i\partial_\mu\mathcal O.$$

Here $\sigma^\mu=(I,-\tau_i)$, $\bar\sigma^\mu=(I,+\tau_i)$ and $\eta=\operatorname{diag}(1,-1,-1,-1)$. The $\tau_i$ are Pauli matrices. For a bosonic chiral $\mathcal O$, the graded Jacobi identity gives

$$\begin{aligned} 2\sigma^\mu_{\alpha\dot\beta}[P_\mu,\mathcal O] &=\{Q_\alpha,[\bar Q_{\dot\beta},\mathcal O]\} +\{\bar Q_{\dot\beta},[Q_\alpha,\mathcal O]\}\\ &=\{\bar Q_{\dot\beta},[Q_\alpha,\mathcal O]\}. \end{aligned}$$

Contract with $\bar\sigma^{\nu\dot\beta\alpha}$. Since $\bar\sigma^{\nu\dot\beta\alpha}\sigma^\mu_{\alpha\dot\beta}=2\eta^{\nu\mu}$, we obtain

$$\boxed{\displaystyle \partial^\nu\mathcal O =\frac{i}{4}\bar\sigma^{\nu\dot\beta\alpha} \{\bar Q_{\dot\beta},[Q_\alpha,\mathcal O]\}.}$$

The ancestor $[Q_\alpha,\mathcal O]$ is odd, which explains the outer anticommutator. The factor $i/4$ follows from our translation and charge normalization; it must not be copied from a source that suppresses those factors.

Now choose a supersymmetric vacuum $|\Omega\rangle$, annihilated by $Q_\alpha$ and $\bar Q_{\dot\alpha}$. Their adjoint relation gives the corresponding annihilation of the bra. Assume the conserved supersymmetry charges and operator products are well defined, with no boundary contribution spoiling the Ward identity.

Let $B$ be a product of bosonic chiral operators at distinct points. For any odd $Y$ inserted separately,

$$0=\langle\bar\delta(YB)\rangle_\Omega =\langle(\bar\delta Y)B\rangle_\Omega -\langle Y\,\bar\delta B\rangle_\Omega =\langle(\bar\delta Y)B\rangle_\Omega.$$

The first equality follows by moving the charge to the bra or ket. The last uses chirality of every factor in $B$. Applying this identity to the boxed ancestor proves

$$\partial_{x_1^\mu} \langle\mathcal O_1(x_1)\cdots\mathcal O_n(x_n)\rangle_\Omega=0$$

away from coincident insertions. This is a Ward identity: a relation between correlation functions implied by a symmetry. It does not say that $\partial_\mu\mathcal O$ vanishes as an ordinary local operator in every correlation function. CDSW, (2.1)–(2.2)

If this chosen infinite-volume vacuum also satisfies cluster decomposition, correlations of widely separated groups approach the product of their expectation values. Move the separated chiral insertions arbitrarily far apart. Their correlator is unchanged, so

$$\langle\mathcal O_1\cdots\mathcal O_n\rangle_\Omega =\prod_{k=1}^n\langle\mathcal O_k\rangle_\Omega.$$

Thus the expectation-value map $\chi_\Omega([\mathcal O])=\langle\mathcal O\rangle_\Omega$ preserves multiplication and sends $[1]$ to $1$. Such a map to $\mathbb C$ is called a character of the ring. Factorization uses both the Ward identity and clustering; averaging different vacua can destroy the latter. Nor does this argument establish a normalizable vacuum when none exists. CDSW, (2.3)

From an auxiliary equation to a polynomial quotient

For neutral chirals with canonical Kähler potential, the classical superfield equation is

$$-\frac14\bar{\mathbb D}^{\,2}\bar\Phi_i+W_i(\Phi)=0, \qquad \bar{\mathbb D}^{\,2} =\bar{\mathbb D}_{\dot\alpha}\bar{\mathbb D}^{\dot\alpha}.$$

The derivatives are those defined in the action lesson. At $\theta=\bar\theta=0$, $\bar{\mathbb D}^{\,2}\bar\Phi_i=-4\bar F_i$, so the equation gives $\bar F_i=-W_i$, checking its sign against the component action.

The barred superspace derivative on this lowest component is a barred supersymmetry variation, up to its fixed normalization. The twice-differentiated expression is therefore exact. Equivalently, the anti-fermion’s barred variation contains $\bar F_i$, and its auxiliary equation identifies that exact component with $-W_i$. Consequently,

$$[W_i]=0$$

in the classical chiral ring. This is the content of the F-term relation, not a claim that the polynomial $W_i$ is identically zero before passing to equivalence classes. CDSW, §2.2, (2.7)

Restricting to scalar polynomials in these neutral fields gives

$$\mathcal R_{\rm cl} =\frac{\mathbb C[x_1,\ldots,x_n]}{(W_1,\ldots,W_n)}.$$

The denominator is an ideal: all sums $\sum_i f_i(x)W_i(x)$ with polynomial coefficients. Two polynomials represent the same class if their difference belongs to this ideal. Writing $x_i$ for a ring generator is convenient, but it does not yet assign that operator a numerical expectation value.

For our first model,

$$K=\bar X X,\qquad W=\frac g3X^3,\qquad g\ne0,$$

$[X]=1$ and $[g]=0$ in engineering mass dimension. The relation is $g x^2=0$, so

$$\mathcal R_{\rm cl}=\mathbb C[x]/(x^2).$$

Every polynomial reduces uniquely to $a+bx$. Multiplication is

$$(a+bx)(c+dx)=ac+(ad+bc)x \quad\text{in }\mathcal R_{\rm cl}.$$

The class $[x]$ is nonzero: a degree-one polynomial cannot be a polynomial multiple of $x^2$. Nevertheless $[x]^2=0$. A nonzero element whose power vanishes is nilpotent.

Compare this with the scalar potential,

$$V=|g|^2|x|^4.$$

Its only zero is the point $x=0$. Every character likewise obeys $\chi([x])^2=0$ in $\mathbb C$, hence $\chi([x])=0$. Numerical evaluation loses the nonzero nilpotent class. Removing all nilpotents gives the reduced ring, here just $\mathbb C$, which is sufficient to describe this point set but contains less information.

An elementary matrix check makes the distinction visible. On the ring’s vector-space basis $(1,x)$, multiplication by $x$ is

$$M_x=\begin{pmatrix}0&0\\1&0\end{pmatrix}, \qquad M_x\ne0,\qquad M_x^2=0.$$

These matrices represent polynomial multiplication; they are not observables on a positive-norm particle Hilbert space. The chiral ring does not include Hermitian conjugation as an internal operation. There is no contradiction with unitarity, and two basis elements do not mean two vacuum points.

Split the repeated root

Deform the same model by a parameter $v$ of mass dimension one:

$$W=g\left(\frac{X^3}{3}-v^2X\right),\qquad v\ne0.$$

Now

$$\mathcal R_v=\mathbb C[x]/(x^2-v^2),\qquad V=|g|^2|x^2-v^2|^2.$$

There are two classical supersymmetric vacua, $x=v$ and $x=-v$. Their fluctuation masses are both $|W''(\pm v)|=2|gv|$, for the complex scalar and Weyl fermion.

A class $a+bx$ is determined by its two values $a+bv$ and $a-bv$. Conversely,

$$a=\frac{f(v)+f(-v)}2,\qquad b=\frac{f(v)-f(-v)}{2v}.$$

Thus $\mathcal R_v$ is the ring $\mathbb C\oplus\mathbb C$ with componentwise multiplication. The elements

$$e_\pm=\frac12\left(1\pm\frac{x}{v}\right)$$

obey $e_\pm^2=e_\pm$, $e_+e_-=0$ and $e_++e_-=1$. Evaluation in the two vacua sends them to $(1,0)$ and $(0,1)$.

The multiplication matrix is now

$$M_x(v)=\begin{pmatrix}0&v^2\\1&0\end{pmatrix}.$$

For $v\ne0$ it has distinct eigenvalues $\pm v$. At $v=0$ it becomes the nonzero nilpotent matrix above. The vector-space dimension stays two, but the distinct vacuum points merge. The ring keeps the repeated-root information that the point set discards. This is a classical deformation, not an instanton correction or a new interacting four-dimensional conformal fixed point.

What changes in the quantum theory?

The Ward argument continues to constrain a quantum theory when supersymmetry, its vacuum and the needed operator definitions satisfy the stated assumptions. Determining the actual quantum ring is a further dynamical task.

In particular, a renormalized composite at coincident points is more than a formal product of bare fields. The separated-insertion derivation does not justify ignoring contact terms in every equation containing composite ancestors or integrated action insertions. In a gauge theory, the Konishi anomaly is a concrete warning: a barred-superderivative relation for a matter bilinear acquires a chiral gauge-field-strength operator. Naively multiplying an elementary equation of motion and declaring every resulting classical relation exact can miss that contribution. This is an anomaly of a matter rotation, not permission for an inconsistent gauge anomaly or a statement that supersymmetry itself has failed. Tong, §7.2.3

Quantum dynamics can also deform classical vacuum constraints. Intriligator–Seiberg, §4.2, (4.14)–(4.15) gives a gauge-theory example. Establishing such a deformation requires information beyond the elementary quotient calculated here.

Even classically, gauge theories require gauge-invariant operators and the appropriate D-flatness and gauge-equivalence analysis. A list of polynomial F equations alone is not a universal recipe for counting physical vacua. Likewise, setting $g=0$ in our neutral model changes the ideal to zero and gives $\mathbb C[x]$; one cannot divide by $g$, then carry the relation $x^2=0$ through that singular change of theory. No superconformal dimension formula is needed for any of these arguments.

Exercises

1. A nilpotent along a vacuum line

Take two neutral canonical chirals with $W=\lambda X^2Y$, $\lambda\ne0$. Calculate the classical polynomial ring and the vacuum point set. Find a normal form for its elements. Does the class of $x$ vanish? Check the small-fluctuation masses at a point $(x,y)=(0,y_0)$.

Hint 1

Use both derivatives of $W$. Reduce separately monomials containing two factors of $x$ and those containing both $x$ and $y$.

Hint 2

To show that $x$ survives, set $y=0$ in any proposed expression $x=A x^2+Bxy$. For the masses, evaluate the two-by-two matrix $W_{ij}$ at the vacuum.

Solution

The equations are $W_X=2\lambda xy$ and $W_Y=\lambda x^2$. Therefore

$$\mathcal R_{\rm cl}=\mathbb C[x,y]/(x^2,xy).$$

Every class has the form $f(y)+a x$, with polynomial $f$ and constant $a$. Its product with $g(y)+b x$ is

$$f(y)g(y)+\bigl[a\,g(0)+b\,f(0)\bigr]x.$$

The class $x$ is nonzero, because setting $y=0$ in an ideal-membership equation would make a degree-one polynomial divisible by $x^2$. It squares to zero and is also annihilated by $y$.

For numerical vacuum values, $x^2=0$ forces $x=0$, while $y$ is arbitrary. The reduced ring is $\mathbb C[y]$. The full ring contains nilpotent information invisible on that whole vacuum line.

At $(0,y_0)$,

$$(W_{ij})=\begin{pmatrix}2\lambda y_0&0\\0&0\end{pmatrix}.$$

The $X$ multiplet has mass $2|\lambda y_0|$ and $Y$ is massless. At $y_0=0$ both quadratic masses vanish. On a patch where $y$ is invertible, $xy=0$ implies $x=0$ even as a ring class. The surviving nilpotent is therefore tied to the special point where this extra mass vanishes; it is not an extra numerical vacuum at every point on the line.

2. A supersymmetric mixture that does not cluster

Assume two supersymmetric clustering vacuum sectors with chiral expectation values $v$ and $-v$, as in the semiclassical split-root model. Consider their equal statistical mixture. Calculate $\langle x\rangle$, $\langle x^2\rangle$ and the widely separated chiral connected correlator. Does the Ward identity fail? Is this expectation map a character?

Hint 1

In each pure sector separately, the two-point function factorizes to $v^2$, including the sector with expectation value $-v$.

Hint 2

Average correlation functions after evaluating them in each sector. Averaging a product is not generally the same as multiplying averages.

Solution

Linearity of the mixture gives

$$\langle x\rangle_{\rm mix}=0,\qquad \langle x^2\rangle_{\rm mix}=v^2,\qquad \langle x(z_1)x(z_2)\rangle_{{\rm mix},c}=v^2$$

for distinct points, even when their separation becomes arbitrarily large. The connected term subtracts $\langle x\rangle_{\rm mix}^2=0$.

Each sector obeys the supersymmetry Ward identity, so their mixture does too. What fails is cluster decomposition. The expectation map is linear and respects the relation $x^2=v^2$, but is not multiplicative: $\chi_{\rm mix}(x^2)\ne\chi_{\rm mix}(x)^2$. This example shows exactly which additional assumption turns position independence into a character. A vacuum count should not treat this statistical mixture as a third pure vacuum.

3. Does heavy-field elimination preserve the relations?

Return to the neutral model

$$W=\frac M2H^2+\frac y2HX^2,\qquad M\ne0,\quad y\ne0.$$

Calculate its classical polynomial ring by eliminating $h$ algebraically. Compare it with the ring from $W_{\rm eff}=-y^2X^4/(8M)$. Then set $M=0$ in the original model and explain why the previous elimination cannot describe the result.

Hint 1

The ideal has two generators, $Mh+yx^2/2$ and $yhx$. Substitute the first relation into the second.

Hint 2

Distinguish an exact isomorphism of classical polynomial quotients at nonzero $M$ from a claim about kinetic terms. At $M=0$, do not use a formula containing $1/M$.

Solution

At nonzero $M$, the first relation gives $h=-yx^2/(2M)$. The second becomes $-y^2x^3/(2M)=0$. Hence

$$\frac{\mathbb C[h,x]}{(Mh+yx^2/2,\;yhx)} \simeq\mathbb C[x]/(x^3).$$

The map sends $x$ to $x$ and $h$ to $-yx^2/(2M)$. Conversely, these substitutions satisfy both original relations, so they define the inverse identification. The quotient has basis $1,x,x^2$ and one numerical vacuum point.

The effective superpotential gives $W_{\rm eff}'=-y^2x^3/(2M)$ and the same ring. This exact algebraic agreement concerns classical chiral relations. It does not improve the momentum or auxiliary approximation of a truncated effective action.

At $M=0$, the original ideal is $(x^2,hx)$, with ring

$$\mathbb C[h,x]/(x^2,hx).$$

Its point set is the entire line $x=0$, $h$ arbitrary. Its elements have normal form $f(h)+a x$, so it retains a nilpotent at the special point $h=0$, just as in exercise 1. The field $H$ must return to the low-energy description. The polynomial ring, the vacuum set and the kinetic effective theory are related pieces of information, but they are not interchangeable.

You can now derive a separated-insertion Ward identity, calculate an F-term quotient, and test whether numerical vacuum values lose information. Classical SQCD applies these questions to a gauge theory: identify its gauge-invariant generators, distinguish classical from quantum relations, and check what makes their expectation values describe actual vacua. Its other preparation is the anomaly lesson.