Sources, currents, and generating functionals

How can a partition function tell us about an electric-like current or a stress tensor? Introduce a field that probes the operator, compute how the theory responds to that field, and then differentiate. The probe is called a source. Its normalization determines the operator normalization, so the signs and factors must be specified before taking derivatives.

We will do this with a complex scalar in Euclidean dimension $d=3$ or $4$. This example has a global $U(1)$ symmetry, no supersymmetry, and no dynamical gauge field. The same source calculus will later apply to supersymmetric theories. Basic QFT and metric variation are sufficient preparation; no three-dimensional spinor conventions are needed.

Couple the theory to two external fields

Let $g_{\mu\nu}$ be a positive-definite metric and $a_\mu$ a real background connection for the scalar’s flavor symmetry. “Flavor” here means an internal symmetry of the matter theory. For a scalar of dimensionless charge $q$, define

$$D_\mu\phi=(\partial_\mu-iq a_\mu)\phi,\qquad (D_\mu\phi)^*=(\partial_\mu+iq a_\mu)\phi^*.$$

Consider the minimally coupled action

$$S_E[\phi;g,a] =\int d^d x\,\sqrt g\, \left[g^{\mu\nu}(D_\mu\phi)^*D_\nu\phi+m^2\phi^*\phi\right], \qquad g=\det(g_{\mu\nu}),\quad m^2\ge0.$$

The expression in brackets is denoted $\mathcal L$. There is no curvature coupling such as $\xi R\phi^*\phi$ in this chosen action.

The fields before the semicolon are integrated over in the quantum theory; those after it are held fixed. We can compare different choices of $g,a$ without integrating over them. A background connection does not become a photon merely because it has a gauge transformation.

Indeed, the simultaneous change

$$\phi'=e^{iq\alpha(x)}\phi,\qquad a'_\mu=a_\mu+\partial_\mu\alpha$$

gives $D'_\mu\phi'=e^{iq\alpha}D_\mu\phi$. It is a redundancy in the description of the sourced theory. Making $a_\mu$ dynamical would additionally require specifying its action and integrating over gauge-equivalent connection configurations with an appropriate gauge prescription.

Vary the flavor connection

Hold the metric and matter fields fixed while varying $a_\mu$. The two differentiated factors contribute

$$\delta_a D_\mu\phi=-iq\phi\,\delta a_\mu,\qquad \delta_a(D_\mu\phi)^*=+iq\phi^*\,\delta a_\mu.$$

Consequently,

$$\delta_a S_E =\int d^d x\,\sqrt g\,j^\mu\delta a_\mu,\qquad j^\mu=iq\left[\phi^*D^\mu\phi-(D^\mu\phi)^*\phi\right].$$

This defines our source-current sign:

$$j^\mu(x)=\frac1{\sqrt{g(x)}}\frac{\delta S_E}{\delta a_\mu(x)}.$$

The functional derivative is defined with the coordinate measure $d^d x$, which explains the factor $1/\sqrt g$. A lower-index source couples to an upper-index current. The plus sign in $\delta S_E=\int\sqrt g\,j^\mu\delta a_\mu$ is part of the convention.

As a separate sign check, apply the position-dependent phase to the matter fields while holding $a_\mu$ fixed. Direct substitution gives $\delta_\alpha\mathcal L=-j^\mu\partial_\mu\alpha$. The source shift supplies $+j^\mu\partial_\mu\alpha$ and cancels it. Thus a definition that calls the coefficient of $+\partial_\mu\alpha$ the Noether current would assign that current the opposite sign. The source coupling tells us which convention is used here.

At zero source in flat space, write the current as $j_0^\mu$. Expanding the action density gives

$$\mathcal L[a]=\mathcal L[0]+a_\mu j_0^\mu+q^2a_\mu a^\mu|\phi|^2, \qquad j^\mu[a]=j_0^\mu+2q^2a^\mu|\phi|^2.$$

The source coupling is not only linear. Its quadratic term will be important when we differentiate twice.

Vary the metric without losing its sign

We hold the covector $a_\mu$ fixed during metric variation. It matters whether the independent variable is $g_{\mu\nu}$ or its inverse:

$$\delta g^{\mu\nu} =-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma}, \qquad \delta\sqrt g =-\frac12\sqrt g\,g_{\mu\nu}\delta g^{\mu\nu}.$$

Vary both the inverse metric in the kinetic term and the volume element. Since $\delta g^{\mu\nu}$ is symmetric,

$$\delta_g S_E =\frac12\int d^d x\,\sqrt g\, \left[(D_\mu\phi)^*D_\nu\phi+(D_\nu\phi)^*D_\mu\phi -g_{\mu\nu}\mathcal L\right]\delta g^{\mu\nu}.$$

Define the Euclidean stress tensor by

$$T_{\mu\nu} =\frac2{\sqrt g}\frac{\delta S_E}{\delta g^{\mu\nu}} =(D_\mu\phi)^*D_\nu\phi+(D_\nu\phi)^*D_\mu\phi -g_{\mu\nu}\mathcal L.$$

Equivalently, the combined variation in covariant metric variables is

$$\delta S_E=\int d^d x\,\sqrt g\left[ j^\mu\delta a_\mu-\frac12T^{\mu\nu}\delta g_{\mu\nu}\right].$$

These equations fix both metric conventions at once. In particular, for $g_{\mu\nu}=\delta_{\mu\nu}+2h_{\mu\nu}$, the first-order action-density change is $-T^{\mu\nu}h_{\mu\nu}$. This agrees with the metric normalization and linear coupling in the three-manifold paper’s equation (2.10), version 3.

The dimensions provide an independent check. With $[\phi]=(d-2)/2$, $[a_\mu]=1$, and $[g_{\mu\nu}]=0$, we find $[j^\mu]=d-1$ and $[T_{\mu\nu}]=d$. Both source variations therefore have the correct action dimension.

The stress tensor also depends on the specified curved-space action. Adding a curvature coupling changes its metric derivative, even if that coupling vanishes on a flat background. This is why specifying only the flat kinetic term does not uniquely choose an improved stress tensor.

Test both responses on a Fourier profile

In flat space, choose a constant connection and the configuration $\phi=Ae^{ik_\mu x^\mu}$, with real $k_\mu,a_\mu$. Define $p_\mu=k_\mu-q a_\mu$. Then

$$\mathcal L=(p^2+m^2)|A|^2,\qquad j^\mu=-2q p^\mu|A|^2,$$ $$T_{\mu\nu} =\left[2p_\mu p_\nu-\delta_{\mu\nu}(p^2+m^2)\right]|A|^2.$$

Differentiating the first expression with respect to $a_\mu$ gives the displayed current, including its minus sign. To check a metric component independently, perturb $g_{11}=1+2h$ while the other diagonal entries remain one. To first order,

$$\sqrt g=1+h,\qquad g^{11}=1-2h.$$

The density changes by $h[(p^2+m^2)-2p_1^2]|A|^2=-hT_{11}$, in agreement with the source definition. No field equation was imposed on this profile; it tests response formulas, not a vacuum.

Differentiate the quantum partition function

Introduce

$$Z[g,a]=\int\mathcal D\phi\,e^{-S_E[\phi;g,a]}, \qquad \mathcal W[g,a]=-\log Z[g,a].$$

Here $\mathcal W$ is a generating functional, also called the dimensionless Euclidean free energy. Use a regulated integral for which differentiation is valid; any source-dependent measure contributions must be included in the action being differentiated. In a continuum theory the resulting composite operators and local terms require the same renormalization prescription.

For any source parameter $\lambda$, differentiating the exponential gives

$$\partial_\lambda Z=-Z\langle\partial_\lambda S_E\rangle, \qquad \partial_\lambda\mathcal W=\langle\partial_\lambda S_E\rangle.$$

Thus

$$\delta\mathcal W =\int d^d x\,\sqrt g \left[\langle j^\mu\rangle\,\delta a_\mu -\frac12\langle T^{\mu\nu}\rangle\,\delta g_{\mu\nu}\right].$$

Taking a derivative means comparing nearby theories in source space. It is not an equation of motion for the source.

A second derivative differentiates both the inserted operator and the normalized probability weight. For any bosonic insertion $O$,

$$\partial_\kappa\langle O\rangle =\langle\partial_\kappa O\rangle -\langle O\,\partial_\kappa S_E\rangle +\langle O\rangle\langle\partial_\kappa S_E\rangle.$$

The last term comes from differentiating $1/Z$. Set $O=\partial_\lambda S_E$ to obtain

$$\partial_\kappa\partial_\lambda\mathcal W =\left\langle\partial_\kappa\partial_\lambda S_E\right\rangle -\left\langle (\partial_\lambda S_E)(\partial_\kappa S_E) \right\rangle_{\!c}.$$

The subscript $c$ means connected: $\langle AB\rangle_c=\langle AB\rangle-\langle A\rangle\langle B\rangle$. The subtraction comes from differentiating the normalization $1/Z$ in an expectation value. The overall minus sign comes from the weight $e^{-S_E}$ and our choice $\mathcal W=-\log Z$.

For the flavor source, at $a=0$ in flat space, the quadratic coupling gives

$$\left.\frac{\delta^2\mathcal W} {\delta a_\mu(x)\delta a_\nu(y)}\right|_{a=0} =2q^2\delta^{\mu\nu}\langle|\phi(x)|^2\rangle \delta^{(d)}(x-y) -\langle j_0^\mu(x)j_0^\nu(y)\rangle_c.$$

The first term is supported only at coincident points. It is a contact term, often called the seagull contribution in this calculation. It comes from differentiating the operator’s explicit source dependence. At separated points the response is minus the connected current correlator; at coincident points, dropping the extra term changes the answer. The coincident expectation and contact contribution must be regulated and renormalized together.

What a supersymmetric background adds

The source method itself required no supersymmetry. Supersymmetry organizes the stress tensor and other currents into related sets of operators, and the corresponding sources must be organized together as well.

Festuccia and Seiberg explain the four-dimensional construction by coupling to off-shell supergravity and then freezing its fields, taking the Planck scale to infinity while keeping the background geometry nontrivial. Their background auxiliary fields are prescribed sources, not fields eliminated through their gravitational equations of motion. See the introduction and the opening of section 2, version 2. Their equation (2.7) explicitly pairs additional background fields with matter operators. This distinction between a fixed source and a dynamical auxiliary field is essential.

In the three-dimensional paper, the bosonic linear couplings include

$$\delta\mathcal L =-T^{\mu\nu}h_{\mu\nu} +j_{(R)}^\mu\left(A_\mu-\frac32V_\mu\right) +j_{(Z)}^\mu C_\mu+J_{(Z)}H.$$

This is equation (7.1). Here $j_{(R)}$ is the R-current, associated with a symmetry that also acts on supercharges; $j_{(Z)}$ and $J_{(Z)}$ are other operators in the same current multiplet. Their detailed construction is a later calculation. For present purposes, identify each coefficient as a source and retain its normalization.

In particular, the linear R-current source is $A_\mu-\tfrac32V_\mu$. The matter covariant derivative in equation (6.6) instead contains $A_\mu-\tfrac12V_\mu$. Other background couplings contribute to the full source response. Reading only the connection inside a matter derivative would miss that distinction.

Exercises

1. Derive the background Ward identity

Assume the regulated partition function of our scalar theory is invariant under $a_\mu\mapsto a_\mu+\partial_\mu\alpha$, and take $\alpha$ to have compact support. Derive the current-conservation identity for its one-point function. Does this require integrating over $a_\mu$?

First hint

Insert $\delta a_\mu=\partial_\mu\alpha$ in the first variation of $\mathcal W$.

Second hint

Integrate by parts using $\nabla_\mu v^\mu=(1/\sqrt g)\partial_\mu(\sqrt g\,v^\mu)$. Compact support removes the boundary term.

Complete solution

Invariance gives

$$0=\delta_\alpha\mathcal W =\int d^d x\,\sqrt g\,\langle j^\mu\rangle\partial_\mu\alpha =-\int d^d x\,\sqrt g\,\alpha\nabla_\mu\langle j^\mu\rangle.$$

Since $\alpha$ is arbitrary in its support, $\nabla_\mu\langle j^\mu\rangle=0$. This follows from a change of matter integration variables and background-source covariance. The connection remains fixed rather than integrated over. The assumed symmetry of the regulated measure matters; this derivation would acquire an additional term if that symmetry were anomalous. We have made no claim about Ward identities with additional charged insertions, which have their own contact terms.

2. Check the trace by a scale change

Take $g_{\mu\nu}=e^{2\sigma}\delta_{\mu\nu}$ with constant $\sigma$, holding $a_\mu$ and the field profiles fixed. Differentiate the action at $\sigma=0$ and compare with $-\int d^d x\,T^\mu{}_\mu$. Give the trace for $d=3$ and $d=4$.

First hint

The measure contributes $e^{d\sigma}$ and the inverse metric contributes $e^{-2\sigma}$. The mass term contains no inverse metric.

Second hint

Write $K=\delta^{\mu\nu}(D_\mu\phi)^*D_\nu\phi$. The action density is $e^{(d-2)\sigma}K+e^{d\sigma}m^2|\phi|^2$.

Complete solution

The derivative is

$$\left.\frac{dS_E}{d\sigma}\right|_0 =\int d^d x\left[(d-2)K+dm^2|\phi|^2\right].$$

Because $\delta g_{\mu\nu}=2g_{\mu\nu}\delta\sigma$, our metric convention gives $dS_E/d\sigma=-\int\sqrt g\,T^\mu{}_\mu$. Therefore

$$T^\mu{}_\mu=(2-d)K-dm^2|\phi|^2.$$

In three dimensions this is $-K-3m^2|\phi|^2$; in four it is $-2K-4m^2|\phi|^2$. Directly tracing the derived stress tensor gives the same answer. This is the minimally coupled tensor chosen in the question; no conformal improvement was assumed. The inverse-metric sign and the volume-element contribution are both needed.

3. A Gaussian with an explicit source contact term

Consider a single real integration variable with

$$S(x;J)=\frac{\beta x^2}{2}+Jx+\frac{cJ^2}{2}, \qquad \beta>0,\qquad Z(J)=\int_{\mathbb R}dx\,e^{-S(x;J)}.$$

Compute $\mathcal W=-\log Z$, $\mathcal W'$, and $\mathcal W''$. Verify the second-response identity using $\partial_JS=x+cJ$. What changes when $c$ changes even though the probability distribution at $J=0$ is unchanged?

First hint

Complete the square in $x$: the Gaussian is centered at $-J/\beta$.

Second hint

The variance is $1/\beta$ and $\partial_J^2S=c$. Retain both terms in the second-response identity.

Complete solution

Completing the square yields

$$Z(J)=\sqrt{\frac{2\pi}{\beta}}\, \exp\!\left[\frac{J^2}{2\beta}-\frac{cJ^2}{2}\right], \qquad \mathcal W(J)=\frac12\log\frac{\beta}{2\pi} +\frac12\left(c-\frac1\beta\right)J^2.$$

Hence $\mathcal W'=(c-1/\beta)J$ and $\mathcal W''=c-1/\beta$. The first derivative equals $\langle x+cJ\rangle$ because $\langle x\rangle=-J/\beta$. The second equals

$$\langle\partial_J^2S\rangle -\langle(\partial_JS)^2\rangle_c =c-\operatorname{Var}(x)=c-\frac1\beta.$$

Changing $c$ shifts the source response but leaves the zero-source distribution and its variance unchanged. With $c=0$, $\mathcal W''$ is negative, exactly as our sign convention predicts. With sufficiently positive $c$, the contact contribution changes that sign. This finite-dimensional example imposes no source gauge symmetry; in a field theory, allowed local source terms must also satisfy its symmetry constraints.

Checkpoint

You can now recover an operator from its source coupling and distinguish a field equation from a source derivative. Before using a second derivative of a partition function as a correlator, identify the choice of $\log Z$, the metric variable, the source normalization, and the explicit second derivative of the action. Those choices determine which connected correlator and contact terms the response contains.

A remaining question is how adding $\xi R|\phi|^2$ changes the stress tensor even where $R=0$. Stress tensors and improvement terms derives that response and explains why two curved-space completions of the same flat action can select different improvements.

Universal observables and scheme dependence asks a complementary question: which changes in a partition function or its contact response can come from local terms containing only the sources? Its comparison also uses fixed points and scaling.

For the round-sphere route, continue with Why minimal coupling is not enough. Its explicit supersymmetry variation shows why a metric source alone generally needs additional partners.