Distributional Correlators, Zero Modes, and Normalization

Noncompact CFT correlators often live in a space of distributions over continuous quantum numbers. A momentum-conserving Dirac delta, its value at zero, and a target-space volume are related but not interchangeable. This page derives the zero-mode distribution, compares Gaussian and finite-box regulators, and states the test-function and limit prescriptions needed to turn formal expressions into well-defined observables.

Required background. Noncompact CFTs and continuous spectra fixes the free-boson state and measure conventions used here. Tempered distributions and Fourier calculus supplies the definition of convergence by pairing with test functions.

Helpful background. Coincident products and contact terms distinguishes distributional extensions in spacetime from conservation deltas in spectral labels.

A zero mode produces a distribution

Use the noncompact free-boson normalization

$$\langle X(z,\bar z)X(0)\rangle=-\log|z|^2, \qquad V_p=:{e^{ipX}}:,$$

and split

$$X=x_0+\widetilde X, \qquad x_0\in\mathbb R, \qquad d\mu_0=dx_0.$$

The constant mode of an $n$-point function gives

$$\int_{\mathbb R}dx_0\, e^{ix_0P}=2\pi\delta(P), \qquad P=\sum_{i=1}^n p_i.$$

The full plane correlator is therefore

$$\left\langle\prod_{i=1}^nV_{p_i}(z_i,\bar z_i)\right\rangle =2\pi\delta\!\left(\sum_i p_i\right) \prod_{i<j}|z_i-z_j|^{2p_ip_j}.$$

For two insertions,

$$\langle V_p(z,\bar z)V_{p'}(0)\rangle =2\pi\delta(p+p')|z|^{-2p^2}.$$

The normalization matches

$$\langle p|p'\rangle=2\pi\delta(p-p'), \qquad \mathbf1=\int_{\mathbb R}\frac{dp}{2\pi}|p\rangle\langle p|.$$

This delta-normalized structure is standard for continuous CFT spectra; see Ribault 2018, §2.2.3, pp. 39–42, and §4.1.3, pp. 101–103.

The correlator is defined through smearing. For smooth test functions $f$ and $g$ for which the integral converges,

$$\begin{aligned} G_{f,g}(z) ={}&\int\frac{dp\,dp'}{(2\pi)^2} f(p)g(p') \langle V_p(z)V_{p'}(0)\rangle\\ ={}&\int_{\mathbb R}\frac{dp}{2\pi} f(p)g(-p)|z|^{-2p^2}. \end{aligned}$$

The qualification on the test functions matters. The bare delta distribution is tempered on Schwartz space. At fixed $|z|<1$, however, the multiplier $|z|^{-2p^2}$ grows like $e^{2|\log|z||p^2}$ and is not a Schwartz multiplier. One may first use compactly supported $C^\infty$ functions in momentum, or wave packets with sufficiently strong Gaussian decay, and enlarge the test space only after proving continuity. Calling the entire momentum-labelled correlator “tempered” without this check is generally too strong.

Gaussian regularization

Regulate the zero-mode integral by

$$e^{-\varepsilon^2x_0^2/4}, \qquad \varepsilon>0.$$

Then

$$\int_{\mathbb R}dx_0\, e^{-\varepsilon^2x_0^2/4}e^{iPx_0} =\frac{2\sqrt\pi}{\varepsilon}e^{-P^2/\varepsilon^2} =2\pi\delta_\varepsilon(P),$$

where

$$\delta_\varepsilon(P) =\frac{1}{\sqrt\pi\,\varepsilon} e^{-P^2/\varepsilon^2}.$$

For every Schwartz function $f$,

$$\begin{aligned} \lim_{\varepsilon\to0^+} \int_{\mathbb R}dP\,\delta_\varepsilon(P)f(P) &=\lim_{\varepsilon\to0^+} \frac1{\sqrt\pi} \int_{\mathbb R}du\,e^{-u^2}f(\varepsilon u)\\ &=f(0). \end{aligned}$$

This is the statement $\delta_\varepsilon\to\delta$ in $\mathcal S'(\mathbb R)$. Pointwise behavior is irrelevant: the peak diverges at $P=0$ and tends to zero at every fixed $P\neq0$.

An exact diagnostic uses the Gaussian test function $f_a(P)=e^{-aP^2}$ with $a>0$:

$$\int_{\mathbb R}dP\, \delta_\varepsilon(P)e^{-aP^2} =\frac{1}{\sqrt{1+a\varepsilon^2}} \xrightarrow[\varepsilon\to0^+]{}1.$$

If a numerical regulator does not reproduce this formula within its stated quadrature and truncation error, its normalization is wrong before any CFT-specific test is attempted.

A finite box and the Kronecker-to-Dirac limit

Alternatively restrict the zero mode to $-L/2\leq x_0<L/2$. Its Fourier kernel is

$$\delta_L(P) =\frac1{2\pi}\int_{-L/2}^{L/2}dx_0\,e^{iPx_0} =\frac{\sin(PL/2)}{\pi P},$$

with $\delta_L(0)=L/(2\pi)$ and $\delta_L\to\delta$ distributionally as $L\to\infty$. If the target is genuinely compact with $X\sim X+L$, its momenta are $p_n=2\pi n/L$ and

$$\int_0^Ldx_0\,e^{i(p_n+p_m)x_0} =L\delta_{n+m,0}.$$

The continuum correspondence is therefore

$$L\delta_{n+m,0} \longleftrightarrow 2\pi\delta(p+p'), \qquad \sum_n\longleftrightarrow\frac{L}{2\pi}\int dp.$$

Both relations must be used together. Replacing the Kronecker delta by a Dirac delta while leaving the sum unchanged creates an extra factor of $L/(2\pi)$. The compact and uncompactified boson zero-mode factors are compared in Di Francesco, Mathieu, and Sénéchal 1997, §§10.1–10.2, pp. 337–343.

The Gaussian and box regulators have different shapes. One can match their peak heights by setting

$$\frac{L}{2\pi}=\frac{1}{\sqrt\pi\varepsilon}, \qquad L=\frac{2\sqrt\pi}{\varepsilon},$$

but that does not make them pointwise equal. Regulator-independent claims must be formulated after smearing and taking the limit, or with an error estimate that controls the difference for the chosen test functions.

Volume division is an observable choice

At total momentum $P=0$, the unregulated zero-mode integral is the target volume

$$V_X=\int_{\mathbb R}dx_0=2\pi\delta(0).$$

There are three distinct quantities one might report:

Quantity	Definition	Appropriate use
Extensive correlator or trace	Retain $V_X$ or its regulator $L$	A finite compact target followed as $L$ grows
Density per target volume	Divide a neutral trace or observable by $L$ before $L\to\infty$	Thermodynamic or modular densities
Momentum-space distribution	Retain $2\pi\delta(P)$ and smear in $P$	Charged correlators and scattering-state normalization

These operations are not interchangeable. Dividing $2\pi\delta_\varepsilon(P)$ by the regulated Gaussian volume $2\sqrt\pi/\varepsilon$ gives $e^{-P^2/\varepsilon^2}$, which tends to zero as a distribution even though it equals one at $P=0$. A per-volume neutral observable is defined by restricting to the neutral sector and dividing the extensive trace, not by dividing a generic charge-conservation delta distribution pointwise.

For shift-invariant quantities, another option is to work directly with the quotient by the constant-mode symmetry. The gauge or quotient measure must then be declared, and charged vertex operators are not observables on that quotient unless accompanied by a compensating insertion.

Global deltas versus contact terms

Two superficially similar distributions have different supports and meanings:

• $\delta(\sum_i p_i)$ lives in momentum-label space and follows from integrating a global zero mode. It is present even when all spacetime insertion points are distinct.
• $\delta^{(2)}(z_i-z_j)$ lives in spacetime and is supported at coincident insertions. It can arise when extending singular products or differentiating Ward identities.

A counterterm can change contact terms without changing the global momentum-conservation delta. Conversely, dividing by a target volume does not remove a spacetime contact term. Any equality that is asserted only for separated points should say so; any integrated Ward identity must keep the allowed contact terms.

A reliable order of operations

For a distribution-valued CFT calculation:

1. declare the zero-mode measure and the normalization of continuum states;
2. introduce a regulator with its dimensions and normalization;
3. pair the expression with a stated test-function class;
4. perform the CFT or spectral integrals where absolute convergence justifies the order;
5. remove the regulator in the weak topology of the chosen distribution space;
6. only then specialize labels, divide by a volume, or take coincident limits when those operations are defined;
7. state any contact-term ambiguity or noncommuting limit that remains.

Fubini’s theorem cannot be invoked for conditionally convergent contour integrals without additional bounds. In particular, taking $P\to0$ before removing a zero-mode regulator produces $\delta_\varepsilon(0)\sim\varepsilon^{-1}$, while smearing first produces the finite value $f(0)$.

The resulting measures enter torus consistency on Nonrational modular consistency and spectral densities.

Exercises

Test the box regulator

Show distributionally that $\delta_L(P)=\sin(PL/2)/(\pi P)$ tends to $\delta(P)$ as $L\to\infty$.

Solution

For a Schwartz function $f$, use the Fourier transform

$$\widehat f(x)=\int_{\mathbb R}dP\,e^{iPx}f(P).$$

Then

$$\int dP\,\delta_L(P)f(P) =\frac1{2\pi}\int_{-L/2}^{L/2}dx\,\widehat f(x).$$

Because $\widehat f$ is Schwartz, the limit is $(2\pi)^{-1}\int_{\mathbb R}dx\,\widehat f(x)=f(0)$ by Fourier inversion.

Identify a noncommuting specialization

For the Gaussian regulator, compare (i) $\lim_{\varepsilon\to0}\delta_\varepsilon(0)$ and (ii) $\lim_{\varepsilon\to0}\int dP\,\delta_\varepsilon(P)f(P)$.

Solution

The pointwise value is

$$\delta_\varepsilon(0)=\frac{1}{\sqrt\pi\varepsilon}\to\infty.$$

The smeared limit is $f(0)$, which is finite. Evaluating a distribution at a point is not a continuous operation; the second limit is the definition of distributional convergence.

References

• Di Francesco, Philippe, Pierre Mathieu, and David Sénéchal. Conformal Field Theory. Graduate Texts in Contemporary Physics. New York: Springer, 1997. DOI.
• Ribault, Sylvain. “Conformal Field Theory on the Plane.” SciPost Physics Lecture Notes 1 (2018). DOI.