Fourier, distributions, and Green functions repair

Fourier analysis turns a translation-invariant differential equation into an algebraic one, but in field theory the resulting multiplier is often singular exactly on the physical modes. Distribution theory gives that singular expression a meaning; boundary, support, or state data then select one Green function from the possible inverses. This lesson develops those three steps as one reliable calculation rather than three formulas to memorize.

Required background. You should be comfortable with integration by parts, constant-coefficient ordinary differential equations, and the distinction between a map and a matrix. If domains and codomains are still easy to lose, begin with Linear and tensor methods repair. No measure-theory survey is required.

Transform conventions determine every sign

We use the site’s Fourier pair in $d$ dimensions,

$$\widetilde f(p)=\int \mathrm d^d x\,e^{+ip\cdot x}f(x), \qquad f(x)=\int\frac{\mathrm d^d p}{(2\pi)^d}\, e^{-ip\cdot x}\widetilde f(p).$$

For a smooth function that decays rapidly enough, integration by parts gives

$$\begin{aligned} \widetilde{\partial_\mu f}(p) &=\int \mathrm d^d x\,e^{ip\cdot x}\partial_\mu f(x)\\ &=-ip_\mu\widetilde f(p). \end{aligned}$$

The omitted surface term must actually vanish. Compact support or Schwartz decay is sufficient; an oscillatory plane wave does not satisfy either condition as an ordinary integrable function. With

$$(f*g)(x)=\int \mathrm d^d y\,f(x-y)g(y),$$

the same convention gives

$$\widetilde{f*g}(p)=\widetilde f(p)\widetilde g(p), \qquad \widetilde{\delta^{(d)}}(p)=1.$$

There is no extra $(2\pi)^d$ in the convolution theorem because the full factor was placed in the inverse transform. If you import a formula with a different exponential sign or a symmetric normalization, rederive these two rules before using it. A systematic development of the transform and its extension beyond integrable functions is given in Duistermaat and Kolk 2010, chs. 15 and 18.

Distributions turn singular formulas into defined objects

Let $\mathcal D(\mathbb R^d)=C_c^\infty(\mathbb R^d)$ be the space of smooth, compactly supported test functions. A distribution $T$ is a continuous linear functional on this space, written $\langle T,\varphi\rangle$. A locally integrable function $f$ defines a distribution through

$$\langle T_f,\varphi\rangle =\int \mathrm d^d x\,f(x)\varphi(x),$$

while the Dirac distribution is defined by

$$\langle\delta^{(d)}(x-a),\varphi\rangle=\varphi(a).$$

Two distributions are equal when they give the same result for every test function. Their support is the smallest closed set outside which every such pairing vanishes. These definitions replace meaningless questions such as “what is $\delta(0)$?” with well-posed questions about the action of a distribution.

Derivatives are transferred to the test function:

$$\langle\partial_\mu T,\varphi\rangle =-\langle T,\partial_\mu\varphi\rangle.$$

For the Heaviside distribution $H$ on the real line,

$$\begin{aligned} \langle H',\varphi\rangle &=-\int_0^\infty \varphi'(x)\,\mathrm d x\\ &=\varphi(0) =\langle\delta,\varphi\rangle, \end{aligned}$$

so $H'=\delta$. Nothing was differentiated pointwise at the jump. This weak derivative is the model for contact terms and Green-function sources.

Fourier transformation is naturally extended to tempered distributions, the continuous functionals on the Schwartz space $\mathcal S(\mathbb R^d)$. That setting includes polynomials, plane waves, delta distributions, and many singular kernels used in perturbative QFT. The familiar differentiation rule $\partial_\mu\mapsto-ip_\mu$ then remains valid by duality. It does not follow that arbitrary products of distributions exist: multiplication by a smooth function is defined, but a product such as $\delta^2$ needs additional data or an extension prescription. See Hörmander 2003, chs. 1–3 for the test-function, distribution, and Fourier framework.

Boundary values remember which side of a pole was chosen

The notation $1/(s\pm i0)$ means a distributional boundary value, not the ordinary function $1/s$. For $\varepsilon>0$,

$$\frac{1}{s\pm i\varepsilon} =\frac{s}{s^2+\varepsilon^2} \mp i\frac{\varepsilon}{s^2+\varepsilon^2}.$$

After pairing with a test function and taking $\varepsilon\to0^+$, the first term approaches the Cauchy principal value and the second approaches a delta distribution. Thus

$$\frac{1}{s\pm i0} =\operatorname{PV}\frac{1}{s}\mp i\pi\delta(s).$$

The two boundary values agree as ordinary functions for $s\neq0$ but differ on the singular set:

$$\frac{1}{s+i0}-\frac{1}{s-i0}=-2\pi i\delta(s).$$

Erasing the prescription therefore erases an on-shell contribution rather than simplifying harmless notation.

A Green function is an inverse plus selecting data

For a constant-coefficient differential operator $P$, a fundamental solution $E$ satisfies

$$P E=\delta$$

in the distributional sense. When the convolution is defined, it produces a solution of $Pu=f$ through $u=E*f$. If $E_1$ and $E_2$ solve the same source equation, then

$$P(E_1-E_2)=0.$$

The source equation alone therefore fixes an inverse only up to homogeneous solutions. The missing information depends on the problem:

Problem	Data that select a Green function
Euclidean equation on a noncompact domain	decay, growth, or integrability at infinity
Boundary-value problem	boundary conditions and treatment of zero modes
Hyperbolic initial-value problem	retarded or advanced support
Vacuum time-ordered correlator	state, ordering, and Feynman pole boundary value

For a compactly supported source $f$, a retarded Green operator obeys $\operatorname{supp}(E_{\mathrm R}f)\subseteq J^+(\operatorname{supp}f)$; an advanced operator uses $J^-(\operatorname{supp}f)$. A Feynman kernel is instead selected by time ordering and vacuum boundary data. It is not a retarded response function. Zero modes add another issue: if $P$ has a kernel, a global inverse may not exist until the source space is restricted or a complementary subspace is chosen.

Worked example: the decaying Euclidean inverse

Take $m>0$ on the real line and seek the solution that decays at both ends:

$$L G(x)\equiv \left(-\frac{\mathrm d^2}{\mathrm d x^2}+m^2\right)G(x) =\delta(x).$$

Fourier transformation gives

$$(p^2+m^2)\widetilde G(p)=1, \qquad \widetilde G(p)=\frac{1}{p^2+m^2}.$$

Inverting the transform, or closing the contour around the pole at $p=-im$ for $x>0$ and at $p=+im$ for $x<0$, yields

$$G(x)=\int\frac{\mathrm d p}{2\pi}\, \frac{e^{-ipx}}{p^2+m^2} =\frac{e^{-m\lvert x\rvert}}{2m}.$$

The formula solves $LG=0$ away from the origin. At the origin its first derivative jumps:

$$G'(0^+)=-\frac12, \qquad G'(0^-)=+\frac12.$$

Integrating the equation across $(-\varepsilon,\varepsilon)$ gives

$$-G'(\varepsilon)+G'(-\varepsilon) +m^2\int_{-\varepsilon}^{\varepsilon}G(x)\,\mathrm d x=1.$$

The integral vanishes in the limit and the jump supplies the unit delta source. Equivalently,

$$\frac{\mathrm d^2}{\mathrm d x^2}e^{-m\lvert x\rvert} =m^2e^{-m\lvert x\rvert}-2m\delta(x)$$

as a distribution. A second decaying fundamental solution would differ by a global solution of $Lh=0$; no nonzero combination of $e^{mx}$ and $e^{-mx}$ decays at both infinities. Operator, source normalization, and decay together give uniqueness.

QFT bridge: the Feynman prescription carries physical data

For the free real scalar in the site’s (+---) convention, define $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$. The vacuum time-ordered two-point function is

$$\Delta_F(x) =\langle0|\mathrm T\phi(x)\phi(0)|0\rangle =\int\frac{\mathrm d^d p}{(2\pi)^d}\, \frac{i\,e^{-ip\cdot x}}{p^2-m^2+i0}.$$

The positive-energy pole lies just below the real $p^0$ axis and the negative-energy pole just above it. That placement produces positive-frequency propagation forward in time and negative-frequency propagation backward in time—the time-ordered vacuum boundary condition. Applying the Klein–Gordon operator checks the normalization:

$$\begin{aligned} (\Box+m^2)\Delta_F(x) &=\int\frac{\mathrm d^d p}{(2\pi)^d}\, e^{-ip\cdot x} \frac{-i(p^2-m^2)}{p^2-m^2+i0}\\ &=-i\delta^{(d)}(x). \end{aligned}$$

Here $(p^2-m^2)/(p^2-m^2+i0)=1$ as a distribution because the delta part is annihilated by $p^2-m^2$. The factor $-i$ matters: $\Delta_F$ is normalized as a correlator, while $i\Delta_F$ is the corresponding kernel normalized to solve $(\Box+m^2)E_F=\delta^{(d)}$.

Retarded and advanced inverses use different boundary values. In momentum space their denominators may be written schematically as

$$p^2-m^2+i0\,p^0 \qquad\text{and}\qquad p^2-m^2-i0\,p^0,$$

placing both poles below or both above the real $p^0$ axis. The first choice vanishes before a compactly supported source acts; the second vanishes after it. All three kernels share the off-shell expression $(p^2-m^2)^{-1}$, but their on-shell distributions and physical questions differ. The free-scalar use of time ordering and $i0$ is developed further in Schwartz 2014, §§6.2 and 14.4.

A reliable five-step calculation

When a propagator or Green kernel appears, use this sequence:

1. Write the transform pair. Derive the derivative multiplier and locate every $(2\pi)^d$.
2. Name the object. State the test-function or source space and whether the formula is an ordinary function, distribution, correlator, or operator kernel.
3. Solve the transformed equation. Include the numerator and source normalization, not just the denominator.
4. Select the inverse. State the boundary, support, ordering, state, or pole data, and check for zero modes.
5. Verify in the original equation. Apply the operator and recover the intended delta source; also test support or boundary behavior.

This sequence separates algebra that can be automated from hypotheses that cannot.

Common pitfalls

Treating a delta distribution pointwise. A delta is defined by its action on test functions. Verify a source through a pairing or a derivative jump, not by substituting the singular point into an ordinary formula.

Calling every reciprocal an inverse. The symbol $1/P(p)$ does not specify boundary conditions, support, state, zero-mode treatment, or even the spaces on which an inverse acts. Add those data before using the word “the.”

Equating Feynman and retarded kernels. Feynman ordering encodes a vacuum in–out boundary value; retarded support encodes causal response. Their poles and on-shell terms differ even when their denominators look identical away from the mass shell.

Multiplying singular distributions formally. Smooth functions multiply distributions, but two singular factors need not have a defined product. For the QFT extension problem, continue to Products, scaling degree, and extensions of singular distributions.

Exercises

1. Find a second weak derivative

Show that $u(x)=\lvert x\rvert$ satisfies $u''=2\delta$ as a distribution.

Solution

For every $\varphi\in C_c^\infty(\mathbb R)$,

$$\langle u'',\varphi\rangle =\langle u,\varphi''\rangle =\int_{-\infty}^0(-x)\varphi''(x)\,\mathrm d x +\int_0^\infty x\varphi''(x)\,\mathrm d x.$$

Compact support removes the terms at infinity. Integrating each integral by parts gives

$$\int_{-\infty}^0(-x)\varphi''(x)\,\mathrm d x=\varphi(0), \qquad \int_0^\infty x\varphi''(x)\,\mathrm d x=\varphi(0).$$

Therefore

$$\langle u'',\varphi\rangle=2\varphi(0) =\langle2\delta,\varphi\rangle,$$

so $u''=2\delta$. The result comes from the jump of $u'$ from $-1$ to $+1$.

2. Recover the information in a pole prescription

Use the boundary-value identity to compute

$$\frac{1}{s+i0}-\frac{1}{s-i0},$$

then show that $s/(s+i0)=1$ as a distribution.

Solution

The two boundary values are

$$\frac{1}{s+i0}=\operatorname{PV}\frac1s-i\pi\delta(s), \qquad \frac{1}{s-i0}=\operatorname{PV}\frac1s+i\pi\delta(s).$$

Subtracting gives

$$\frac{1}{s+i0}-\frac{1}{s-i0}=-2\pi i\delta(s).$$

For any test function $\varphi$,

$$\left\langle s\operatorname{PV}\frac1s,\varphi\right\rangle =\left\langle\operatorname{PV}\frac1s,s\varphi\right\rangle =\int_{-\infty}^{\infty}\varphi(s)\,\mathrm d s,$$

while $s\delta(s)=0$. Hence

$$\frac{s}{s+i0}=1.$$

The second result explains why applying the differential operator recovers a delta source, while the first shows that the sign of $i0$ still changes the kernel on its singular support.

3. Build a retarded oscillator Green function

For $E>0$, find the retarded fundamental solution of

$$\left(\frac{\mathrm d^2}{\mathrm d t^2}+E^2\right)g(t)=\delta(t).$$

Verify the source and identify the pole placement of its Fourier transform.

Solution

Retarded support requires $g(t)=0$ for $t<0$. The homogeneous solution for $t>0$ and the unit derivative jump at the source give

$$g_{\mathrm R}(t)=\theta(t)\frac{\sin(Et)}{E}.$$

Because $\sin(Et)/E$ vanishes at $t=0$,

$$g_{\mathrm R}'(t)=\theta(t)\cos(Et),$$

and differentiating once more gives

$$g_{\mathrm R}''(t)=\delta(t)-E\theta(t)\sin(Et).$$

Thus $g_{\mathrm R}''+E^2g_{\mathrm R}=\delta$. With the transform $\widetilde g(p^0)=\int \mathrm d t\,e^{ip^0t}g(t)$,

$$\widetilde g_{\mathrm R}(p^0) =-\frac{1}{(p^0)^2-E^2+i0\,p^0}.$$

Equivalently, it is the boundary value of $-[\,(p^0+i\varepsilon)^2-E^2\,]^{-1}$ as $\varepsilon\to0^+$. Both poles lie below the real axis. For $t<0$, the inverse-transform contour closes above and encloses no poles, which is exactly the required retarded support. The advanced solution $g_{\mathrm A}(t)=-\theta(-t)\sin(Et)/E$ puts both poles above instead.

Re-check and return

Without consulting the worked examples, take either the decaying Euclidean kernel or the retarded oscillator and produce a short derivation containing:

• the Fourier pair and derivative rule you used;
• the test-function or source space;
• the transformed equation with its full normalization;
• the condition that selects the inverse;
• a distributional source check; and
• a comparison with one other inverse, naming the changed support, boundary, state, or pole data.

Mark the capability demonstrated when the transform signs, delta normalization, and selecting data all agree. Mark it uncertain when the algebra is correct but the test-function meaning or choice of inverse remains implicit. Mark it not yet demonstrated when a pointwise manipulation replaces the source check or the pole prescription can be erased without changing your explanation.

Then retry only the Fourier/distribution section of the mathematics diagnostic. If it is demonstrated, continue to Complex and asymptotic methods repair or return to Core QFT; the next direct application is Canonical quantization of the free scalar.

References

• J. J. Duistermaat and J. A. C. Kolk, Distributions: Theory and Applications, Birkhäuser, 2010, doi:10.1007/978-0-8176-4675-2.
• Lars Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, second edition, Springer, 2003, doi:10.1007/978-3-642-61497-2.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940.