Fourier Transforms

This table collects continuous Fourier-transform pairs and operational rules used in wave mechanics. Every entry is tied to a declared exponential sign and normalization; changing convention changes factors of $2\pi$, factors of $\hbar$, and sometimes signs.

For derivations and functional-analytic context, use the canonical Fourier Transform page. For a guided comparison of physics conventions, use Fourier Transform Conventions.

Scope

Included here:

• the asymmetric mathematical convention in the wavenumber $k$;
• the unitary position–momentum convention in the momentum $p$;
• common transform pairs for functions and tempered distributions;
• differentiation, translation, scaling, convolution, and Parseval rules;
• the position–momentum operator dictionary;
• the multidimensional extension.

Discrete Fourier transforms, Fourier series, numerical FFT ordering, and boundary-adapted transforms require their own grids and normalization conventions and are outside this table.

Mathematical Convention

Unless a row is explicitly labeled otherwise, define

$$F(k)=\mathcal F[f](k) =\int_{-\infty}^{\infty} f(x)e^{-ikx}\,dx$$

and

$$f(x)=\mathcal F^{-1}[F](x) =\frac{1}{2\pi} \int_{-\infty}^{\infty} F(k)e^{ikx}\,dk.$$

The exponent uses angular wavenumber $k$, not cyclic spatial frequency. The inverse transform carries the full $1/(2\pi)$ factor.

For ordinary manipulations, one may begin with Schwartz functions $f\in\mathcal S(\mathbb R)$, where differentiating and integrating by parts are well controlled. The Plancherel theorem extends the transform to $L^2(\mathbb R)$, and duality extends it to tempered distributions $\mathcal S'(\mathbb R)$.

The delta normalization consistent with this convention is

$$\int_{-\infty}^{\infty}e^{ikx}\,dk =2\pi\delta(x).$$

Position–Momentum Convention

Position and momentum wavefunctions are related by the unitary pair

$$\phi(p) =\frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi(x)\,dx$$

and

$$\psi(x) =\frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{ipx/\hbar}\phi(p)\,dp.$$

Consequently,

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx = \int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp.$$

If $\psi$ is normalized, its dimensions are

$$[\psi]=L^{-1/2}, \qquad [\phi]=P^{-1/2}.$$

The connection with the mathematical transform is

$$\phi(p) =\frac{1}{\sqrt{2\pi\hbar}}\, F\left(\frac{p}{\hbar}\right), \qquad p=\hbar k, \qquad dp=\hbar\,dk.$$

Convention Comparison

Name	Forward transform	Inverse transform
Asymmetric angular-wavenumber	$F(k)=\int f(x)e^{-ikx}\,dx$	$f(x)=\frac{1}{2\pi}\int F(k)e^{ikx}\,dk$
Symmetric angular-wavenumber	$\widehat f(k)=\frac{1}{\sqrt{2\pi}}\int f(x)e^{-ikx}\,dx$	$f(x)=\frac{1}{\sqrt{2\pi}}\int\widehat f(k)e^{ikx}\,dk$
Cyclic frequency	$\overline f(\nu)=\int f(x)e^{-i2\pi\nu x}\,dx$	$f(x)=\int\overline f(\nu)e^{i2\pi\nu x}\,d\nu$
Quantum momentum	$\phi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int\psi(x)e^{-ipx/\hbar}\,dx$	$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int\phi(p)e^{ipx/\hbar}\,dp$

For the first two rows,

$$\widehat f(k)=\frac{F(k)}{\sqrt{2\pi}}.$$

For cyclic frequency, $k=2\pi\nu$. For momentum, $p=\hbar k$. These variable changes bring Jacobians as well as renamed arguments.

Transform Pairs

The table below uses the asymmetric mathematical convention.

Elementary and distributional pairs

$f(x)$	$F(k)$	Conditions
$\delta(x-a)$	$e^{-ika}$	Tempered distribution
$\delta^{(n)}(x-a)$	$(ik)^n e^{-ika}$	Distributional derivative
$1$	$2\pi\delta(k)$	Tempered distribution
$e^{ik_0x}$	$2\pi\delta(k-k_0)$	Tempered distribution
$\cos(k_0x)$	$\pi[\delta(k-k_0)+\delta(k+k_0)]$	Tempered distribution
$\sin(k_0x)$	$-i\pi[\delta(k-k_0)-\delta(k+k_0)]$	Tempered distribution
$\operatorname{PV}(1/x)$	$-i\pi\,\operatorname{sgn}(k)$	Principal-value distribution

Decaying functions

$f(x)$	$F(k)$	Conditions
$e^{-ax^2}$	$\sqrt{\pi/a}\,e^{-k^2/(4a)}$	$\operatorname{Re}a>0$; consistent square-root branch
$e^{-a\lvert x\rvert}$	$\frac{2a}{a^2+k^2}$	$a>0$
$\frac{1}{x^2+a^2}$	$\frac{\pi}{a}e^{-a\lvert k\rvert}$	$a>0$
$\frac{a}{\pi(x^2+a^2)}$	$e^{-a\lvert k\rvert}$	Normalized Cauchy profile, $a>0$

Rectangular and band-limited pairs

Define

$$\operatorname{sinc}u\equiv \begin{cases} \dfrac{\sin u}{u},&u\ne0,\\[4pt] 1,&u=0, \end{cases}$$

and let $\operatorname{rect}(x/L)$ equal $1$ for $\lvert x\rvert<L/2$ and $0$ for $\lvert x\rvert>L/2$. Endpoint values do not affect the ordinary integral.

$f(x)$	$F(k)$	Conditions
$\operatorname{rect}(x/L)$	$L\,\operatorname{sinc}(kL/2)$	$L>0$
$\frac{\sin(Kx)}{\pi x}$	$\mathbf 1_{\lbrace\lvert k\rvert<K\rbrace}$	Distributional endpoint value $1/2$ at $\lvert k\rvert=K$
$\frac{1}{L}\operatorname{rect}(x/L)$	$\operatorname{sinc}(kL/2)$	Unit integral

The rectangle–sinc pair is a quick diagnostic for convention errors: the argument must be dimensionless, and the value at $k=0$ must equal the integral of the original function.

Operational Rules

Let $F=\mathcal F[f]$ and $G=\mathcal F[g]$. The rules assume sufficient decay and regularity or a valid distributional interpretation.

Operation in position space	Transform in wavenumber space	Comment
$\alpha f+\beta g$	$\alpha F+\beta G$	Linearity
$f^{(n)}(x)$	$(ik)^nF(k)$	Boundary terms vanish or are distributional
$x^n f(x)$	$i^n\,d^nF/dk^n$	Multiplication becomes differentiation
$f(x-a)$	$e^{-ika}F(k)$	Translation
$e^{ik_0x}f(x)$	$F(k-k_0)$	Modulation
$f(ax)$	$\frac{1}{\lvert a\rvert}F(k/a)$	Real $a\ne0$
$f(-x)$	$F(-k)$	Reflection
$f^\ast(x)$	$F^\ast(-k)$	Complex conjugation
$(f*g)(x)$	$F(k)G(k)$	Convolution as defined below
$f(x)g(x)$	$\frac{1}{2\pi}(F*G)(k)$	Product–convolution duality

The convolution definitions are

$$(f*g)(x) =\int_{-\infty}^{\infty} f(y)g(x-y)\,dy$$

and

$$(F*G)(k) =\int_{-\infty}^{\infty} F(q)G(k-q)\,dq.$$

Parseval and Plancherel

For the asymmetric convention,

$$\int_{-\infty}^{\infty} f^\ast(x)g(x)\,dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} F^\ast(k)G(k)\,dk.$$

Setting $g=f$ gives

$$\int\lvert f(x)\rvert^2\,dx =\frac{1}{2\pi} \int\lvert F(k)\rvert^2\,dk.$$

The factor disappears in the symmetric convention and in the unitary position–momentum convention.

Position–Momentum Operator Dictionary

Under the unitary quantum convention:

Position-space object	Momentum-space image
$\psi(x)$	$\phi(p)$
$x\psi(x)$	$i\hbar\,\partial_p\phi(p)$
$x^n\psi(x)$	$(i\hbar)^n\partial_p^n\phi(p)$
$-i\hbar\,\partial_x\psi(x)$	$p\phi(p)$
$(-i\hbar\,\partial_x)^n\psi(x)$	$p^n\phi(p)$
$\psi(x-a)$	$e^{-iap/\hbar}\phi(p)$
$e^{ip_0x/\hbar}\psi(x)$	$\phi(p-p_0)$
$(f*g)(x)$	$\sqrt{2\pi\hbar}\,\phi(p)\chi(p)$
$f(x)g(x)$	$\frac{1}{\sqrt{2\pi\hbar}}(\phi*\chi)(p)$

In the last two rows, $\phi$ and $\chi$ are the unitary momentum transforms of $f$ and $g$, and the momentum convolution uses

$$(\phi*\chi)(p) =\int_{-\infty}^{\infty} \phi(q)\chi(p-q)\,dq.$$

The operator correspondences

$$x\longleftrightarrow i\hbar\frac{\partial}{\partial p}, \qquad p\longleftrightarrow p$$

are representation statements, not substitutions between classical numbers. Their domains and boundary behavior matter.

Continuum normalization

With

$$\langle x\vert p\rangle =\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar},$$

one has

$$\langle p\vert p'\rangle=\delta(p-p'), \qquad \int_{-\infty}^{\infty} \lvert p\rangle\langle p\rvert\,dp=I.$$

A plane wave is therefore delta-normalized and is not an element of $L^2(\mathbb R)$.

Gaussian Width Check

Consider the normalized packet

$$\psi(x) =\frac{1}{(2\pi\sigma_x^2)^{1/4}} \exp\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} +\frac{ip_0x}{\hbar} \right].$$

Its transform is

$$\begin{aligned} \phi(p) =\left( \frac{2\sigma_x^2}{\pi\hbar^2} \right)^{1/4} &\exp\left[ -\frac{\sigma_x^2(p-p_0)^2}{\hbar^2} \right] \\ &\times \exp\left[ -\frac{i(p-p_0)x_0}{\hbar} \right]. \end{aligned}$$

The probability densities have standard deviations

$$\Delta x=\sigma_x, \qquad \Delta p=\frac{\hbar}{2\sigma_x}, \qquad \Delta x\,\Delta p=\frac{\hbar}{2}.$$

The factors of $2$ depend on whether a quoted width belongs to the amplitude or the probability density. Writing the full Gaussian before comparing widths avoids that ambiguity.

Multidimensional Extension

In $d$ dimensions, the asymmetric mathematical convention is

$$F(\boldsymbol k) =\int_{\mathbb R^d} f(\boldsymbol x)e^{-i\boldsymbol k\cdot\boldsymbol x}\,d^d x$$

and

$$f(\boldsymbol x) =\frac{1}{(2\pi)^d} \int_{\mathbb R^d} F(\boldsymbol k)e^{i\boldsymbol k\cdot\boldsymbol x}\,d^d k.$$

The main differential rules become

$$\begin{aligned} \mathcal F[\nabla f] &=i\boldsymbol k F, & \mathcal F[\boldsymbol x f] &=i\nabla_{\boldsymbol k}F, \\ \mathcal F[\nabla^2f] &=-\lVert\boldsymbol k\rVert^2F, & \mathcal F[\delta^{(d)}(\boldsymbol x-\boldsymbol a)] &=e^{-i\boldsymbol k\cdot\boldsymbol a}. \end{aligned}$$

The unitary position–momentum kernel is $(2\pi\hbar)^{-d/2}e^{i\boldsymbol p\cdot\boldsymbol x/\hbar}$, with $d^dp=\hbar^d\,d^dk$.

Distributional Reading

Several rows cannot be interpreted as absolutely convergent integrals. Instead, a tempered distribution $T$ is transformed by duality against a Schwartz test function. This gives precise meaning to entries such as

$$1\longleftrightarrow2\pi\delta(k), \qquad \operatorname{PV}\frac1x \longleftrightarrow -i\pi\operatorname{sgn}(k).$$

Distributional differentiation preserves the same rule $\mathcal F[T'](k)=ik\,\mathcal F[T](k)$. Principal values, delta terms, and endpoint half-values should not be discarded merely because they occur on a set of ordinary measure zero; they encode the distribution itself.

How to Use This Table

1. Identify the transform pair, exponential sign, and transformed variable.
2. Check whether the source uses $k$, cyclic frequency, or momentum $p$.
3. Translate both the argument and the integration measure.
4. Determine whether the entry is an ordinary function, an $L^2$ equivalence class, or a distribution.
5. Apply operational rules from the same normalization convention.
6. Check dimensions, the value at zero, inverse transformation, and Parseval normalization.

Common Mistakes

• Mixing $p$ and $k$ without using $p=\hbar k$ and $dp=\hbar\,dk$.
• Taking transform pairs from one normalization and convolution rules from another.
• Reversing the sign of a translation or modulation phase.
• Forgetting the absolute value in the scaling rule for real negative $a$.
• Treating constants and plane waves as square-integrable functions on the full line.
• Dropping principal-value or delta contributions in distributional transforms.
• Confusing amplitude width with probability-density standard deviation.
• Using cyclic frequency $\nu$ while retaining an angular-frequency derivative rule.
• Applying integration by parts without checking boundary terms or distributional contributions.

Verification

The table is editorially maintained rather than generated. Each pair can be checked through at least one of the following:

• direct integration in its convergence domain;
• differentiation, scaling, or translation from a known pair;
• inverse transformation against a test function;
• the value $F(0)=\int f(x)\,dx$ when the integral exists;
• Parseval normalization;
• dimensional analysis in the position–momentum convention.

The rectangle–sinc, Gaussian–Gaussian, and Cauchy–exponential pairs provide independent checks of normalization and scaling. Last reviewed: 2026-08-19.

Exercises

Exercise 1: Translating from wavenumber to momentum

Let $F(k)=\int\psi(x)e^{-ikx}\,dx$. Derive the relation between $F(k)$ and the unitarily normalized momentum wavefunction $\phi(p)$, including the change of measure.

Solution

Set $p=\hbar k$, so $k=p/\hbar$ and $dp=\hbar\,dk$. Comparing definitions gives

$$\phi(p) =\frac{1}{\sqrt{2\pi\hbar}}\, F\left(\frac{p}{\hbar}\right).$$

Then

$$\begin{aligned} \int\lvert\phi(p)\rvert^2\,dp &= \frac{1}{2\pi\hbar} \int \left\lvert F\left(\frac{p}{\hbar}\right) \right\rvert^2dp \\ &= \frac{1}{2\pi} \int\lvert F(k)\rvert^2\,dk \\ &= \int\lvert\psi(x)\rvert^2\,dx. \end{aligned}$$

The second line uses $dp=\hbar\,dk$, and the last uses Parseval in the asymmetric convention.

Exercise 2: Translation sign

Under the mathematical convention, find the transform of $g(x)=f(x-a)$. Then use the answer to determine how translating a state by positive $a$ changes its momentum-space phase.

Solution

Substitute $y=x-a$:

$$\begin{aligned} \mathcal F[g](k) &=\int f(x-a)e^{-ikx}\,dx \\ &=\int f(y)e^{-ik(y+a)}\,dy \\ &=e^{-ika}F(k). \end{aligned}$$

For momentum, $k=p/\hbar$, so

$$\phi(p)\longmapsto e^{-iap/\hbar}\phi(p).$$

This is consistent with the active translation operator $T(a)=e^{-iap/\hbar}$.

Exercise 3: Gaussian uncertainty

Read the position and momentum variances from the Gaussian pair above and verify the minimum-uncertainty product.

Solution

The position density is proportional to

$$\exp\left[ -\frac{(x-x_0)^2}{2\sigma_x^2} \right],$$

so its variance is $\sigma_x^2$. The momentum density is proportional to

$$\exp\left[ -\frac{2\sigma_x^2(p-p_0)^2}{\hbar^2} \right] = \exp\left[ -\frac{(p-p_0)^2}{2\sigma_p^2} \right],$$

which gives

$$\sigma_p^2=\frac{\hbar^2}{4\sigma_x^2}.$$

Therefore

$$\Delta x\,\Delta p =\sigma_x\frac{\hbar}{2\sigma_x} =\frac{\hbar}{2}.$$

Canonical Links

• Fourier Transform
• Fourier Transform Tables for QM
• Inverse Fourier Transform
• Plancherel and Parseval Theorems
• Convolution
• Poisson Summation Formula
• Fourier Transform Conventions
• Delta Function
• Distributions
• Distributional Derivatives
• Principal-Value Distributions
• Gaussian Wave Packets

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• E. M. Stein and R. Shakarchi, Fourier Analysis: An Introduction, Princeton University Press, 2003.
• M. J. Lighthill, Introduction to Fourier Analysis and Generalised Functions, Cambridge University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.