Fourier Transform Conventions

The default position–momentum convention uses symmetric factors of $(2\pi\hbar)^{-1/2}$ in one dimension. The signs are chosen so that

$$\langle x|p\rangle =\frac{1}{\sqrt{2\pi\hbar}} \exp\left(\frac{ipx}{\hbar}\right)$$

and the momentum operator in position representation is

$$P=-i\hbar\frac{d}{dx}.$$

Other Fourier conventions are equally valid. A page using another convention must state the exponential signs, normalization factors, variables, and integration measures before combining formulas.

One-Dimensional Position–Momentum Pair

Define

$$\psi(x)=\langle x|\psi\rangle, \qquad \phi(p)=\langle p|\psi\rangle.$$

The inverse transform from momentum components to position components is

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

The forward transform is

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

The words “forward” and “inverse” are conventional labels. The equations, signs, and measures are authoritative.

Origin in Bra-Ket Completeness

Position and momentum generalized eigenkets obey

$$\begin{aligned} \int dx\,|x\rangle\langle x|&=I,\\ \int dp\,|p\rangle\langle p|&=I. \end{aligned}$$

Inserting momentum completeness gives

$$\begin{aligned} \psi(x) &=\langle x|\psi\rangle\\ &=\int dp\, \langle x|p\rangle \langle p|\psi\rangle, \end{aligned}$$

which yields the inverse transform. Taking the complex conjugate of $\langle x|p\rangle$ gives

$$\langle p|x\rangle =\frac{1}{\sqrt{2\pi\hbar}} \exp\left(-\frac{ipx}{\hbar}\right),$$

and inserting position completeness yields the forward transform.

This derivation fixes the two signs together. Reversing both signs defines an equivalent convention if all associated operator formulas are translated.

Normalization and Plancherel Identity

The symmetric convention is unitary:

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

For a normalized state, both integrals equal one. Therefore

$$\begin{aligned} [|\psi(x)|^2]&=L^{-1},\\ [|\phi(p)|^2] &=(\text{momentum})^{-1}. \end{aligned}$$

in one dimension. The two wavefunctions do not have the same physical dimensions because their probability measures differ.

The canonical mathematical treatment, including hypotheses for the transform and its unitary extension, is Fourier Transform.

Delta Normalization

The plane-wave normalization implies

$$\langle p|p'\rangle=\delta(p-p').$$

Indeed,

$$\begin{aligned} \langle p|p'\rangle &=\int dx\, \langle p|x\rangle \langle x|p'\rangle\\ &=\frac{1}{2\pi\hbar} \int dx\, \exp\left( \frac{i(p'-p)x}{\hbar} \right)\\ &=\delta(p-p'). \end{aligned}$$

The last equality is distributional. Plane waves are generalized eigenstates, not square-integrable wavefunctions. See Distributions for the canonical delta-function framework.

Operator Representations

The chosen exponential signs determine the differential representations.

Acting on a plane wave,

$$-i\hbar\frac{d}{dx} \exp\left(\frac{ipx}{\hbar}\right) =p\exp\left(\frac{ipx}{\hbar}\right),$$

so

$$(P\psi)(x) =-i\hbar\frac{d\psi}{dx}.$$

In momentum representation, multiplication by $x$ becomes differentiation:

$$(X\phi)(p) =i\hbar\frac{d\phi}{dp},$$

subject to the appropriate domains and boundary behavior.

The transform also exchanges derivatives and multiplication:

$$\begin{aligned} \mathcal F_p\left[ -i\hbar\frac{d\psi}{dx} \right](p) &=p\,\phi(p),\\ \mathcal F_p[x\psi](p) &=i\hbar\frac{d\phi}{dp}. \end{aligned}$$

Here $\mathcal F_p$ denotes the forward position-to-momentum transform defined above, not an independent convention.

Three Spatial Dimensions

Use

$$\langle\mathbf r|\mathbf p\rangle =\frac{1}{(2\pi\hbar)^{3/2}} \exp\left( \frac{i\mathbf p\cdot\mathbf r}{\hbar} \right).$$

Then

$$\psi(\mathbf r) =\frac{1}{(2\pi\hbar)^{3/2}} \int d^3p\, e^{i\mathbf p\cdot\mathbf r/\hbar} \phi(\mathbf p),$$

and

$$\phi(\mathbf p) =\frac{1}{(2\pi\hbar)^{3/2}} \int d^3r\, e^{-i\mathbf p\cdot\mathbf r/\hbar} \psi(\mathbf r).$$

Normalization is

$$\int d^3r\,|\psi(\mathbf r)|^2 = \int d^3p\,|\phi(\mathbf p)|^2.$$

In $d$ dimensions, replace $3$ by $d$ and use $(2\pi\hbar)^{-d/2}$.

Wave Number Versus Momentum

Wave number and momentum are related by

$$p=\hbar k, \qquad dp=\hbar\,dk.$$

Define a unitary wave-number transform by

$$\begin{aligned} \psi(x) &=\frac{1}{\sqrt{2\pi}} \int dk\,e^{ikx}\widehat\psi(k),\\ \widehat\psi(k) &=\frac{1}{\sqrt{2\pi}} \int dx\,e^{-ikx}\psi(x). \end{aligned}$$

Comparison with the momentum transform gives

$$\widehat\psi(k) =\sqrt{\hbar}\,\phi(\hbar k),$$

or equivalently,

$$\phi(p) =\frac{1}{\sqrt{\hbar}}\, \widehat\psi\left(\frac{p}{\hbar}\right).$$

The square-root Jacobian preserves norm:

$$\int dk\,|\widehat\psi(k)|^2 = \int dp\,|\phi(p)|^2.$$

Replacing $k$ by $p/\hbar$ only in the exponential is incomplete; the measure and wavefunction normalization must also change.

Energy and Time

For time evolution, the sign convention is chosen to match $e^{-iEt/\hbar}$. A symmetric time–energy transform may be written

$$g(t) =\frac{1}{\sqrt{2\pi\hbar}} \int dE\, e^{-iEt/\hbar}\widetilde g(E),$$

with inverse

$$\widetilde g(E) =\frac{1}{\sqrt{2\pi\hbar}} \int dt\, e^{iEt/\hbar}g(t).$$

This transform does not by itself make time an observable represented by a self-adjoint operator canonically conjugate to every Hamiltonian. It is a Fourier relation between a time-dependent function and an energy-frequency variable.

With angular frequency $\omega=E/\hbar$,

$$\widehat g(\omega) =\sqrt{\hbar}\,\widetilde g(\hbar\omega)$$

under the corresponding symmetric $d\omega/\sqrt{2\pi}$ convention.

Periodic Domains and Discrete Momentum

On an interval of length $L$ with periodic boundary conditions,

$$k_n=\frac{2\pi n}{L}, \qquad p_n=\hbar k_n, \qquad n\in\mathbb Z.$$

A convenient orthonormal position-space mode is

$$\langle x|n\rangle =\frac{1}{\sqrt L}e^{ik_nx}.$$

Integrals over continuous momentum become sums over discrete modes. In a large-volume limit,

$$\sum_n \longrightarrow \frac{L}{2\pi} \int dk =\frac{L}{2\pi\hbar} \int dp$$

in one dimension. The density-of-states factor depends on boundary conditions, dimension, and normalization.

Do not combine continuum delta normalization with box-normalized discrete states without an explicit conversion.

Asymmetric Transform Conventions

A common mathematical convention is

$$\begin{aligned} \widehat f(k) &=\int dx\,e^{-ikx}f(x),\\ f(x) &=\frac{1}{2\pi} \int dk\,e^{ikx}\widehat f(k). \end{aligned}$$

This convention places all normalization in the inverse transform. It is not wrong; its transform amplitude differs from the symmetric unitary amplitude by a factor of $\sqrt{2\pi}$.

Another convention reverses both exponential signs. Translation requires replacing $k$ by $-k$ or complex conjugating the kernel as appropriate. Never change a sign in one transform without changing its inverse and derivative rules.

Correlation and Many-Body Transforms

Many-body and response calculations often transform both space and time. Conventions may differ because authors choose:

• real time or imaginary time;
• ordinary frequency, angular frequency, or energy;
• $e^{-i\omega t}$ or $e^{+i\omega t}$ in the forward transform;
• volume-normalized sums or continuum integrals;
• lattice momentum in the first Brillouin zone;
• retarded, advanced, time-ordered, or Matsubara correlators.

Use Structure Factors, Green Functions in Many-Body QM, and Retarded and Advanced Response for their local declarations. The position–momentum convention on this page does not silently determine every spacetime-transform convention.

Discrete Fourier Transforms

A discrete Fourier transform introduces separate choices:

• sample spacing and total interval;
• normalization split between forward and inverse transforms;
• ordering of positive and negative frequencies;
• angular frequency versus ordinary frequency;
• treatment of endpoint duplication;
• aliasing and windowing.

Library defaults are not universal mathematical conventions. Record the library, transform direction, normalization mode, and frequency ordering. See Fast Fourier Transform for the numerical treatment.

Quantum Fourier Transform owns the separate finite-register unitary convention, including its sign, normalization, basis-index order, inverse, and optional output reversal; the continuum position–momentum convention on this page does not determine those choices automatically.

Common Mistakes

• Mixing an exponential sign from one convention with normalization from another.
• Omitting $\hbar$ in $e^{ipx/\hbar}$ without declaring natural units.
• Treating $\phi(p)$ and $\widehat\psi(k)$ as the same function.
• Changing variables from $k$ to $p$ without transforming the measure and amplitude.
• Assuming position- and momentum-space wavefunctions have the same physical dimensions.
• Treating a delta-normalized plane wave as a normalized Hilbert-space state.
• Using $P=+i\hbar\,d/dx$ with the plane-wave convention on this page.
• Confusing energy $E$, angular frequency $\omega$, and ordinary frequency $f$.
• Replacing a box momentum sum by an integral without the density-of-states factor.
• Trusting an FFT array’s index order without mapping indices to physical frequencies.

Exercises

Exercise 1: Momentum-operator sign

Use the inverse transform to show that multiplication by $p$ in momentum space corresponds to $-i\hbar\,d/dx$ in position space.

Solution

Differentiate the inverse transform:

$$\begin{aligned} -i\hbar\frac{d\psi}{dx} &= \frac{-i\hbar}{\sqrt{2\pi\hbar}} \int dp\, \frac{ip}{\hbar} e^{ipx/\hbar}\phi(p)\\ &= \frac{1}{\sqrt{2\pi\hbar}} \int dp\, e^{ipx/\hbar} p\phi(p). \end{aligned}$$

The last line is the inverse transform of $p\phi(p)$, so

$$(P\psi)(x) =-i\hbar\frac{d\psi}{dx}.$$

Exercise 2: Translate from wave number to momentum

Suppose $\widehat\psi(k)$ is normalized by

$$\int dk\,|\widehat\psi(k)|^2=1.$$

Find the normalized momentum amplitude $\phi(p)$.

Solution

With $p=\hbar k$ and $dp=\hbar\,dk$, define

$$\phi(p) =\frac{1}{\sqrt{\hbar}}\, \widehat\psi\left(\frac{p}{\hbar}\right).$$

Then

$$\begin{aligned} \int dp\,|\phi(p)|^2 &= \int dp\, \frac{1}{\hbar} \left| \widehat\psi\left(\frac{p}{\hbar}\right) \right|^2\\ &= \int dk\,|\widehat\psi(k)|^2 =1. \end{aligned}$$

Exercise 3: Recover the momentum delta function

Evaluate

$$\frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} dx\, e^{i(p'-p)x/\hbar}$$

as a distribution.

Solution

Use

$$\frac{1}{2\pi} \int dx\,e^{iqx} =\delta(q)$$

with $q=(p'-p)/\hbar$. Since

$$\delta\left(\frac{p'-p}{\hbar}\right) =\hbar\,\delta(p'-p)$$

for $\hbar>0$, the prefactor gives

$$\frac{1}{2\pi\hbar} \int dx\, e^{i(p'-p)x/\hbar} =\delta(p'-p).$$

Exercise 4: Box sum to continuum integral

For periodic length $L$, show that the spacing of allowed momenta is $\Delta p=2\pi\hbar/L$ and derive the one-dimensional replacement for a smooth large-volume sum.

Solution

The allowed momenta are

$$p_n=\frac{2\pi\hbar n}{L}.$$

Adjacent values differ by

$$\Delta p=\frac{2\pi\hbar}{L}.$$

A Riemann sum therefore becomes

$$\begin{aligned} \sum_n F(p_n) &\longrightarrow \frac{1}{\Delta p} \int dp\,F(p)\\ &=\frac{L}{2\pi\hbar} \int dp\,F(p). \end{aligned}$$

The replacement assumes that $F$ varies slowly on the momentum spacing and that the large-volume limit is appropriate.

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.
• R. N. Bracewell, The Fourier Transform and Its Applications, 3rd ed., McGraw-Hill, 2000.