Fourier Transform

The Fourier transform resolves a function into complex exponential modes. In quantum mechanics, those modes are generalized momentum eigenfunctions, so the transform changes a state from its position wavefunction to its momentum wavefunction.

The idea is simple; the bookkeeping is not. Signs, factors of $2\pi$, factors of $\hbar$, normalization measures, and function-space assumptions all matter. This page fixes one convention and derives the rules needed to use it.

Position–momentum convention

For a one-dimensional wavefunction, define

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

The inverse transform is

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

The same normalization appears in both directions, and the signs in the exponents are opposite. This is the unitary momentum convention used throughout the quantum-mechanics pages.

The exponent $px/\hbar$ is dimensionless. If a normalized position wavefunction has units

$$[\psi]=[\text{length}]^{-1/2},$$

then its momentum wavefunction has units

$$[\phi]=[\text{momentum}]^{-1/2}.$$

Consequently, both probability elements are dimensionless:

$$\lvert\psi(x)\rvert^2dx, \qquad \lvert\phi(p)\rvert^2dp.$$

The convention is recorded compactly in Fourier Transform Conventions.

Why the modes represent momentum

The kernel is built from plane waves,

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

They satisfy the formal eigenvalue equation

$$-i\hbar\frac{d}{dx} \langle x\vert p\rangle = p\langle x\vert p\rangle.$$

Therefore,

$$\phi(p) = \langle p\vert\psi\rangle$$

is the amplitude for momentum $p$. The inverse transform is the generalized basis expansion

$$\psi(x) = \int_{-\infty}^{\infty} \langle x\vert p\rangle \langle p\vert\psi\rangle\,dp.$$

Plane waves are not square-integrable on the full line. They are generalized eigenvectors normalized by delta functions, not physical normalized states by themselves. The representation-theoretic interpretation is developed in Position and Momentum Representations.

Unitarity and probability

The Fourier transform preserves inner products:

$$\begin{aligned} I_{12} &= \int_{-\infty}^{\infty} \psi_1(x)^*\psi_2(x)\,dx\\ &= \int_{-\infty}^{\infty} \phi_1(p)^*\phi_2(p)\,dp. \end{aligned}$$

In particular,

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

A normalized state therefore remains normalized after transformation. This also shows why the $p$-space measure and prefactor cannot be changed independently.

For the theorem on $L^2$, including the extension from well-behaved functions, see Plancherel and Parseval Theorems.

Inversion and the delta kernel

Insert the forward transform into the inverse:

$$\begin{aligned} \psi(x) &= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty}dx'\,\psi(x')\\ &\quad\times \int_{-\infty}^{\infty}dp\, e^{ip(x-x')/\hbar}. \end{aligned}$$

Reconstruction follows from

$$\frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{ip(x-x')/\hbar}\,dp = \delta(x-x').$$

This identity is distributional. The inner integral is not an ordinary convergent improper integral. Its role is to reproduce a test function after integration over $x'$. The reconstruction theorem and its hypotheses have a canonical home in Inverse Fourier Transform; the generalized kernel belongs in Delta Function.

Function-space meaning

Several levels of interpretation are useful:

Schwartz functions. If $\psi$ and all its derivatives decrease faster than every inverse power, the transform and inverse are ordinary smooth functions. Differentiation, multiplication, and integration by parts are especially safe. The transform maps the Schwartz space to itself.

Integrable functions. If $\psi\in L^1(\mathbb R)$, the transform is bounded and continuous and tends to zero as $\lvert p\rvert\to\infty$. Inversion requires additional hypotheses or an appropriate limiting interpretation.

Square-integrable functions. If $\psi\in L^2(\mathbb R)$, the transform is defined by unitary extension. A pointwise integral need not exist everywhere, but the $L^2$ transform is unambiguous up to equality almost everywhere.

Tempered distributions. Plane waves, delta functions, constants, and many Green-function kernels are transformed distributionally. Algebraic rules still work when interpreted through test functions. See Distributions.

One should state which level is being used instead of treating every transform as the same kind of integral.

Translation and momentum shift

Let $\mathcal F_\hbar[\psi]=\phi$. Translating the function by $a$ gives

$$\mathcal F_\hbar \left[ \psi(x-a) \right](p) = e^{-ipa/\hbar}\phi(p).$$

A spatial shift changes only the momentum-space phase, so it does not change the momentum probability density.

Multiplication by a plane-wave phase shifts momentum:

$$\mathcal F_\hbar \left[ e^{ip_0x/\hbar}\psi(x) \right](p) = \phi(p-p_0).$$

Thus translation in one representation becomes phase modulation in the conjugate representation, while phase modulation becomes translation.

Scaling

For nonzero real $a$,

$$\mathcal F_\hbar \left[ \psi(ax) \right](p) = \frac{1}{\lvert a\rvert} \phi\left(\frac{p}{a}\right).$$

If the scaled state is normalized as

$$\psi_a(x) = \sqrt{\lvert a\rvert}\,\psi(ax),$$

then

$$\phi_a(p) = \frac{1}{\sqrt{\lvert a\rvert}} \phi\left(\frac{p}{a}\right).$$

Compressing a wavefunction in position therefore broadens it in momentum. This reciprocal scaling is the geometric core of the position–momentum uncertainty relation.

Differentiation and multiplication

Assuming the boundary term vanishes, integration by parts gives

$$\mathcal F_\hbar \left[ \frac{d\psi}{dx} \right](p) = \frac{ip}{\hbar}\phi(p).$$

Therefore the position-space momentum operator becomes multiplication:

$$\mathcal F_\hbar \left[ -i\hbar\frac{d\psi}{dx} \right](p) = p\phi(p).$$

Differentiating the transform with respect to $p$ gives the complementary identity

$$\mathcal F_\hbar \left[ x\psi(x) \right](p) = i\hbar\frac{d\phi}{dp}.$$

Thus the basic operator correspondences are

$$\begin{aligned} \hat p &=-i\hbar\frac{d}{dx} \quad\longleftrightarrow\quad p,\\ \hat x &=x \quad\longleftrightarrow\quad i\hbar\frac{d}{dp}. \end{aligned}$$

These formulas involve operator domains. A square-integrable wavefunction need not have a square-integrable derivative, and integration by parts is not valid without controlling its boundary behavior.

Products and convolutions

Define convolution in position by

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

With the symmetric momentum convention,

$$\mathcal F_\hbar[f*g](p) = \sqrt{2\pi\hbar}\, \mathcal F_\hbar[f](p) \mathcal F_\hbar[g](p).$$

Multiplication in position becomes convolution in momentum:

$$\begin{aligned} \mathcal F_\hbar[fg](p) &= \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \mathcal F_\hbar[f](p')\\ &\qquad\times \mathcal F_\hbar[g](p-p')\,dp'. \end{aligned}$$

The prefactors change with convention. Derivations and applications to response kernels belong in Convolution.

Momentum and wave-number conventions

Mathematics and wave physics often use wave number $k$ instead of momentum:

$$\widetilde\psi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx}\psi(x)\,dx.$$

Because $p=\hbar k$, the normalized amplitudes are related by

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

This factor is a Jacobian, not a matter of taste:

$$\lvert\phi(p)\rvert^2dp = \left\lvert\widetilde\psi(k)\right\rvert^2dk.$$

Another common convention puts no prefactor on the forward $k$ transform and $1/(2\pi)$ on the inverse. All are valid when used consistently. Before borrowing a formula, record:

1. whether the spectral variable is $p$ or $k$;
2. the sign in the forward exponential;
3. both normalization factors;
4. the integration measure.

Changing only one of these produces dimensionally or numerically incorrect results.

Worked example: a Gaussian packet

Let

$$A_x = \left( \frac{1}{2\pi\sigma_x^2} \right)^{1/4}$$

and define the normalized packet

$$\psi(x) = A_x e^{-(x-x_0)^2/(4\sigma_x^2)} e^{ip_0x/\hbar}.$$

Its transform is

$$\begin{aligned} \phi(p) &= A_p e^{-\sigma_x^2(p-p_0)^2/\hbar^2}\\ &\quad\times e^{-i(p-p_0)x_0/\hbar}, \end{aligned}$$

where

$$A_p = \left( \frac{2\sigma_x^2}{\pi\hbar^2} \right)^{1/4}.$$

The probability distributions have standard deviations

$$\sigma_p = \frac{\hbar}{2\sigma_x}, \qquad \sigma_x\sigma_p = \frac{\hbar}{2}.$$

The center $x_0$ appears only in the momentum-space phase, while the carrier momentum $p_0$ shifts the momentum distribution. A Gaussian saturates the position–momentum uncertainty bound. The dynamical packet is developed in Gaussian Wave Packets, and the inequality itself in Position–Momentum Uncertainty.

Three dimensions

For $\mathbf x,\mathbf p\in\mathbb R^3$,

$$\phi(\mathbf p) = \frac{1}{(2\pi\hbar)^{3/2}} \int_{\mathbb R^3} e^{-i\mathbf p\cdot\mathbf x/\hbar} \psi(\mathbf x)\,d^3x,$$

with inverse

$$\psi(\mathbf x) = \frac{1}{(2\pi\hbar)^{3/2}} \int_{\mathbb R^3} e^{i\mathbf p\cdot\mathbf x/\hbar} \phi(\mathbf p)\,d^3p.$$

The transform factorizes into Cartesian one-dimensional transforms when the function does. In curvilinear coordinates, angular-momentum and radial decompositions can be more useful than a direct Cartesian transform.

Periodic boxes and discrete modes

On a periodic interval of length $L$, the allowed wave numbers and momenta are

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

The integral transform becomes a Fourier series. As $L\to\infty$, the spacing

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

tends to zero and sums approach integrals with the corresponding density of states. The finite-volume normalization must be converted together with the measure; a normalized box mode does not literally become a normalized plane wave on the full line.

See Periodic Functions and Fourier Series for the discrete theory.

Sampled transforms and FFT grids

Suppose a uniform grid has $N$ samples,

$$x_j=x_0+j\Delta x, \qquad L=N\Delta x.$$

The compatible momentum spacing is

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

and the Nyquist scale is

$$p_{\mathrm{Ny}} = \frac{\pi\hbar}{\Delta x}.$$

Two independent approximations are present:

• finite $\Delta x$ limits the resolvable momentum range and can cause aliasing;
• finite $L$ limits momentum resolution and effectively windows or periodizes the wavefunction.

FFT libraries also choose an array ordering and discrete normalization. One must supply the physical factors $\Delta x$, $\Delta p$, $\hbar$, and any shift of the zero-frequency bin. See Fast Fourier Transform for the implementation workflow.

A reliable transform workflow

1. Write the forward and inverse pair before calculating.
2. Check that every exponential has a dimensionless argument.
3. Identify the function space or distributional interpretation.
4. Track units of the transformed amplitude and its measure.
5. Derive shift, derivative, or convolution factors from the chosen pair instead of memory.
6. Test the result with inversion, norm preservation, and a limiting case.
7. For numerical work, check both window-size and grid-spacing convergence.

These checks catch most convention errors before they reach a physical prediction.

Common mistakes

• Mixing $p$ and $k$ without using $p=\hbar k$ and the amplitude Jacobian.
• Using the same sign in both forward and inverse transforms.
• Losing a factor of $2\pi$, $\hbar$, $\Delta x$, or $\Delta p$.
• Assigning the same physical units to $\psi(x)$ and $\phi(p)$.
• Treating a plane wave as a normalized state on the full line.
• Applying integration by parts when the boundary term or derivative is not controlled.
• Reading an $L^2$ equality as pointwise equality everywhere.
• Using ordinary integrals for delta functions or constant functions without a distributional interpretation.
• Assuming a narrow sampled peak is resolved merely because it appears on an FFT plot.
• Forgetting that an FFT assumes periodic continuation of the sampled array.

Exercises

1. Assuming sufficient decay, derive the Fourier transforms of $\psi'(x)$ and $x\psi(x)$. Use them to identify the momentum-space actions of $\hat p$ and $\hat x$.

Solution

For the derivative,

$$\begin{aligned} \mathcal F_\hbar[\psi'](p) &= \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar}\psi'(x)\,dx\\ &= \frac{ip}{\hbar} \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar}\psi(x)\,dx\\ &= \frac{ip}{\hbar}\phi(p), \end{aligned}$$

where the endpoint term was set to zero. Therefore,

$$\mathcal F_\hbar[\hat p\psi](p) = p\phi(p).$$

Next,

$$\frac{d\phi}{dp} = -\frac{i}{\hbar} \mathcal F_\hbar[x\psi](p),$$

so

$$\mathcal F_\hbar[x\psi](p) = i\hbar\frac{d\phi}{dp}.$$

Hence $\hat p$ is multiplication by $p$ in momentum space, while $\hat x$ is $i\hbar\,d/dp$.

2. If $\mathcal F_\hbar[\psi]=\phi$, find the transform of

   $$\chi(x) = e^{ip_0x/\hbar}\psi(x-a).$$

Solution

First translate:

$$\mathcal F_\hbar[\psi(x-a)](p) = e^{-ipa/\hbar}\phi(p).$$

Multiplication by $e^{ip_0x/\hbar}$ then replaces $p$ by $p-p_0$ in this result. Therefore,

$$\mathcal F_\hbar[\chi](p) = e^{-i(p-p_0)a/\hbar} \phi(p-p_0).$$

The translation changes phase, and the modulation shifts the probability density:

$$\left\lvert \mathcal F_\hbar[\chi](p) \right\rvert^2 = \lvert\phi(p-p_0)\rvert^2.$$

3. Transform the centered normalized Gaussian

   $$\psi(x) = \left( \frac{1}{2\pi\sigma_x^2} \right)^{1/4} e^{-x^2/(4\sigma_x^2)}.$$

   Read off its momentum standard deviation.

Solution

Use the Gaussian integral

$$\int_{-\infty}^{\infty} e^{-ax^2+bx}\,dx = \sqrt{\frac{\pi}{a}} e^{b^2/(4a)}.$$

This formula holds for $\operatorname{Re}a>0$.

Here $a=1/(4\sigma_x^2)$ and $b=-ip/\hbar$. Substitution gives

$$\phi(p) = \left( \frac{2\sigma_x^2}{\pi\hbar^2} \right)^{1/4} e^{-\sigma_x^2p^2/\hbar^2}.$$

Therefore,

$$\lvert\phi(p)\rvert^2 \propto \exp\left( -\frac{2\sigma_x^2p^2}{\hbar^2} \right).$$

Comparing with $\exp[-p^2/(2\sigma_p^2)]$ gives

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

Both transforms are normalized, and $\sigma_x\sigma_p=\hbar/2$.

4. Let $\widetilde\psi(k)$ use the symmetric wave-number convention. Derive the relation between $\widetilde\psi(k)$ and $\phi(p)$, then verify that normalization is unchanged.

Solution

The two definitions are

$$\begin{aligned} \widetilde\psi(k) &= \frac{1}{\sqrt{2\pi}} \int e^{-ikx}\psi(x)\,dx,\\ \phi(p) &= \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar}\psi(x)\,dx. \end{aligned}$$

Set $k=p/\hbar$. The integrals then agree, while the prefactors differ by $1/\sqrt{\hbar}$:

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

Since $dp=\hbar\,dk$,

$$\begin{aligned} \int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp &= \int_{-\infty}^{\infty} \frac{1}{\hbar} \lvert\widetilde\psi(k)\rvert^2 \hbar\,dk\\ &= \int_{-\infty}^{\infty} \lvert\widetilde\psi(k)\rvert^2\,dk. \end{aligned}$$

The amplitude factor and measure Jacobian cancel exactly.

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.
• R. N. Bracewell, The Fourier Transform and Its Applications, 3rd ed., McGraw–Hill, 2000.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, World Scientific, 1998.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.