advanced30 min studyLesson 362 in Physics

Momentum representation and Fourier duality

Transform amplitudes between position and momentum with consistent units, normalization, and continuous-spectrum conventions.

Before you begin

On this page

The question for today

What you’ll understand

Apply a unitary Fourier convention to position and momentum amplitudes.
Explain why phase gradients shift momentum without shifting position density.
Distinguish continuous plane-wave distributions from discrete normalized periodic modes.

Position and momentum wavefunctions are different coordinate descriptions of the same state. They are related by a unitary Fourier transformation, so probability is conserved even though its density has different units in each representation. A narrow shape in position generally requires many momentum components. The connection concerns amplitudes and their phases; Fourier transforming a position probability density does not, in general, produce the momentum probability density.

Choose a convention and retain its factors

Take momentum eigenfunctions on the line proportional to exp(ipx/ℏ), delta normalized in momentum. The Fourier kernel then includes the inverse square root of two pi hbar. Momentum coordinates have the inverse square root of momentum as their amplitude units. Parseval's relation makes the total position and momentum probabilities equal. Switching from momentum p to wavenumber k=p/ℏ changes both the measure and the amplitude normalization, so factors of hbar cannot be dropped without redefining the represented function.

In momentum representation, momentum acts by multiplication and position acts as a derivative with an i hbar factor on a suitable domain. This reverses the convenient roles they have in position representation. No physical measurement or dynamical rotation is required to change coordinates. An active unitary operation changes a state relative to fixed observables; a passive representation change transforms state coordinates and observable matrices together, preserving all their paired predictions.

\phi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{-ipx/\hbar}\psi(x)\,dx,\qquad \psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{ipx/\hbar}\phi(p)\,dp

Here x is in m, p in kg m/s, and ℏ in J s. The transform is unitary on L² with the corresponding measures; plane-wave kernels are generalized eigenfunctions. The formulas are full-line identities, with the usual L² interpretation when pointwise integrals do not converge absolutely.↔ Scroll wide equations horizontally.

A phase ramp is a momentum translation

Multiplying a position wavefunction by exp(ip0x/ℏ) leaves its position density fixed because the factor has unit magnitude. In the Fourier integral it replaces p by p−p0, translating the momentum distribution. Relative phase across position therefore contains measurable information missing from the position density alone. Two preparations with identical spatial detection histograms can have different mean momenta and evolve differently under the same free Hamiltonian.

A Gaussian makes the scale tradeoff exact. Its spatial amplitude width parameter is not its position standard deviation: squaring the amplitude narrows the Gaussian exponent. Fourier transformation gives another Gaussian whose momentum width varies inversely with the spatial width. For this unchirped minimum-uncertainty family, the product of standard deviations equals hbar divided by two. Adding a quadratic phase can increase momentum variance while preserving the same position density, so the equality does not follow from a Gaussian position histogram alone.

\psi(x)=\frac{e^{-x^2/(2a^2)+ip_0x/\hbar}}{(\pi a^2)^{1/4}}\ \Longrightarrow\ \phi(p)=\left(\frac{a^2}{\pi\hbar^2}\right)^{1/4}e^{-a^2(p-p_0)^2/(2\hbar^2)},\quad \Delta X=\frac{a}{\sqrt2},\quad \Delta P=\frac{\hbar}{\sqrt2a}

a>0 is a length, p0 is a momentum, and the displayed family is normalized on the line with no quadratic phase. Δ denotes standard deviation, not a support width. Its uncertainty product is ℏ/2; generic states need not saturate that bound.↔ Scroll wide equations horizontally.

Spectrum and geometry determine sum or integral

A full-line plane wave has constant modulus and infinite norm. It represents a continuous-spectrum generalized eigenstate, useful inside integrals but unavailable as a unit-normalized preparation. A normalizable packet superposes these generalized eigenfunctions with a square-integrable amplitude. Approximating a packet by a very long nearly uniform region does not make its edges or its finite momentum spread disappear at finite size.

On a circle of circumference L, periodicity instead permits normalized plane-wave modes with discrete momenta 2πℏn/L. A Fourier series replaces the full-line integral. Changing the boundary conditions changes the allowed spectral problem rather than merely choosing a different notation. The distinction is especially important when a formal derivative expression is called momentum: its domain and physical geometry decide whether discrete modes, generalized modes, or another self-adjoint realization is appropriate.

Work it through

Widths of a phase-shifted Gaussian

Use the displayed Gaussian with a=2.0 nm and $p_0=3.0\hbar/\mathrm{nm}$ . Find the two standard deviations, mean momentum, and probability of measuring momentum above p0.

The position density is proportional to exp(−x²/a²), giving $\Delta X=a/\sqrt2=\sqrt2$ nm. The parameter a itself is larger than the standard deviation, so using a directly would introduce a factor of square root two.
The momentum density is centred at p0 with $\Delta P=\hbar/(\sqrt2a)=\hbar/(2\sqrt2\,\mathrm{nm})$ . Their product is exactly ℏ/2, consistent with an unchirped normalized Gaussian.
The phase ramp centres the momentum distribution at $\langle P\rangle=3.0\hbar/\mathrm{nm}$ . Its symmetric continuous Gaussian has half its probability above that centre, so P(P>p0)=1/2 despite the nonzero mean.

The result

The standard deviations are 1.414 nm and 0.354 ℏ/nm, the mean is 3.0 ℏ/nm, and the above-mean momentum probability is 1/2. The phase shift does not alter the position density.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Phase hidden from one basis

Reset and set both controls to Mixing angle θ=90° and Relative phase φ=0°. Predict equal zero/one probabilities and Bloch x=1. For a fresh comparison, Reset and set both controls to θ=90° and φ=180°; the basis probabilities remain equal while Bloch x becomes −1. This finite two-mode analogy isolates phase information that a single basis histogram misses.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forIdentical probabilities in one basis do not establish identical states. The lab shows pure two-level geometry and coherent phase precession, not a position–momentum Fourier transform or a continuous packet.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The factor has unit magnitude at each x, preserving position density. Its x-dependent phase changes the Fourier kernel so that φ(p) becomes φ(p−p0). Only a constant phase is global; a finite packet retains its momentum spread.

Your turn

Try it yourself

On a periodic circle of circumference L, prepare $(|n=1\rangle+i|n=-1\rangle)/\sqrt2$ using normalized momentum modes. Find momentum probabilities, mean, and variance.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The mode momenta are ±2πℏ/L. Square the coefficients before summing moments; a zero mean need not imply a definite zero value.

Compare with the solution

Each coefficient has squared magnitude 1/2, so the outcomes +2πℏ/L and −2πℏ/L are equally likely. Their mean is zero and their second moment is $(2\pi\hbar/L)^2$ , which is also the variance. This state is not the n=0 momentum eigenstate: that state has zero variance. The relative factor i affects interference in a complementary representation but does not change these diagonal momentum probabilities.

Keep the idea

Take it with you

Fourier duality transforms amplitudes with a convention-dependent measure and normalization.
A spatial phase gradient can change momentum statistics without changing position density.
Normalizable periodic modes and generalized full-line plane waves belong to different spectral geometries.

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