Thinking in Waves Episode 9 / Part IV · Constraints imposed on waves

Signal processing got there before quantum mechanics

The uncertainty principle is
the bandwidth theorem times \(\hbar\) For any wave, \(\Delta t\cdot\Delta\omega\ge1/2\) ── a theorem of signal processing, with no physics in it.
Measure it and a Gaussian gives exactly 0.5000; everything else is larger.
Substitute \(E=\hbar\omega\) and you have Heisenberg's inequality.

Prerequisites: Episode 1Computation: kenshou/fukakutei.js

Last time causality constrained a filter's real and imaginary parts.
This time, the theorem constraining its time width and frequency width. And this one has a famous face.

01First, measure it without quantum mechanics

The Gabor limit $$\Delta t\cdot\Delta\omega\ \ge\ \frac12$$

A short pulse necessarily has a wide band ── that is the whole theorem

Build various waveforms and measure the standard deviations on the time and frequency sides.

Waveform\(\Delta t\)\(\Delta\omega\)\(\Delta t\cdot\Delta\omega\)
Gaussian (σ=1)0.70710.70710.5000★ minimum
Gaussian (σ=3)2.12130.23570.5000★ minimum
Triangle (width 4)0.63250.86610.5478
Two-sided exponential (τ=1)0.70711.00030.7073
Rectangle (width 4)1.154426.3330.40(diverges)

Conclusion of §01

Only the Gaussian reaches 0.5; everything else is strictly larger.

★ Not once does \(\hbar\) appear in this calculation. The same inequality holds for sound, water waves and radio.

※ The rectangle's 30.4 actually diverges ── the sinc spectrum's tail falls only as \(1/\omega\), so the variance integral does not converge. Another way of saying "a waveform with corners has unbounded bandwidth".

02Now multiply by \(\hbar\)

All quantum mechanics adds is two correspondences.

Wave units → particle units $$E=\hbar\omega,\qquad p=\hbar k$$ $$\Delta t\cdot\Delta\omega\ge\tfrac12\ \longrightarrow\ \boxed{\Delta t\cdot\Delta E\ge\tfrac\hbar2},\qquad \Delta x\cdot\Delta k\ge\tfrac12\ \longrightarrow\ \boxed{\Delta x\cdot\Delta p\ge\tfrac\hbar2}$$

The heart of §02

The content of the uncertainty principle is a Fourier theorem.
All \(\hbar\) does is convert "wave units" into "particle units".

ConstantWhat it convertsUnitsNote
\(c\)space ↔ timem/sEpisode 4; a defined value
\(\hbar\)wave ↔ particleJ·s★ this episode; defined since 2019
\(k_B\)temperature ↔ energyJ/Kdefined since 2019

All three are exchange rates, and all three became defined values. Episode 4's "dimensionful is bookkeeping" is at work here too.

03A common misconception ── it is not measurement disturbance

ExplanationCorrect?Reason
Because measurement disturbs×Heisenberg's microscope is intuitive but not the essence
Because instruments are imperfect×a perfect instrument still obeys it, as long as it is a wave
Because it is a wave○★ measured without quantum mechanics in §01

The evidence is §01 itself

There is no \(\hbar\) in that calculation. All quantum mechanics contributed is the single fact that particles are also waves; the inequality itself comes from tools Fourier already had in 1822.
The surprising thing was never the inequality — it was that particles are waves.

04Comparison with measurement ── natural linewidth

The lifetime \(\tau\) of an excited state becomes the width of a spectral line ── a wave that lasts only a finite time has width.

Natural linewidth $$\Gamma=\frac{\hbar}{\tau}$$
TransitionLifetime \(\tau\)\(E\) [eV]\(\Gamma\) [eV]\(\Gamma/E\)
Hydrogen 2p → 1s1.6 ns10.2\(4.1\times10^{-7}\)\(4.0\times10^{-8}\)
Sodium D line16 ns2.1\(4.1\times10^{-8}\)\(2.0\times10^{-8}\)
Fe-57 (Mössbauer)141 ns14 400\(4.7\times10^{-9}\)\(3.24\times10^{-13}\)

§04 ── gravity was measured with this width

Fe-57 has \(\Gamma/E=3.2\times10^{-13}\). Absorb the \(\gamma\) ray recoil-free and that width becomes usable directly.

In 1959 Pound and Rebka used it to measure the gravitational redshift:

$$\frac{gh}{c^2}=\frac{9.807\times22.5}{(3\times10^8)^2}=2.46\times10^{-15}$$

That is 0.76 % of the linewidth. A shift smaller than the width is still measurable if you track the line's centre (the position of a hill's summit is known far better than the hill is wide).
★ The bandwidth theorem fixes the width, not the precision of the centre.

05The commutation relation is Fourier conjugacy too

Multiplication and differentiation do not commute $$[x,\ d/dx]f = x f' - (xf)' = xf' - f - xf' = -f\quad\Longrightarrow\quad [x,\ d/dx]=-1$$ $$p=-i\hbar\frac{d}{dx}\quad\Longrightarrow\quad \boxed{[x,p]=i\hbar}$$
VariableConjugatePhysical quantityInequality
time \(t\)angular frequency \(\omega\)energy \(E=\hbar\omega\)\(\Delta t\Delta E\ge\hbar/2\)
position \(x\)wavenumber \(k\)momentum \(p=\hbar k\)\(\Delta x\Delta p\ge\hbar/2\)
angle \(\varphi\)integer \(m\)angular momentum \(L=\hbar m\)subtle, because of periodicity

The commutation relation restates "position and spatial frequency are Fourier conjugates". Here too, \(\hbar\) is only fixing the units.

06But \(\Delta E\Delta t\) is a different animal

InequalitySituationCharacter
\(\Delta x\Delta p\ge\hbar/2\)both are operatorsderivable exactly from the commutator
\(\Delta E\Delta t\ge\hbar/2\)time is not an operator★ needs interpretation to mean anything

The heart of §06 ── the wave view is the wider one

The "\(\Delta t\)" in \(\Delta E\Delta t\) means the time for a state to change, or an excited state's lifetime, or the time spent measuring ── it differs by context.

★ In wave language (§01), time and frequency are symmetric axes. Move to quantum mechanics and only one becomes an operator, creating the asymmetry ──
a sign that the "see it as a wave" view is broader than quantum mechanics.

07Why is the Gaussian minimal?

The reason is one line $$\text{the Fourier transform of a Gaussian is a Gaussian}$$

The same shape on both sides ── maximal symmetry

\(\sigma_t\) (set)measured \(\Delta t\)measured \(\Delta\omega\)product
0.50.353551.414210.50000
1.00.707110.707110.50000
2.01.414210.353550.50000
5.03.535530.141420.50000

However you vary \(\sigma\), the product is exactly 0.5. You can watch "narrow one side and the other widens by exactly as much" happen numerically.

Exercises

  1. Estimate the bandwidth of a 10 fs ultrashort laser pulse.
    Show answer
    \(\Delta\omega\ge1/(2\Delta t)=1/(2\times10^{-14})=5\times10^{13}\) rad/s.
    In frequency, \(\Delta\nu\ge8\times10^{12}\) Hz ── against a centre wavelength of 800 nm (\(\nu=3.7\times10^{14}\) Hz) that is at least 2 % of bandwidth.
    Which is why ultrashort pulses are always whitish. A monochromatic ultrashort pulse cannot be made.
  2. From this episode, give the evidence that "uncertainty is due to measurement disturbance" is wrong.
    Show answer
    That \(\hbar\) never appears in §01.
    With no measurement, no observer and no quantum mechanics — just an FFT of some waveforms — \(\Delta t\Delta\omega\ge1/2\) came out. Since the same inequality holds for sound waves, quantum disturbance cannot be the cause.
    (Disturbance-based uncertainty is formalised separately as Ozawa's inequality, a different statement from Heisenberg's.)
  3. In wave language, why do longer-lived levels give narrower lines?
    Show answer
    Because a wave that lasts longer has a better-defined frequency. \(\Delta\omega\ge1/(2\Delta t)\): the larger \(\Delta t\) (how long the wave lasts), the smaller the floor on \(\Delta\omega\).
    Fe-57's 141 ns is ten times longer than an atomic transition and its energy is a thousand times higher, so the relative width \(\Gamma/E\) falls to \(10^{-13}\) ── which is the resolution of Mössbauer spectroscopy.

Summary The surprise was never the inequality

For any wave, \(\Delta t\cdot\Delta\omega\ge1/2\). Measured: a Gaussian gives exactly 0.5000, a triangle 0.548, a two-sided exponential 0.707, a rectangle diverges. No \(\hbar\) anywhere in that calculation.

Substitute \(E=\hbar\omega\) and \(p=\hbar k\) and you have Heisenberg's inequality. \(\hbar\) merely converts wave units into particle units ── the same role as \(c\) (space↔time) and \(k_B\) (temperature↔energy), and all three are now defined values.

It survives comparison with measurement: the natural linewidth \(\Gamma=\hbar/\tau\). Fe-57's \(\Gamma/E=3.2\times10^{-13}\) is the resolution of Mössbauer spectroscopy, and Pound and Rebka used it to measure the gravitational redshift (0.76 % of the linewidth).

The commutator \([x,p]=i\hbar\) follows from one line — multiplication and differentiation do not commute ── a restatement of Fourier conjugacy.
Only \(\Delta E\Delta t\) is different, because time is not an operator. In wave language time and frequency are symmetric, so that asymmetry was born in the move to quantum mechanics ── the wave view is broader.

The uncertainty principle is a property of waves times \(\hbar\).
The surprise was not the inequality but the single fact that particles are waves.

This document is Episode 9 of the series "Thinking in Waves". The computation is in kenshou/fukakutei.js. On the numbers: width is defined as the standard deviation weighted by the intensity \(|f|^2\) ── change the definition and the numbers change (using full width at half maximum gives \(\Delta t\Delta\omega\ge2\ln2\approx1.39\), for instance). The rectangle's 30.40 depends on grid and window length; the true value diverges. The lifetimes are published values, and real spectral lines are usually dominated by Doppler or collisional broadening (the Mössbauer effect is special because it suppresses recoil and Doppler at once). The Pound–Rebka experiment was a precision affair with round-trip measurements and temperature corrections; our estimate shows only the underlying principle. Disturbance-based uncertainty is formalised separately by Ozawa's inequality; what this episode rejects is only the claim that Heisenberg's inequality originates in disturbance. ── To print, use your browser's Print → Save as PDF.

Print / PDF: ⌘+P (Ctrl+P on Windows). "Show answer" opens the solutions.