Heisenberg Uncertainty Principle

Equivalent forms

\Delta E \cdot \Delta t \geq \frac{\hbar}{2}

An inequality that limits what can be known about reality itself.

Symbol	Name	SI unit	Dimension	Range
$Δx$	Position uncertainty Standard deviation of position	m	L	1e-15 – 0.000001
$Δp$	Momentum uncertaintyoutput Standard deviation of momentum	kg·m/s	M·L·T⁻¹	1e-30 – 1e-20

Unit systems

Symbol	SI	CGS	Imperial
$Δx$	m	cm	ft
$Δp$	kg·m/s	g·cm/s	slug·ft/s

Where it holds

Universal in non-relativistic and relativistic quantum mechanics for any pair of conjugate observables ([Â,B̂

] = i\hbar )

. The energy-time form is heuristic and requires care because t is a parameter, not an operator.

Dimensional analysis

$[\Delta x]\cdot [\Delta p] = L \cdot M\cdot L\cdot T^{-1} = M\cdot L^{2}\cdot T^{-1} = [\hbar ] \checkmark$

Discovery

Werner Heisenberg · 1927

Heisenberg derived the principle while developing matrix mechanics, formalizing quantum indeterminacy.

Try this

Can you know exactly where an electron is AND how fast it's moving?

An electron's position is known to within 1 nm. What is the minimum uncertainty in its velocity?

Research status: stable

Real-world applications

Zero-point energy of atoms and molecules — atoms cannot sit still even at $T = 0$ .
Stability of matter — electrons can't fall into nuclei because of position-momentum trade-off.
Squeezed light in gravitational-wave detectors (LIGO) and quantum sensing.
STM and quantum dot design — tunneling and confinement energies.

Common misconceptions

It is NOT a measurement disturbance principle — it's a property of the quantum state itself, even when no measurement is made.
$\Delta x$ $and \Delta p$ are statistical standard deviations, not 'errors' or 'measurement precision'.
It does not say 'observation creates reality' — it constrains what *can* be specified simultaneously.

Experimental verification

Single-slit electron diffraction directly demonstrates the trade-off: narrowing the slit (smaller

\Delta x)

widens the diffraction pattern (larger

\Delta p)

. Modern squeezed-light experiments at LIGO exploit the inequality to suppress quantum noise in one quadrature by inflating the other.

Derivation

From the canonical commutator [x̂, p̂

] = i\hbar and

the Cauchy–Schwarz inequality applied

to \psi in

Hilbert space, one derives the Robertson inequality

\sigma

_

A \sigma _B

≥ |⟨[Â,B̂]⟩|/2.

Substituting [x̂,p̂

] = i\hbar

gives

\sigma _x \sigma _p \ge \hbar /2

.

The saturation case is the Gaussian wavepacket.

Limiting cases

\Delta x \to 0

⟶

\Delta p \to \infty

Perfect localization requires infinite momentum spread — a position eigenstate has all momenta equally.

\Delta p \to 0

⟶

\Delta x \to \infty

A perfect plane wave has definite momentum but is delocalized over all space.

\Delta x\cdot \Delta p = \hbar /2

⟶ Saturation (Gaussian wavepacket)Equality is achieved only for minimum-uncertainty Gaussian states; all others lie above the bound.

What if…

What

if \hbar

were as large as

1 J\cdot s

?

Macroscopic objects would obviously delocalize. A 1 kg ball localized to 1 m would have $\Delta v \ge 0.5\,\mathrm{m/s}$ — quantum effects would dominate everyday life.

What if you tried to confine an electron to a nuclear-size region

(\sim 1\,\mathrm{fm})

?

$\Delta p \ge \hbar /(2\times 10^{-15}) \approx 5\times 10^{-20} kg\cdot m/s$ , requiring relativistic energies $\sim 100\,\mathrm{MeV}$ . This is why electrons cannot live inside nuclei $(\beta$ -decay creates them).

What

if \Delta x\cdot \Delta p = 0

were allowed?

We could specify exact phase-space points, recovering classical mechanics. Atoms would collapse, lasers wouldn't lase, and chemistry would not exist as we know it.

1

Electron localized to 1 nm

Given ·

Δx:: 1e-9
m e:: 9.109e-31

Find · minimum

\Delta v

Steps

$\Delta p \ge \hbar /(2\Delta x) = 1.055\times 10^{-34} / (2\times 10^{-9}) \approx 5.27\times 10^{-26} kg\cdot m/s$
$\Delta v = \Delta p/m_e = 5.27\times 10^{-26} / 9.109\times 10^{-31} \approx 5.79\times 10^{4} m/s$
Locking an electron in a 1 nm box forces $\sim 58\,\mathrm{km/s}$ velocity spread.

Result ·

\Delta v \ge \hbar /(2\,\mathrm{m}_e \Delta x) \approx 5.79\times 10^{4} m/s

2

1 g particle localized to 1 μm

Given ·

Δx:: 0.000001
m:: 0.001

Find · minimum

\Delta v

Steps

$\Delta p \ge 1.055\times 10^{-34} / (2\times 10^{-6}) \approx 5.27\times 10^{-29} kg\cdot m/s$
$\Delta v = 5.27\times 10^{-29} / 10^{-3} \approx 5.3\times 10^{-26} m/s$
Utterly negligible — uncertainty is invisible at macroscopic mass.

Result ·

\Delta v \ge \hbar /(2m\Delta x) \approx 5.3\times 10^{-26} m/s

De Broglie Wavelength→Schrödinger Equation (Time-Independent)→