Heisenberg Uncertainty Principle

Equivalent forms

\Delta E \, \Delta t \geq \hbar/2

A single inequality that forbids classical determinism.

Symbol	Name	SI unit	Dimension	Range
$Delta_x$	Position uncertainty Standard deviation of position	m	L	1e-18 – 0.001
$Delta_p$	Momentum uncertaintyoutput Standard deviation of momentum	kg*m/s	MLT^-1	1e-30 – 1e-18
$hbar$	Reduced Planck constant h / (2*pi)	J*s	ML^2T^-1	1.054571817e-34 – 1.054571817e-34

Unit systems

Symbol	SI	CGS	Imperial
$Delta_x$	m	cm	ft
$Delta_p$	kg*m/s	g*cm/s	lb*ft/s
$hbar$	J*s	erg*s	ftlbfs

Where it holds

Holds for any quantum state and any pair of non-commuting observables with [A,

B] = i

*hbar. Generalized form: Delta_A *

\Delta _B \ge

|<[A,B]>|/2.

Dimensional analysis

[L] * [M*L*T^-1] \to [M*L^2*T^-1]

hbar

\to [M*L^2*T^-1]

Both sides are action.

\checkmark

Discovery

Werner Heisenberg · 1927

Heisenberg derived this limit from matrix mechanics; Kennard and Robertson later formalized it as a variance inequality.

Try this

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

Quantum mechanics says no — the product of position and momentum uncertainties has a hard lower bound. How precisely can you locate an electron confined to 1 Å?

Research status: stable

Real-world applications

Zero-point motion of atoms in crystals
Squeezed-light interferometry (LIGO gravitational wave detection)
Natural linewidth of atomic transitions (energy-time form)
Fundamental limit on electron microscope resolution

Common misconceptions

The uncertainty is NOT due to measurement disturbance — it is an intrinsic property of the state
It does NOT mean 'observer affects reality' in a mystical sense
The factor of 2 matters: the lower bound is hbar/2, not h or hbar

Experimental verification

Single-slit diffraction patterns, electron microscopy resolution limits, and precision measurements of squeezed light (LIGO) all confirm the relation. Ozawa-type uncertainty refinements measured directly with neutron spin experiments.

Derivation

Start from [x,

p] = i

*hbar.

For any state |psi>, the Robertson inequality gives Delta_A *

\Delta _B \ge

|<[A,B]>|/2.

Substituting [x,

p] = i

*hbar yields

\Delta _x * \Delta _p \ge

hbar/2.

A Gaussian wavefunction saturates the bound.

Limiting cases

\Delta _x \to 0

⟶

\Delta _p \to \infty

Perfectly localized particle has completely undetermined momentum.

\Delta _x \to \infty

⟶

\Delta _p \to 0

Plane-wave momentum eigenstate is completely delocalized in space.

Gaussian wave packet⟶

\Delta _x*\Delta _p =

hbar/2 (equality)The minimum-uncertainty state — saturates the bound.

What if…

What if hbar

\to 0

?

Classical limit — position and momentum can be known simultaneously with arbitrary precision.

What if we measure Delta_x to 10 fm (nuclear scale)?

$\Delta _p \ge 5.27e-21\,\mathrm{kg}*m/s$ ; Delta_v for an electron exceeds c, signalling relativistic QM is needed.

What if we use a non-minimum uncertainty state?

Product Delta_x*Delta_p > hbar/2 strictly — e.g., higher excited states of the harmonic oscillator.

1

Electron in a 1 Angstrom box

Given ·

Delta x:: 1e-10

Find · Delta_p_min, Delta_v_min

Steps

$\Delta _p \ge$ hbar / $(2 * \Delta _x) = 1.054571817e-34 / 2e-10$
$\Delta _p \ge 5.27e-25\,\mathrm{kg}*m/s$
$\Delta _v = \Delta _p / m_e = 5.27e-25 / 9.1093837015e-31 \approx 5.79e_{5} m/s$

Result ·

\Delta _p \ge 5.27e-25\,\mathrm{kg}*m/s

;

\Delta _v \ge 5.79e_{5} m/s

2

Energy-time: atomic level lifetime

Given ·

Delta t:: 1e-8

Find · Delta_E_min

Steps

$\Delta _E \ge$ hbar / (2 * Delta_t)
$= 1.054571817e-34 / 2e-8$
$\approx 5.27e-27\,\mathrm{J}$

Result ·

\Delta _E \ge 5.27e-27\,\mathrm{J} \approx 3.3e-8\,\mathrm{eV}

(natural linewidth)

de Broglie Wavelength→Time-Dependent Schrödinger Equation→Particle in a 1D Box→