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beginner · Physics · The Uncertainty Principle

What Uncertainty Means

One of the most striking results in all of physics is that quantum mechanics forbids, in principle, knowing two particular pairs of physical quantities with arbitrary precision at the same time. This is not a statement about imperfect instruments — it is a fundamental feature of nature encoded in the mathematical structure of quantum states.

Standard deviations in quantum mechanics

In quantum mechanics, a physical observable AA is represented by a Hermitian operator A^\hat{A}. When a system is in state ψ|\psi\rangle, the expected value of a measurement is

A=ψA^ψ.\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle.

The spread of outcomes around that mean is captured by the standard deviation

ΔA=A2A2,\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2},

which is zero only if ψ|\psi\rangle is an eigenstate of A^\hat{A}. In every other state there is irreducible spread — not because the measurement is rough, but because the observable genuinely does not have a sharp value in that state.

The Heisenberg uncertainty principle

Heisenberg (1927) identified a special pair: position xx and momentum pp. Their operators satisfy the canonical commutation relation

[x^,p^]=i,[\hat{x},\, \hat{p}] = i\hbar,

where =h/2π1.055×1034 Js\hbar = h/2\pi \approx 1.055 \times 10^{-34}\ \text{J}\cdot\text{s} is the reduced Planck constant. A rigorous derivation using the Cauchy–Schwarz inequality on Hilbert space (found in any standard text) shows that for any state:

ΔxΔp    2.\Delta x \cdot \Delta p \;\geq\; \frac{\hbar}{2}.

This is the Heisenberg uncertainty principle. The product of the position standard deviation and the momentum standard deviation can never fall below /2\hbar/2, no matter what state the particle is in and no matter how carefully you measure.

What the principle is NOT saying

A common misconception is that the uncertainty principle is about the disturbance caused by measurement — the idea that shining a photon on an electron to locate it unavoidably kicks it. While that picture has a grain of truth for some measurement schemes, the real content of ΔxΔp/2\Delta x \cdot \Delta p \geq \hbar/2 is more radical: the spread Δx\Delta x and Δp\Delta p are properties of the quantum state itself before any measurement happens. A particle prepared in a state with very narrow position spread (Δx\Delta x small) necessarily has a wide momentum spread (Δp\Delta p large), and vice versa.

A numerical feel for the bound

To see how small /2\hbar/2 is in everyday terms, consider a grain of sand of mass m1 μg=109 kgm \approx 1\ \mu\text{g} = 10^{-9}\ \text{kg}. If its position is known to Δx=1 nm=109 m\Delta x = 1\ \text{nm} = 10^{-9}\ \text{m}, the momentum uncertainty is at least

\approx 5.3 \times 10^{-26}\ \text{kg}\cdot\text{m/s}.$$ The corresponding velocity uncertainty $\Delta v = \Delta p / m \approx 5 \times 10^{-17}\ \text{m/s}$ is utterly unmeasurable — hence quantum uncertainty is irrelevant for macroscopic objects. For an **electron** ($m_e \approx 9.1 \times 10^{-31}\ \text{kg}$) confined to a hydrogen atom ($\Delta x \sim 10^{-10}\ \text{m}$), the same formula gives $\Delta v \gtrsim 6 \times 10^5\ \text{m/s}$ — about 0.2% of the speed of light. Quantum mechanics is essential at atomic scales. ## Other conjugate pairs Position and momentum are not the only pair subject to an uncertainty relation. Energy and time satisfy the analogous bound $$\Delta E \cdot \Delta t \;\geq\; \frac{\hbar}{2},$$ which governs the natural linewidth of atomic transitions: a short-lived excited state ($\Delta t$ small) emits a photon with a broad spread of energies ($\Delta E$ large). Spin components along perpendicular axes obey similar relations arising from their non-commuting operators. The unifying theme is the commutator: whenever $[\hat{A}, \hat{B}] \neq 0$, there is an irreducible lower bound on $\Delta A \cdot \Delta B$. The next lessons in this module make that connection precise.

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