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Heisenberg Uncertainty Principle Calculator

Compute the minimum uncertainty allowed by quantum mechanics with Δx·Δp ≥ ħ/2 — enter the position uncertainty to get the minimum momentum uncertainty, or solve the other way.

 

Formula

$$ \Delta x\,\Delta p \ge \dfrac{\hbar}{2} \qquad \Delta p_{min}=\dfrac{\hbar}{2\,\Delta x} \qquad \Delta x_{min}=\dfrac{\hbar}{2\,\Delta p} $$
ħ = 1.0546×10⁻³⁴ J·s (reduced Planck constant)

Worked example

Confining an electron to an atom-sized Δx = 1×10⁻¹⁰ m forces a momentum uncertainty of at least \( \Delta p = \hbar/(2\Delta x) = (1.0546\times10^{-34})/(2\times10^{-10}) \approx 5.3\times10^{-25}\ \text{kg·m/s} \).

How it works

Heisenberg's uncertainty principle sets a fundamental limit: the more precisely a particle's position is known, the less precisely its momentum can be, and vice versa. The product of the uncertainties can never fall below ħ/2: Δx·Δp ≥ ħ/2.

This is not a measurement flaw but a property of quantum systems. The calculator returns the minimum uncertainty in one quantity given the other, using the reduced Planck constant ħ = 1.0546×10⁻³⁴ J·s.

Frequently asked questions

What is the Heisenberg uncertainty principle?

It states that the position and momentum of a particle cannot both be known exactly: Δx·Δp ≥ ħ/2. Reducing the uncertainty in one increases it in the other.

What is the value of ħ (h-bar)?

The reduced Planck constant ħ = h/2π ≈ 1.0546×10⁻³⁴ J·s. It sets the scale of the smallest possible uncertainty product.

Does the uncertainty principle come from imperfect instruments?

No. It is a fundamental property of quantum systems, not a limitation of measurement tools. Even with perfect instruments the product Δx·Δp cannot be less than ħ/2.

Why don't we notice it in everyday life?

Because ħ is extraordinarily small. For macroscopic objects the required uncertainties are far too tiny to detect; the principle only matters at atomic and subatomic scales.

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