Heisenberg Uncertainty Principle (original) (raw)

Last Updated : 8 May, 2026

The Heisenberg Uncertainty Principle is an important concept in quantum mechanics that explains the limitations in measuring the exact position and motion of tiny particles such as electrons. According to this principle, it is impossible to determine both the exact position and the exact momentum (velocity × mass) of a moving particle at the same time.

heisenberg_s_principle

where:

Reason for the Impossibility of Measuring Position and Momentum Simultaneously

**Heisenberg Uncertainty Principle Equation

The Heisenberg Uncertainty Principle can be expressed using mathematical equations that show the relationship between the uncertainties in different physical quantities of microscopic particles.

The equations are:

**1. Position and Momentum Uncertainty Equation

This equation states that the uncertainty in position and uncertainty in momentum cannot both be very small at the same time.

\Delta x \, \Delta p \geq \frac{h}{4\pi}

where:

**2. Energy–Time Uncertainty Equation

Another form of the uncertainty principle relates energy and time. This equation means that the uncertainty in the energy of a system and the uncertainty in time cannot both be very small simultaneously.

It is given by:

\Delta E \, \Delta t \geq \frac{h}{4\pi}

where:

**3. Equation in Terms of Velocity

Since momentum is given by:

p = mv

The uncertainty in momentum becomes:

Δp = mΔv

Substituting this in the main equation:

\Delta x .\, m \Delta v \geq \frac{h}{4\pi}

or

\Delta x \, \Delta v \geq \frac{h}{4\pi m}

Heisenberg’s γ-ray Microscope

The Gamma-ray microscope is a thought experiment used to explain the Heisenberg Uncertainty Principle. It was proposed by Werner Heisenberg to show why it is impossible to determine the exact position and momentum of an electron at the same time.

**Basic Idea

To observe a very small particle such as an electron, we must use a microscope. The accuracy with which we can see the particle depends on the wavelength of the light used in the microscope.

Thus, both quantities cannot be measured exactly at the same time. Since electrons are extremely small, ordinary visible light is not sufficient to locate them accurately. Therefore, very short wavelength radiation such as gamma rays is required. This leads to the concept of the gamma-ray microscope.

**Working

heisenberg

The gamma-ray microscope experiment proposed by Werner Heisenberg explains why the exact position and momentum of an electron cannot be measured at the same time.

The working process can be understood step-by-step from the diagram:

**1. Emission of Gamma Rays

**2. Passing Through Narrow Slits

**3. Interaction with the Electron

**4. Detection on the Screen

**5. Disturbance of Electron Momentum

Gamma-ray photons have very high energy and momentum. When the photon strikes the electron:

**6. Uncertainty

Applications of Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle is used in several fields of physics, technology, and even philosophy nowadays. Here are some notable applications:

Solved Examples

**Example 1: A particle's position is known with an uncertainty of Δx=10⁻¹⁰ meters. Calculate the minimum uncertainty in its momentum according to the Heisenberg Uncertainty Principle.

**Solution:

According to the Heisenberg Uncertainty Principle:

Given:

Δx = 10-10 meters

Using the uncertainty principle equation.

Δp ≥ ℏ/2Δx

Δp ≥ (1.05×10-34m2kg/s) / (2×10-10m)

Δp ≥ (1.05×10-34m) / (2×10-10) kg m/s

Δp ≥ 5.25×10-25 kg m/s

So, the minimum uncertainty in momentum is Δp ≥ 5.25×10-25 kg m/s.

**Example 2: The position of an electron is known to an accuracy of 10⁻¹⁰ m. What is the minimum uncertainty in its velocity?.(Given h = 6.63×10⁻³⁴ JS and me = 9.1×10⁻³¹ kg)

**Solution:

We know Heisenberg's Uncertainty Principle

Δx⋅Δp ≥ h/2

Where:

Given Δx=10-10 m, h=6.63×10-34, and find Δp=?.

Δx Δp ≥ h/4π

Δp ≥ 6.63×10-34/4π×10-10 kg m/s

Δp≥5.28 x 10-25kg m/s

This is the minimum uncertainty in momentum. Now, using the relation p=mv, we can find the minimum uncertainty in velocity. Since

9.1×10-31 kg for an electron.

Δv = 5.28×10-25 / 9.1×10-31

Δv ≈5.8×105 m/s

So, the minimum uncertainty in the velocity of the electron is 5.8×105 m/s.