Heisenberg Uncertainty Principle

1.0What Is the Uncertainty Principle?

The Heisenberg Uncertainty Principle, also known as the Principle of Uncertainty, states that it's impossible to know the exact location and velocity of a subatomic particle simultaneously. This means that the more precisely we determine the particle's position, the less accurately we can measure its velocity, and vice versa. The uncertainty principle is a fundamental concept in quantum mechanics and has significant implications for our understanding of the behavior of atoms and subatomic particles.

2.0Introduction

Particles sometimes act like waves, and we find them more often where these waves are strongest. But when these waves are really wavy, we can't pin down their exact size. This uncertainty about the wave's size affects how fast the particle moves.

So, if we know exactly where a particle is, we can't be sure how fast it's going. And if we know its speed really well, we can't pinpoint its exact position. It's like a trade-off: the better we know one thing, the fuzzier the other thing gets. This is just how particles roll in the quantum world! That’s why we are going to discuss The Principle of Uncertainty, in detail.

3.0Origin of the Heisenberg Uncertainty Principle

In 1927, German physicist Werner Heisenberg introduced the Uncertainty Principle, stating that the position and momentum of any particle cannot be precisely measured at the same time. 

There's a limit to how accurately both these properties can be known simultaneously, setting a minimum threshold for the product of their uncertainties. This principle extends beyond position and momentum to energy and time uncertainties, revealing a minimum limit for their product as well. These limitations stem from the wave-like behaviors fundamental to the quantum description of nature.

4.0Heisenberg Uncertainty Principle Formula

Mathematical Expression of Heisenberg Uncertainty Principle:

The Heisenberg Uncertainty Principle formula is often expressed mathematically as:

$\Delta x.\Delta p\ge \frac{h}{4\pi }$

Where:

• Δx represents the uncertainty in position.
• Δp represents the uncertainty in momentum.
• h is the Planck’s constant, and its value is, equal to 6.6260715×10^-34 joule seconds 

Since momentum, p = mv, Heisenberg uncertainty principle formula can be alternatively written as,

$\Delta x\cdot m\Delta v\ge \frac{h}{4\pi }$

So above equation can define Heisenberg uncertainty relation, Applying the Heisenberg principle to an electron in an orbit of an atom, with h = 6.6260715×10^-34 and 

m= 9.11 ×10^-31Kg  (9.11 ×10^-28 gm.) 

$\begin{array}{rl}\Delta \mathrm{x}\cdot \Delta \mathrm{v}\ge & \frac{6.624\times 10^{-27}}{4\pi \mathrm{m}}\ge \frac{6.624\times 10^{-27}}{4\times 3.14\times 9.10\times 10^{-28}}\\ & =0.57\mathrm{erg}\text{ sec per gram approximately }\end{array}$

If the electron's position is precisely measured to its size (10^-10 meters), the error in measuring its velocity will be equal to or larger than 10⁶ meters, which is equivalent to 1000 kilometers.

Which shows that Heisenberg's principle applies to only dual-natured microscopic particles and not to macroscopic particles whose wave nature is minimal.

5.0Questions Related to Heisenberg’s Principle

Question.1 Calculate the uncertainty in the position of a particle when the uncertainty in momentum is

(a) 1 × 10^–3 g cm s^–1                                  

(b) zero.

Solution:

(a) Given Δp = 1 × 10^–3 g cm s^–1, h = 6.62 × 10^–27 erg s, 𝝅 = 3.142

According to Heisenberg’s uncertainty principle

$\Delta \mathrm{x}\cdot \Delta \mathrm{p}\ge \frac{\mathrm{h}}{4\pi }\text{ or }\Delta \mathrm{x}\ge \frac{\mathrm{h}}{4\pi }\cdot \frac{1}{\Delta \mathrm{p}}\ge \frac{6.62\times 10^{-27}}{4\times 3.142}\times \frac{1}{10^{-3}}\ge 0.527\times 10^{-24}\mathrm{cm}$        

(b)   When the value of Δp = 0, the value of Δx will be infinity.

Question.2 Calculate the a cricket ball of mass 150 g if the uncertainty in its position is of the order of 1Å (h = 6.6 × 10^–34 kg m² s^–1).

Solution:

$\Delta \mathrm{x}.\mathrm{m}\Delta \mathrm{v}=\frac{\mathrm{h}}{4\pi }\text{ or }\Delta \mathrm{v}=\frac{\mathrm{h}}{4\pi \Delta \mathrm{x}.\mathrm{m}}=\frac{6.6\times 10^{-34}}{4\times 3.143\times 10^{-10}\times 0.150}=3.499\times 10^{-24}{\mathrm{ms}}^{-1}$