 # Heisenberg uncertainty principle and its application

Introduction

Heisenberg uncertainty principle and its application Is established by de-Broglie. L. De Broglie established the fact that all material particles in motion have wave like characteristics. This dual nature character gives rise to question . In this article we are going to discuss about Heisenberg uncertainty principle and its application . According to Heisenberg , It is impossible to determine simultaneously the position and momentum of electron with 100% accuracy.If, however , an experiment is designed for the exact determination of one of them the determination of other will become completely indeterminant or uncertain and vice versa. ## theory of Heisenberg uncertainty principle and its application

### 1) uncertainty in position and momentum-

Let Δx is uncertainty in position of a particle and ΔPx is uncertainty in momentum (in x-direction) then , the product of these can never be smaller then h/2 .

Mathematically, the uncertainty principle is written as

ΔxΔPx ≥  h/2

Where h is called Planck’s constant . Whose value is

h = h/2π = 6.62 Χ 10 ¯³⁴/ 2×3.14 = 1.054 × 10⁻³⁴

From above equation it is clear that if the momentum of particle is known accurately, the uncertainty in momentum is zero. I.e. ΔPx = 0 , then  Δx is infinite. Which means that According to uncertainty principle the particle is represented by a wave of constant amplitude extending over all space.In other words , the particle can be localised any where between −∞  to  +∞.  It is impossible to know both position and momentum of particle at the same time with accuracy.

In classical mechanics , h is zero i.e value of Planck’s constant is zero . However, in quantum mechanics Planck’s constant is not zero.

### 2) uncertainty in energy and time-

According to Heisenberg , it is impossible that to determinen energy and time of a moving particle simultaneously,

∆E∆t ≥ h /2

### 3) uncertainty in Angel and momentum-

According to Heisenberg, it is impossible to determine angle and momentum of a moving particle simultaneously, with 100% accuracy.

∆θ∆J ≥ h /2

4) for microscopic particle uncertainty always →0

#### Experiment based on Heisenberg uncertainty principle

##### 1) conceptual gamma ray microscope for the determination of position of an electron- When a photon of certain wavelength λ and initial momentum P (h/λ) after scattering enters into field of microscope. It may be anywhere within the angle 2θ . So that uncertainty in momentum is given by

∆P = [Psin θ ]₋θ+Θ

∆P = 2Psin θ

Also, ∆P =2h sinθ/λ

By resolving power of Microscope

∆x ≈ λ/2 sinθ

Now multiply above two equation

∆x∆P = 2h sinθ/λ . λ/2 sinθ

Hence,∆x∆P = h

∆x∆P ≈ h >h/2

If a particle is moving with velocity v and it is found in the distance ∆x then uncertainty in time ∆t,

∆t =∆x /v

Hence kinetic energy E = 1mv²/2

∆E = ∆(mv²/2)

Or, ∆E = mv∆v

∆E =v ∆(mv)

And,∆E =v ∆P

∆t∆E= (∆x /v ). (v ∆P)

∆t∆E= ∆x∆P> h/2

##### Diffraction of electron by a single slit –

Now as an another example let us consider the wave nature of a particle. A narrow beam of electrons of momentum p falls on a slit AB of the width ‘a’ . The de-Broglie wavelength assciatied with these moving electrons is equal to h/p.the electron passes through the slit,but the point at which it passes through the slit is not known.therefore the uncertainty in the position of the electron in the y-direction will be of the order of the width of the slit.

Condition of the diffraction minima –

First diffraction minima due to single slit

dsinθ =nλ

∆ysinθ=λ

∆y = λ/sin θ

Uncertainty in momentum is

∆py= [psinθ]⁺θ₋θ

Or,

∆py= psinθ−(−psinθ)

∆py=2psinθ

By multiplying above two equations

∆py∆y=(  λ/sin θ).(2h sinθ/λ)

∆py∆y= 2h > h/4π

###### Application of Heisenberg uncertainty principle

1)Non existent of electron in nucleus :-

According to Heisenberg principle,

∆x∆P ≥h/2 (1)

∆x ≈ 5× 10⁻¹⁵ m (2)
Possibility of position of electron in nucleus
Then from first equation ,

∆p =h/2 ∆x∆p =h/2(5×10⁻¹⁵m)

∆P= 1.054×10⁻³⁴/2×(5×10⁻¹⁵)

∆p≈ 1.054×10⁻²⁰kgm/sec

This is the minimum momentum of electron from mass energy relation-

E= √p²c²+m₀c⁴

Emin = √p²minc²+m₀c⁴

Rest,mass energy of electron is of order 0.511mev which is very less then p²minc²

p²minc²>> m₀²c⁴

Emin = pmin c
Emin= 1.054×10⁻²⁰×3 ×10⁸joule
Emin= 20mev
But β-decay experiment shows that emitted electron dosn’t have more then two theree mev energy that is electron cannot exist in nucleus.

2) ground state energy of harmonic oscillator-

Total energy of a particle which is oscillating SHM

,E=K+U
E=(mv²/2 )+ (kx²/2)
E=(P²/2m) +(mω²x²)
Minimum energy of oscillator,
Emin= (∆P²/2m )+ (mω²(∆x²)/2)
Put in equation
Emin = (h/2∆x)²×1/2m + (mω²∆x²)/2

Emin=(h²/8∆x²m)+ mω²∆x²/2       (a)

For minimum energy ,

dEmin /d ∆x =0 = d/d∆x (/8m∆x² + mω²(∆x²)/2)0=h/8m(-2/∆x³)+mω²(2∆x)/2

In order to solve this equation we get
∆x²= h /2mω
Put in equation(a)

Emin =  hω/2

3)natural width of spectrum lines-

When an electron or atom transits from lower State to upper energy state, it remains 10⁻⁸Sec in the excited state and come back to the lower state after emitting E-M radiations. So that we have
,∆t = 10⁻⁸sec
from Heisenberg uncertainty principle ∆E∆t ≈ h

∆E=/10⁻⁸

∆E= 10⁻²⁶joule
But we know

E=hν

∆E = h∆v
∆v= ∆E/h
∆v = 10⁻²⁶/6.6×10⁻³⁴
∆v=1.5 ×10⁻⁷/sec