The document discusses the uncertainty principle proposed by Werner Heisenberg in 1927. It explains that in the quantum world, one cannot measure a particle's properties like position and momentum simultaneously with perfect precision due to unavoidable uncertainty introduced during measurement. This uncertainty is inherent in nature and not due to limitations of measurement techniques. The document then derives and proves the mathematical relations for position-momentum, energy-time, and angular momentum-angle uncertainty principles. It also lists some applications of the uncertainty principle like explaining the non-existence of electrons within atomic nuclei.

1. JAI NARAYAN VYAS UNIVERSITY,
JODHPUR
SUBJECT: - QUANTUM MECHANICS
SUBMITTED TO: - SUBMITTED BY:-
DR.S.S. MEENA ARCHNA SHARMA
MSC 1ST SEM
PHYSICS

3. THE UNCERTAINTY
PRINCIPLE
 In the latter part of the 19th century,
Newtonians mechanics, Maxwell equation
of electromagnetism and
thermodynamics were thought no basic
laws were left to discover.
 But in 1927, The Werner Heisenberg
proposed a very interesting principle of
far reaching importance, known as The
principle of indeterminacy or The
uncertainty principle.

4. HISENBERG REALISED
THAT……
 In the world of very small particles,
one cannot measure any property
of a particle without interacting
with it in some way.
 This introduces an unavoidable
uncertainty into the result.
 One can never measure all the
properties exactly.
 These uncertainties are inherent in
the physical world and have nothing
to do with skill of the observer.

5. WAVE PACKET AND THE
UNCERTAINTY PRINCIPLE:-
 In quantum mechanics, a particle is described by wave
packet, which represents and symbolises all about particle
and moves with group velocity.
 According to bohr’s probability interpretation the particle
may be found anywhere within the wave packet. This
implies that the position of the particle is uncertain within
the limits of the wave- packet.
 Moreover the wave packet has a velocity spread and hence
there is uncertainty about the velocity or momentum of the
particle.
 This means it is impossible to know where within the wave-
packet , the particle is and what is its exact velocity or
momentum.

7. CONTINUED……
 By plank’s law E=hc/λ, a photon with a
short wavelength has a large energy.
Thus it would impact a large kick to the
electron.
 But to determine its momentum
accurately, electron must only be given a
small kick.
This means using light of long
wavelength.

8. FUNDAMENTAL TRADE OFF…
Use light with short wavelength:-
Accurate measurement of position but not
momentum.
Use light with long wavelength:-
Accurate measurement of momentum but not
position.

9. THE UNCERTAINTY RELATIONS
WITH PROOF:-
1) Position – Momentum Relation:-
Let us consider a particle which is moving along x- axis having momentum
Px.
According to de-Broglie this particle associated with the wave packet
wavelenlength
λ = h/Px.
Px = h/λ = h.k
ΔPx = h.Δk = ħ/2лΔk……(1)
The uncertainty in the position of the particle is equal to the length of the
wave packet.
Δx = L = 2л/ΔK……….(2)
Equation (1) * equation(2)=
ΔPx .Δx = (ħ/2л) .Δk. (2л/Δk)
ΔPx.Δx = ħ
ΔPx.Δx ≥ħ

10. 2. ENERGY-TIME RELATION:-
Let us consider a particle which is moving along x- axes.
This particle cover a distance Δx in timeΔt.
So the velocity of the particle V = Δx/Δt
Δt = Δx/V……(1)
The energy of particle is function of momentum E=F(P)
So any change in energy due to the change in momentum
ΔE = (∂E/∂P)ΔP……..(2)
Multiply the equation (1) and (2)
ΔE.Δt= (Δx/v)(∂E/∂P)ΔP
ΔE.Δt= (ΔX.ΔP/V) (∂E/∂P)
As ∂E/∂P = Vg = V
Δt. ΔE = (ΔX. ΔP/V).V
Δt. ΔE = ΔX. ΔP
SINCE ΔX.ΔP ≥ ħ
SINCE
ΔE.Δt ≥ ħ

11. 3. ANGULAR MOMENTUM - ANGLE
RELATION:-
Let L is the angular momentum of the particle.
The energy of the rotating particle will be E= (1/2)Iω2
I is the moment of inertia. Then Lo = ΔΦ/Δt
Any change in this energy
ΔE = ½*I*2*Lo*Δω
ΔE = (I.Δω).ω
ΔE = ΔL.ω
ΔE = Δh.(ΔΦ/Δt) {Iω = L; IΔω = ΔL}
ΔE.Δt = ΔL.ΔΦ [ΔE.Δt ≥ ħ]
NOTE:- Because ħ is soo small, these uncertainties are not
observable in normal everyday situations.
ΔL. ΔΦ ≥ ħ

12. APPLICATIONS OF UNCERTAINTY PRINCIPLE:-
The principle of uncertainty explains a large
number of
Facts which could not be explained by classical
ideas.
Some of its applications are:-
1. The non existence of the e- in the nucleus.
2. The radius of the bohr's first orbit.
3. Light quanta.
4. Minimum energy of a harmonic oscillator.
5. Bohr’s atomic energy levels and uncertainty
principle.
6. Energy of particle in one- dimensional box.

13. 1). THE NON EXISTENCE OF ELECTRON
IN NUCLEUS:-
The diameter of nucleus of any atom is of the order of 10-14m. If any electron is confined within the nucleus then the
uncertainty in its position (Δx) must not be greater than 10-14m.
According to Heisenberg’s uncertainty principle, equation (1.27)
Δx Δp > h / 2π
The uncertainty in momentum is
Δp > h / 2πΔx , where Δx = 10-14m
Δp > (6.63X10-34) / (2X3.14X10-14)
i.e. Δp > 1.055X10-20 kg.m /s --------------(1.30)
This is the uncertainty in the momentum of electron and then the momentum of the electron must be in the same
order of magnitude. The energy of the electron can be found in two ways one is bynon relativistic method and
the other is by relativistic method.
Non-Realistic method:-
The kinetic energy of the electron is given by,
E = p2/ 2m
p is the momentum of the electron = 1.055X10-20 kg.m /s
m is the mass of the electron = 9.11X10-31kg
∴ E = (1.055X10-20)2/ (2X9.11X10-31) J
= 0.0610X10-9J
= 3.8X108eV
The above value for the kinetic energy indicates that an electron with a momentum of 1.055X10-20 kg.m /s and mass
of 9.11X10-31kg to exist with in the nucleus, it must have energy equal to orgreater than this value. But the
experimental results on β decay show that the maximum kinetic an electron can have when it is confined with in
the nucleus is of the order of 3 – 4 Mev. Therefore the free electrons cannot exist within the nucleus

14. REALISTIC METHOD:-
According to the theory of relativity the energy of a particle is given by
E = mc2 = (m0c2)/(1-v2/c2)1/2 ------- (1.31)
Where m0 is the particle’s rest mass and m is the mass of the particle with velocity v.
Squaring the above equation we get,
E2 = (m0
2c4)/ (1-v2/c2)= (m0
2c6)/ (c2-v2) ------- (1.32)
Momentum of the particle is given by p = mv = (m0v)/ (1-v2/c2)1/2
And p2 =(m0
2v2)/ (1-v2/c2)
= (m0
2v2c2)/ (c2-v2)
then p2c2 = (m0
2v2c4)/ (c2-v2) ------------(1.33)
Subtract equation (1.33) from (1.32)
E2 - p2c2 = {(m0
2c4) (c2-v2)}/ (c2-v2)
= (m0
2c4)
or E2 = p2c2 + m0
2c4 (1.34)
= c2(p2 + m0
2c2)
Substituting the value of momentum from eq(1.30) and the rest mass as = 9.11X10-31kg we get the kinetic energy of the
electron as
E2 > (3X108)2 (0.25 X 10-40 + 7.469 X 10-44 )
The second term in the above equation being very small and may be neglected then we get
E > 1.5X10-12 J
Or E > 9.4 MeV
The above value for the kinetic energy indicates that an electron with a momentum of 1.055X10-20 kg.m /s and mass of
9.11X10-31kg to exist with in the nucleus it must have energy equal to or greater than this value. But the experimental
results on β decay show that the maximum kinetic an electron can have when it is confined with in the nucleus is of the
order of 3 - 4 Mev. Therefore the electrons cannot exist within the nucleus

15. Thank
You  ..