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Recap
Heisenberg uncertainty relations

p x x 
2

E t 
2
 The
product of the uncertainty in momentum
(energy) and in position (time) is at least as
large as Planck’s constant
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Conjugate variables
(Conjugate observables)

{px,x}, {E,t} are called
conjugate variables
 The conjugate
variables cannot in
principle be measured
(or known) to infinite
precision
simultaneously
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Example

The speed of an electron is measured to have a
value of 5.00 x 103 m/s to an accuracy of
0.003%. Find the uncertainty in determining the
position of this electron

SOLUTION
Given v = 5.00  103 m/s; (v)/v = 0.003%
By definition, p = mev = 4.56 x 10-27 Ns;
p = 0.003% x p = 1.37x10-27 Ns
Hence, x ≥ h/4pp = 0.38 nm




p = (4.56±1.37)10-27 Ns
x = 0.38 nm
x
0
x
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Example

A charged p meson has rest energy of 140 MeV and a
lifetime of 26 ns. Find the energy uncertainty of the p
meson, expressed in MeV and also as a function of its
rest energy


Solution
Given E = mpc2 = 140 MeV, t = 26 ns.
E ≥h/4pt = 2.0310-27J
= 1.2710-14 MeV;
E/E = 1.2710-14 MeV/140 MeV = 910-17


“Now you
see it”
Exist only for
E ±E
t = 26 ns
“Now you
DONT”
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Example
estimating the quantum effect on a macroscopic particle

Estimate the minimum uncertainty velocity of a billard ball
(m ~ 100 g) confined to a billard table of dimension 1 m
Solution
For x ~ 1 m, we have
p ≥h/4px = 5.310-35 Ns,
 So v = (p)/m ≥ 5.310-34 m/s
 One can consider v = 5.3x10-34 m/s
(extremely tiny) is the speed of the
billard ball at anytime caused by quantum
effects
 In quantum theory, no particle is
absolutely at rest due to the Uncertainty
Principle
v = 5.310-34 m/s
A billard ball of
100 g, size ~ 2 cm
1 m long billard
table
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A particle contained within a
finite region must has some
minimal KE

One of the most dramatic consequence of the
uncertainty principle is that a particle confined
in a small region of finite width cannot be
exactly at rest (as already seen in the
previous example)

Why? Because…

...if it were, its momentum would be precisely
zero, (meaning p = 0) which would in turn
violate the uncertainty principle
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What is the Kave of a particle in a
box due to Uncertainty Principle?




We can estimate the minimal KE of a particle confined in
a box of size a by making use of the UP
Uncertainty principle requires that p ≥ (h/2p)a (we
have ignored the factor 2 for some subtle statistical
reasons)
Hence, the magnitude of p must be, on average, at least
of the same order as p: | p |  p
Thus the kinetic energy, whether it has a definite value or
not, must on average have the magnitude
K ave
2
 p2 

p )
2
 
= 

~ 2m ~ 2ma 2
2
m

ave
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Zero-point energy
p 

p )

 
K ave = 

2
~
~
 2m  av 2m 2ma
2
2
2
This is the zero-point energy, the minimal
possible kinetic energy for a quantum
particle confined in a region of width a
a
Particle in a box of size a can never be at rest
(e.g. has zero K.E) but has a minimal KE Kave (its
zero-point energy)
We will formally re-derived this result again when
solving for the Schrodinger equation of this system
(see later).
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Recap





Measurement necessarily involves interactions between
observer and the observed system
Matter and radiation are the entities available to us for
such measurements
The relations p = h/l and E = hn are applicable to both
matter and to radiation because of the intrinsic nature of
wave-particle duality
When combining these relations with the universal
waves properties, we obtain the Heisenberg uncertainty
relations
In other words, the uncertainty principle is a necessary
consequence of particle-wave duality
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Introductory Quantum
mechanics
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Probabilistic interpretation
of matter wave
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A beam of light if pictured as monochromatic wave (l, n)
l
Intensity of the light beam is
A=1
unit
area
I =  0c E 2
A beam of light pictured in terms of photons A = 1
unit
area
E=hn
Intensity of the light beam is I = Nhn
N = average number of photons per unit time crossing unit area
perpendicular to the direction of propagation
Intensity = energy crossing one unit area per
unit time. I is in unit of joule per m2 per second
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Probability of observing a photon
Consider a beam of light
 In wave picture, E = E0 sin(kx–wt),
electric field in radiation
 Intensity of radiation in wave
picture is I =  0 c E 2

On the other hand, in the photon
picture, I = Nhn
 Correspondence principle: what is
explained in the wave picture has to
be consistent with what is explained
in the photon picture in the limit
2
Ninfinity:
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I =  0c E = Nhn

Statistical interpretation of radiation




The probability of observing a photon at a point in unit
time is proportional to N
2
However, since Nhn =  0c E
the probability of observing a photon must also  E 2
This means that the probability of observing a photon at
any point in space is proportional to the square of the
averaged electric field strength at that point
Prob (x)  E 2
Square of the mean of the square of
the wave field amplitude
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What is the physical interpretation
of matter wave?

we will call the mathematical representation of the de Broglie’s wave /
matter wave associated with a given particle (or an physical entity) as
The wave function, Y


We wish to answer the following questions:
Where is exactly the particle located within x? the locality of a particle
becomes fuzzy when it’s represented by its matter wave. We can no
more tell for sure where it is exactly located.
 Recall that in the case of conventional wave physics, |field amplitude|2 is
proportional to the intensity of the wave). Now, what does |Y |2
physically mean?
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