Quantum
Quantum
Quantum
1
The physical values or motion of a
macroscopic particles can be observed
directly. Classical Mechanics can be
applied to explain that motion.
Only probability of future state can be predicted using
known laws of force and equations of quantum mechanics.
Waves and Particles
Particles:
A particle or matter has mass and it is located at a
some definite point and it is specified by its mass,
velocity, momentum and energy.
Waves and Particles : What do we mean by them?
Material Objects:
Ball, Car, person, or point like objects called particles.
They can be located at a space point at a given time.
They can be at rest, moving or accelerating.
Falling Ball
Ground level
Waves and Particles: What do we mean by them ?
Common types of waves:
Ripples, surf, ocean waves, sound waves, radio waves.
Need to see crests and troughs to define them.
Waves are oscillations in space and time.
Direction of travel, velocity
Updown
oscillations
Wavelength ,frequency, velocity and oscillation size defines waves
Particles and Waves: Basic difference in behaviour
When particles collide they cannot pass through each other !
They can bounce or they can shatter
Before collision After collision
Another after
collision state
shatter
Waves and Particles Basic difference:
Waves can pass through each other !
As they pass through each other they can enhance or cancel
each other
Later they regain their original form !
Wavelength Frequency
Waves and Particles:
Spread in space and time
Waves
Can be superposed – show
interference effects
Pass through each other
Localized in space and time
Particles
Cannot pass through each other
they bounce or shatter.
Prince Louis-Victor
de Broglie
Won the Nobel Prize in Physics in 1929
From 1 & 2 mc 2 = h
c
λ
h
λ=
mc
h
λ=
p
h
λ=
Hence the de Broglie’s wave length 2mE
de Broglie wavelength associated with
electrons:
Let us consider the case of an electron of rest mass m0 and
charge ‘ e ‘ being accelerated by a potential
1 V volts.
m0 v = eV
2
2
If ‘v ‘ is the velocity 2eV
v=
attained by the electron m0
due to acceleration
h h m0
λ= ⇒λ =
m0 v m0 2eV
0
The de Broglie wavelength 12.26
λ= A
V
Characteristics of Matter waves:
Lighter the particle, greater is the wavelength
associated with it.
Lesser the velocity of the particle, longer the
wavelength associated with it.
For v = 0, λ=∞ , i.e.,wave becomes indeterminate
and if v=∞ then λ=0 This shows that matter waves
are generated by the motion of particles .These
waves are produced whether the particles are
charged particles or they are uncharged .
This fact reveals that these waves are not
electromagnetic but a new kind of waves .
The velocity of matter wave is greater than the
velocity of light
It can be proved that the matter waves E = hϑ
travel faster than light.
E = mc 2
We know that
hϑ = mc 2
mc 2
ϑ=
The wave velocity (ω) is given by h
w = ϑλ
Substituting for λ we get
mc 2
w=( )λ
As the particle velocity v cannot h
exceed velocity of light c, ω is h
λ=
greater than mv
velocity of light. mc 2 h
w=( )
h mv
c2
w=
v
There are two experimental evidences
Principle:
Based on the concept of wave nature of matter
fast moving electrons behave like waves.
Hence accelerated electron beam can be used for
diffraction studies in crystals.
Experimental Arrangement
Nickel
Target
Anode
filament cathode G
Faraday
cylinder
S
Circular scale
c
Galvanometer
G
Working
The target crystal can be rotated about an axis
parallel to the direction of incident electron beam. A
sensitive galvanometer connected to the detector.
First of all , the accelerating potential V is given a low
value and the crystal is set at any azimuth(ө),Now
the Faraday cylinder is moved to various position on
the scale S and galvanometer current is noted for
each position.
44v
Incident beam
48v
Incident beam
50
54v
Incident beam
60v
Obseravations
With increasing potential, the bump moves
upwards
The ‘bump’ becomes most prominent in the
curve for 54volts electrons at ө=500
At higher potentials, the bumps gradually
disappear.
The bump in its most prominent state verifies
the existence of electron waves
Result:
12 .26
λ= A0
V
According 12 .26
to experiment
λ = we have
=1.67diffracted
A0 beam at
ө=500.The 54
corresponding angle of incidence relative the
family of Bragg plane
250
650
250
650 Diffracted
beam
This is in excellent agreement with
the experimental value.
EXPERIMENTAL ARRANGEMENT:
cathode
slit Anode Gold foil Photo
graphic
plate
Diffraction pattern.
Observations
This pattern is similar as produced by X-ray
diffraction powder method .
To be sure that this pattern is due to the
electrons or due to X-rays generated by the
electrons in their passage through the foil, the
cathode rays in the discharge tube are deflected
by magnetic field
The observed rings can only be interpreted by
considering that the diffraction pattern of
incoming beam is due to the diffractions of
electrons by the foil.
As the diffraction pattern can only be produced
by waves and not by the particles
conclusions
Thomson experiment clearly demonstrated
the existence of matter waves because the
diffraction can be produced by the waves
R
Incident Gold foil
electron radius
beam θ
B θ c
A o
θ L
Brage plane
Tan 2θ = R / L
If θ is very small 2θ = R / L
2θ = R / L ………. (1)
2d sin θ = nλ
if , θ ..is..very.small
2d (θ ) = nλ
for., n = 1
λ
d= .......(2)
2θ
frm,.eq n .....(1)
L
d = λ..............(3)
r
According to de Broglie’ s wave equation
h
λ= ( for., electron)
2m0 eV
where., m0 ..is..a., relatavestic..mass
n
from., eq (3)
L h
d= ( )
r 2m0 eV
d = 4.08 A0
The value of ‘d’ so obtained agreed well with
the values using X-ray techniques.
∂ 2ψ 4π 2υ 2
(4)
=− 2 ψ
∂t 2 λ
υ
ν =
λ
Substituting the value of
∂ 2ψ from eq .(3)in eq.
2
(2).weget ∂t
4π υ
2 2
υ ∇ψ =− 2 ψ
2 2
λ
4π 2
∇ 2ψ + 2 ψ = 0 (5)
λ
h
λ=
p
Now de Broglie relation h
λ= .......... .(1)
mv
4π 2 2 2
∇ ψ + 2 mυ ψ =0
2 (6)
h
If E and V the total and potential energies of the particle respectively,
Then its Kinetic energy is given by
1
mυ 2 + V = E
2
1
mυ 2 = E − V
2
m 2υ 2 = 2m( E − V ) (7)
From eqs(6)&(7) we have
4 π 2
∇ 2ψ + 2 × 2m( E − V )ψ = 0
h
8π 2
m
∇ ψ + 2 ( E − V )ψ = 0
2
h (8)
Eq (8) is known as Schrödinger time independent wave
equation
h
=
2π
Substituting in eq (8) the Schrödinger wave equation
can be written as
2m
∇ ψ + 2 ( E − V )ψ = 0
2
(9)
Eq (9) can also be expressed in the following way:
2 2
∇ ψ + ( E − V )ψ = 0
2m
2 2
∇ ψ − Vψ = − Eψ
2m
2 2
− ∇ + V ψ = Eψ
2m
H ψ = Eψ
Physical significance of the wave
function:
∫ dxdydz = 1
ψ ψ*
-∞
+∞ 2
∫ψ
-∞
dxdydz = 1
d 2ψ
** It must be continuous : For Schrödinger equation
d x2 must
be finite everywhere .This is possible
dψonly when has
discontinuity at any boundary wheredxpotential changes .This
implies Ψ must be continuous across the a boundary
∞ ∞
m
V V
A B
X=0 X=a
Let us consider the case of a particle of mass m
moving along x-axis between the two rigid walls A
and B at x=0 and x=a
d 2ψ 8π 2 m
2
+ ( E − V )ψ = 0
(2) dx h
As V=0 between the walls ,hence the equation has the
following form
d 2ψ 8π 2 m (3)
+ 2 Eψ = 0
dx 2 h
8π 2 m
E = k 2 4
h2
Let
then eq(3) takes the the form
d 2ψ 5
2
+ k 2
ψ =0
dx
ψ ( x) = A sin kx + B cos kx 6
Where A and B are two constants
The values of these constants can be obtained by
applying the boundary condition of the problem
Ψ2represents the probability of finding the particle
at any instant. The particle cannot penetrate the
walls.hence Ψ=0 at x=0 and Ψ=0 at x=a. These are
the boundary condition .Apply boundary condition
we have
From Eq(4)
nπx
a
∫ A sin dx =1
2 2
a
nπx
a
A2 ∫ sin 2 dx = 1
0
a
2 nπ x
a
1
A ∫ 1 − cos
2
dx = 1
o
2 a
a
A 2
a 2nπx
x −
2πn Sin =1
a
0
2
A2 a 2 2
= 1orA = orA =
2
2 a a
2 nπ x
ψ ( x) = sin ..............(11)
a a
Eq (11)gives the wave function of the particle enclosed in
infinitely deep potentialwell.
The normalized wave functions Ψ1,Ψ2,and Ψ3 together with
the probability
ψ 1 , ψ densities
2 2
and ψ
2
2 3 areplotted
Ψ(x)
For n = 1: |Ψ(x)|²
x
For n = 2:
Ψ(x) |Ψ(x)|²
x
ψ n = 2 / a sin
nπx n2h2
a En =
8ma 2
E3=9h2 / 8ma2
n =3
E2=4h2/8ma2
n=2
√ (2 / a)
E1=h2 / 8ma2
n=1
X=0 a/2 X=a
According to classical mechanics the probability distribution
is constant .Ψn2=0at x=0 and x=a. At particular point , the
probability of the particle being present is different for
different quantum numbers .
ψ1
2
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