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Quantum Mechanics
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Introduction
1. Stability of an atom
2. Spectral series of Hydrogen atom
3. Black body radiation
There are a few phenomenon which the classical mechanics
failed to explain.
Max Planck in 1900 at a meeting of German Physical Society read his paper “On the theory of the Energy distribution law of the Normal Spectrum”. This was the start of the revolution of Physics i.e. the start of Quantum Mechanics.
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Quantum Mechanics
Quantum Physics extends that range to the region of small dimensions.
It is a generalization of Classical Physics that includes classical laws as special cases.
Just as ‘c’ the velocity of light signifies universal constant, the
Planck's constant characterizes Quantum Physics.
sec.10625.6
sec.1065.634
27
Jouleh
ergh−
−
×=×=
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Quantum Mechanics
1. Photo electric effect
2. Black body radiation
3. Compton effect
4. Emission of line spectra
It is able to explain
The most outstanding development in modern science was the conception of Quantum Mechanics in 1925. This new approach was highly successful in explaining about the behavior of atoms, molecules and nuclei.
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Photo Electric Effect
The emission of electrons from a metal plate when illuminated by light or any other radiation of any wavelength or frequency (suitable) is called photoelectric effect. The emitted electrons are called ‘photo electrons’.
V
Evacuated Quartz tube Metal
plate
Collecting plate
Light
^^^^^^^^ A
+_
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Photo Electric Effect
Experimental findings of the photoelectric effect
1. There is no time lag between the arrival of light at the metal surface and the emission of photoelectrons.
2. When the voltage is increased to a certain value say Vo, the photocurrent reduces to zero.
3. Increase in intensity increase the number of the photoelectrons but the electron energy remains the same.
3I
Voltage
Photo Current
2I
I
Vo
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Photo Electric Effect
4. Increase in frequency of light increases the energy of the electrons. At frequencies below a certain critical frequency (characteristics of each particular metal), no electron is emitted.
Voltage
Photo Current
v1
v2
v3
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Einstein’s Photo Electric Explanation
The energy of a incident photon is utilized in two ways
1. A part of energy is used to free the electron from the atom known as photoelectric workfunction (Wo).
2. Other part is used in providing kinetic energy to the emitted electron .
2
2
1mv
2
2
1mvWh o +=ν
This is called Einstein’s photoelectric equation.
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If , no photoelectric effect
maxKEWh o +=ν
maxKEhh o += νν
)(max ohKE νν −=
oνν <
ooo
hchW
λν ==
o
ooo A
eVWW
hc
)(
12400==λ
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It is in form of . The graph with on y-axis and on x-axis will be a straight line with slope
oo hheV νν −=
oV If is the stopping potential, then
)(max ohKE νν −=
e
h
e
hV oo
νν −=
cmxy +=eh
oVν
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Photons
Einstein postulated the existence of a particle called a photon, to explain detailed results of photoelectric experiment.
λν hchE p ==
Photon has zero rest mass, travels at speed of light
Explains “instantaneous” emission of electrons in photoelectric effect, frequency dependence.
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Compton Effect
When a monochromatic beam of X-rays is scattered from a material then both the wavelength of primary radiation (unmodified radiation) and the radiation of higher wavelength (modified radiation) are found to be present in the scattered radiation. Presence of modified radiation in scattered X-rays is called Compton effect.
electron
scattered photon
recoiled electron
νhE =
c
hp
ν=
'' νhE =
v
θφ φcosmv
φsinmv
incident photon
θνcos'
c
h
θνsin'
c
h
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From Theory of Relativity, total energy of the recoiled electron with v ~ c is
22 cmKmcE o+==
Similarly, momentum of recoiled electron is
22 cmmcK o−=
2
22
2
1cm
cv
cmK o
o −−
=
−
−= 1
1
122
2
cvcmK o
221 cv
vmmv o
−=
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Now from Energy Conversation
φθννcos
1cos'
22 cv
vm
c
h
c
h o
−+=
−
−+= 1
1
1'
22
2
cvcmhh oνν (i)
From Momentum Conversation
(ii) along x-axis
φθνsin
1sin'
022 cv
vm
c
h o
−−= (iii) along y-axis
and
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Rearranging (ii) and squaring both sides
φθνν 222
222
cos1
cos'
cv
vm
c
h
c
h o
−=
− (iv)
φθν 222
222
sin1
sin'
cv
vm
c
h o
−=
(v)
Rearranging (iii) and squaring both sides
Adding (iv) and (v)
22
22
2
222
1cos'2'
cv
vm
c
h
c
h
c
h o
−=−
+
θνννν
(vi)
From equation (i)
221
'
cv
cmcm
c
h
c
h oo
−=+− νν
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On squaring, we get
Subtracting (vi) from (vii)(vii)
22
22
2
222
22
1)'(2
'2'
cv
cmhm
c
hcm
c
h
c
h ooo −
=−+−+
+
νννννν
0)'(2)cos1('2
2
2
=−+−− ννθννohm
c
h
)cos1('2
)'(22
2
θνννν −=−c
hhmo
)cos1('
)'(2
θνννν −=−c
hmo
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But
is the Compton Shift.
λν c=
)cos1(''
11 θλλλλ
−=
− hcmo
and'
'λ
ν c= So,
)cos1(''
' θλλλλ
λλ −=
− hcmo
)cos1(' θλλλ −=∆=−cm
h
o
λ∆
θIt neither depends on the incident wavelength nor on the scattering material. It only on the scattering angle i.e.
is called the Compton wavelength of the electron and its value is 0.0243 Å.cm
h
o
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Experimental Verification
Monochromatic X-ray Source
photon
θ
Graphite target
Bragg’s X-ray Spectrometer
1. One peak is found at same position. This is unmodified radiation
2. Other peak is found at higher wavelength. This is modified signal of low energy.
3. increases with increase in .λ∆ θλ∆
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0.0243 (1- cosθ) Å=−=∆ )cos1( θλcm
h
o
=∆ maxλ
λ∆
So Compton effect can be observed only for radiation having wavelength of few Å.
Compton effect can’t observed in Visible Light
is maximum when (1- cosθ) is maximum i.e. 2.λ∆
0.05 Å
For 1Å ~ 1%
λ∆=λ
=λ
For 5000Å ~ 0.001% (undetectable)
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Pair Production
When a photon (electromagnetic energy) of sufficient energy passes near the field of nucleus, it materializes into an electron and positron. This phenomenon is known as pair production.
In this process charge, energy and momentum remains conserved prior and after the production of pair.
PhotonNucleus (+ve)
−e
+e
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The rest mass energy of an electron or positron is 0.51 MeV (according to E = mc2).
The minimum energy required for pair production is 1.02 MeV.
Any additional photon energy becomes the kinetic energy of the electron and positron.
The corresponding maximum photon wavelength is 1.2 pm. Electromagnetic waves with such wavelengths are called gamma rays . )(γ
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Pair Annihilation
When an electron and positron interact with each other due to their opposite charge, both the particle can annihilate converting their mass into electromagnetic energy in the form of two - rays photon.γ
γγ +→+ +− ee
Charge, energy and momentum are again conversed. Two - photons are produced (each of energy 0.51 MeV plus half the K.E. of the particles) to conserve the momentum.
γ
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From conservation of energyγν 22 cmh o=
Pair production cannot occur in empty space
In the direction of motion of the photon, the momentum is conserved if
θνcos2p
c
h =
θθchν
θcospθcosp
p
p
−e
+e
here mo is the rest mass and2211 cv−=γ
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γvmp o=Momentum of electron and positron is
(i)
1cos ≤θ1<cvBut
θν cos2cph =
Equation (i) now becomes
θγν cos2 cvmh o=
θγν cos2 2
=c
vcmh o
and
γν 22 cmh o<
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But conservation of energy requires that
γν 22 cmh o=
Hence it is impossible for pair production to conserve both the energy and momentum unless some other object is involved in the process to carry away part of the initial photon momentum. Therefore pair production cannot occur in empty space.
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Wave Particle Duality
Light can exhibit both kind of nature of waves and particles so the light shows wave-particle dual nature.
In some cases like interference, diffraction and polarization it behaves as wave while in other cases like photoelectric and compton effect it behaves as particles (photon).
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De Broglie Waves
Not only the light but every materialistic particle such as electron, proton or even the heavier object exhibits wave-particle dual nature.
De-Broglie proposed that a moving particle, whatever its nature, has waves associated with it. These waves are called “matter waves”.
Energy of a photon is
νhE =
For a particle, say photon of mass, m
2mcE =
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Suppose a particle of mass, m is moving with velocity, v then the wavelength associated with it can be given by
hvmc =2
λhc
mc =2
mc
h=λ
mv
h=λp
h=λ or
(i) If means that waves are associated with moving material particles only.
∞=⇒= λ0v
(ii) De-Broglie wave does not depend on whether the moving particle is charged or uncharged. It means matter waves are not electromagnetic in nature.
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Wave Velocity or Phase Velocity
When a monochromatic wave travels through a medium, its velocity of advancement in the medium is called the wave velocity or phase velocity (Vp).
kVp
ω=
where is the angular frequency
and is the wave number.
πνω 2=
λπ2=k
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Group Velocity
So, the group velocity is the velocity with which the energy in the group is transmitted (Vg).
dk
dVg
ω=
The individual waves travel “inside” the group with their phase velocities.
In practice, we came across pulses rather than monochromatic waves. A pulse consists of a number of waves differing slightly from one another in frequency.
The observed velocity is, however, the velocity with which the maximum amplitude of the group advances in a medium.
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Relation between Phase and Group Velocity
dk
dVg
ω= )( pkVdk
d=
dk
dVkVV p
pg +=
( )λπλπ
2
2
d
dVVV ppg +=
( )λλ 1
1
d
dVVV ppg +=
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λλd
dVVV ppg −=
−
+=λ
λλ
d
dVVV ppg
2
11
λλd
dVpSo, is positive generally (not always).
pg VV <⇒
In a Dispersive medium Vp depends on frequency
≠k
ω
generally
i.e. constant
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λλd
dVVV ppg −=
0=⇒λddVp
pg VV =⇒
In a non-dispersive medium ( such as empty space)
=k
ω constant pV=
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Phase Velocity of De-Broglie’s waves
According to De-Broglie’s hypothesis of matter waves
mv
h=λ
wave numberh
mvk
πλπ 22 == (i)
If a particle has energy E, then corresponding wave will have frequency
h
E=ν
then angular frequency will beh
Eππνω 22 ==
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Dividing (ii) by (i)
(ii)h
mc22πω =
mv
h
h
mc
k ππω
2
2 2
×=
v
cVp
2
=
But v is always < c (velocity of light)
(i) Velocity of De-Broglie’s waves (not acceptable) cVp >
(ii) De-Broglie’s waves will move faster than the particle velocity (v) and hence the waves would left the particle behind.
)( pV
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Group Velocity of De-Broglie’s waves
The discrepancy is resolved by postulating that a moving particle is associated with a “wave packet” or “wave group”, rather than a single wave-train.
A wave group having wavelength λ is composed of a number of component waves with slightly different wavelengths in the neighborhood of λ.
Suppose a particle of rest mass mo moving with velocity v then associated matter wave will have
h
mc22πω = andh
mvk
π2= where221 cv
mm o
−=
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and
On differentiating w.r.t. velocity, v
22
2
1
2
cvh
cmo−
= πω221
2
cvh
vmk o
−= π
( ) 23
221
2
cvh
vm
dv
d o
−= πω
(i)
( ) 23
221
2
cvh
m
dv
dk o
−= π (ii)
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Wave group associated with a moving particle also moves with the velocity of the particle.
o
o
m
vm
dk
dv
dv
d
ππω2
2. =
Dividing (i) by (ii)
gVvdk
d ==ω
Moving particle wave packet or wave group≡
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Davisson & Germer experiment of electron diffraction
• If particles have a wave nature, then under appropriate conditions, they should exhibit diffraction
• Davisson & Germer measured the wavelength of electrons• This provided experimental confirmation of the matter waves
proposed by de Broglie
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Davisson and Germer Experiment
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0φ =
090φ =
Current vs accelerating voltage has a maximum (a bump or kink noticed in the graph), i.e. the highest number of electrons is scattered in a specific direction.
The bump becomes most prominent for 54 V at φ ~ 50°
Inci
dent
Bea
m
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According to de Broglie, the wavelength associated with an electron accelerated through V volts is
o
AV
28.12=λ
Hence the wavelength for 54 V electron
o
A67.154
28.12 ==λ
From X-ray analysis we know that the nickel crystal acts as a plane diffraction grating with grating space d = 0.91 Å
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ooo
652
50180 =
−=θ
Here the diffraction angle, φ ~ 50°
The angle of incidence relative to the family of Bragg’s plane
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From the Bragg’s equation
which is equivalent to the λ calculated by de-Broglie’s hypothesis.
θλ sin2d=o
oo
AA 65.165sin)91.0(2 =××=λ
It confirms the wavelike nature of electrons
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Electron Microscope: Instrumental Application of Matter Waves
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Resolving power of any optical instrument is proportional to the wavelength of whatever (radiation or particle) is used to illuminate the sample.
An optical microscope uses visible light and gives 500x magnification/200 nm resolution. Fast electron in electron microscope, however, have much shorter wavelength than those of visible light and hence a resolution of ~0.1 nm/magnification 1,000,000x can be achieved in an Electron Microscope.
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Heisenberg Uncertainty Principle
It states that only one of the “position” or “momentum” can be measured accurately at a single moment within the instrumental limit.
It is impossible to measure both the position and momentum simultaneously with unlimited accuracy.
or
uncertainty in position
uncertainty in momentum
→∆x→∆ xp
then
2
≥∆∆ xpx π2
h=∴
The product of & of an object is greater than or equal to2
xp∆x∆
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If is measured accurately i.e. x∆ 0→∆x ∞→∆⇒ xp
Like, energy E and time t.
2
≥∆∆ tE
2
≥∆∆ θL
The principle applies to all canonically conjugate pairs of quantities in which measurement of one quantity affects the capacity to measure the other.
and angular momentum L and angular position θ
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Determination of the position of a particle by a microscope
i
Incident Photon
ScatteredPhoton
Recoiled electron
Suppose we want to determine accurately the position and momentum of an electron along x-axis using an ideal microscope free from all mechanical and optical defects.
The limit of resolution of the microscope is
ix
sin2
λ=∆
here i is semi-vertex angle of the cone of rays entering the objective lens of the microscope.
is the order of uncertainty in the x-component of the position of the electron.
x∆
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The scattered photon can enter the microscope anywhere between the angular range +i to –i.
We can’t measure the momentum of the electron prior to illumination. So there is uncertainty in the measurement of momentum of the electron.
The momentum of the scattered photon is (according to de-Broglie)
λh
p =
Its x-component can be given as
ih
px sin2
λ=∆
The x-component of the momentum of the recoiling electron has the same uncertainty, (conservation of momentum)xp∆
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The product of the uncertainties in the x-components of position and momentum for the electron is
ih
ipx x sin
2
sin2.
λλ ×=∆∆
This is in agreement with the uncertainty relation.
2.
>=∆∆ hpx x
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Order of radius of an atom ~ 5 x10-15 m
then
If electron exist in the nucleus then
Applications of Heisenberg Uncertainty Principle
(i) Non-existence of electron in nucleus
mx 15max 105)( −×=∆
2
≥∆∆ xpx
2)()( minmax
=∆∆ xpx
120min ..101.1
2)( −−×=
∆=∆ smkg
xpx
MeVpcE 20== ∴ relativistic
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Thus the kinetic energy of an electron must be greater than 20 MeV to be a part of nucleus
Thus we can conclude that the electrons cannot be present within nuclei.
Experiments show that the electrons emitted by certain unstable nuclei don’t have energy greater than 3-4 MeV.
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But
Concept of Bohr Orbit violates Uncertainty Principle
m
pE
2
2
=
2.
≥∆∆ px
m
ppE
∆=∆m
pmv∆= pt
x ∆∆∆=
pxtE ∆∆=∆∆ ..
2.
≥∆∆ tE
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According to the concept of Bohr orbit, energy of an electron in a orbit is constant i.e. ΔE = 0.
2.
≥∆∆ tE
∞→∆⇒ t
All energy states of the atom must have an infinite life-time.
But the excited states of the atom have life–time ~ 10-8 sec.
The finite life-time Δt gives a finite width (uncertainty) to the energy levels.
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Two-slit Interference Experiment
Laser
Source
Slit
Slit Detector
Rate of photon arrival = 2 x 106/sec
Time lag = 0.5 x 10-6 sec
Spatial separation between photons = 0.5 x 10-6 c = 150 m
1 meter
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– Taylor’s experiment (1908): double slit experiment with very dim light: interference pattern emerged after waiting for few weeks
– interference cannot be due to interaction between photons, i.e. cannot be outcome of destructive or constructive combination of photons
⇒ interference pattern is due to some inherent property of each photon - it “interferes with itself” while passing from source to screen
– photons don’t “split” – light detectors always show signals of same intensity
– slits open alternatingly: get two overlapping single-slit diffraction patterns – no two-slit interference
– add detector to determine through which slit photon goes: ⇒ no interference
– interference pattern only appears when experiment provides no means of determining through which slit photon passes
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Double slit experiment – QM interpretation
– patterns on screen are result of distribution of photons
– no way of anticipating where particular photon will strike
– impossible to tell which path photon took – cannot assign specific trajectory to photon
– cannot suppose that half went through one slit and half through other
– can only predict how photons will be distributed on screen (or over detector(s))
– interference and diffraction are statistical phenomena associated with probability that, in a given experimental setup, a photon will strike a certain point
– high probability ⇒ bright fringes
– low probability ⇒ dark fringes
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Double slit expt. -- wave vs quantum
• pattern of fringes:
– Intensity bands due to variations in square of amplitude, A2, of resultant wave on each point on screen
• role of the slits:
– to provide two coherent sources of the secondary waves that interfere on the screen
• pattern of fringes:
– Intensity bands due to variations in probability, P, of a photon striking points on screen
• role of the slits:
– to present two potential routes by which photon can pass from source to screen
wave theory quantum theory
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Wave function
ψψψ *|| 2 =
The quantity with which Quantum Mechanics is concerned is the wave function of a body.
|Ψ|2 is proportional to the probability of finding a particle at a particular point at a particular time. It is the probability density.
Wave function, ψ is a quantity associated with a moving particle. It is a complex quantity.
Thus if iBA+=ψ iBA−=*ψ222222 *|| BABiA +=−==⇒ ψψψ
then
ψ is the probability amplitude.
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Normalization
τd|Ψ|2 is the probability density.
The probability of finding the particle within an element of volume
τψ d2||
Since the particle is definitely be somewhere, so
1|| 2 =∫∞
∞−
τψ d
A wave function that obeys this equation is said to be normalized.
∴ Normalization
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Properties of wave function
1. It must be finite everywhere. If ψ is infinite for a particular point, it mean an infinite large
probability of finding the particles at that point. This would violates the uncertainty principle.
2. It must be single valued. If ψ has more than one value at any point, it mean more than
one value of probability of finding the particle at that point which is obviously ridiculous.
3. It must be continuous and have a continuous first derivative everywhere.
zyx ∂∂
∂∂
∂∂ ψψψ
,, must be continuous
4. It must be normalizable.
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Schrodinger’s time independent wave equation
One dimensional wave equation for the waves associated with a moving particle is
From (i)
ψ is the wave amplitude for a given x. where
A is the maximum amplitude.
λ is the wavelength
ψλπψ
2
2
2
2 4−=∂∂x
(ii)
)0,( and ),()(
2)( x
ipxEt
i
AetxAetx λπ
ψψ ===−−
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vm
h
o
=λ
2
22
2
1
h
vmo=⇒λ 2
2
21
2
h
vmm oo
=
22
21
h
Kmo=λ
where K is the K.E. for the non-relativistic case
(iii)
Suppose E is the total energy of the particle
and V is the potential energy of the particle
)(21
22VE
h
mo −=λ
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This is the time independent (steady state) Schrodinger’s wave equation for a particle of mass mo, total energy E, potential energy V, moving along the x-axis.
If the particle is moving in 3-dimensional space then
Equation (ii) now becomes
ψπψ)(2
42
2
2
2
VEmhx o −−=
∂∂
0)(2
22
2
=−+∂∂ ψψ
VEm
xo
0)(2
22
2
2
2
2
2
=−+∂∂+
∂∂+
∂∂ ψψψψ
VEm
zyxo
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For a free particle V = 0, so the Schrodinger equation for a free particle
02
22 =+∇ ψψ E
mo
0)(2
22 =−+∇ ψψ VE
mo
This is the time independent (steady state) Schrodinger’s wave equation for a particle in 3-dimensional space.
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Schrodinger’s time dependent wave equation
)( pxEti
Ae−−
= ψ
Wave equation for a free particle moving in +x direction is
(iii)
ψψ2
2
2
2
p
x−=
∂∂
where E is the total energy and p is the momentum of the particle
Differentiating (i) twice w.r.t. x
(i)
2
222
xp
∂∂−=⇒ ψψ (ii)
Differentiating (i) w.r.t. t
ψψ
iE
t−=
∂∂
tiE
∂∂=⇒ ψψ
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For non-relativistic case
Using (ii) and (iii) in (iv)
(iv)
ψψψV
xmti +
∂∂−=
∂∂
2
22
2
E = K.E. + Potential Energy
txVm
pE ,
2
2+=
ψψψ Vm
pE +=⇒
2
2
This is the time dependent Schrodinger’s wave equation for a particle in one dimension.
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Linearity and Superposition
2211 ψψψ aa +=
If ψ1 and ψ2 are two solutions of any Schrodinger equation of a system, then linear combination of ψ1 and ψ2 will also be a solution of the equation..
Here are constants
Above equation suggests:
21& aa
is also a solution
(i) The linear property of Schrodinger equation
(ii) ψ1 and ψ2 follow the superposition principle
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21 ψψψ +→ Then
Total probability will be2
212 |||| ψψψ +==P
due to superposition principle
)()( 21*
21 ψψψψ ++=))(( 21
*2
*1 ψψψψ ++=
1*22
*12
*21
*1 ψψψψψψψψ +++=
1*22
*121 ψψψψ +++= PPP
21 PPP +≠ Probability density can’t be added linearly
If P1 is the probability density corresponding to ψ1 and P2 is the probability density corresponding to ψ2
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Expectation values
dxxf∫∞
∞−
= 2||)( ψ
Expectation value of any quantity which is a function of ‘x’ ,say f(x) is given by
for normalized ψ
Thus expectation value for position ‘x’ is
>< )(xf
dxx∫∞
∞−
= 2||ψ>< x
Expectation value is the value of ‘x’ we would obtain if we measured the positions of a large number of particles described by the same function at some instant ‘t’ and then averaged the results.
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dxx∫=1
0
2||ψ Solution
>< x
1
0
42
4
= xa
Q. Find the expectation value of position of a particle having wave function ψ = ax between x = 0 & 1, ψ = 0 elsewhere.
dxxa ∫=1
0
32
>< x4
2a=
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Operators
ψψpi
x =
∂∂
(Another way of finding the expectation value)
For a free particle
An operator is a rule by means of which, from a given function we can find another function.
)( pxEti
Ae−−
= ψ Then
Here
xip
∂∂= ^
is called the momentum operator
(i)
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ψψEi
t −=
∂∂ Similarly
Here
tiE
∂∂=
^
is called the Total Energy operator
(ii)
Equation (i) and (ii) are general results and their validity is the same as that of the Schrodinger equation.
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Uximt
i +
∂∂=
∂∂ 2
2
1
If a particle is not free then
This is the time dependent Schrodinger equation
^^^
.. UEKE +=^
^2^
2U
m
pE
o
+=⇒
UU =∴^
Uxmt
i +∂∂−=
∂∂
2
22
2
ψψψU
xmti +
∂∂−=
∂∂
2
22
2
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If Operator is Hamiltonian
Then time dependent Schrodinger equation can be written as
Uxm
H +∂∂−=2
22^
2
ψψ EH =^
This is time dependent Schrodinger equation in Hamiltonian form.
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Eigen values and Eigen function
Schrodinger equation can be solved for some specific values of energy i.e. Energy Quantization.
ψψα a=^
Suppose a wave function (ψ) is operated by an operator ‘α’ such that the result is the product of a constant say ‘a’ and the wave function itself i.e.
The energy values for which Schrodinger equation can be solved are called ‘Eigen values’ and the corresponding wave function are called ‘Eigen function’.
then ψ is the eigen function of
a is the eigen value of
^
α^
α
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Q. Suppose is eigen function of operator then find the eigen value.
The eigen value is 4.
Solution.
xe2=ψ2
2
dx
d
2
2^
dx
dG =
2
2^
dx
dG
ψψ = )( 22
2xe
dx
d=
xeG 2^
4=ψ
ψψ 4^
=G
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Particle in a Box Consider a particle of rest mass mo enclosed in a one-dimensional
box (infinite potential well).
Thus for a particle inside the box Schrodinger equation is
Boundary conditions for Potential
V(x)= 0 for 0 < x < L
∞ { for 0 > x > L
Boundary conditions for ψ
Ψ = 0 for x = 0
{ 0 for x = L
0222
2
=+∂∂ ψψ
Em
xo
x = 0 x = L
∞=V ∞=V
particle
0=V
0=∴V inside (i)
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Equation (i) becomes
p
h=λk
π2=
pk =⇒
Emo2=
22 2
Emk o=⇒
(k is the propagation constant)
(ii)
022
2
=+∂∂ ψψ
kx
(iii)
General solution of equation (iii) is
kxBkxAx cossin)( +=ψ (iv)
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Equation (iv) reduces to
Boundary condition says ψ = 0 when x = 0
(v)
0.cos0.sin)0( kBkA +=ψ
1.00 B+= 0=⇒ B
kxAx sin)( =ψ Boundary condition says ψ = 0 when x = L
LkAL .sin)( =ψ
LkA .sin0 =0≠A 0.sin =⇒ Lk
πnLk sin.sin =⇒
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Put this in Equation (v)
(vi)L
nk
π=
L
xnAx
πψ sin)( =
When n # 0 i.e. n = 1, 2, 3…., this gives ψ = 0 everywhere.
πnkL =
Put value of k from (vi) in (ii)
22 2
Emk o=
2
22
Em
L
n o=
π
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Where n = 1, 2, 3….
Equation (vii) concludes
om
kE
2
22=⇒ 2
22
8 Lm
hn
o
= (vii)
1. Energy of the particle inside the box can’t be equal to zero.
The minimum energy of the particle is obtained for n = 1
2
2
1 8 Lm
hE
o
= (Zero Point Energy)
If momentum i.e.01 →E 0→ 0→∆p∞→∆⇒ x
But since the particle is confined in the box of dimension L.
Lx =∆ max
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Thus zero value of zero point energy violates the Heisenberg’s uncertainty principle and hence zero value is not acceptable.
2. All the energy values are not possible for a particle in
potential well.
Energy is Quantized
3. En are the eigen values and ‘n’ is the quantum number.
4. Energy levels (En) are not equally spaced.
n = 1
n = 3
n = 2
3E
1E
2E
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Using Normalization conditionL
xnAxn
πψ sin)( =
1sin0
22 =∫ dxL
xnA
L π
1|)(| 2 =∫∞
∞−
dxxnψ
12
2 =
L
AL
A2=⇒
The normalized eigen function of the particle are
L
nx
Lxn
πψ sin2
)( =
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Probability density figure suggest that:
1. There are some positions (nodes) in the box that will never be occupied by the particle.
2. For different energy levels the points of maximum probability are found at different positions in the box.
|ψ1|2 is maximum at L/2 (middle of the box)
|ψ2|2 is zero L/2.
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Eigen functionzyx ψψψψ =
L
zn
L
yn
L
xnAAA zyxzyx
πππψ sinsinsin=
L
zn
L
yn
L
xn
Lzyx πππψ sinsinsin
23
=
2
2222
8)(mL
hnnnE zyx ++=
Particle in a Three Dimensional Box
zyx EEEE ++= Eigen energy