Document

```QUANTUM MECHANICS FOR
NANOTECHNOLOGY I
EEE5425 Introduction to Nanotechnology
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Classical Mechanics
z
A classical particle is what we think of an ordinary
object (ball, car etc.) .
v(t)
A classical particle with mass m occupies a definite
position in space r(t) at a time t like:
T
Where a, b, c are unit vectors along x, y, z
coordinates, respectively.
r(t)
y
x
r(t)=ax(t) + by(t) + cz(t)
If the particle is moving along a trajectory T it has a
definite velocity v=dr(t)/dt
definite momentum p= mv
definite acceleration a=d2r (t)/d2t
Classical particles obey Newtonian mechanics: F=m[d2r(t)/dt2]
In classical physics, physical quantities such as position and momentum can , in principle, be
measured with absolute certainty.
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New Observations
About the turn of the century, there were many experimental and natural phenomena
that could not be explained by classical (Newtonian) mechanics.
1) The frequency spectrum of black body radiation (Max Planck, Nobel Prize 1918).
2) Photoelectric effect: Photo-emission of electrons from metals, waves acting like
particles! (A. Einstein, Nobel Prize 1921)
3) The characteristic line spectra of atoms (Niels Bohr, Nobel Prize 1922)
4) Particles (like billiard balls) could behave like waves interference, diffraction, DavisonGermer experiment.)
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black body is an object that absorbs all light that falls on it. Because no light is reflected or
transmitted, the object appears black when it is cold. If the black body is hot, these
properties make it an ideal source of thermal radiation. If a perfect black body at a certain
temperature is surrounded by other objects in thermal equilibrium at the same
temperature, it will on average emit exactly as much as it absorbs, at every wavelength.
Max Planck applied quantization to the tiny
oscillators that were thought to exist in the walls
of the cavity. He assumed that the energy of
these oscillators was limited to a set of discrete,
integer multiples of a fundamental unit of
energy, E, proportional to the oscillation
frequency ν:
He derived
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Photoelectric Effect
Consider monochromatic light is incident on the surface of a
metal plate in a vacuum. The electrons in the metal absorb
energy from the light, and some of the electrons receive
enough energy to be ejected from the metal surface into the
vacuum.
The maximum energy of electrons Em can be found by placing
another electrode to create an electric field in between. The
potential necessary to retard all electron flow between the
plates gives the energy Em.
Em  h  q
h: Planck constant (=6.63x10-34 J.s=4.14x10-15 eV.s)
n: frequency
q: electron charge (=1.6x10-19 coulomb)
q: metal work function (Joules or eV)
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Photoelectric Effect
For a particular frequency of light incident on the sample, a maximum energy Em is
observed for the emitted electrons. The resulting plot of Em vs. n is linear, with a slope
equal to Planck’s constant.
Em  h  q
Planck was right!!!
Light energy is contained in discrete units rather than
in a continuous distribution of energies. The quantized
units of light energy can be considered as localized
packets of energy, called photons.
Einstein’s interpretation of photoelectric based on Planck’s hypothesis is considered to be
the birth of Quantum Mechanics.
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Wave-Particle Duality
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De Broglie Hypothesis
Louis-Victor-Pierre-Raymond, 7th duc de Broglie ( 1892 – 1987) was a
French physicist and a Nobel laureate.
He proposed that particles of matter (such as electrons) could manifest a
wave character in certain experiments just like light manifested the
discrete units of energy called photons. His hypothesis completed the
concept of duality.
p  k
h
h
de Broglie wavelengt h  l  
p mv
Remembering
Momentum:
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c= l f
,
E=h f , k=2p / l , w  2p f
p  k
Energy:
EEE5425 Introduction to Nanotechnology
and
ħ = h /2p
E  w
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Quantum Mechanics
What is quantum mechanics?
Quantum mechanics is the study of matter and radiation at an atomic level
where particles and waves can be described in a similar way .
If classical physics is wrong, why do we still use it?
For everyday things, which are much larger than atoms and much slower than the
speed of light, classical physics does an excellent job. Plus, it is much easier to use
than either quantum mechanics or relativity (each of which require an extensive
amount of math).
What is the importance of quantum mechanics?
The following are among the most important things which quantum mechanics
can describe while classical physics cannot:
-Discreteness of energy
-The wave-particle duality of light and matter
-Quantum tunneling
-The Heisenberg uncertainty principle
-Spin of a particle
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Wavepackets -1
Quantum particles (light, electrons, bowling balls, etc) can be thought of as quantized
bundles of energy E=ħω having wave-like properties (frequency ω and wavelength λ) and
particle-like properties (momentum p) that are interrelated –the so- called wave-particle
duality.
A typical plane wave is described by
 (t , x)  Ae j (wt kx )
that extends over a region of (or all of) a space rather than being localized to single point. It
implies that quantum particles will not be localized at a single point like classical particles.
Viewed from a distance large compared to its de Broglie wavelength, en electron appears like
a particle. Viewed from a distance small compared to its wavelength, usually atomic
dimensions, the “spread” of electron becomes evident.
One way to model this dual behavior is with a wavepacket which is a wave that is both
propagating and localized in space and time.
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EEE5425 Introduction to Nanotechnology
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Wavepackets -2
 (t , x)  Ae j (wt kx ) and recall basic
Consider a single frequency plane wave
relationships: c=λf, f=ω/2π which leads ω=ck.
For a particle with mass m and only kinetic energy
1 2 p 2 (k ) 2
E  w  mv 

2
2m
2m
Such that
k 2
w (k ) 
2m
Relationships between frequency and wavenumber such as above are called dispersion
relations.
The phase velocity of the plane wave is the velocity of a constant phase (and amplitude in this
case) planar wavefront. Therefore taking the derivative of the phase wrt time and setting
equal to 0 gives:
dx
w  k  w  kvp  0
dt
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Yielding the phase velocity: v p 
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w
k
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Wavepackets -3

Now, instead of a single plane wave, consider
 (t , x)   a(k )e j (w ( k )t kx ) dk

of which the integrand represents plane waves of varying amplitudes and wavenumbers.
The integration is simply a summation of those planewaves. Assume:
a(k) = 1,
a(k) = 0,
a(k)
k0 – Δk ≤ k ≤ k0 +Δk
elsewhere
k
k0 -Δk k0 k0 +Δk
With this form of a(k), one can interpret the integral as a summation of waves with wave
numbers within some Δk range of a give value k0. For photon in free space with ω=ck the
integral becomes:
 (t , x)  
k0  k
k0  k
e
 jk ( ct  x )
 jk0 ( ct  x )
dk  2ke
sin( k (ct  x))
k (ct  x)
which is wave packet moving with velocity c (note that vp= ω/k=c) and having an envelope
proportional to
sin( k (ct  x))
sinc (k (ct  x)) 
k (ct  x)
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Wavepackets -4
In the previous discussion, the wavepacket did not change its shape as it propagated. Often,
On needs to consider a dispersion relation that is more complicated than the simple linear
dependence. In general this leads to the wavepacket changing shape (generally spreading
out ) as it propagates. In addition, in this case the phase velocity is not the velocity of prime
interest.
To examine this phenomenon it is convenient to expand ω(k) in a Taylor’s series around the
center wavenumber k=k0 obtaining:
w
w (k )  w (k0 ) 
k
1  2w
(k  k0 ) 
2 k 2
k  k0
(k  k0 ) 2  ...
k  k0
 w0   (k  k0 )   (k  k0 ) 2  ...
Assuming that it is sufficient to keep only the first two terms:
 ( x, t )  e
e
where vp=ω0/k0.
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 jk0 ( v p t  x )
 jk0 ( v p t  x )

k 0  k
k 0  k
2k
e  j ( k k0 )(t  x ) dk
sin( k (t  x))
k (t  x)
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Wavepackets -5
The velocity of the envelope is not the phase velocity but α which is called the group velocity:

w
k
 vg
k  k0
Therefore:
 ( x, t )  e
 jk0 ( v p t  x )
2k
sin( k (vg t  x))
k (v g t  x)
In this case, the wavepacket moves through space and time as localized bundle of
approximate width Δk(vgt-x) = π/2 that is centered at the point (vgt-x) = 0.
That is starting at t=0, the wavepacket is centered at x=0 and a t a given time t the
wavepacket is centered at the point x= vgt and occupies a spatial extent Δx= vgt – π /2Δk.
In reality rather than the abrupt amplitude function a(k) used in above discussion, amore
physically realistic function is used, typically a Gaussian:
a(k )  e
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
( k  k0 ) 2
2 k 2
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Probability and the Uncertainty Principle
It is impossible to describe with absolute precision events involving
individual particles on the atomic scale.
Instead, we must speak of the average values (expectation values)
of position, momentum, and energy of a particle such as an
electron.
The theory describes the probabilistic nature of events involving
small particles (atoms, electrons, elementary particles). The fact is
that such quantities as the position and momentum of an electron
do not exist apart from a particular uncertainty.
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EEE5425 Introduction to Nanotechnology
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Heisenberg Uncertainty Principle
Werner Heisenberg (1901 –1976) was a German theoretical physicist.
He made contributions to quantum mechanics, nuclear physics,
quantum field theory, and particle physics. Heisenberg, along with Max
Born and Pascual Jordan, set forth the matrix formulation of quantum
mechanics in 1925. Heisenberg was awarded the 1932 Nobel Prize in
Physics.
The magnitude of uncertainty to determine physical quantities (position,
momentum, energy) is described by the Heisenberg uncertainty principle (also
known as the principle of indeterminacy)
In any measurement of the position and momentum of a
particle, the uncertainties in the two measured quantities
will be related by
(Δx).(Δpx) ≥ ħ/2
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Heisenberg Uncertainty Principle –2
Similarly,
The uncertainties in an energy measurement will be
related to the uncertainty in the time at which the
(ΔE).(Δt) ≥ ħ/2
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Heisenberg Uncertainty Principle –3
Example:
What is the uncertainty in velocity for an electron in a 1 &Aring; radius orbital in which
the positional uncertainty is 1% of the radius?
x  11010 (m)  0.01  11012 (m)
 1
h
6.626 1034 (J.s)
 23
p 



5
.
28

10
(kg.m/s)
12
2 x 4px 4p 110 (m)
p 5.28 1023 (kg.m/s )
8
v 


0
.
6

10
(m/s )
31
m
9.1110 (kg)
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EEE5425 Introduction to Nanotechnology
Huge!
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Heisenberg Uncertainty Principle –4
Example:
What is the uncertainty in position for a 80 kg student walking across campus at
1.3 m/s with an uncertainty in velocity of 1%.
p  m  v  80(kg)  0.013(m/s )  1.04(kg.m/s )
 1
h
6.626 1034 (J.s)
x 


 5.07 1035 (m)
2 p 4pp 4p 1.04(kg.m/ s)
Uncertainty in position of a student is very small –we know where you are!
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EEE5425 Introduction to Nanotechnology
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Probability–1
The uncertainty principle brings us to an idea that we cannot properly speak of
the position of an electron, but must look for the probability of finding an
electron at a certain position.
Thus one of the important results of quantum mechanics is that a probability
density function can be obtained for a particle in a certain environment, and this
function can be used to find the expectation value of important quantities such
as position, momentum, and energy.
For this purpose, it is common to define a probability density function P(x) which
describes the probability of finding a particle within a certain volume.
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EEE5425 Introduction to Nanotechnology
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Probability–2
The probability of finding the particle in a range from x to (x+ dx) is P(x)dx.
Since the particle will be somewhere, the probability to find it in some point
within region (-∞,∞) must be 1:

 P( x)dx  1

if the function P(x) is properly chosen –normalized.
To find the average value of a function of x(f(x)), we need only multiply the value
of that function in each increment dx by the probability (P(x)) of finding the
particle in that dx and sum over all range of x:

Average value of f(x):
f ( x) 
 f ( x) P( x)dx

where P(x) is normalized.
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The Schr&ouml;dinger Wave Equation -1
Erwin Rudolf Josef Alexander Schr&ouml;dinger ( 1887 – 1961) was an Austrian
physicist who achieved fame for his contributions to quantum mechanics,
especially the Schr&ouml;dinger equation, for which he received the Nobel Prize
in 1933. In 1935, after extensive correspondence with personal friend
Albert Einstein, he proposed the Schr&ouml;dinger's cat thought experiment.
Basic postulates
Postulate 1:
Each particle in a physical system is described by a wave function
Ψ(r,t)=Ψ(х,у,z,t)
This function and its partial space derivative (∂ψ/∂x + ∂ψ/∂y + ∂ψ/∂z) are
continuous, finite, and single valued.
Wave function can be interpreted as probability amplitude
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The Schr&ouml;dinger Wave Equation -2
Postulate 2:
In dealing with classical quantities such as energy E and momentum p, we must
relate these quantities with abstract quantum mechanical operators defined in
the following way (one-dimensional case).
Classical variable
Quantum operator
x
x
f(x)
p(x)
f(x)
 
j x
 

j t
E
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The Schr&ouml;dinger Wave Equation -3
Postulate 3:
The probability of finding a particle with wave function Ψ in the volume
(dx&times;dy&times;dz) is (Ψ*Ψ)dx&times;dy&times;dz.
(Ψ* is the complex conjugate of Ψ, obtained by reversing the sign of each j.
Thus, (ejx)*=e-jx).
The product Ψ*Ψ is normalized so that

*

 dxdydz  1

and the average value (or expectation value) ⟨Q⟩ of any variable Q is
calculated from the wave function by using the quantum operator Qop
defined in postulate 2:

Q    *Qop dxdydz  1

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The Schr&ouml;dinger Wave Equation -Example
Example:
Given a plane wave Ψ =Aejkx, what is the expectation value of px ?


Remember:
Q    *Qop dxdydz  1
P

Numerator
px 


*

 dx

Denominator
   *dx
   * Pdx




jk x
*  jk x x    
A
 e  j x  Ae x dx

 A
→



 
j x
*

 Pdx
2
 jk x
e x

jk x jkx x
e dx
j

jk x x
*  jk x x
A
e
Ae
dx

 A k x  dx

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px 

 A
2
 jk x x jk x x
e
 e dx
A k x  dx

A

a
2

 dx

A k x  dx
2
 lim a 
a
a
A
2
 dx
a
p x  k x


2

2

 A
2
 dx

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The Schr&ouml;dinger Wave Equation -4
The classical equation for the energy of a particle: Ekin+ Epot= Etot
Kinetic energy Ekin:
Potential energy Epot:
Ekin
mV 2 p 2


2
2m
E pot

 U (r )
(U - potential energy)
2
p
 U  Etot
2m
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Th The Schr&ouml;dinger Wave Equation -5
p2
 U  Etot
2m
In quantum mechanics we have to use the operator form for variables
momentum p and energy E (postulate 2);
the operators are allowed to operate on the wave function Ψ.
Classical variable
Quantum operator
x
x
f(x)
f(x)
 
j x
p(x)
E
p2
 U  Etot
2m
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
 
j t




 2  2  (r , t )
 (r , t )

 2  U (r ) (r , t )  
2m r
j t
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The Schr&ouml;dinger Wave Equation -6
We can rewrite the equation using conventional notation:
2
2
2






2  2  2  2
x
y
z




 2  2  (r , t )
 (r , t )

 2  U (r ) (r , t )  
2m r
j t
Schr&ouml;dinger wave equation
2 2
 

   U  
2m
j t
The wave function Ψ in the Schr&ouml;dinger wave equation includes both space and
time dependencies.
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The Schr&ouml;dinger Wave Equation -7
2 2
 

   U  
2m
j t
where Ψ(r,t) = Ψ(х,у,z,t) is a function of space coordinates and time;
U(r) –potential energy (field). (It may have very complex form)
Wave functions are solutions to the Schrodinger wave equation. The wave
function, Ψ(х, t) describes physical state of the particle, such as its momentum,
energy etc. and also where the particle is (in terms of probability ).
This is quite complex differential equation –typically it is very difficult to solve it
and find Ψ(r,t).
In many cases, it is possible to solve the wave equation by breaking it into two
equations by the technique of separation of space coordinates and time variables.
Let Ψ(х,t) be represented by the product ψ(х)&times;φ(t)
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The Schr&ouml;dinger Wave Equation -8
Let Ψ(х,t) be represented by the product ψ(х)&times;φ(t):
Ψ(x,t) = ψ(х)&times;φ(t)
Substituting this product in the Schr&ouml;dinger wave equation
 2  2 ( x, t )
 ( x, t )


U
(
x
)

(
x
,
t
)


2m x 2
j t
we have:
2
 2 ( x)

 (t )

 (t )

U
(
x
)

(
x
)

(
t
)



(
x
)
2m
x 2
j
t
Now the variables can be separated!
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The Schr&ouml;dinger Wave Equation -9
Separation of variables allows us to derive two independent equations:
(1) the time-dependent Schr&ouml;dinger equation in one dimension:
 (t ) jE

 (t )  0
t

(2) the time-independent Schr&ouml;dinger equation.
To derive time-independent Schr&ouml;dinger equation, we have to recall quantum
operator to determine energy:
 

j t
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The Schr&ouml;dinger Wave Equation -10
2
 2 ( x)

 (t )

 (t )

U
(
x
)

(
x
)

(
t
)



(
x
)
2m
x 2
j
t
Using quantum energy operator:
 
E
j t
2
 2 ( x)

 (t )
 U ( x) ( x) (t )   ( x) E (t )
2
2m
x
Thus, eliminating time dependent function (t) we have:
 2  2 ( x)

 U ( x) ( x)   ( x) E
2
2m x
 2 ( x) 2m
 2 [ E  U ( x)] ( x)  0
2
x

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The Schr&ouml;dinger Wave Equation -11
The time-independent (stationary) Schr&ouml;dinger equation:
 2 ( x) 2m
 2 [ E  U ( x)] ( x)  0
2
x

Solution of this equation, wave function ψ(x), describes a particle in stationary
state.
Constant E corresponds to the energy of the particle when particular solutions
are obtained, such that a wave function ψn corresponds to a particle energy En.
This equation is the basis of wave mechanics. From it we can determine the
wave functions for particles in various simple systems.
For calculations involving electrons, the potential term U(x) usually represents
electrostatic or magnetic field.
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EEE5425 Introduction to Nanotechnology
33
The Schr&ouml;dinger Wave Equation -1
 2 ( x) 2m
 2 [ E  U ( x)] ( x)  0
2
x

It is quite difficult to find solutions
to the Schr&ouml;dinger wave equation
for most realistic potential fields
U(x).
&copy; Nezih Pala npala@fiu.edu
The simplest problem is the
potential energy well with infinite
boundaries - “particle in a box”.
EEE5425 Introduction to Nanotechnology
34
Free Electrons -1
As a first approximation of solving Schrodinger’s equation, consider a free electron in an
infinite space. By “free electron”, we mean that there is no potential energy variation to
influence the particle, i.e. U(x)=U0 (where U0 can be zero, the important thing is that U is
constant).
For solid materials, the most common source of potential is the atomic lattice, where the
potential energy between an electron with charge q and ionized atom of charge –q is given
(in one dimension)
U(x)
x
1 (q )(  q )
1D model of
U ( x) 
potential due to an
4p0 | x |
atom.
}
Other sources of potential could be, for example, other electrons. Here it will be assumed
that there is a single electron and no other particles.
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EEE5425 Introduction to Nanotechnology
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Free Electrons -2
Consider 1D Schrodinger’s equation
 2  ( x ) 2m
 2 [ E  V0 ] ( x)  0
2
x

A typical solution for such a 2nd order differential equation is
( x)  Ae jkx  Be  jkx
with
2m( E  V0 )
k 
2
2
The parabolic relationship between wave vector k and energy E is shown in the figure.
Putting in the time variation we have the complete solution for a
free electron:
E


( x, t )  Ae jkx  Be  jkx e  jEt / 
V0
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k
Such a solution is called a plane wave solution since the surfaces
of constant amplitude and phase are plane waves. The terms with
the constants A and B represent the forward and backward
traveling plane waves.
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Free Electrons -3
Recall the two concepts of velocity: Phase velocity vp= ω/k and the group velocity vg= δω/δ k.
Although they were derived from a consideration of wavepackets, they can be taken as
possible definitions of wave velocities. For Schrodinger’ s equation presented previously,
setting V0=0 for convenience and using E=ħω, we obtain
vp 
w
k

k
p

2m 2m
Recalling that solutions of Schrodinger’s equation should agree with classical physics in the
classical limit, in order to see if the phase velocity agrees with our classical notion of velocity,
we equate p=mv for classical electron to obtain
vp 
v
2
Therefore, the phase velocity does not yield a reasonable value for the electron’s velocity.
However the group velocity is
vg 
w k p mv

 
v
k
m m m
Therefore, as concepts of velocity is the more meaningful. For a classical electromagnetic
plane wave in free space, k= ω/c and vp=vg=c
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EEE5425 Introduction to Nanotechnology
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Particle in a 1D Potential Well -2
U ( x)  0 for 0  x  L

U ( x)   for x  0 and x  L
 2 ( x) 2m
 2 [ E  U ( x)] ( x)  0
2
x

(U(x) = 0) Schr&ouml;dinger time-independent (stationary) equation:
 2 ( x) 2m
 2 E ( x)  0
2
x

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EEE5425 Introduction to Nanotechnology
for 0 &lt; x &lt; L
38
Particle in a 1D Potential Well -3
This Schr&ouml;dinger equation for free particle (electron) of mass minside onedimensional well (U(x) = 0)
 2 ( x) 2m
 2 E ( x)  0
2
x

for 0 &lt; x &lt; L
has general solution: ψ(x) = Ae jkx + Be -jkx
Shape of the potential well dictates the boundary conditions: the solution (ψ(x))
must have zero values at the walls of the well (x = 0 and x= L). This is due to the
fact that probability to find particle outside the well must be zero: |ψ(x)|2=0 for
any x &lt; 0 and x &gt; L.
Thus, the boundary conditions:
&copy; Nezih Pala npala@fiu.edu
ψ(0)=0; ψ(L)=0
EEE5425 Introduction to Nanotechnology
39
Particle in a 1D Potential Well -4
General solution of the equation ψ(x) = Ae jkx + Be -jkx
Remembering
e &plusmn;jθ = cosθ &plusmn; jSinθ
ψ(x) = ASin(kx) + BCos(kx)
We must examine boundary conditions to choose a solution:
Solution should satisfy boundary condition: ψ(x=0) = 0
At x = 0 we have:(1) Asin(k0) = 0 ; (2) Bcos(k0) ≠0
ψ(x) = Asin(kx) satisfies boundary condition - ψ(0)=0
and ψ(x) = Bcos(kx) –does not.
The solution of the equation is ψ(x) = Asin(kx)
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EEE5425 Introduction to Nanotechnology
40
Particle in a 1D Potential Well -5
Thus, the solution of the Schr&ouml;dinger equation for free particle inside onedimensional potential well is ψ(x) = Asin(kx)
Parameter k can be found by substituting solution ψ(x) = Asin(kx) into
equation
2
  ( x ) 2m
 2 E ( x)  0
2
x

for 0 &lt; x &lt; L
 2 A sin( kx) 2m
 2 EA sin( kx)  0
2
x

 Ak 2 sin( kx) 
2mE
k  2 0

2
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2m
EA sin( kx)  0
2

d sin( x)
 cos( x)
dx
d cos( x)
  sin( x)
dx
2mE
k
2
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41
Particle in a 1D Potential Well -6
On the other hand, according to the boundary conditions ψ(x) has to be zero at
x= 0 and x= L.
k must then be some integer multiple of π/L:
np
k
L
where n=1,2,3, …
2mE
nπ
k
and k 

2

L
np 
En 
2
2mL
2
&copy; Nezih Pala npala@fiu.edu
2
2
2mE nπ

2

L
n 2p 2 2
 En 
2mL2
This formula shows what values of energy the
particle in the potential well may have.
The energy is quantized!!!
The integer n is called a quantum number.
EEE5425 Introduction to Nanotechnology
42
Particle in a 1D Potential Well -7
In order to find amplitude A of the wave function ψ(x) = Asin(kx),
–the solution of the Schr&ouml;dinger equation, –we have to use Postulate 3 stating
that probability to find particle anywhere from -∞ to ∞ is 1. Actually, the
probability to find particle in the region from 0 to L is 1:

L

0
*
*

(
x
)

(
x
)
dx



 ( x) ( x)dx  1
1
1
Using formula from table of integrals :  (sin x) 2 dx  x  sin 2 x  C
2
4

2
2
  np 
  np   np 
2 L
 dx  0 A sin  L x  dx  A np 0 sin  L x  d  L x 
L
*
L
2
A L  1  np  1  np

x   sin  2


np  2  L  4  L
2
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


L
0
A2 L  1  np  A2 L

L  


np  2  L 
2
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43
Particle in a 1D Potential Well -8

*

 ( x) ( x)dx  1
Based on Postulate 3:

A2 L
1  A 
2
2
L
Conclusion:
Free particle of mass m in one-dimensional potential well (U(x) = 0) is described
by Schr&ouml;dinger time-independent (stationary) equation:
 2 ( x) 2m
 2 E ( x)  0
2
x

&copy; Nezih Pala npala@fiu.edu
with solution
EEE5425 Introduction to Nanotechnology
2
np
 n ( x) 
sin
x
L
L
44
Particle in a 1D Potential Well -9
Summary
Free particle of mass m in one-dimensional potential well (U(x) = 0) is described by
wave functions
2
np
 n ( x) 
sin
x
L
L
For each allowable value of n the particle may have only certain value of energy
given by
n 2p 2  2
En 
2mL2
This formula describes energy spectrum of the particle in the potential well
•The energy of a particle in potential well is quantized.
•The integer n is called a quantum number
•The particular wave function ψn(x) and corresponding to it energy En describe
the quantum state of the particle.
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EEE5425 Introduction to Nanotechnology
45
Particle in a 1D Potential Well -10
np 
En 
2mL2
2
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EEE5425 Introduction to Nanotechnology
2
2
46
Particle in a 1D Potential Well -11
2
np
 n ( x) 
sin
x
L
L
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EEE5425 Introduction to Nanotechnology
47
Particle in a 1D Potential Well -12
Wave function ψn(x).
&copy; Nezih Pala npala@fiu.edu
Probability of finding a particle at a
position x inside the well is
proportional to |ψn(x)|2
EEE5425 Introduction to Nanotechnology
48
Particle in a 1D potential well -13
3
Probability to find electron in the
interval from x = 2 to x = 3 is
  ( x)
2
2
 0.194  20%
2
10
  ( x)
2
2
0
5.5
Probability to find electron in the
interval from x = 4.5 to x = 5.5 is
  ( x)
2
2
 0.0000065  0.00065%
4.5
10
  ( x)
2
2
0
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EEE5425 Introduction to Nanotechnology
49
Particle in a 1D potential well – Summary
•Schr&ouml;dinger stationary equation for particle in the infinitely deep potential well:
 2 ( x) 2m
 2 E ( x)  0
x 2

for 0  x  L
•Boundary condition: ψ(x=0) = 0; ψ(x=L) = 0
2mE
k
2
•To satisfy boundary conditions, k must be some integral multiple of π/L:
•General solution: ψ(x) = Asin(kx) where
k n
p
L
2mE nπ

2

L
(n  1,2,3,...)
n 2p 2 2
 En 
2mL2
•Probability of the particle existence within the well (0 &lt; x&lt; L) must be 1:

2
*

(
x
)

(
x
)
dx

1

A


L

2
np
 n ( x) 
sin
x
L
L
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EEE5425 Introduction to Nanotechnology
n 2p 2  2
En 
2mL2
50
Tunneling –1
The wave functions are relatively easy to obtain for the potential well with
infinite walls, since the boundary conditions force wave function ψn to be zero
at the walls.
Such shape of potential well models quite unrealistic situation.
A finite potential well is more appropriate model of cases existing in real world.
In this case, process of quantum mechanical tunneling of an electron through a
barrier of finite height and thickness may take place.
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EEE5425 Introduction to Nanotechnology
51
Tunneling –2
Solution ψn(x) of corresponding Schr&ouml;dinger
equation has nonzero value inside the
barrier and beyond it.
Exponential decrease of
probability inside barrier
|ψn(x)|2≠0 beyond barrier
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EEE5425 Introduction to Nanotechnology
52
Tunneling –3
When the barrier width and height is not infinite, the boundary conditions do
not force ψ to zero at the barrier. Instead, we must use the condition that ψ and
its slope dψ/dx are continuous at each boundary of the barrier (postulate 1).
Thus ψ must have a nonzero value within the barrier and also on the other side.
Since ψ has a value to the right of the barrier, ψ*ψ exists there also, implying
that there is some none-zero probability of finding the particle beyond the
barrier. The particle does not go over the barrier! –particle’s total energy is less
than the barrier height U0.
The mechanism by which the particle &quot;penetrates&quot;
the barrier is called tunneling.
It is impossible to explain effect of tunneling using classical concept. Quantum
mechanical tunneling is bound to the uncertainty principle.
Tunneling is important only over very small dimensions, but it can be of great
importance in the conduction of electrons in solid-state devices: p-n junctions,
field-effect transistors.
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EEE5425 Introduction to Nanotechnology
53
Schr&ouml;dinger equation in 3D -1
2 2
 

   U  
2m
j t
For 3D
2  2
2
2  
 
 2  2  2 r , t   U  

2m  x y
z 
j t
Let us write the wave function in the form of


j (( k x  k y  k z ) wt )
j ( k .r wt )
r , t   A  e
 A e x y z
To separate time and space variables:

  jwt
r , t   r e
Also meaning that
where

j (k xk y k z )
 r   A  e x y z

 r   x ( x)y ( y )z ( z )
Now let’s look at the particle in a box problem again but this time in 3D!
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Schr&ouml;dinger equation in 3D -2

  jwt
r , t   r e
Lz
Boundary conditions:
  x  0     x  Lx   0
Lx
  y  0    y  Ly   0
Ly
  z  0     z  Lz   0
We know the solution for x coordinate:
x x  
&copy; Nezih Pala npala@fiu.edu
2
sin knx x
Lx
where
np
k nx 
Lx
EEE5425 Introduction to Nanotechnology
nx  1,2,3...
55
Schr&ouml;dinger equation in 3D -3
Similarly
Lz
Ly
Lx
x  Lx , y, z   A  sin( k x Lx )  e
j ( k y y kz z )
0
Is true of and only if
nxp
k nx 
Lx
nx  1,2,3...
Repeating the same procedure for y and z, we conclude:

 r  
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 n xp
8
sin 
Lx L y Lz
 Lx
  n yp   n z p
x  sin 
y  sin 


  Ly   Lz
EEE5425 Introduction to Nanotechnology

z 

56
Schr&ouml;dinger equation in 3D -4
Allowed energy levels are given by:
Lz
E
Lx
Ly
 2 (k n2x  k n2y  k n2z )
2m
p n
nz2

(  2  2)
2 m L L y Lz
2
2
2
x
2
x
n y2
Also known as “quantum states”. Pauli exclusion principle: Each unique combination of nx,
ny, nz can only have two electrons (spin up, spin down).
States with different quantum numbers but the same energy (e.g. (1,2,3) and (3,1,2) are
called degenerate and the number of states having the same energy is called the
degeneracy.
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EEE5425 Introduction to Nanotechnology
57
Schr&ouml;dinger equation in 3D -5
If we think of a cube of material of side L and we compress (squeeze) the material ,
then L decreases and the energy levels increase. Thus electrons in the material
must increase their energy, and this energy gain comes from the work done by
squeezing the material. The resulting pressure is called Pauli pressure, since the
Pauli exclusion principle keeps multiple electrons (more than two) from occupying
the same energy level. This pressure partially accounts for the resistance to
squeezing of materials with high electron concentration.
Lastly, we should emphasize that we have solved the time independent
Schrodinger’s equation to obtain the possible (i.e. allowed electron states. What
state an electron actually occupies will depend on other factors such as
temperature, the presence of other electrons and other energy sources.
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EEE5425 Introduction to Nanotechnology
58
Periodic Boundary Conditions -1
Rather than the boundary conditions we used, it is more realistic to use periodic boundary
conditions that result in traveling rather than standing wave solutions. Periodic boundary
conditions emulate an infinite solid, rather than a finite region and are given by
 x, y, z    x  Lx , y, z ,  x, y, z    x, y  Ly , z ,  x, y, z    x, y, z  Lz 
where
  1
 r  
L L L
 x y z
 



k  a x k x  a y k y  a z k z , k  |k |
2n p
kx  x ,
Lx
ky 
2n yp
Ly
, kz 
and
1/ 2
 ikr
 e




k 2  k x2  k y2  k z2 
2me E
2
2n zp
,
Lz
nx,y,z=0,&plusmn;1, &plusmn;2,… Note that now we have the index 2nx,y,z instead of nx,y,z and that both
positive and negative values of k are allowed to account for waves moving in opposite
directions.
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EEE5425 Introduction to Nanotechnology
59
Periodic Boundary Conditions -2
Allowed energy levels are given by
2 p n
nz2
E
(  2  2)
m
L Ly Lz
2
2
2
x
2
x
n y2
And for a box having equal sides L,
2 2p 2 2
2
2
E
(
n

n

n
x
y
z)
2
mL
A more general form
E
 2p 2
mL2
(nx2  n y2  nz2 )
Represents either hard wall case for α=1/2 and the periodic boundary condition case for
α=2 .
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EEE5425 Introduction to Nanotechnology
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Finite Potential Well -1
V=V0
V=V0
I
x=-L
II
V=0
V=0 -L ≤ x ≤ L,
V=V0 x &lt; L, x &gt; L
III
x=L
This also approximates the influence of an ionized atom
on electron.
Now let us think about how classical mechanics would treats this problem:
Introduce an electron having total energy E&lt;V0 into the potential well. The electron would
stuck in the well, classically. Outside of the well, the electron’s total energy would still be E
since there is no source of energy for the electron. Therefore, if the electron were outside
the well we would have
E = EK + EP
= EK + V0 &lt; V0
where EK is the kinetic energy and EP is the potential energy. So
EK &lt; 0
which indicates that the electron has negative kinetic energy. According to classical physics,
this cannot occur, and therefore, classically, the electron must be in the well.
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EEE5425 Introduction to Nanotechnology
61
Finite Potential Well -2
However, quantum mechanically, there is some probability that the particle will be found
 2 d 2

 
 V ( x) x   Ex 
2
 2m dx

We solve this equation separately in the three regions and connect the three solutions
together by applying boundary conditions at the interfaces.
In the region I we have:
V=V0
V=V0
I
x=-L
II
V=0
III
x=L
 2 d 2


 (V0  E ) 1 x   0
2
 2m dx

1 x   Aek1x  Be  k1x
with
k12 
2m(V0  E )
2
However, the wavefunction should be finite as x → -∞ and assuming that E &lt; V0, then
B=0
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EEE5425 Introduction to Nanotechnology
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Finite Potential Well -3
V=V0
V=V0
I
II
V=0
For region II
III
x=-L
x=L
 2 d 2


2 x   0

E
2
 2m dx

2 x  C sin k2 x  D cos k2 x with k 22 
For region III
2mE
2
 2 d 2


3 x   0

(
V

E
)
0
2
 2m dx

3 x   Fek3 x  Ge k3 x
with
2m(V0  E )
2
k 

k
1
2
2
3
However, the wavefunction should be finite as x → +∞ and assuming that E &lt; V0, then
F=0
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Finite Potential Well -4
In summary we obtain
1 x   Aek1x
2 x   C sin k 2 x  D cos k 2 x
3 x   Ge k3 x
with
k 22 
2mE
,
2

k12  k32 
x  -L
-L  x  L
xL
2m(V0  E )
2
The boundary conditions at the interfaces between regions I and II and regions II and III are
that the wave function and first derivative must be continuous. Therefore
1  x   L   2  x   L 
1 x   L   2  x   L 
2  x  L   3  x  L 
2  x  L   3  x  L 
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Finite Potential Well -5
Ae  k1L  C sin k 2 L  D cos k 2 L
k1 Ae  k1L  Ck2 cos k 2 L  Dk
Adding and subtracting the first two
equations and the last two equations
and remembering
C sin k 2 L  D cos k 2 L  Ge k3 L
k12  k32
Ck2 cos k 2 L  Dk 2 sin k 2 L  Ge k3 L
Dividing out the exponentials from this
set of equations leads to the following
two transcendental equations:
k1  k 2 tan k 2 L
 k1  k 2 cot k 2 L
( A  G )e  k1L  2 D cos k 2 L
(G  A)e  k1L  2C sin k 2 L
k1 ( A  G )e  k1L  2k 2C cos k 2 L
k1 ( A  G )e  k1L  2k 2 D sin k 2 L
The only unknown in above equations is energy eigenvalue E (note that k’s re functions of E
and V0 . Due to the nature of these two equations thee energy can not be solved in closed
form. It is necessary to use graphical or numerical methods.
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Finite Potential Well -6
A simple graphical solution method is as follows. First note that
k12  k 22 
2m(V0  E ) 2mE 2mV0
 2 
2


2
2
2
mV
L
0
(k1 L) 2  (k 2 L) 2 
2
Which is the equation of a circle in the k1L – k2L plane. We can also plot k1L  k2 L tan k2 L
In the same plane. Obviously the intersections will be the desired (discrete) solutions for
energy En.
k1L
Only intersections in the upper-half plane are valid
since k1 &lt; 0 would cause the wavefunctions Ψ1 and Ψ3
to become infinitely large as |x| →∞.
k2L
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Finite Potential Well -7
k1L
The radius of the circle is
2
2
2
mV
L
2
mV
L
0
0
(k1 L) 2  (k2 L) 2 

r

2
2
k2L
So that for very small V0 or L, there is only one solution. As V0 or L increase, the radius of
the circle and so more discrete states will exist, although for any finite V0 and L there will
be finite number of solutions. In the limit V0 →∞ a countable infinity of discrete
solutions will exist in agreement with the infinite well problem.
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Finite Potential Well -8
We assumed that E&lt; V0 and discrete energy
values were obtained. The electron is mot likely
to be found in the well, although it can also be
found outside the well with decreasing
probability as we move away form the well.
Other solutions exist for E&gt; V0 corresponding to
a continuum of allowed energy values. In this
case presence of the well merely perturbs the
electrons wave function. Far from the well we
expect the wavefunction to correspond to a
plane wave, since in these locations the electron
is essentially free.
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Parabolic Well – Harmonic Oscillator -1
Parabolic potential has importance in modeling of many quantum heterostructures. In this
case the potential profile is given as
V ( x) 
1 2
Kx
2
Which describes a classical harmonic oscillator in analogous to mass on a spring which
gives rise to harmonic motion x(t) =Acos ωt where the ω2=K/m.
For the harmonic potential, Schrodinger equation is
 2 d 2 1 2 2 
 
 w0 mx x   Ex 
2
 2m dx 2

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Parabolic Well – Harmonic Oscillator -2
Which has the solution of
 mw0 1/ 2   mw0 x 2
 x   Cn H n 
 xe 2
   
Where Cn is a constant
Where Hn are Hermite polynomials:
H 0 ( x)  1
 1  mw0 
Cn  


n
 p 2 n!   
1/ 4
H1 ( x)  2 x
H 2 ( x)  4 x 2  2
Energy levels are found to be
1

En   n  w0 ,
2

n  1,2,3...
And equally spaced according to the index n.
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Parabolic Well – Harmonic Oscillator -3
Solution wavefunctions for parabolic well.
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Triangular Well
V
Triangular wells are used to model junctions between two
materials.
V=Cx
The potential profile is
V(x)=
{
V=∞
V=Cx
x&lt;0
x&gt;0
x
The solution of Schrodinger’s equation with the triangular
potential is fairly complicated and the resulting wavefunctions
are expressed in terms of Airy functions.
The energy levels are given as
1/ 3
   3

En  
  pC 
 2m   2

2/3
1

n  
4

2/3
,
n  1,2,3...
Note that for a rectangular well En ~ n2, for a parabolic well En ~ n and for a
triangular well En ~ n2/3
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The Hydrogen Atom - 1
Finding the wave functions for the
hydrogen atom requires a solution of
the Schr&ouml;dinger wave equation in three
dimensions for a coulombic potential
field.
Since the problem is spherically
symmetric, the spherical coordinate
system is used in the calculation.
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The Hydrogen Atom -2
Spherical polar coordinates
The radial dependence of the potential suggests that we should from Cartesian
coordinates to spherical polar coordinates.
r = interparticle distance
(0 ≤ r ≤ ∞)
θ = angle from z-axis to “x-yplane”
(0 ≤ θ ≤ π)
φ = rotation in “x-yplane”
(0 ≤ φ ≤ 2π)
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The Hydrogen Atom -3
Potential term U(x,y,z) in the Schr&ouml;dinger equation written in rectangular coordinate
system must be replaced by U(r,θ,φ), representing the Coulomb potential the spherical
(polar) coordinate system which the electron experiences in the vicinity of the proton.
The Coulomb potential varies only with r in spherical coordinates:
U (r )  
q2
4p0 rn
Then, the Schr&ouml;dinger equation:
2m
  (r ,  ,  )  2 ( E  U (r ,  ,  ))  (r ,  ,  )  0

2
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The Hydrogen Atom -4
Coulomb potential which the electron experiences in the vicinity of the proton:
q2 1
U (r , ,  )  U (r )  
4p0 r
Now, variables in the Schr&ouml;dinger equation can be separated:
 (r , ,  )  R(r )( )( )
This wave function of an electron in hydrogen atom ψ(r,θ,) is a product of three parts.
Three separate solutions must be obtained for:
•the r -dependent equation
•the θ -dependent equation
•the  -dependent equation
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The Hydrogen Atom -5
As in the simple potential well problem discussed before, each of the three hydrogen
atom equations gives a solution which is quantized.
Thus we would expect a quantum number to be associated with each of the three
parts of the wave equation.
Let’s analyze only the  -dependent equation obtained after separation of variables in
the Schr&ouml;dinger equation written in polar coordinate system:
d 
2

m
0
2
d
2
where m is a quantum number associated with coordinate  .
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The Hydrogen Atom -6
d 2
2

m
0
2
d
The solution to this equation is: Φm()=Aejm
where A can be evaluated by the normalization condition, as we have seen before:
2p
*

 m ( ) m ( )d  1
0
2p
2p
0
0
A2  e  jm e jm d A2  d  2pA2
Thus
2pA  1
2
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
1
A
2π
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The Hydrogen Atom -7
d 2
2

m
0
2
d
the  -dependent wave equation
 m ( )  Ae jm 
1

jm


(

)
e
1

m
A
2p

2p

the  -dependent wave function
Since values of  repeat every 2π radians, Φ() should repeat also. This occurs if m is an
integer, including negative integers and zero:
The wave functions for the φ-dependent equation are quantized with the
following selection rule for the quantum numbers:
m = ..., -3, -2, -1,0, +1, +2, +3,...m
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The Hydrogen Atom -8
Applying similar logics, the functions R(r) and Θ(θ) can be obtained, each
being quantized by its own selection rule.
For the r -dependent equation, the quantum number n can be any positive integer (not
zero).
For the θ -dependent equation the quantum number l can be zero or a positive integer.
However, there are some interrelationships among the equations which
restrict the various quantum numbers used with a single wave function
Ψnlm(r,θ,) = Rn(r) Θl(θ) Φm()
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The Hydrogen Atom -9
These restrictions are summarized as follows:
Principal quantum number:
n = 1, 2, 3, ...
Azimuthal quantum number:
l = 0, 1, 2, ..., (n-1)
Magnetic quantum number:
m = -l, ..., -2, -1, 0, +1, +2, ..., +l
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The Hydrogen Atom -10
Energy depends only on the principle quantum number n and is given by:
En  
q
4
8 h n
2
0
2
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2
For n=1 the ground state:
q 4
E1   2 2  13.6eV
8 0 h
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The Hydrogen Atom -11
In addition to the three quantum numbers arising from the three parts of the wave
equation, there is an important quantization condition on the spin of the electron.
Investigations of electron spin employ the theory of relativity as well as quantum
mechanics; therefore, we shall simply state that the intrinsic angular momentum s of
an electron with Ψnlm specified is

s
2
That is, in units of ħ, the electron has a spin of
&frac12;, and the angular momentum produced by
this spin is positive or negative depending on
whether the electron is “spin up” or “spin
down”.
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The Hydrogen Atom -12
Summary
Each allowed energy state of the electron in the hydrogen atom
is uniquely described by the set of four quantum numbers: n, l,
m, and s.
Using these four quantum numbers, we can identify the various states which the
electron can occupy in a hydrogen atom.
The number n, called the principal quantum number, specifies the “orbit” of the
electron in Bohr terminology.
There is considerable fine structure in the energy levels about the Bohr orbits, for
example:
•an electron with n= 1 can have only l= 0 and m= 0, but there are two spin states
allowed.
•for n= 2, l can be 0 or 1, and m can be -1, 0, or +1 with two spins for each state.
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The Hydrogen Atom -13
These four quantum numbers with the selection rules precisely
describe the structure of energy states in hydrogen atom.
Electron in a hydrogen atom can occupy only one of a large
number of excited states including the lowest (ground) state.
Energy differences between the various states properly account for the
observed lines in the hydrogen spectrum.
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Larger Atoms
The structure of energy states and quantum numbers just discussed arise
from solutions to the hydrogen atom problem.
Question: How can we extend the knowledge of hydrogen atom
energy structure on description of more complex atoms?
Answer: The quantum number selection rules are valid for more
complicated structures. They can be used to describe the
arrangement of atoms in the periodic table of chemical elements.
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Pauli Exclusion Principle -1
Wolfgang Ernst Pauli (1900 – 1958) was an Austrian theoretical
physicist noted for his work on spin theory, and for the discovery
of the exclusion principle underpinning the structure of matter and
the whole of chemistry.
In multi-electron system only one electron may occupy a specific
discrete energy level.
or in other words:
No two electrons can have the same set of quantum numbers n, l, m, s.
or in other words:
Only two electrons can have the same three quantum numbers n, l, m,
and those two must have opposite spin.
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Pauli Exclusion Principle -2
Pauli exclusion principle is basic to the electronic structure of all atoms in the
periodic table
Quantum numbers to n= 3
and allowable states for
the electron in a hydrogen
atom.
The first four columns
show the various
combinations of quantum
numbers allowed by the
selection rules.
The last two columns
indicate the number of
allowed states
(combinations of n, l, m,
and s) for each l (sub-shell)
and n (shell, or Bohr orbit).
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The Periodic Table –1
In the first electronic shell (n= 1), l can be only zero since the maximum value of
l is always n-1.
Similarly, m can be only zero since m runs from the negative value of l to the
positive value of l:
First electronic shell: n= 1; l= 0; m= 0; s= &plusmn;&frac12; ψ100
Two electrons with opposite spin can fit in this ψ100 state; therefore, the first
shell can have at most two electrons.
For the helium (He) atom (atomic number Z= 2) in the ground state, both
electrons will be in the first Bohr orbit (n= 1), both will have l= 0 and m= 0, and
they will have opposite spin.
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The Periodic Table –2
Convention in the naming of the l sub-shells:
l
Notation
0
s
1
p
2
d
3
f
4
g
This convention was created by early spectroscopists who referred to the first
four spectral groups as:
sharp, principal, diffuse, and fundamental.
Alphabetical order is used beyond f.
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The Periodic Table –3
With this convention for l, we can describe each electron state as follows:
6 electrons in the 3p subshell
6
3p
n=3
Principal quantum number
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l=1
Azimuthal quantum number
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Ex: Silicon Atom -1
Example:
The total electronic configuration for Si (Z= 14) in the ground state is:
Si:1s2 2s2 2p6 3s2 3p2
Number of electrons: 2+2+6+2+2=14
The total electronic configuration for Ne (Z= 10) in the ground state is:
Ne:1s2 2s2 2p6
Number of electrons: 2+2+6=10
In Ne, all the three sub-shells (1s, 2s, and 2p) are completely filled with maximum
possible number of electrons forming a closed shell (typical of the inert elements).
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Ex: Silicon Atom -2
Example:
The total electronic
configuration for Si (Z= 14) in
the ground state is:
1s2 2s2 2p6 3s2 3p2
Number of electrons:
2+2+6+2+2=14=Z
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Ex: Silicon Atom -3
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Ex: Silicon Atom -4
The p subshell can hold up to 6 electrons, but in the case of Si has only 2. Interestingly, in a
Si crystal when we bring individual atoms very close together, the s- and p-orbitals overlap
so much that they lose their distinct character, and lead to four mixed sp3 orbitals. The
negative part of the p orbital cancels the s-type wavefunction, while the positive part
enhances it, thereby leading to a “directed” bond in space. These “hybridized” sp3 orbitals
point symmetrically in space along the 4 tetragonal directions
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Electronic configurations of atoms
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The Periodic Table –1
In 1869 Mendeleev and Lothar Meyer (Germany) published nearly
identical classification schemes for elements known to date.
The periodic table is base on the similarity of properties and reactivities
exhibited by certain elements.
Later, Henri Moseley (England, 1887-1915) established that each
elements has a unique atomic number (Z), which is how the current
periodic table is organized.
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The Periodic Table –2
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```