Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
CHAPTER 4
Solutions Of Schrodinger Equations For Simple Systems.
4.1 Introduction
For non-relativistic quantum mechanics the basic equation to be solved is the Schrodinger equation.
Like Newton's laws, the Schrodinger equation must be written down for a given situation of a
quantum particle moving under the influence of some external forces, although it turns out to be
easier to frame this in terms of potential energies instead of forces. However, unlike Newton's laws,
the Schrodinger equation does not give the trajectory of a particle, but rather the wave function of
the quantum system, which carries information about the wave nature of the particle, which allows
us to only discuss the probability of finding the particle in different region of space at given
moment of time.
4.2 Free Particle in One Dimension
A “free" particle refers to a particle that has no external forces acting upon it, in other words the
potential energy is constant V0. The state of such a free particle is represented by its wave function
ψ(x).
A “free" particle refers to a particle that has no external forces acting upon it, in other words the
potential energy is constant V0. The state of such a free particle is represented by its wave function
ψ(x).
Consider a particle of mass m moving along positive x-axis. Particle is said to be free if it is not
under the influence of any field or force. Therefore for a free particle potential energy can be
considered to be constant or zero. The Schrodinger wave equation for a free particle is given by.
d 2
8 2 m
 E  0 

2
2
dx
h
d 2
8  2 mE

  0
2
2
dx
h
2
d 
  2  0
2
dx
8  2 mE
2
where 

h2
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 0
1
1
2
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
The solution of the equation 1 is of the form
Where A and B are unknown constants to be determined. Since there are no boundary conditions A, B
and  can have any values.
Energy of the particle is given by
Since there is no restriction on  there is no restriction on E.
Therefore energy of the free particle is not quantized. i.e., free particle can have any value of
energy.
4.3 Particles in a One Dimensional Box
The simplest problem to treat in quantum mechanics is that of a particle of mass m constrained to
move in a one-dimensional box of length a (“particle in a box”). The potential energy V (x) is
taken to be zero for 0 < x < L and infinite outside this region. The infinite potential can be treated
as a boundary condition (i.e., the wave function must be zero outside 0 < x < L). Such a bound
potential will lead to quantized energy levels. In general, either a bound potential or a suitable
boundary condition is required for quantization.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
Time Independent Schrödinger Equation
  2 d 2 ( x)
 V ( x)  E
2m dx2
(4)
Applying boundary conditions,
  2 d 2 ( x)
  *  E
2m dx2
Region I and III
 
2
d 2 ( x )
2mE


 ( x)
2
2
dx
(5)
0
(6)
Region II:
  2 d 2 ( x)
 E
2m dx2
d 2 ( x) 2m
 
 2 E
dx 2

This is similar to the general differential equation:
2
2
2
d  ( x)

k 
dx
  A sin kx  B cos kx
(7)
(8)
(9)
(10)
Applying boundary conditions:
a) x = 0 ψ = 0
0  A sin 0k  B cos0k
0  0  B *1
 B  0
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
b) x=L ψ =0
0  Asin kL
ButA  0
kL  n
nx
L
But what is ‘A’?
Thus, wave function:
 II  A sin
(11)
L
Normalizing wave function:
(12)
2
 ( A sin kx) dx  1
0
(13)
L
sin 2kx 
2x
A  
 1
2
4
k

0
Normalizing wave function:
Calculating Energy Levels:
L
 ( A sin kx) dx  1
2
0
k2 
2mE
2
L
 x sin 2kx 
A  
1
4k  0
2
2
n 

sin 2
L

2 L
L
A  
1
n 
2


4
L 

2 L 
A   1
2
A
k 2 2
E
2m
k 2h2
E
2m4 2
( 
h
)
2
n 2 2 h 2
E 2
L 2m4 2
2
L
Thus normalized wave function is:
2
nx
 II 
sin
L
L
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Thus Energy is:
n2h2
E
8mL2
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
EXAMPLE 4-1 Consider an electron in a one-dimensional box of length 258pm.
a) What is the zero-point energy (ZPE) for this system? and for a mole of such systems?
b)What electronic speed classically corresponds to this ZPE? Compare to the speed
of light.
4.4 Rotational Motion
Rotational motion is an important topic in chemical systems as it will be used to describe the
rotational motion of gas phase molecules and electronic motion in atoms and molecules. The
model problems presented in this topic will be the basis for modeling rotational motion.
4.5 Particle on a Ring
Consider a particle of mass m confined to a circle with a constant radius r as shown in Figure 4.1. The
potential energy anywhere on the circle is defined as zero. The Hamiltonian, Ĥ, for the particle in
Cartesian coordinates is given below.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
The motion of the particle is not separable between the x and y-axes. The problem can be made
separable by transforming the coordinates from Cartesian to polar coordinates. In polar coordinates,
the variables become the radius of gyration, r, and the angle, ɸ of the particle from the origin.
x= rcosɸ
y = rsinɸ
The moment of inertia, I, of the particle, is equal to its mass times the square of the radius of
gyration r. The Hamiltonian in Equation (15) is very similar to the Hamiltonian for the onedimensional Particle-in-a-Box problem.
by using Euler's formula
The constants A and B are normalization constants, and the constant k will be determined by the
boundary conditions for the particle. The Schrödinger equation and the energy eigenvalue becomes:
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
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The symbol for k is traditionally ml known as the magnetic quantum number when referring to
electronic states. The possible values for ml are as follows:
ml =0, 1, 2,
The allowed energy values become:
The final task is to normalize the wavefunction. Since the value of ml can be both positive and
negative integers, the wavefunction can be reduced to just one exponential term.
(20)
The limits of integration for normalization will be from 0 to 2 since this covers the entire circular
path.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
4.6 Particle on a Sphere
Particle on a sphere is one of the models that describe rotational motion. A single particle travels
on the surface of the sphere. Unlike particle in a box, the particle on a sphere requires angular
momentum, L.
There is a vector that contains direction of the axis of rotation. The magnitude of angular
momentum of the particle that travels around the sphere can be defined as:
L=p.r
(21)
where

p is linear momentum is the result of the mass and velocity of object (p=mv)

r is the radius of the sphere
The faster a particle travels in a sphere the higher the angular momentum. In other words, if we
increase the velocity of a particle we get an increase in angular momentum. Therefore this required
stronger torque to bring the particle to stop. Particle of mass is not restrict to move anywhere on
the surface of the sphere radius. The potential energy of the particle on a sphere is zero because the
particle can travel anywhere on the surface of the sphere without a preference in location; the
particle on the sphere is infinity. Furthermore, the wave-function needs to satisfy two cyclic
boundary conditions which are passing over the poles and around the equator of the sphere
surrounding the central point. Using the Schrödinger equation we are able to find the energy of the
particle: The orientation of a linear rigid rotator is completely specified by the two angles
and ,
so rigid-rotator wave functions depend upon only these two variable. The rigid-rotator wave
functions are customarily denoted by Y(θ,
), so the Schrödinger equation for a rigid rotator reads.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
Spherical Polar Coordinates = most efficient way to describe position of particle on
sphere
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
The Wave Functions of a Rigid Rotator Are Called Spherical Harmonics
Example. Rotational energy levels
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
4.7. Harmonic Oscillator
The vibrational motion of a diatomic molecule can be approximated as a harmonic oscillator. In
this topic, we will first study a classical harmonic oscillator and then present and discuss the
energies and the corresponding wave functions of a quantum mechanical harmonic oscillator.
simple harmonic oscillator is a system with a continuously varying potential. There are several
reasons for studying this problem in detail. First, the quantum-mechanical harmonic oscillator
plays an essential role in our understanding of molecular vibrations, their spectra, and their
influence on thermodynamic properties. We will use the quantum-mechanical energies to describe
the infrared spectrum of a diatomic molecule and learn how to determine molecular force constants
from vibrational spectra. Then we will model the rotational motion of a diatomic molecule by a
rigid rotator. We will discuss the quantum-mechanical energies of a rigid rotator and show their
relation to the rotational spectrum of a diatomic molecule. We will use the rotational spectrum of a
diatomic molecule to determine the bond length of the molecule.
4.8 Classical harmonic oscillator
A Harmonic Oscillator Obeys Hooke's Law.
Consider a mass m connected to a wall by a spring as shown in Figure 4.3. Suppose further that no
gravitational force is acting on m so that the only force is due to the spring.
FIGURE 4.3
A mass connected to a wall by a spring. If the force acting upon the mass is directly proportional to
the displacement of the spring from its undistorted length, then the force law is called Hooke's law.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
If we let l0 be the equilibrium, or undistorted, length of the spring, then the restoring force must be
some function of the displacement of the spring from its equilibrium length. Let this displacement
be denoted by x = l - l0, where l is the length of the spring. The simplest assumption we can make
about the force on m as a function of the displacement is that the force is directly proportional to
the displacement and to write.
F= -k(l - l0) = -kx
(24)
Let's look at the total energy of a harmonic oscillator. The force is given by Equation 24. Recall
from physics that a force can be expressed as the negative derivative of a potential energy or that
The frequency of motion (ν) is given by
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
28
Squaring the both side of equation 28
29
30
)
31
Schrodinger equation become
32
32
,
The solution of above equation for the wave function is complex and it is expressed
in terms of what are known as Hermit polynomials Hn(y)
n = 0,1,2,3,4
Nn is normalized constant
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
35
The Harmonic-Oscillator Wave Functions Involve Hermite Polynomials
The first seven eigenfunction of harmonic oscillator
The energy eigen values
The allowed energy value of harmonic oscillators is given by
E is vibrational energy
v is vibrational frequency
n is vibrational quantum number
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
4.9 Quantum Mechanical Treatment of the Harmonic Oscillator
The diatomic molecule can be modeled as harmonic oscillator oscillates
around center of mass μ
If we model the potential energy function of a diatomic molecule as a harmonic
oscillator, and the vibrational energy levels of the diatomic molecule are given by
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
FIGURE 4.4
The energy levels of a quantum-mechanical harmonic oscillator.
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
Note that the energy levels are equally spaced, with a separation w or hv. This uniform spacing
between energy levels is a property peculiar to the quadratic potential of a harmonic oscillator.
Note also that the energy of the ground state, the state with n = 0, is ~ h v and is not zero as the
lowest classical energy is. This energy is called the zero-point energy of the harmonic oscillator.
1. As n increase E increase
2. Energy levels are equally spaced by hv
3. Minimum possible energy =1hv/2 is called the Zero-Point Energy (ZPE)
4. Wave functions extend beyond separations allowed by classical mechanics
4.10. Vibration of Diatomic Molecules
Figure 4.5. Two masses connected by a spring, which is a model used to describe the vibrational
motion of a diatomic molecule.
A diatomic molecule can make a transition from one vibrational energy state to another by
absorbing or emitting electromagnetic radiation whose observed frequency satisfies the Bohr
frequency condition.
Harmonic-oscillator model allows transitions only between adjacent energy states, so that we have
the condition that ∆n = ± 1. Such a condition is called a selection rule.
For absorption to occur, ∆ n = + 1 and so
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
where the tilde indicates that the units are cm-1. Furthermore, because successive energy states of a
harmonic oscillator are separated by the same energy, ∆E is the same for all allowed transitions.
For example, for H35Cl, ṽobs is 2.886 x 103 cm-1 and so, the force constant of H35Cl
is
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Chapter Four: Solutions of Schrodinger Equations for Simple Systems
Quantum Chemistry
EXAMPLE
Exercise
1. A 1H35Cl molecule has a force constant of 516 Nm-1. Calculate the
vibrational stretching frequency
2. Hydrogen bromide (1H81Br) absorbs infrared radiation of wavenumber
2649.7 cm-1. Calculate the force constant of the 1H81Br bond. (The molar
masses of 1H and 81Br are 1.008 g mol-1, and 80.916 g mol-1, respectively.)
3. The infrared absorption spectrum of 1H35Cl has its strongest band at
8.65×1013 Hz.
(1) Calculate the force constant
(2) Calculate the zero-point vibrational energy
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