11/2/2023
PHY 412 Introductory Quantum
Mechanics
continue from last lecture…….Particle in a One-Dimensional Box
Normalization of the wave function
Normalization means that the integral of the square of the
wavefunction (probability density) over all space is equal to one.
As we mentioned earlier, the quantity
finding the particle in the small volume
Lecture 06: The Schrödinger Equation
,Particle in a Box and The
Postulates of Quantum Mechanics
∗
gives the probability of
.
Since the particle must be found inside the box, the total probability
of finding the particle inside the box must equal 1.
∗
=1
Because the wave function doesn’t include imaginary numbers, we
have ∗ = , and because the particle is limited to move in one
direction, we have
= .
In addition, because the particle is limited to move inside the box
between = 0 and = , we can put the integration limits between 0
and .
2
Particle in a One-Dimensional Box
Particle in a One-Dimensional Box
Using these facts, we get
=
ψ∗
But we have sin 2
Substituting for
2
=
−
sin 2
4
=
sin
=
sin
=1
⇒
using the relation
sin
=
−
2
2
=
And when substitute for
sin 2
4
=
2
−
sin
4
,
3
=1
ℎ
=
=1
1
2
2
=
in the equation of the wave function, we get
we get
2
sin 0 = 1
4
= 0 for any value of . therefore,
, we get
=
−0−
2
sin
And this is the final form of the normalized wave function of the
particle in a one dimensional box.
4
Particle in a One-Dimensional Box
Particle in a One-Dimensional Box
Orthogonality of wave functions
Using this realtion
This is an important mathematical characteristic of Eigenfunctions. This
mathematical rule indicates that when a number of Eigenfunctions are
different solutions with different eigenvalues of a given equation, these
Eigenfunctions must be orthogonal.
This rule is expressed
mathematically as:
=0,
2
=
2
sin
2
sin
sin
=
sin −
sin +
−
2( − )
2( + )
we get
=
≠
1
2
( −
)
sin
( −
)
−
( +
)
sin
( +
)
=0
Because
and
are real numbers, their sum or difference is also a real
number, and sin
= 0 when = 0, 1, 2, 3, … ,therefore the above integral
must equal zero.
This means that the two wave function equations
and
for the particle
in the one dimensional box are orthogonal.
It must also be noticed that the orthogonality conditions apply only
when the eigenfunctions are real solutions to the equation and not
approximate solutions.
For a particle in a one dimensional box
=
sin
sin
sin
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Particle in a One-Dimensional Box
Particle in a One-Dimensional Box
Energy and wave function of a particle in a box
2. The highest probability of finding the particle is in the middle of the
distance between two nodes. The highest probability of finding the
particle for = 1 is at /2, and for = 2 is at /4 and 3 /4, and so on.
For a particle moving in a one dimensional box, we have shown that
=
2
and
sin
=
ℎ
8
3. The orthogonality of the wave functions is
clear. From the figure we can see that
We noticed from the graph that:
=
2 take zero at the same
1. Both
and
certain points. These points are called
nodes. At the nodes, the wave function of
the particle vanishes and there is zero
probability of finding the particle. The
number of nodes for a given value of
equals ( − 1) and the number of nodes
increases as then value of
increases.
This is in consistence with the de Broglie
hypothesis ( = ℎ/ ). As the kinetic energy
increases the wave length decreases and
the number of nodes increases.
=−
−
−
=−
−
−
The integration over all values of for the
two equations between = 0 and = /2 is
equal but different in sign for the same
integral between = /2 and = . This
means that the integration∫
= 0 is
equal to zero.
⇓
7
Particle in a One-Dimensional Box
8
Particle in a One-Dimensional Box
4.
is inversely proportional to and : as the mass of the
particle and or the size of the box get smaller the higher the
energy of the particle.
Correspondence principle
“classical mechanics emerges from quantum mechanics at high
quantum numbers.”
Note that in the particle in one dimensional box problem when the
size of the box becomes very large or when the mass of the particle
gets very large, the quantity
becomes very large compared to
ℎ (
≫ ℎ ) and the energy of the system becomes
continuum and we go back from quantum physics to classical physics.
This is always the case. When sizes get larger, the quantum physics
problem converges with classical physics. This is called the
correspondence principle. So it might be said that the classical
physics laws are special case of quantum chemistry laws.
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Correspondence principle
The most probable position of the particle is near the
= 1 state but that the
center of the box for the
probability density becomes more uniformly distributed
as increases.
The probability density for higher energy levels (e.g. at
= 20) is fairly uniformly distributed from 0 to . the
particle tends to behave classically in the limit of
→ ∞ each position
large quantum numbers (as
is equally probable).
|
|
→
The probability distribution
of a particle in a box.
0
→
10
Particle in a 1-D Box Model and the Real World
Particle in a 1-D Box Model and the Real World
The particle in a box model can be applied to electrons moving freely (
electrons) in a molecule.
According to the Pauli exclusion principle each energy level can hold only two
electrons (with opposite spins), and so the four electrons fill the first two levels.
The first excited state of this system of four
electrons is that which has one electron
elevated from the = 2 state to the = 3
state. The energy needed to make a transition
from the = 2 state to the = 3 state is
Consider butadiene, H2C=CH-CH=CH2, which has four
Butadiene has an absorption band at 217 nm for the 1st
Although butadiene
simplicity that the four
electrons.
→
∗
transition.
is not a linear molecule, we will assume for
electrons in butadiene move along a straight line.
The length of this straight line can be estimated as
2 C=C bond (2 x 135 pm) + C- C bond (154 pm) + the distance of a carbon
atom radius at each end (2 x 77.0 pm= 154pm), giving a total distance of
578 pm (5.78 Å or 578 x 10-12 m).
The allowed
electronic energies are given by
=
is the electron mass.
ℎ
8
∆ =
=
ℎ
(3 − 2 )
8
the mass of the electron is 9.109 x 10-31 kg,
and the length of the box is taken to be 578
pm, or 578 x 10-12 m.
∆ =
= 1, 2, 3, …
11
−
ℎ
∆
1
ℎ :
= 1 × 10
HOMO: Highest Occupied Molecular Orbital
LUMO: Lowest Unoccupied Molecular Orbital
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Particle in a 1-D Box Model and the Real World
∆ =
(6.626 × 10
J. s ) 5
8(9.109 x 10−31 kg )(578 × 10
)
= 9.02 x 10−19 J
and
=
(6.626 × 10
J. s )(2.998 × 10
9.02 x 10−19 J
= 2.20 × 10
1
(
220
/ )
Postulates of Quantum Mechanics
Any theory is usually built on a number of postulates. A postulate is a
theoretical assumption that has no prove but reasonable and found to
be consistent with observation. A postulates can play a key rule in
deriving the theory.
The following are some of the quantum mechanics postulates:
First postulate
The state of a system is fully described by its state function or wave
function (often written as wavefunction)
, that depends on the
coordinates , , of the particle(s) and on time .
)
ℎ :
= 1 × 10
For a system consisting of
described by a wave function
The causes of discrepancy between experimental value (217
)
and calculated value (220
) is the approximations that have been
made.
13
≡ ( 1,
where
1, and
1,
2, and
4,
5,
particles then this system can be fully
which is given by
2,
3,
4, … ,
3
, )
3 are the spatial coordinates ( 1,
1, 1)
and 6 are the spatial coordinates ( 2,
2 and so on. Whereas is the time.
for particle
2, 2)
for particle
14
Postulates of Quantum Mechanics
Postulates of Quantum Mechanics
When the function
is independent of time the function is said to be
stationary wave function. Fortunately, many applications of quantum
mechanics to chemistry use stationary wavefunction.
There are requirements for a wave function to be acceptable.
These requirements are:
The probability of finding particle lies in the volume element , where
refers to the region between and +
, and + , and and
+
is given by
∗
Now, because the particle has to be found some where in space, we must
have
∗
∗
found in one space at a given time.
b. The wave function and its first derivative must be continuous
since a particle can not transfer from one space to another
without passing in the path connecting them.
c. The wave function must be finite.
=1
The notation "all space" here means that we integrate over all possible
values of , , and .
The quantity
a. The wave function must be single valued since a particle can be
is called the probability density.
d. The wave function must be quadratically integrable.
The wave function that fulfil these requirements is said to be well
behaved wave function.
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16
Postulates of Quantum Mechanics
Postulates of Quantum Mechanics
Second postulate
To every measurable physical property (observable) in classical mechanics
there is a linear Hermitian operator in quantum mechanics. The value of
the physical property for a system can be found when the corresponding
operator operates on its wave function.
Physical observables and their corresponding quantum operators.
The linear and Hermitian operator have been defined earlier when we
discussed operators. However we add here that a Hermitian operator also
fulfil these requirements:
Classical mechanics
Name
Symbol
=
(
=
(
)∗
− ℏ( ̂
Kinetic energy
ℏ
−
2
Total energy
For all well behaved functions.
)∗
Angular momentum
17
Multiply by
Linear momentum
and
∗
Operation
̂
Position
( )
Potential energy
∗
Quantum mechanics
Symbol
=
+ ( )
( )
=
+
+ ̂
(
+
+
+
Multiply by
( )
−
ℏ
2
)
)
( )
+
+
+ ( )
− ℏ(
−
)
− ℏ(
−
)
− ℏ(
−
)
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Postulates of Quantum Mechanics
Postulates of Quantum Mechanics
How to derive an operator of a property?
To derive the operator of a physical property, follow the next steps:
a. Write the equation of the desired property using the classical
physics laws in terms of the coordinates, momentum and time.
b. Leave time and coordinates with no change.
c. When using the cartesian coordinates, replace the momentum
by the operator
Now, we make the replacement
 , where q ≡( x, y, z )
 i
q
pˆ x  i
pˆ x2   2
T=
2
x 2
Likewise
2
2
and
pˆ z2   2 2
y 2
z
And substitution in the operator of the kinetic energy, we get
pˆ 2y    2
Example:
If we want to calculate the kinetic energy of the system, the corresponding
operator is and from classical physics we have

x
 2   2
2
 2 
2 2
Tˆ  




2m  x 2 y 2 z 2 
2m
The operator
is called the Laplacian operator (read as “del squared”).
2
2
2
1 2 m 2 v 2 Px + Py + Pz
mv =
=
2
2m
2m
≡
+
+
19
20
Postulates of Quantum Mechanics
Applications of Postulates of Quantum Mechanics
Third postulate
If is an eigenfunction of the operator , then in any measurement of the
observable associated with the operator , the only values that will ever be
observed are the eigenvalues , which satisfy the eigenvalue equation
Lets, for example, operate on the wave function for the particle in
one dimensional box by the operator of the linear momentum in the
direction,
=
= −ℏ
Fourth postulate
If is not an eigenfunction of the operator , we still can calculate the
average or expected value of that property using the relation
=−ℏ
expected value =
=
∗
∫
If the wave function
is a normalized , then the average value of the
observable corresponding to is given by
What if we operate by
cos
?
2
d 
d2

pˆ x2    i    2 2
dx 
dx

∗
=
2
sin
It is clear that the function here is not an eigenfunction of the
operator .
∗
∫
2
21
Applications of Postulates of Quantum Mechanics
2
=−ℏ
= ℏ
= ℏ
It is clear that the function
and the eigenvalue is
ℏ
2
cos
a
 2
d  2
 nx 
 nx  
pˆ x   
sin 
sin 
  i 
  dx
dx  a
 a 
 a 
0  a
sin
=
Applications of Postulates of Quantum Mechanics
Because the function
is not an eigenfunction of the operator ,
we can calculate the expected value of
as the fourth postulate
states.
∫ ∗
∗
=
=
∫ ∗
sin
2
=−ℏ
22
2 2ℏ2
 i 
2
is an eigenfunction of the operator
.
Using the relation

23
2  n  a
 nx 
 nx 

  sin 
 cos 
 dx
a a 0
 a 
 a 
sin bx cos bx dx 
sin 2 bx
2b
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Applications of Postulates of Quantum Mechanics
2  nπ  2a
Pˆx  i  
a  a  nπ
⇒
a
The importance of the particle in a 3-D box model arises from the idea that
it shows the technique used to solve Schrodinger equation when it is a
function of more than one variable.
 2  nx 
sin  a 

 0

2  nπ  2a
sin 2 nπ  sin 2 0  0
 i  
a  a  nπ
Therefore, the expectation value of
This technique is called the separation of variables in which the total
equation is separated into as many equations as there variables.
is zero.
For the particle in three dimensional box, the wave function is a function of
the three variables , and
= ( , , )
It was found from the early discussion that:
n 2 2  2
p x2 
a2
Taking the square root of this equation gives
px  
n 2 2  2
a2
 
Particle in a Three-Dimensional Box
Using the separation of variables technique, This equation can be divided
into three equations, equation for each variable, and the original equation
will be the product of these three equations.
n
a
, ,
=
Thus for a large number of measurements of , we will have half of the
results with
= ℏ/ , and the other half with
= − ℏ/ .
Therefore, the average value of the results will be zero.
25
The solution of the equation of each variable can be found separately, and
the solution of the original equation can then be found.
26
Particle in a Three-Dimensional Box
Particle in a Three-Dimensional Box
For the particle in three dimensional box, suppose we have a particle
of mass moving inside a box of dimensions , and .
Now, we want to solve the Schrodinger equation for this system
The potential energy inside the box is zero and infinity any where
else outside the box. In this case, the particle is confined to lie within
a box.
Note that solving the Schrödinger equation means finding the right
form of the Hamiltonian ( ), the wave function ( ) and the energy
( ) of the system.
, ,
, ,
=0
=∞
0<
0<
0<
ℎ
<
<
<
=
We know that when the potential energy is zero, the Hamiltonian in
three dimensions , and is given by
h2   2
2
2 
h2
Hˆ   2  2  2  2    2  2
8 m  x
y
z 
8 m
ℎ
Now the equation we want to solve would be
The potential term can be treated by boundary conditions (i.e.,
infinite potential implies that the wavefunction must be zero there).
27
Particle in a Three-Dimensional Box

=
and , we can write
=
and
=
28
Likewise, we can write
To solve this equation, we use the separation of variables technique.
We will try to make the function
, , a product of three
functions
,
, and
where each function of the three
functions is a function of only one variable that is separate of the
other two variables.
, , =
=
is independent of
h2   2
2
2 

 ( x, y, z )  E ( x, y, z )


8 2 m  x 2 y 2 z 2 
Particle in a Three-Dimensional Box
h2   2
2
2 

 ( x, y, z )  E ( x, y, z )


8 2 m  x 2 y 2 z 2 
Now, because

29
=
when we substitute by this into the total equation, we get
ℎ
−
+
+
=
8
If we divide both sides be
, we get
ℎ
1
1
1
−
+
+
=
8
We note that each one of the three terms inside the square brackets
depends only on one variable and totally independent of the other two.
If the particle moves only on the coordinate , then the energy of the
other two directions is constant and we can solve for the energy in the
direction ( ).
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Particle in a Three-Dimensional Box
Particle in a Three-Dimensional Box
In this case we can write
The previous equation is just similar to the equation of a particle in
one dimensional box, therefore, the solution is known and given by
ℎ
−
1
=
8
Similarly, when the particle is moving in the other directions
ℎ 1
ℎ 1
−
=
−
=
8
8
or .
+
+
=
Now, we can solve each equation of the above three equations
separately.
If we start with the equation in the x direction, it can be written in
the form
ℎ
−
=
8
31
Particle in a Three-Dimensional Box
( , , )=
sin
+
=
For the case of a cubic box
=
ℎ
+
2
sin
=
=
2
,
= 1, 2, 3, …
ℎ
8
,
= 1, 2, 3, …
= ), if
=
=
= 1 then,
is normalized and so is the total function.
∗
=
=
3. The three states
+
= 1·1·1 =1
,
, and
=
Particle in a Three-Dimensional Box
References
This means that it is possible to have different states with different quantum
numbers taking the same energy. This is what is called degeneracy. Some other
examples are:
English References:
=
ℎ
8
32
33
=
= 1, 2, 3, …
and
2. Each function , ,
= , the total energy is give by
8
=
,
=
8
And this is called the zero point energy for the particle in a cubic box.
This level,
, is said to be nondegenerate.
= 1, 2, 3, …
= 1, 2, 3, …
= 1, 2, 3, …
+
sin
ℎ
8
ℎ
The total energy is given as
ℎ
=
8
=
Notice the following:
1. For a cubic box where ( =
= 1, 2, 3, …
= 1, 2, 3, …
= 1, 2, 3, …
sin
sin
Particle in a Three-Dimensional Box
The total wave function equation is then given as
8
2
Likewise,
It is also clear that
=
=
are equal in energy
6ℎ
=
=
8
34
:‫اﻟﻣراﺟﻊ اﻟﻌرﺑﯾﺔ‬
McQuarrie, Donald A. “Quantum Chemistry.” 2nd Ed., University Science Books, Mill Valey, CA, 1983.
=
11ℎ
=
8
Levine, Ira N. “Quantum Chemistry.” 7th Ed., Pearson Education, 2014.
=
9ℎ
=
8
Sherrill, C. D. “A Brief Review of Elementary Quantum Chemistry.” 2001
http://vergil.chemistry.gatech.edu/notes/quantrev/quantrev.html
Atkins, P. W. “Molecular Quantum Mechanics.”, 2nd Ed., Oxford University Press. Oxford 1983.
Undergraduate Quantum Chemistry by Jussi Eloranta (2018).
Degeneracy – different wave functions with the
same energy.
Reflects symmetry of the box.
When
=
Cockett, Martin; Doggett, Graham, “Maths for Chemists.” RSC publishing, 2012.
Cunningham, Allan; Whelan, Rory “Maths for Chemists.” University of Birmingham, University of
Leeds 2014.
= , the lowest levels have the following degeneracy factors:
35
36
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