ADVANCE QUANTUM
MECHANICS
Instructor: Prof. Amir Al-Q’rawee
E- mail:
University of
Karbala
M.Sc. Course
2016-2017
1

Course Description
This course for graduate students. The students are assumed
undergraduate-level quantum mechanics and basic
mathematics background. It aims to help students gain
advanced foundation in quantum mechanics.
Text book:
Quantum Mechanics L.I. SCHIFF
Quantum Mechanics J. POWELL
Modern Quantum Mechanics J. J. Sakurai
2

THE COURSE WILL COVER THE FOLLOWING SUBJECTS
An advanced undergraduate level introduction to the
principles, formalism and results of quantum mechanics;
including: historical background, Schrödinger equations,
particle in box, harmonic oscillator, hydrogen atom,
angular momentum, Hilbert space, Dirac notation,
introduction to Approximation methods for stationary
states, and scattering theory.
Objectives:
1- To achieve an understanding of the theory of quantum
mechanics, and an ability to apply that theory correctly to
important physical systems.
2- To became aware of the necessity for quantum methods in the
analysis of physical systems of atomic and nuclear physics.
3

The Wave Function
•The Schrödinger Equation
•The Statistical Interpretation
•Probability
•Normalization
•Momentum
•The Uncertainty Principle
4

Wave function
One of the most profound and mysterious principles in all
of physics is the Born Rule, named after Max Born. In
quantum mechanics, particles don’t have classical properties
like “position” or “momentum”; rather, there is a wave
function that assigns a (complex) number, called the
“amplitude,” to each possible measurement outcome. Q.M
says that every object in the universe is associated with a
mathematical expression that encodes in it every property (its
charge, mass, location, energy...) that it is possible to know about
the object. This math expression is called the object’s wave
function (psi) ψ.
5

The Schrödinger has two "forms",
Time dependent Schrödinger wave equation:
Time independent Schrödinger equation:
Born's Statistical Interpretation
We have seen that matter must be consider to have wave-
like properties in order to explain experimental data, but
what is the nature of these waves? What is "waving"? Born
postulated that the wave function, , that describes a
particle's behavior is related to the probability of finding the
particle by:
The Schrödinger Equation
6

Where   2 is the complex square, or *  and * is the
complex conjugate of the (perhaps complex) wave function. This
tells us that the wave function itself does not represent the
probability, but a probability amplitude, and that the information
contained in  only represents the probability that one would
measure a certain dynamical quantity, but cannot give pre-
determined results in the same way that the deterministic (if you
know) the initial conditions, you can say exactly where the
particle will be at a later point in time), quantum mechanics only
tells you statistical information about what the possible
measurements will be. This interpretation, although since born
out by much experimentation, caused much debate in the history
of quantum mechanics, and continues to prick at our intuition.
b)
and
a
between
t
at time
particle
the
finding
of
y
probabilit
(
)
,
(
Ψ
2


b
a
dx
t
x
7

What exactly is the wave function.
Born came up with a statistical interpretation of the wave function,
which says that gives the probability of finding the particle at
point x, at time t, or more precisely,
The particle would be relatively likely to be found near A, and unlikely to be found near B. The shaded
area represents the probability of finding the particle in the range dx.







t
at time
)
(
and
between
particle
the
finding
of
y
probabilit
)
,
(
Ψ
2
dx
x
x
dx
t
x
8

 The probability P(r,t)dV to find a particle
associated with the wave function Ψ(r,t)
within a small volume dV around a point in
space with coordinate r at some instant t is
 P(r,t) is the probability density
 For one-dimensional case
dV
t
dV
t
P
2
)
,
(
)
,
( r
r 

dx
t
x
dV
t
x
P
2
)
,
(
)
,
( 

Here |Ψ(r,t)|2 = Ψ*(r,t)Ψ(r,t)
Probability
Y
Z
X
r
dV
9

The wave function contains information about where the
particle is located, its square being a probability density. A
wave function must be "well behaved", in other words it
should be defined and continuous everywhere. In addition
it must be square-integrable, meaning:






dx
t
x
2
)
,
(
Ψ
10

Properties of a valid wave function:
Boundary conditions
1) In order to avoid infinite probabilities, the wave
function must be finite everywhere.
2) In order to avoid multiple values of the probability,
the wave function must be single valued.
3) For finite potentials, the wave function and its
derivative must be continuous. This is required
because the second-order derivative term in the wave
equation must be single valued. (There are exceptions
to this rule when V is infinite.)
4) In order to normalize the wave functions, they must
approach zero as x approaches infinity.
11

The wave function itself is complex, but 2 =  
(where  is the complex conjugate of ) is real and non
negative – as a probability must be.
* = ||2 = “Probability distribution function”
||2 dV =
For a stationary state,
• * is independent of time
• * = |(x,y,z)|2
probability of finding a particle near a
given point x,y,z at a time t
A wave function  must:
Be single valued
Be continuous
Be differentiable
Be integrable
12











dx
t
x
P
dx
t
x
P
everywhere
b
a
ab
2
2
)
,
(
)
,
(
For one-dimensional case, the
probability of funding the
particle in the arbitrary interval
a ≤ x ≤ b is
Example:
Let two functions  and  be defined for 0 ≤ 𝑥 ≤ ∞.
Explain why (𝑥) = 𝑥 can not be a wave function but
could be a valid wave function.
Solution:
Both functions are continuous and defined on the interval of
interest. They are both single valued and differentiable. However,
consider the integral of 𝑥:
2
)
(
Φ x
e
x 










0 0
3
3
0
2
3
)
(
x
dx
x
dx
x
ψ
13

Given that, (𝑥) = 𝑥 is not square integrable over this range it cannot
be a valid wav function, On the other hand:
Therefore, is square integrable so we can say it’s well
behaved over the interval of interest. This makes it a valid candidate
wave function.







0
2
2
0
2
8
)
(
Φ
π
dx
e
dx
x x
2
)
(
Φ x
e
x 

Example:
Suppose that a certain probability distribution is given for
1 ≤ 𝑥 ≤ 3. Find the probability that 5/2 ≤ 𝑥 ≤ 3.
Solution:
If this were a wave function, there would be about a 5.5 percent chance of finding
the particle between 5/2 ≤ 𝑥 ≤ 3.
3
1
4
9
)
(
x
x
p 
 
















3
2
5
3
2
5
2
3
055
.
0
200
11
25
4
9
1
8
9
)
2
(
4
9
1
4
9 x
dx
x
p
14

3
1
4
9
)
(
x
x
p 
 
















3
2
5
3
2
5
2
3
055
.
0
200
11
25
4
9
1
8
9
)
2
(
4
9
1
4
9 x
dx
x
p
Example:
Suppose that a certain probability distribution is given for
1 ≤ 𝑥 ≤ 3. Find the probability that 5/2 ≤ 𝑥 ≤ 3.
Solution:
If this were a wave function, there would be about a 5.5 percent chance of finding
the particle between 5/2 ≤ 𝑥 ≤ 3.
Since ̋ Ψ*(𝑥,𝒕) Ψ(𝑥,𝒕) 𝑑𝑥 ̋ is a probability, then the sum of all probability must
equal one, Thus
This is the Normalization Condition and is very useful in solving
problems.
1
)
,
,
,
(
Ψ
)
,
,
,
(
Ψ
or
1
)
,
(
Ψ
)
,
(
Ψ 
 









dV
t
z
y
x
t
z
y
x
dx
t
x
t
x
Normalizing the wave function means you find the exact form of ψ
15

Expectation values
Although particle is never in a definite location, it is more likely to
be in one location than others. We have seen that is the probability
density of a measurement of a particle's displacement yielding the
value at time .
Suppose that we made a large number of independent measurements
of the displacement on an equally large number of identical systems.
In general, measurements made on different systems will yield
different results. However, from the definition of probability, the
mean of all these results is simply
16

The expectation value of x2 is given by
And the stander deviation, or uncertainty, in x is given by
Although particle is never in a definite location, it is more likely to be
in one location than others, if any potential V is active.
The definition of a weighted average position:


space
all
dx
x
p
x
x
2
2
)
(
2
2
Δ x
x
x 

:
)
(
)
(
)
(
e
Us
)
(
)
(
x
ψ
x
ψ
x
p
dx
x
p
dx
x
xp
x











17

)
(
)
(
)
(
)
(











dx
x
ψ
x
ψ
dx
x
ψ
x
x
ψ
x
This is the “expectation value of 𝑥”
By convention, place 𝑥 between ’s
If  has been normalized, this
denominator is.
Examples: Lets assume tha wave function of QM particle is of an
observable ̋𝑥 ̋ is given by:
We can normalize Ψ to get the constant A:




 


otherwise
0
1
1
-
)
,
(
Ψ
x
Ax
t
x
18

 
 



x
x
dx
x
A
dx
t
x
t
x
1
1
2
2
1
or
)
,
(
Ψ
)
,
(
Ψ
1
Then we can calculate the expectation value of 𝑥, <𝑥> as:
Note: That the probability Ψ*Ψ of observing the QM particle is 0 at 𝑥 = 0 but
multiple measurements will average to net weighted average measurement of
0.



















1
1
2
3
2
3
dx
x
x
x
x
2
3
3
2
3
1 2
1
1
3
2





A
A
x
A
0
x
4
1
2
3
)
(
2
3
1
1
4
1
1
3














 x
dx
x
x
19

As the previous case illustrates, sometimes the “average value” of an
observable is not all we need to know. Particularly, since the integrand
of our <𝑥> equation was odd, the integrand was 0. (Looking at
whether a function is evev or odd can often save calculation effort).
We may like to also know the “stander deviation” or the variance
(stander deviation squared) of the observable around the average
value. In statistics, the standard deviation is:
And since <𝑥>, all we need is:
2
2
x
x
σ 

 


 


1
1
2
2
2
3
or
)
,
(
Ψ
)
,
(
Ψ xdx
xx
dx
t
x
t
x
x
5
3
10
3 1
1
5
2



x
x
775
.
0
5
3
0
5
3
2
2





 x
x
σ
20

Expectation Value
for Momentum of a Free Particle
 



































dx
x
x
x
i
p
dx
x
x
i
x
dx
x
p
x
p
)
(
)
(
)
(
)
(
)
(
ˆ
)
(
*
*
*








Generally
  
 
       p
k
dx
Ae
Ae
k
dx
Ae
ik
i
Ae
p
dx
Ae
x
i
Ae
p
A
dx
Ae
Ae
dx
x
Ae
x
ikx
ikx
ikx
ikx
ikx
ikx
ikx
ikx
ikx














































*
*
*
*
2
n
integratio
of
limits
as
0
where
,
1
)
(
with
)
( 

Free Particle
Similarly, we could also calculate other higher order moments of
distributions such as skewness, curtosis, etc …
21

Q) A particle is described by the wave function
calculate the expectation values of momentum and energy.
Solution: The normalization condition
Determines N up to an arbitrary phase factor. We choose N to be real.
Then
constant
0
;
1
)
,
( 2


 

b
e
e
N
t
r t
i
br 
1
)
,
(
2
3


 dV
t
r
R
      










0 0
2
0
4
2
2
3
2
2
sin
 




 dr
d
d
e
r
e
dV
e
e
e
e
N br
t
i
br
R
t
i
br





0
3
4
2
8b
4

 dr
e
r br
22

 Operators:
 Every classical obtained dynamical variable can be replaced
by an "Operator" that "acts on the wave function". An
operator is merely the mathematical rule used to describe a
certain mathematical operation. For example, the "X "
derivative operator" is defined as "d/dx". The wave function is
said to be the operators, operand, i.e. what is being acted on.
The following table lists the corresponding of the classical
dynamical variables and their corresponding quantum
mechanical operator:
 An operator, , operates on a wave function, ψ , to produce
an observable, O
 An operator, , is a mathematical entity which transforms one
function into another.
23

The operator x represents position, and the operator
represents momentum, in quantum mechanics; to calculate
expectation values, we sandwich the appropriate operator
between  and , and integrate. The fact is, all such
quantities can be written in terms of position and
momentum. Kinetic energy, for example, is
m
p
mv
T
2
2
1 2
2


24

2
)
)(
(



 x
px
1. The limitations imposed by the uncertainty principle have
nothing to do with quality of the experimental equipment
2. The uncertainty principle does imply that one cannot
determine the position or the momentum with arbitrary
accuracy
 It refers to the impossibility of precise knowledge about
both: e.g. if Δx = 0, then Δ px is infinity, and vice versa
3. The uncertainty principle is confirmed by experiment, and
is a direct consequence of the de Broglie’s hypothesis
The Uncertainty Principles
25

Energy and time, for instance, form a pair of
complementary variable. Their simultaneous measurement
must obey the time-energy uncertainty relation
Estimate the uncertainty in the position of a) a neutron moving at
(5 x 106 m/s) and b) a 50Kg person moving at (2 m/s).
This distance is comparable to the size of a nucleus.
An uncertainty of this magnitude is beyond human detection,
therefore can be neglected. So the position and momentum
uncertainty are important for microscopic systems, but negligible for
macroscopic systems.
2



 t
E
26

Probability current density is conserved
If a particle is not being created or destroyed it's integrated
probability always remains constant (=1 for a normalized wave
function). However, if the particle is moving, we can define a
"probability current density" and a probability continuity equation
that describes the particles movement through a Gaussian surface
(analogous to electromagnetic).
Probability current density equation
0






J
t



electric current density
electric charge density
27

J → probability current density and ρ is the probability density ρ = ψ* ψ
(∂ρ ∕ ∂t) → electric charge density electric current density
We need an analogous expression to describe
probability density ρProb and
probability current JProb which can flow in space but remain
conserved.
Assume ρProb and Jprob involve ψ somehow, but in an unspecified
function.
Plan:
1. Use the only equation we have for ψ: the Schrödinger Equation
2. Manipulate (‫معالجة‬ ) it to get the form


 J

28

Convert 1− d Schrödinger Eq to 3 - d :
Form [ψ *• (Sch Eq)] - [ψ • (Sch Eq)*]⇒
÷ by iћ and collect terms :
 
   
  real)
is
V
if
(0
2
0
2
2
2























 





























mi
t
i
V
V
mi
t
t
V is real because, Imaginary potentials do cause probability not to
be conserved.
29

     
 
Prob
Prob
0
2
J
mi
t











 













 


Q) A particle is described by the wave function
Calculate the current density when r2 = x2 + y2 +z2. How does the
probability current density behave for large r.
 
constant
0
;
)
( 



a
r
e
r
r
a
ik
30

Mathematical Structure of Quantum Mechanics
CHAPTER TWO
 The Hilbert Space
 Dirac Notation
 Operators
 Hermitian Adjoint
 Commutator Algebra
31

Hilbert space is the mathematical foundation used for
quantum mechanics. This formalism is based on the basic
ideas of vector analysis, with functions taking the role of
vectors and each vector can have finite or infinite number of
elements. A Hilbert space ℋ consists of a set of vectors ψ, ϕ,
χ,….. and a set of scalars a, b, c,….. which satisfy the
following four properties:
(a) ℋ is a linear space
The properties of a linear space. A linear vector space consists of
two sets of elements and two algebraic rules: A vector space is a
set of objects called vectors ( |A〉, |B〉, |C〉,...) and a set of
numbers called scalars (a, b, c,...) along with a rule for vector
addition and a rule for scalar multiplication.
The Hilbert Space
32

If the scalars are real, we have a real vector space; if the
scalars are complex, we have a complex vector space. The
set must be closed under vector addition and scalar
multiplication.
Vector addition must have these properties:
• The sum of any 2 vectors is a vector: |A〉 + |B〉 = |C〉
• Vector addition is commutative and associative:
|A〉 + |B〉 = |B〉 + |A〉 and |A〉 + (|B〉 + |C〉) = (|A〉 +| B〉) + C〉
• There exists a zero vector |0〉 such that : |A〉 + |0〉 = |A〉 for
any vector |A〉
• For every vector |A〉 there is an inverse vector |-A〉 such
that |A〉 + |-A〉 = |0〉
33

Scalar multiplication must have these properties:
• The product of a scalar and a vector is another vector:
b |A〉 = |C〉
• It is distributive with respect to vector addition and scalar
addition:
a (|A〉 + |B〉) = a |B〉 + a |A〉 and (a + b) |A〉 = a |A〉 + b |A〉
• It is associative with respect to ordinary scalar
multiplication: a (b |A〉) = (a b) |A〉
• Multiplication by the scalars 0 and 1 yields the expected: 0
|A〉 = |0〉 and 1 |A〉 = |A〉
34

The scalar product of an element ψ with another element ϕ
is in general a complex number, denoted by (ψ,ϕ), where
(ψ,ϕ) = complex number. Because the scalar product is a
complex number, the quantity (ψ,ϕ), is generally not equal
to (ϕ,ψ), : (ψ,ϕ) = ψ*ϕ while (ϕ,ψ) = ϕ*ψ, The scalar
product satisfies the following properties:
The scalar product of ψ with ϕ is equal to the complex
conjugate of the scalar product of ϕ with ψ:
   
 


 ,
,
The scalar product of ϕ with ψ is linear with respect to the
second factor if ψ =aψ1 + bψ2:
     
2
1
2
1 ,
,
, 





 b
a
b
a 


(b) ℋ has a defined scalar product that is strictly positive
35

and antilinear with respect to the first factor if ϕ = aϕ1 + bϕ2
(aϕ1 + bϕ2,ψ) = a*(ϕ1,ψ) + b*(ϕ2,ψ) .
The scalar product of a vector ψ with itself is positive real
number:
where the equality holds only for ψ = 0.
(c) ℋ is separable:
(d) ℋ is complete :
  ,
0
,
2

 


36

Question:
Check whether the following sets of functions are linearly
independent or dependent on the real x-axis.
x
e
x
h
x
x
g
x
f
a 2
2
)
(
,
)
(
,
4
)
(
) 


3
2
)
(
,
)
(
,
)
(
) x
x
h
x
x
g
x
x
f
b 


Dirac Notation
In QM, the state of a system is completely characterized by
a state vector.
Dirac developed general notation to describe these state
vectors which was independent of basis, today this is called
Dirac notation.
37

Dirac bra/ket notation














d




2
1
Ket: ψ always denotes a column vector, e.g.
Bra: ψ always denotes a row vector that is the
conjugate transpose of ψ, e.g. [ *
1 *
2  *
d ]
Bra-ket
Dirac denoted the scalar (inner) product by the symbol 〈 ∣ 〉,
which he called a a bra-ket. For instance, the scalar
product (ψ,ϕ) is denoted by the bra-ket 〈 ψ ∣ ϕ 〉:
Note: When a ket (or bra) is multiplied by a complex
number, we also get a ket (or bra).
38

〈 ψ ∣ ϕ 〉 = the inner product
The inner product satisfies 〈 ψ ∣ ϕ 〉 = 〈 ϕ ∣ ψ 〉*
∣ψ〉〈ϕ∣ = the outer product
Properties of kets, bras, and bra-kets
Every ket has a corresponding bra
To every ket ∣ψ〉, there corresponds a unique bra 〈ψ∣ and
vice versa: ∣ψ〉 ⟷ 〈ψ∣
a∣ψ〉 + b∣ϕ〉 ⟷ a*〈ψ∣ + b*〈ϕ∣
Where a and b are complex numbers. The following
common notation: ∣aψ〉 = a∣ψ〉, 〈aψ∣ = a*〈ψ∣
39

Properties of the scalar product
In quantum mechanics, since the scalar product is a
complex number. We must be careful to distinguish a scalar
product from its complex conjugate; 〈ψ∣ϕ〉 is not the same
thing as 〈ϕ∣ψ〉 : 〈ϕ∣ψ〉* = 〈ψ∣ϕ〉
the scalar product 〈ϕ∣ψ〉 is given by
    3
,
, dr
t
r
t
r 


 


This property becomes clearer : 〈ϕ∣ψ〉* = 〈ψ∣ϕ〉
   
      r
d
t
r
t
r
r
d
t
r
t
r 3
3
,
,
,
, 




 










40

additional properties of the scalar product:
〈ψ∣𝑎1ψ1+𝑎2ψ2〉 = 𝑎1〈ψ∣ψ1〉 + 𝑎2〈ψ∣ψ2〉 ,
〈𝑎1ϕ1 + 𝑎2ϕ2∣ψ〉 = 𝑎1*〈ϕ1∣ψ〉 + 𝑎2*〈ϕ2∣ψ〉 ,
〈𝑎1ϕ1 + 𝑎2ϕ2∣b1ψ1+b2ψ2〉 = 𝑎1*b1〈ϕ1∣ψ1〉 + 𝑎1*b2〈ϕ1∣ψ2〉
+ 𝑎2*b1〈ϕ2∣ψ1〉 + 𝑎2*b2〈ϕ2∣ψ2〉
Orthogonal states
Two kets, ∣ψ〉 and ∣ϕ〉, are said to be orthogonal if they have
vanishing scalar product
Bra and Ket orthogonal if scalar product equals zero.
0
B A 
41

2 Bras or 2 Kets orthogonal;
and orthogonal if
0
B A
B A 
and orthogonal if
0
B A
B A 
Two states of a dynamical system are orthogonal if the
vectors representing them are orthogonal.
Orthonormal states
Two kets, ∣ψ〉 and ∣ϕ〉, are said to be orthonormal if they
are orthogonal and if each one of them has a unit norm:
〈ψ∣ϕ〉 = 0 , 〈ψ∣ψ〉 = 1 , 〈ϕ∣ϕ〉 = 1
42





 dx
x
x
dx
x
x )
(
)
( 





The integration ψ*(x) ψ(x) over all values of x will yield 1
if ψ(x) is normalized.
Forbidden quantities
If ∣ψ〉 and ∣ϕ〉 belong to the same vector (Hilbert) space,
products of the type ∣ψ〉 ∣ϕ〉 and 〈ψ∣〈ϕ∣ are forbidden.
They are nonsensical.
43

Operators
Definition of an operator: An operator  is a mathematical
rule that when applied to a ket ∣ψ〉 transforms it into
another ket ∣ψ'〉 of the same space and when it acts on a
bra 〈ϕ∣ transforms it into another bra 〈ϕ’∣ :
Â∣ψ〉 = ∣ψ'〉 , 〈ϕ∣Â = 〈ϕ'∣
Operators stand to the left of kets and to the right of bras.
A similar definition applies to wave functions:
),
(
)
(
ˆ r
r
A 
 
 )
(
ˆ
)
( r
A
r



 

consider some simple operators:
The Identity Operator:- The simplest operator of all is the
identity operator, which does nothing to a ket or Unity
operator 
 
I
ˆ
44

Outer Product
The outer product between a ket and a bra is written as
follows: ∣ψ〉〈ϕ∣
(∣ψ〉〈ϕ∣)∣χ〉 = ∣ψ〉〈ϕ∣χ〉 = 𝛂∣ψ〉
inner product(complex number)
Products of operators:
The product of two operators ,the operators do
not commute.
The product of operators is associative:
A
B
B
A ˆ
ˆ
ˆ
ˆ 
C
B
A
C
B
A
C
B
A ˆ
)
ˆ
ˆ
(
)
ˆ
ˆ
(
ˆ
ˆ
ˆ
ˆ 

45

Linear operators
An operator that obeys the distributive law
That is, an operator  is linear if, for any vectors ∣ψ1〉 and
∣ψ2〉 and any complex numbers 𝑎1 and 𝑎2, we have:
Â(𝑎1∣ψ1〉 + 𝑎2∣ψ2〉) = 𝑎1Â∣ψ1〉 + 𝑎2Â∣ψ2〉 ,
and
(〈ψ1∣𝑎1 + 〈ψ2∣𝑎2)Â = 𝑎1〈ψ1∣Â + 𝑎2〈ψ2∣Â .
The expectation or mean value  of an operator  with
respect to a state ∣ψ〉 is defined by:



 A
A
ˆ
ˆ  
 A
A ˆ
ˆ 
46

Hermitian Adjoint
Hermitian Conjugate or adjoint of an operator  to be
A†. Since the inner product is just a complex number, we can
form the complex conjugate which is given by relation
〈ψ∣ϕ〉 = 〈ϕ∣ψ〉*. When an operator is present inside the
inner product this becomes
〈ψ∣†∣ϕ〉 = 〈ϕ∣Â∣ψ〉*
DEFINITION: Forming the Adjoint of a General Expression
1. Replace any constants by their complex conjugates
2. Replace kets by their associated bras, and bras by their
associated kets (∣ψ〉)† = 〈ψ∣ , (〈ψ∣)† = ∣ψ〉
3. Replace each operator by its adjoint
4. Reverse the order of all factors in the expression
47

The adjoint operation also has the following properties:
(†)† = Â
(𝑎Â)†=𝑎*†
(Âⁿ)†=(†)ⁿ
48

The expectation or mean value Â of an operator  with
respect to a state |ψ is defined by
The quantity |ψ| is a linear operators in Dirac's notation
Since is a complex number.
49

Example:
(a) Discuss the hermiticity of the operators
( + †),i(( + †), and ( - †).
(b) Find the Hermitian adjoint of
f(Â) =(1 + i  + 3 Â2)(1-2i  - 9 Â2)/(5 + 7 Â).
(c) Show that the expectation value of a Hermitian
operator is real and that of an anti-Hermitian
operator is imaginary.
Solution:
(a)The operator † is Hermitian regardless of
whether or not  is Hermitian,
Since
Similarly the operator i( - †) is also Hermitian; but
i( + †) is anti-Hermitian, since
[i( + †)]† = -i( + †).
A
A
B ˆ
ˆ
ˆ 

50

(b) Since the Hermitian adjoint of an operator function f(Â)
is given by f †(Â) = f *(†),
We can write
(c) From  = † or 〈ψ∣†∣ϕ〉 = 〈ϕ∣Â∣ψ〉*
we immediately infer that the expectation value of a
Hermitian operator is real, for it satisfies the following
property: 〈ψ∣†∣ϕ〉 = 〈ϕ∣Â∣ψ〉*
that is, if † =  then 〈ϕ∣Â∣ψ〉 is real. Similarly, for anti-
Hermitian operator,
we have
which means that is purely imaginary number. 51

Find the Hermitian Conjugate of
x


52

Commutator Algebra
3a – 5b + 7a original (given) statement
3a + 7a – 5b Commutative Property
(3a + 7a) – 5b Associative Property
a(3 + 7) – 5b Distributive Property
a(10) – 5b simplification (3 + 7 = 10)
10a – 5b Commutative Property
53

Commutative
Laws: a + b = b + a
a × b = b × a
Associative Laws:
(a + b) + c = a + (b + c)
(a × b) × c = a × (b × c)
Distributive Law:
a × (b + c) = a × b + a × c
54

The commutator of two operator  and , denoted by
, is defined by
And anticommutator , is defined by
Two operators are said to commute if their commutator is
equal to zero and hence
Any operator commutes with itself:
55

A
B
B
A
B
A ˆ
ˆ
ˆ
ˆ
]
ˆ
,
ˆ
[ 

0
]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
[ 
 A
B
B
A
0
]
ˆ
,
ˆ
[ 
A
A
]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
[
]
ˆ
ˆ
,
ˆ
[ C
A
B
A
C
B
A 


]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
ˆ
[ C
B
C
A
C
B
A 


]
ˆ
,
ˆ
[
ˆ
ˆ
]
ˆ
,
ˆ
[
]
ˆ
ˆ
,
ˆ
[ C
A
B
C
B
A
C
B
A 

]
ˆ
,
ˆ
[
ˆ
ˆ
]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
ˆ
[ C
B
A
B
C
A
C
B
A 

Consider the following properties:
]
ˆ
,
ˆ
[
]
ˆ
,
ˆ
[ A
B
B
A 

Distributive
Antisymmetric
linearity
56

If two operators are Hermitian and their product is also
Hermitian, these operators commute:
Operator commute with scalar: an operator  commute
with any scalar b: [Â,b⦌=0
57

Example:
(a) Show that the commutator of two Hermitian operators
is anti-Hermitian.
(b) Evaluate the commutator
Solution:
(a) If  and are Hermitian, we can write
that is, the commutator of  and is anti-Hermitian
]
ˆ
]
ˆ
,
ˆ
[
,
ˆ
[ D
C
B
A
58

(b) Using the distributivity relation, we have
D
C
B
A
D
A
C
B
D
C
B
A ˆ
]]
ˆ
,
ˆ
[
,
ˆ
[
]
ˆ
,
ˆ
][
ˆ
,
ˆ
[
]
ˆ
]
ˆ
,
ˆ
[
,
ˆ
[ 

59

CHAPTER THREE
The Time-Independent Schrodinger Equation
 Stationary States
 The Infinite Square Well
 The Free Particle
 The Finite Square Well
 The Harmonic Oscillator
 The Delta-Function Potential
60

The idea of "stationary states" was first introduced by Bohr as
a name given to those states of hydrogen atom for which the
orbits that the electron occupied were stable, i.e. the electron
remained in the same orbital state for all time. This idea is now
no longer appropriate, the notation of a stationary state remains a
valid one. It is a term that is now used to identify those states of
a quantum system that do not change in time. This is not to say
that stationary state is one for which "nothing happens" – there
is still a rich collection of dynamical physics to be associated
with such a state-but a stationary state turns out to be one for
which the probabilities of outcomes of a measurement of any
property of the system is the same no matter at what time the
measurement is made.
stationary states
A stationary state is called stationary because the system remains in
the same state as time elapses. The Hamiltonian is unchanging in time
61

The starting point is to let the time-dependent Schrodinger equation
for a particle of mass m moving in a time-dependent potential V(r,t)
can be written as follows:
Now, let us consider the particular case of time-independent
potentials: V(r,t) = V(r). In this case the Hamiltonian operator will
also be time independent, and hence the Schrodinger equation will
have solutions that are separable, i.e. solutions that consist of a
product of two functions, one depending only on r and the other only
on time:
For separable solutions we have
  )
(
)
(
, t
f
x
t
x 


,
dt
df
t





2
2
2
2
dx
d
x





Schrodinger Equation
62

And the Schrodinger equation
Dividing by f
Now the left side is a function of t alone, and the right side is a
function of x alone. The only way this can possibly be true is if
both sides are in fact constant; this constant, which we denote
by E , has the dimensions of energy.
f
V
f
dx
d
m
dt
df
i 

 

 2
2
2
2


V
dx
d
m
dt
df
f
i 

 2
2
2
1
2
1 



)
(
)
(
t
Ef
dt
t
df
i 

63

and
or
Separation of variables has turned a potential differential equation
into two ordinary differential equations. The first of these is easy to
solve
The second term in the S.E is called the time-independent
Schrödinger equation:
,
f
iE
dt
df


 E
V
dx
d
m


 2
2
2
1
2






E
V
dx
d
m


 2
2
2
2

)
(
)
(
ˆ x
E
x
H 
 
If a Hamiltonian does not depend on time, then the
solution of the Schrödinger equation can be expressed in
terms of eigenfunctions of the Hamiltonian
Stationary wave function
64

This particular solution of the Schrodinger equation for time-
independent potential is called a stationary state. Why is this is called
stationary? The reason is obvious: the probability density is stationary,
i.e., it does not depend on time:
Note that such a state a precise value for the energy, E = ħ.
In summary, stationary states, which are given by the solutions of
exist only for time-independent potentials. The set of energy levels
that are solutions to this equation are called the energy spectrum of
the system. The states corresponding to discrete and continuous
spectra are called bound and unbound states, respectively.



E
V
dx
d
m


 2
2
2
2

65

One-Dimensional Problems
Properties of One-Dimensional Motion
To study the dynamical properties of a single particle moving in a
one-dimensional potential, let consider a potential V(x). One such
potential is display in Fig.(1); it is finite at x = , V(-) = V1 and
V(+) = V2 with V1 smaller than V2, and it has a minimum, Vmin.
Fig.(1) Shape of
general potential
66

Discrete Spectrum(Bound States)
Bound states occur whenever the particle cannot move to infinity.
That is, the particle is confined or bound at all energies to move within
a finite and limited region of space. The Schrodinger equation in this
region admits only solutions that are discrete. The infinite square well
potential and harmonic oscillator are typical examples that display
bound states.
In the potential of Fig.(1), the motion of the particle is bounded
between the classical turning points x1 and x2 when the particle's
energy lies between Vmin and V1:
The states corresponding to this energy range are called bound states.
They are defined as states whose wave functions are finite (or zero) at
x  .
67

 In a one-dimensional problem the energy levels of a bound state
system are discrete and not degenerate.
 The wave function n(x) of a one-dimensional bound state system
has n nodes if n = 0 corresponds to the ground state and (n-1) nodes if
n = 1 corresponds to the ground state.
Continuous Spectrum (Unbound States)
Unbound states occur in those cases where the motion of the system
is not confined; a typical example is the free particle. For the
potential displayed in Fig.(1) there are two energy ranges where the
particle's motion is finite: V1  E  V2 and E  V2.
68

Mixed Spectrum
Potentials that confined the particle for only some energies give rise to
mixed spectra; the motion of the particle for such potentials is
confined for some energy values only. For instance, for the potential
display in Fig.(1), if the energy of the particle is between Vmin  E 
V1, the motion of the particle is confined (bound) and its spectrum is
discrete, but if E  V2, the particle's motion is unbounded and its
spectrum is continuous (if V1  E  V2 , the motion is unbounded only
along the x = - direction). Other typical examples where mixed
spectra are encountered are the finite square well potential.
Nondegenerate spectrum
Since the eigenstates of the parity operator have definite parity, the
bound eigenstates of a particle moving in a one-dimensional
symmetric potential have definite parity; they are either even or odd:
 (-x) =  (x)
69

Degenerate spectrum
If the spectrum of the Hamiltonian corresponding to a symmetric
potential is degenerate, the eigenstates are expressed only in terms of
even and odd states. That is, the eigenstates do not have definite parity.
The Free Particle: Continuous States
This is the simplest one-dimensional problems; it corresponds to
V(x)=0 for any value of x. In this case the Schrodinger equation is
given by
Where k2 = 2mE/ ħ2 , k being the wave number
The most general solution is a combination of two linearly independent
plane waves
0
)
(
)
(
)
(
2
2
2
2
2
2
2












 x
k
dx
d
x
E
dx
x
d
m




ikx
ikx
k e
A
e
A
x 

 

)
(

70

Where A+ and A- are two arbitrary constants. The complete wave function
is thus given by the stationary state.
Since  = E/ħ = ħk2/2m. The first term represents a wave traveling to
the right, while the second term represents a wave traveling to the left.
The intensities of these waves are given by A+2 and A-2, respectively.
We should note that the waves +(x,t) and -(x,t) are associated,
respectively, with a free particle traveling to the right and to the left
with well-defined momenta and energy: p = ħk, E=ħ2k2/2m. We
will comment on the physical implications of this in a moment. Since
there are no boundary conditions, there are no restriction on k or on E
; all values yield solutions to the equation.
)
(
)
(
)
,
( t
kx
i
t
kx
i
k e
A
e
A
t
x 

 



 

)
2
(
)
2
(
2
2
m
t
k
kx
i
m
t
k
kx
i
e
A
e
A






 

71

There are three Physical subtleties: First, the probability densities
corresponding to either solutions
p(x,t) = (x,t)2 = A2
are constant, for they depend neither on x nor on t. This is due to the
complete loss of information about the position and time for a state
with definite values of momentum, p = ħk, E=ħ2k2/2m. This is a
consequence of Heisenberg’s uncertainty principle: when the
momentum and energy of a particle are known exactly, ∆p = 0 and
∆E=0, there must be total uncertainty about its position and time:
∆x and ∆t.
The second subtlety pertains to an apparent discrepancy between the
speed of the wave and the speed of the particle it is supposed to
represent. The speed of the plane waves (x,t) is given by
72

On the other hand, the classical speed of the particle is given by
Third, the wave function is not normalizable:
The solutions (x,t) are thus unphysical; physical wave function
must be square integrable. The problem can be traced to this: a free
particle cannot have sharply defined momenta and energy. According
to above subtleties the Schrodinger equation that are physically
acceptable cannot be plane waves. Can construct physical solutions
by means of a linear superposition of plane waves. The answer is
provided by wave packets.
73

Where  (k), the amplitude of the wave packet, is given by the
Fourier transform of  (x,0) as
initial wave function ψ (x,0) at time t = 0
The wave packet solution cures and avoids all subtleties raised
above. First, the momentum, the position and the energy of the
particle are no longer known exactly; only probabilistic outcomes
are possible. Second, the wave packet and particle travel with the
same speed vg=p/m, called the group speed or the speed of the whole
packet. Third, the wave packet is normalizable.
Group velocity
phase velocity
74

Example 1. A free particle moves along x-
direction. Set up its Schrödinger equation.
Solution: The motion of free particle has
kinetic energy only, so we have:
  0
2
2
1
2
1
2
2
2
2



 p
x
x
k E
m
P
m
mv
m
mv
E
Substituting
x
i
Px



  into above equation, we get
2
2
2
2
2
2
ˆ
dx
d
x
H 
 





75

)
(
)
(
ˆ r
E
r
H



 
Substituting the above Hamiltonian
operator into the following equation
The schrödinger equation could be obtained
0
2
2 2
2
2
2
2
2




 




 mE
dx
d
E
dx
d
m
  0
)
(
2
2
2
2


 

x
V
E
m
dx
d

76

The Potential Step: Unbounded states
Another simple problem consists of a particle that is free everywhere,
but beyond a particular point, say x = 0, the potential increases
sharply(i.e. it becomes repulsive or attractive). A potential of this type
is called a potential step as shown in Fig.(2):
Two cases:
Case E  V0
The particles are free for x  0. Consider a particle of energy E
moving in region in which the potential energy is the step function.
What happened when a particle moving from left to right encounters
the step? The classical answer is simple:
All particles with E < Vo will be reflected back.
All particles with E > Vo will be pass through into zone II.
An unbound state occurs when the energy is sufficient to take the
particle to infinity, E > V().
77

78

the particle moves with a speed v = √2E/m. As the particles enter the
region x  0, where the potential now is V = V0, they slowdown to
momentum 2m(E-V0) ; they will then conserve this momentum as
they travel to the right.
as example of potential step is a charged particle moves
along the axis of cylindrical electrodes held at different
voltages. If the total energy E of the electron is GREATER
than the work function of the metal, V0, when the electron
reaches the end of the wire, it will…
either be reflected or transmitted with some probability.
Wave function splits into two pieces – transmitted part and
reflected part.
e
79

Fig.(2) Potential step and propagation directions of the incident,
reflected, and transmitted waves, plus their probability densities (x)2
when E V0.
V(x)=
0
V(x)=
V0
Since the particles have sufficient energy to penetrate into the region
x  0, there will be total transmission: all the particles will emerge to
the right with a smaller kinetic energy E – V0. This is then a simple
scattering problem in one dimension.
80

The probabilities of reflection and transmission can be
calculated by solving the Schrödinger equation in each
region of space and comparing the amplitudes of
transmitted waves and reflected waves with that of the
incident wave.
where and . The most
general solutions to these two equations are waves:
(x < 0)
(x ≥ 0)
2
2
1 /
2 
mE
k  2
0
2
2 /
)
(
2 
V
E
m
k 

x
ik
x
ik
x
ik
x
ik
De
Ce
x
Be
Ae
x
2
2
1
1
)
(
)
(
2
1








81

Since no wave is reflected from the region x  0 to the left, then D = 0
and the complete wave function is thus given by:
x < 0
x ≥ 0
Using the boundary conditions of the wave function at x = 0
to determine the constants. Since both the wave function and
its first derivative are continuous at x = 0.
















)
(
2
)
(
)
(
1
2
1
1
)
(
)
(
)
,
(
t
x
k
i
t
i
t
x
k
i
t
x
k
i
t
i
Ce
e
x
Be
Ae
e
x
t
x







 is continuous: 1(0) = 2(0) ⟹ A + B = C
is continuous: ⟹ A – B =
x


x
x 



 )
0
(
)
0
( 2
1 

C
k
k
1
2
82

The reflection and transmission coefficient, R and T, is
defined by:
To calculate R and T
2
1
incident
)
(
)
(
)
(
)
(
2
A
m
k
dx
x
ψ
d
x
ψ
dx
x
ψ
d
x
ψ
m
i
J i
i
i
i












 

2
2
d
transmitte
2
1
reflected , C
m
k
J
B
m
k
J





2
2
1
2
2
2
,
A
C
k
k
T
A
B
R 

83

,
2
,
;
)
(
,
2
1
1
2
1
2
1
2
1
A
k
k
k
C
A
k
k
k
k
B
C
k
B
A
k
C
B
A









   2
2
2
1
2
1
2
2
2
2
1
2
2
1
1
4
4
,
)
1
(
)
1
(
)
(
)
(














k
k
k
k
T
k
k
k
k
R
where  = k2/k1 = √1-V0/E . The sum of R and T is equal
to 1. When E gets smaller and smaller, T also gets smaller
and smaller so that when E = V0 the transmission coefficient
T becomes zero and R = 1. On the other hand, when
E » V0, we have κ = 1-V0/E  1; hence R = 0 and T = 1.
84

Case E  V0
The Schrodinger equation is given by:
.(2) Potential step and propagation directions of the incident, reflected, and
transmitted waves, plus their probability densities (x)2 when E  V0.
(x ≥ 0)
)
(
2
2
2
2
2
x
k
dx
d











85

2
0
2
2 /
)
(
2 
E
V
m
k 


Where . This equation’s solution is
(x ≥ 0)
Since the wave function must be finite everywhere, and
since the term diverges when x ⟹ , the constant D
has to be zero. Thus, the complete wave function is
x < 0,
x≥ 0.
x
k
x
k
De
Ce
x 2
2
)
(
2






x
k
e 2





 







)
(
)
(
)
(
2
1
1
)
,
(
t
x
k
i
t
x
k
i
t
x
k
i
Ce
Be
Ae
t
x



86

Thus , the reflected coefficient is given by
1
2
2
2
1
2
2
2
1
2
2







k
k
k
k
A
B
R
The Potential Barrier and Well
Consider a beam of a particles of mass m and energy E
approaching a potential barrier of height V0 and the
potential everywhere else is zero.











.
,
0
,
0
0,
,
0
)
( 0
a
x
a
x
V
x
x
V
87

Case E  V0 unbound states(continuum states)
We will first consider the case when the energy is greater
than the potential barrier (figure 3).
The wave function will consist of an incident wave, a
reflected wave, and a transmitted wave.
The potentials and the Schrödinger wave equation for the
three regions are as follows:
0
2
0
)
Region(
0
)
(
2
)
Region(0
0
2
0
0)
Region(
3
2
2
3
2
2
0
2
2
2
2
0
1
2
2
1
2




















E
m
dx
d
V
a
x
V
E
m
dx
d
V
V
a
x
E
m
dx
d
V
x



88




















,
,
)
(
,
0
,
)
(
,
0
,
)
(
)
(
1
2
2
1
1
3
2
1
a
x
Ee
x
a
x
De
Ce
x
x
Be
Ae
x
x
x
ik
x
ik
x
ik
x
ik
x
ik




The corresponding solutions are:
E
V(x)
V0
Particle
x
0 ɑ
Fig.(3) Potential barrier and propagation directions of the incident, reflected, and
transmitted waves, plus their probability densities (x)2 when E  V0 89

where and .The constant
B, C, D, and E can be obtained in terms of A from the boundary
conditions: (x) and d/dx must be continuous at x = 0 and x = ɑ,
respectively:
These equation yield
Solving for E, we obtain
2
1 /
2 
mE
k  2
0
2 /
)
(
2 
V
E
m
k 

.
)
(
)
(
),
(
)
(
,
)
0
(
)
0
(
(0),
(0)
3
2
3
2
2
1
2
1
dx
a
d
dx
a
d
a
a
dx
d
dx
d












),
(
)
( 2
1 D
C
ik
B
A
ik
D
C
B
A 





a
ik
a
ik
a
ik
a
ik
a
ik
a
ik
Ee
ik
De
Ce
ik
Ee
De
Ce 1
2
2
1
2
2
1
2 )
(
, 





1
2
2
1
2
2
1
2
1 ]
)
(
)
[(
4 2
2
1 






a
ik
a
ik
a
ik
e
k
k
e
k
k
Ae
k
k
E
.
)]
sin(
)
(
2
)
cos(
4
[
4 1
2
2
2
2
1
2
2
1
2
1
1 



 a
k
k
k
i
a
k
k
k
Ae
k
k
a
ik
90

91

The Case E  V0: Tunneling: bound states
Now we consider the situation where classically the particle does not
have enough energy to surmount the potential barrier (figure3).
Classically, we would expect total reflection: every particle that arrives
at the barrier (x = 0) will be reflected back; no particle can penetrate
the barrier, where it would have a negative kinetic energy. Quantum
mechanical predictions differ sharply from their classical counterparts,
for the wave function is not zero beyond the barrier. The solutions of
the Schrödinger equation in the three regions yield expression that are
similar to



















,
,
)
(
,
0
,
)
(
,
0
,
)
(
)
(
1
2
2
1
1
3
2
1
a
x
Ee
x
a
x
De
Ce
x
x
Be
Ae
x
x
x
ik
x
ik
x
ik
x
ik
x
ik




92

E
V(x)
V0
Particle
x
0 ɑ
+
Fig.(3) Potential barrier and propagation directions of the incident, reflected, and
transmitted waves, plus their probability densities (x)2 when E  V0
Except that should be replaced with
x
ik
x
ik
De
Ce
x 2
2
)
(
2




x
k
x
k
De
Ce
x 2
2
)
(
2




93




















,
,
)
(
,
0
,
)
(
,
0
,
)
(
)
(
1
2
2
1
1
3
2
1
a
x
Ee
x
a
x
De
Ce
x
x
Be
Ae
x
x
x
ik
x
k
x
k
x
ik
x
ik




where and The behavior
of the probability density corresponding to this wave function is
expected, as display in Fig.(3), to be oscillatory in the regions x  0
and x  a, and exponentially for 0  x  a. To find the reflection and
transmission coefficients,
2
2
1 /
2 
mE
k  .
/
)
(
2 2
0
2
2 
E
V
m
k 

2
2
2
2
,
A
E
T
A
B
R 

94

we need only to calculate B and E in terms of A. The continuity
conditions of the wave function and its derivative at x = 0 and x = a
yield
The last two equations lead to the following expressions for C and
D:
Inserting these two expression into the two above equations and
dividing by A,
),
(
)
( 2
1 D
C
k
B
A
ik
D
C
B
A






a
ik
a
k
a
k
a
ik
a
k
a
k
Ee
ik
De
Ce
k
Ee
De
Ce
1
2
2
1
2
2
1
2 )
(
,






   a
k
ik
a
k
ik
e
k
k
i
E
D
e
k
k
i
E
C 2
1
2
1
2
1
2
1
1
2
,
1
2






















95

Solving these two equations for B/A and E/A, we obtain:
96

Thus, the coefficients R and T become
Rewrite R in terms of T as
97

Since cosh2(k2a) = 1 + sinh2(k2a) can reduce T to
Note T is finite. This means that the probability of the transmission
of the particle into the region x  a is not zero.
98

The Infinite Square Well Potential
The Asymmetric Square Well
Consider a particle of mass m confined to move inside an
infinitely deep asymmetric potential well
V(x) is infinite outside the region 0 ≤ x ≤ a, the wave
function of the particle must be zero outside the boundary.
The solutions only inside the well.
99

V   V  
V(x)
0 a
0
V 
The solutions are,
The wave function vanishes at the walls, ψ(0) = ψ(a) = 0, the
condition ψ(0) = 0 gives B = 0, while ψ(a) = Asin(ka) = 0
gives kna = n (n = 1,2,3,……)
This condition determines the energy
)
cos(
)
sin(
)
(
)
( kx
B
kx
A
x
e
B
e
A
x ikx
ikx






 


,.....)
3
,
2
,
1
(
2
2
2
2
2
2
2
2


 n
n
ma
k
m
E n
n



100

Normalisation of the wave functions gives:
So that:
Example: A typical diameter of a nucleus is about 10-14m.
Use the infinite square well potential to calculate the
transition energy from the first excited state to the ground
state for a proton confined to the nucleus.
Solution:
The first excited state energy is found to be E2 = 4E1 =
8MeV, and the transition energy is ΔE =E2 – E1 = 6MeV.
This is a reasonable value for proton in the nucleus.
  



a
n
n
n
n
n
a
A
dx
x
k
A
dx
0
2
2 2
,
1
)
(
sin


)
sin(
2
)
( x
k
a
x n
n 

MeV
c
ma
c
E
n
ma
En 0
.
2
2
,
2
2
2
2
2
2
1
2
2
2
2




 

101

A particle in an infinite square-well potential has
ground-state energy 4.3eV. (a) Calculate and sketch the
energies of the next three levels, and (b) sketch the
wave functions on top of the energy levels.
a
102

A particle in a box according to Newton's laws of classical mechanics (A), and according to
the Schrödinger equation of quantum mechanics (B-F). In (B-F), the horizontal axis is
position, and the vertical axis is the real part (blue) and imaginary part (red) of the
wavefunction
103

Why there is no state with zero energy for square well
potential, if the particle has zero energy, it will be at rest
inside the well, and this violates Heisenberg's uncertainty
principle.
Remember Heisenberg uncertainty relation
Particle is confined in box, so x ~ a.
Must be an uncertainty in momentum
Since momentum cannot be zero, minimum
energy must be of order
E
p
m ma
min
( )
 
 2 2
2
2 2

Generally, the lowest energy in a quantum system is called
the zero point energy.
Zero-point energy
104

Wave function Probability distribution function
a a
105

Example (zero point):
106

107

The Symmetric Potential Well
What happens if the potential become symmetric?
First, the energy spectrum to remain unaffected by this
translation, since the Hamiltonian is invariant under spatial
translation, as it contains only a kinetic part, it commutes
with the particle's momentum, [H,P] = 0. The energy
spectrum is discrete and nondegenerate.
Second, for symmetric potentials, V(-x) = V(x), the wave
function of bound states must be either even or odd. The
wave function corresponding for above potential can be
written as follows: 108

That is, the wave functions corresponding to odd quantum
numbers n = 1,3,5,….. are symmetric, ψ(-x) = ψ(x), and
those corresponding to even numbers n = 2,4,6,…. are
symmetric, ψ(-x) = -ψ(x).
cos (-x) = cos(x) and sin(-x) = -sin(x)
n
odd
n
even
109

The Finite Square Well
The infinite potential is an oversimplification which can
never be realised. A more realistic finite square well.
Consider a particle of mass m moving in the following
symmetric potential:
The two physically interesting cases are E  V0 and E V0.
We expect the solutions to yield a continuous doubly-
degenerate energy spectrum for E  V0 and discrete
nondegenerate spectrum for 0  E V0.
110

Fig.(4) Finite square well potential and propagation directions of the incident,
reflected and transmitted waves when E  V0 and 0  E V0.
111

The Scattering Solutions (E > V0)
Classically, if the particle is initially incident from left with
constant momentum √2m(E-V0) it will speed up to √2mE
between –a/2≤ x ≤ a/2 and then slow down to its initial
momentum in the region x > a. All the particles that
come from the left will be transmitted, non will be
reflected back; therefore T = 1 and R = 0.
Quantum mechanically, and as we did for step and
barrier potentials, we can verify that we get a finite
reflection coefficient. The solution is straightforward to
obtain. The wave function has an oscillating pattern in
all three regions.
112

The Bound state Solution (0 < E < V0)
Classically, when E < V0 the particle is completely
confined to the region -a/2 ≤ x ≤ a/2; it will bounce
back and fourth between x = -a/2 and x = a/2 with
constant momentum p = √2mE.
Quantum mechanically, the solution are particularly
interesting for they are expected to yield a discrete
energy spectrum and wave functions that decay in the
two regions x < -a/2 and x > a/2, but oscillate in -a/2 ≤
x ≤ a/2. In these three regions, the Schrödinger
equation can be written as:
113

,
2
1
0
)
(
,
2
1
2
1
-
0
)
(
,
2
1
0
)
(
3
2
1
2
2
2
2
2
2
2
1
2
1
2
2



























































a
x
x
k
dx
d
a
x
a
x
k
dx
d
a
x
x
k
dx
d



where k1 = √2m (V0-E)/ ħ and k1 = √2mE/ ħ.
Eliminating the physically unacceptable solutions which
grow exponentially for large values of ∣x∣, we can write
the solution to this Schrödinger equation in the regions
x < -a/2 and x> a/2 as follows:
114

).
2
1
(
)
(
),
2
1
(
)
(
1
1
3
1
a
x
De
x
a
x
Ae
x
x
k
x
k








Since the bound state eigenfunctions of symmetric one-
dimensional Hamiltonian are either even or odd under space
inversion, the solutions are then either antisymmetric (odd)















,
2
/
,
2
/
2
/
),
sin(
2
/
,
)
(
1
1
2
a
x
De
a
x
a
x
k
C
a
x
Ae
x
x
k
x
k
a

115

or symmetric (even)
To determine the eigenvalues, we need to use the
continuity conditions at x = ± a/2. The continuity of the
logarithmic derivative, (1/a(x) )da(x)/dx, of a(x) at x
= ± a/2 yields















,
2
/
,
2
/
2
/
),
cos(
2
/
,
)
(
1
1
2
a
x
De
a
x
a
x
k
B
a
x
Ae
x
x
k
x
k
s

116

Solve these equations to obtain:
function
parity
odd
.
2
cot 1
2
2 k
a
k
k 









Similarly, the continuity of (1/s(𝑥))ds(𝑥)/d𝑥 at 𝑥 = ± 𝒂/2 gives
function
parity
even
.
2
tan 1
2
2 k
a
k
k 








..)
(1,2,3,...
2
π
n
αn 
Since αn
2 = m𝒂2En / (2h2) we see that we recover the energy
expression for the infinite well
2
2
2
2
2
2
n
ma
π
E
π
n
α n
n




117

)
2
/(
and
)
2
/(
where
,
cot
and
tan
with
of
on
intersecti
by the
given
are
they
:
potential
well
sequare
finit
for the
solutions
Graphical
Fig.(5)
2
2
2
2
n
2
2
n
2
2

 o
n
n
n
n
n V
ma
R
E
ma
R 


 





118

In the limiting case V0→∞, the circle’s radius R also becomes infinite,
and hence the function
infinite
become
cot
and
tan
both
when
because
,
2
/
asymptotes
at the
,
cot
and
tan
cross
will n
2
2
n
n
o
n
n
n
n
n
V
n
R














...)
1,2,3,....
(n
cot
..)
0,1,2,3,..
(n
2
1
2
tan
n 
















n
n
n
n
n
states.
bound
other
two
yields
cot
on with
intersecti
its
and
0,2,
n
states,
bound
two
yields
tan
with the
circle
larger
the
of
on
intersecti
the
whereas
0,
n
boundstate
one
only
yields
tan
curve
with the
circle
small
the
of
on
intersecti
The
n
n
n
n
n
n









119

When
is.
small
how
matter
no
on)
intersecti
one
(i.e.
state
bound
one
least
at
always
is
e
that ther
Note
states.
bound
of
number
the
greater
the
hence
and
R,
of
value
larger the
the
well,
e
broader th
and
deeper
The
.
)
2
/(
since
well,
the
of
width
the
and
depth
on the
depends
in turn
which
R,
of
size
on the
depends
solutions
of
number
The
2
2
o
o
o
V
V
ma
R
a
V 

2
0
2
2
0
or
2
0
ma
V
R













there is only one bound state corresponding to n = 0 as shown in Fig. ;
this state-the ground state-is even. Then, and when
2
2
2
0
2
2
or
2 ma
V
R









there are two bound states; an even state (the ground state)
corresponding to n = 0 and the first odd state corresponding to n = 1.
Now, if
120

2
2
2
0
2
2
2 2
2
3
2
or
2
3
ma
V
ma
R
















there exist three bound states; an even state (the ground state)
corresponding to n = 0, the first excited state (odd state),
corresponding to n = 1, and the second excited state (even state),
which corresponds to n = 2. In general, the well width at which n
states are allowed is given by
2
2
2
2
0
2
2
or
2
n
ma
V
n
R











121

Example:
Find the number of bound states and the corresponding energies for
the finite square well when: (a) R=1, and (b) R = 2
Solution:
equations
ing
correspond
the
of
solutions
numerical
The
ly.
respective
1,
n
and
0
n
to
correspond
they
;
cot
and
tan
with
4
of
on
intersecti
the
from
resulting
states
bound
two
are
there
2
R
When
b)
(
).
/(
1
.
1
yields
which
,
73909
.
0
)
2
/(
relation
by the
determined
is
energy
ing
correspond
the
Thus,
.
73909
.
0
by
y
numericall
given
is
cos
of
solution
The
.
-
1
with
tan
of
on
intersecti
by the
given
is
energy
ing
correspond
The
0.
n
to
s
correspond
state
bound
This
.
since
state
bound
one
only
is
there
,
1
)
2
/(
when
Fig.5
(a)From
1
1
0
0
2
0
2
2
0
2
2
0
2
0
0
2
2
0
0
0
2
2
























ma
E
E
ma
R
V
ma
R
o
n
o



122

,
22
.
7
9
.
1
2
,
12
.
2
03
.
1
2
are
energies
ing
correspond
The
ly.
respective
,
9
.
1
and
03
.
1
yield
,
sin
4
4
tan
,
cos
4
4
tan
2
2
1
2
1
2
1
2
2
0
2
0
2
0
1
0
2
1
1
2
2
1
1
1
2
0
0
2
2
0
0
0
ma
E
E
ma
ma
E
E
ma





































123

The Harmonic Oscillator
The harmonic oscillator is one of those few problems that
are important to all branches of physics. It provides a
useful model for a variety of vibrational phenomena that
are encountered, for instance, in classical mechanics,
electrodynamics, statistical mechanics, solid state, atomic,
nuclear, and particle physics. In quantum mechanics, it
serves as an invaluable tool to illustrate the basic concepts
and the formalism.
Classical treatment :
and the solution is
Where is the angular frequency of oscillation .
2
2
dt
x
d
m
kx
F 


)
cos(
)
sin(
)
( t
B
t
A
t
x 
 

124

A particle oscillating in a harmonic potential
125

its graph is a parabola.
V
x
Turning point
Kinetic energy
zero; potential
energy max. Turning
point
Turning
point
Classical particle can never
be past turning point.
Particle can be stationary at bottom of well,
know position, x = 0; know momentum, p = 0.
126

Which can be reduced to
where x0 = ћ/m is a constant that has dimensions of
length.
0
)
(
2
)
(
4
0
2
2
2
2










 x
x
x
mE
dx
x
d



The analytic method: this approach consists in using the
power series method to solve the following differential
Schrodinger equation:
)
(
)
(
2
1
)
(
2
2
2
2
2
2
x
E
x
x
m
dx
x
d
m








127

The solution of differential equation is expressed in terms
of some special functions, the Hermit polynomials.
The energy eigenvalues which are discrete or quantized:
Wave functions that are physically acceptable and satisfy
that Schrödinger equation are given:
where Hn(y) are nth order polynomials called Hermite
polynomials:
,...)
2
,
1
,
0
(
2
1










 n
n
En 

)
(
!
2
1
)
(
0
2
0
2
0
2
x
x
H
e
x
n
x n
x
x
n
n




128

129

130

From this relation it is easy to calculate the first few
polynomials:
2
2
2
2
)
1
(
)
( y
n
y
n e
dy
d
e
y
H 


En
Hn(y)
n
½ ħ
3/2 ħ
5/2 ħ
7/2 ħ
9/2 ħ
11/2 ħ
1
2y
4y2 – 2
8y3 – 12y
16y4 - 48y2 +12
32y5 -160y3 +120y
0
1
2
3
4
5
131

We will deal with the physical interpretations of the harmonic
oscillator results when we study the second method.
Algebraic method:
The Hamiltonian operator can be expressed in terms of the operators


















 









m
P
m
m
m
P
H
2
2
2
2
2
2
2
2
2
  

 

 m
q
m
P
p
q
p
H 



 ˆ
ˆ
and
ˆ
ˆ
ˆ
ˆ
2
ˆ 2
2
and then introduce two-non-Hermitian, dimensionless operators:
   
p
i
q
a
p
i
q
a ˆ
ˆ
2
1
ˆ
,
ˆ
ˆ
2
1
ˆ †




 
p
q
i
p
q
q
p
i
p
q
i
p
q
p
i
q
p
i
q
a
a ˆ
,
ˆ
2
)
2
ˆ
2
ˆ
(
2
1
)
ˆ
ˆ
ˆ
ˆ
2
ˆ
2
ˆ
(
2
1
)
ˆ
ˆ
)(
ˆ
ˆ
(
2
1
ˆ
†
ˆ 









    i
p
p
m
m
p
q 











 ˆ
,
ˆ
1
ˆ
1
,
ˆ
ˆ
,
ˆ


 

132

Hence
therefore
where is known as the number operator or occupation
number operator, which is clearly Hermitian.
N̂
133

  







 
 a
H
a
a
H
a ,
,
Can also show
Now, we know that energy eigenstates of H must also be eigenstates
of N, so we’ll denoted those (normalized) orthogonal eigenstates by
∣n〉, such that N∣n〉=n∣n〉 and H∣n〉 = En∣n〉 with En = ħ(n+1/2).
They only constraint on n that we know right now is that it must
be real number. Since we have
134

It shall be useful to know a few mathematical relations:
The Matrix Representation of Various Operators
Her we look at the matrix representation of several operators
in the N-space.
In the matrix form, represented in the |n  basis,
135

136

In quantum mechanics the delta potential is a potential
well mathematically described by the Dirac delta function -
a generalized function. Qualitatively, it corresponds to a
potential which is zero everywhere, except at a single
point, where it takes an infinite value. This can be used to
simulate situations where a particle is free to move in two
regions of space with a barrier between the two regions.
For example, an electron can move almost freely in a
conducting material, but if two conducting surfaces are put
close together, the interface between them acts as a barrier
for the electron that can be approximated by a delta
potential.
The Delta-Function Potential
137

The Dirac delta function, δ(x), is defined informally as
follows:
δ(x-a) would be a spike of area 1 at the point a.
If you multiply δ(x-a) by an ordinary function
f(x), it’s the same as multiplying by f(a):
f(x) δ(x-a) = f(a)δ(x-a)
δ(x)
x
because the product is zero anyway except
at the point a. In particular,
138

139

Let’s consider a potential of the form
V (x) = -αδ (x)
where α is some constant. This potential has a delta function well located
at x = 0 (so was the infinite square well). Solving Schrödinger’s equation
with delta function potential
This potential yields both bound states (E < 0) and scattering states
(E > 0).
V (x)
x
x=0
E >0
E <0
140

First consider bound states, away from x = 0, V (x) = 0 , so
(E is negative, by assumption, so k is real and positive).
energy)
negative
0
(
)
(
)
(
2 2
2
2



 E
x
E
dx
x
d
m



)
(
2
)
( 2
2
2
2



k
x
mE
dx
x
d




2

mE
k


negative
real
+
The general solution is  (x) = A e -kx + B e kx
x
Potential going up at x=0
x>0
B=0
X<0
A=0
 2(x) = B e -kx
 1(x) = A e kx
What happen in
the boundary
condition
141

Boundary condition matching
1-  is always continuous, and
2- d/dx is continuous except at points where the potential [V(x) ]is
finite.
ψ continuous
1(x=0) = 2(x=0)
A = B
d/dx is continuous
d/dx at the boundary condition
x
 2
 1
x=0
0
0 



x
kx
x
kx
Be
Ae
1
1
)
(
)
(
)
(
2 2
2
2
x
E
x
V
x
x
m









-αδ (x)
142

)
(
)
(
)
(
2 2
2
2














 x
E
x
V
x
x
m




















 dx
x
E
dx
x
x
dx
x
x
m
)
(
)
(
)
(
)
(
2 2
2
2






0
)
0
(
2
2








dx
d
m

0
)
0
(
2
2










m
dx
d
0
0
1
0
2



dx
d
dx
d 

)
0
(
2
2



m
Bke
Bke kx
kx



 
B
A 
kx
Be
m
Bk
Bk 



 2
2


(0)
143

kx
Be
m
Bk
Bk 



 2
2


B
m
Bk
Bk 2
2






1
2
2
2

m
k



 2
2

m
k



mE
k
2


2
2
2
m
E












0
0
)
(
2
2
x
e
x
e
B
x x
m
x
m





144

Normalized bound state solution
(x)
x



2
)
(
2
2




m
E
e
m
x
x
m




145

In the region x > 0, V (x) is again zero, and the general
solution is of the form,  (x) = F e-kx + G ekx
which reduces as G ekx  as x
 (x) = F e-kx , (x > 0).
1-  is always continuous, and
2- d/dx is continuous except at points where the
potential [V(x) ]is finite.
In this case the first boundary condition F = B, so
The general solution is  (x) = A e -kx + B e kx ,
but the first term blows up as x  -, so A = 0:
 (x) = B e kx , (x < 0).
146

The delta function must determine the discontinuity
in the derivative of  , at x = 0. The idea is to
integrate the Schrodinger equation, from -ϵ to +ϵ ,
and then take the limit as ϵ  0:
The right hand side becomes zero as ϵ  0 , then:
147

k
(x)
k e-
kx
k
ekx
Bound state wave function for delta function
potential.
0 x
148

or you can write it in another way
Ordinarily, the limit on the right is again zero, and hence
d/dx is continuous. But when V(x) is infinite at the
boundary, that argument fails. In particular, if
V(x) = -αδ(x), then:
149

and hence ∆(d/dx) = -2Bk. And (0) = B.
So the equation says
The corresponding energy is
Finally, we normalize 
150

Evidently the delta-function well, regardless of its
“strength” α, has exactly one bound state.
The wave function can be written as:
There is only one bound state for any given value of α.
151

Scattering States:
Scattering states have E  0 , for x < 0, away from x = 0,
V(x)=0
so the Schrödinger equation becomes
The general for x < 0 is :
 = A e ikx + B e –ikx . At this point we cannot
eliminate either the A or B term.
From the region away from the delta function
0
1 

 
x
Be
Ae ikx
ikx

0
2 

 
x
Ge
Fe ikx
ikx

152

Boundary condition matching
 is continuous
1(0) = 2(0)
A + B = F + G
0
1 

 
x
Be
Ae ikx
ikx

0
2 

 
x
Ge
Fe ikx
ikx

d/dx at the boundary condition
)
(
)
(
)
(
2 2
2
2
x
E
x
V
x
x
m









)
0
(
2
)
(
)
(
)
(
2 2
2
2
2







 m
dx
d
x
E
x
V
x
x
m





















)
(
0
1
1 B
A
ik
dx
d
Be
Ae ikx
ikx



  

153

)
(
0
2
2 G
F
ik
dx
d
Ge
Fe ikx
ikx



  

)
(
2
)
(
)
( 2
B
A
m
B
A
ik
G
F
ik 







k
m
2
2


 
)
2
1
(
)
2
1
( 
 i
B
i
A
G
F 




2 equations and 4 unknowns
0
1 

 
x
Be
Ae ikx
ikx

0
2 

 
x
Ge
Fe ikx
ikx

B
A
G
F 


)
2
1
(
)
2
1
( 
 i
B
i
A
G
F 




154

x=0
A
B
F
G
Initial conditions ⇒Ics = scattering from the left ⇒G=0
A is incident wave
Reflection and Transmission
A
i
F
A
i
i
B







1
1
,
1
A
i
F
A
i
i
B







1
1
,
1
2
2
2
2
2
2
2
1
1
,
1 








A
F
T
A
B
R
1
T
R
2
1
1
,
2
1
1
2
2
2
2






E
m
T
m
E
R

 
 155

For higher energy the reflection coefficient decreases, that means
more penetrated. In transmission coefficient the dominator is very
small and the transmission coefficient approach to one
Delta function barrier
2
1
1
2
1
1
2
2
2
2
E
m
T
m
E
R








V(x)= -
aδ(x)
a<
0
Classically:
a > 0 (well) T = 1 and R =
0
a < 0 (barrier) T = 0 and R =
1
Q.M : is the same
Ex; Tunneling
156

For x  0 we have (x) = F e ikx + G e –ikx and at x
= 0 (x) is continuous so we can write A + B = F + G
.
Differentiating (x) we find that:
Going back to our previous integration of the Schrodinger
equation, we have the result that
For the scattering states (0) = A + B and we have
157

Let , divide by ik and collect the A and B
terms on one side and we get F – G = A (1 + 2iβ) – B(1-
2iβ) .
What do the terms in the
Schrodinger equation
Represent?
A e ikx represents a wave
incident from the left and
F e ikx is a wave travelling
from the potential well to the right. 0
V(x)
0 x
Feikx
Ge-
ikx
Aeikx
Be-ikx
Scattering from a delta –function well 158

Suppose G e –ikx is zero, then B e –ikx represents a wave
reflected from the potential well, moving to the left. we put
G = 0 and solve the two equations A + B = F + G and
F – G = A (1+2iβ) – B (1-2iβ) .
We can define reflection and transition coefficient, R and
T which are the probabilities of reflection and transmission
from the potential well.
and
159

and R + T = 1
In terms of energy we can write:
The higher energy, the greater the probability of
transmission (which seems reasonable).
160

CHAPTER FOUR
Angular Momentum
Commutation Relations
The Uncertainty Relations for Angular Momentum
161

Angular momentum is as important in classical
mechanics and quantum mechanics is the motion of
a particle in a central force field where the potential
function V depends only on the distance r between
the particle and the origin of the coordinate system.
It is particularly useful for studying the dynamics of
systems that move under influence of spherically
symmetric, or central, potentials. The orbital
angular momenta of these systems are conserved.
Additionally , angular momentum plays a critical
role in the description of molecular rotations, the
motion of electron in atoms, and the motion of
nucleons in nuclei. 162

In classical mechanics the angular momentum of a particle
with momentum P and position r is defined as cross
product of position vector and momentum vector
In quantum mechanics the classical vectors
become operators. More precisely, they give us triplets of
operators:
p
r
L





L
p
r



and
,
)
ˆ
,
ˆ
,
ˆ
(
)
ˆ
,
ˆ
,
ˆ
(
)
ˆ
,
ˆ
,
ˆ
(
z
y
x
z
y
x
L
L
L
L
p
p
p
p
z
y
x
r






163

  x y z
x y z p p p
 
   
L
x y z
x y z
p p p

i j k
L
The Cartesian component of are
L̂
x z y
y x z
z y x
L yp zp
L zp xp
L xp yp
 
 
 
ˆ
ˆ
ˆ
x
y
z
L i y z
z y
L i z x
x z
L i x y
y x
 
 
  
 
 
 
 
 
  
 
 
 
 
 
  
 
 
 
x
y
z
z
x
y
y
z
x
p
y
p
x
L
p
x
p
z
L
zp
p
y
L
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ






i x y z
d d d
dx dy dz
 
i j k
L
164

Clearly, angular momentum does not exist in a one-dimensional space.
We should mention that the components
Commutation relations:
,
ˆ
of
square
the
and
,
ˆ
,
ˆ
,
ˆ L
L
L
L z
y
x

2
2
2
2 ˆ
ˆ
ˆ
ˆ
z
y
x L
L
L
L 


      have
we
,
ˆ
,
ˆ
,
ˆ
,
ˆ
,
ˆ
,
ˆ
since
and
ˆ
and
,
ˆ
,
ˆ
do
so
and
commute
mutually
ˆ
and
,
ˆ
,
ˆ
Since


 i
p
z
i
p
y
i
p
x
p
p
p
z
y
x
z
y
x
z
y
x



      y
x
z
x
z
y
z
y
x L
i
L
L
L
i
L
L
L
i
L
L ˆ
ˆ
,
ˆ
,
ˆ
ˆ
,
ˆ
,
ˆ
ˆ
,
ˆ 

 


165

We will use a " ladder operator " technique, introduce
raising and lowering operators, 
 L
L and
,
The different components of L do not commute with each
another, but they do commute with the squared magnitude of
the angular momentum vector:
      0
ˆ
,
ˆ
ˆ
,
ˆ
ˆ
,
ˆ 2
2
2


 z
y
x L
L
L
L
L
L
2 2
2
ˆ
, , , 0
x y z
L L L L L L
   
 
  
 
   
   
y
x L
i
L
L ˆ
ˆ
ˆ 


)
ˆ
ˆ
(
2
1
ˆ
,
)
ˆ
ˆ
(
2
1
ˆ



 


 L
L
i
L
L
L
L y
x
166

In addition, satisfy
      



 


 L
L
L
L
L
L
L
L z
z
ˆ
ˆ
,
ˆ
,
ˆ
2
ˆ
,
ˆ
,
0
ˆ
,
ˆ2


These relations leads to
z
z L
L
L
L
L ˆ
ˆ
ˆ
ˆ
ˆ 2
2


 
 
Which in turn yield
2
2 ˆ
)
ˆ
ˆ
ˆ
ˆ
(
2
1
ˆ
z
L
L
L
L
L
L 

 



167

The different components of angular momentum do not
commute
• Lx, Ly and Lz are not compatible observables
• They do not have simultaneous eigenfunctions (except when
L = 0)
• We can not have perfect knowledge of any pair at the same
time
But, the different components all commute with L2
• L2 and each component are compatible observables
• We can find simultaneous eigenfunctions of L2 and one
component
168

The Uncertainty relations for angular momentum
Recalling the generalized uncertainty relation for two operators
   
 
2
2
2
2
,













i
B
A
B
A
 
B
A
b
a ,
2
1



 
2 2
2
A A A
    
2 2
2
B B B
  
  
i
p
x x 
ˆ
,
ˆ
169

We can write down uncertainty relation for the
components of angular momentum using the
commutators. For example, we find
 
z
z
y
x
y
x L
i
L
i
i
L
L
L
L
2
2
2
, 






  z
y
x L
i
L
L 

,
170

CHAPTER FIVE
Quantum Mechanics In Three Dimensions
•Schrodinger Equation in Spherical Coordinates
•Separation of Variables
• The Angular Equation
• The radial Equation
• The Box Potential
• The Hydrogen Atom
171

The generalization to three dimensions is straightforward.
Schrodinger's equation says
the Hamiltonian operator H is obtained from classical
energy
by the standard prescription (applied now to y and z, as
well as x):
or
Schrodinger Equation in Spherical Coordinates
172

for short. Thus
where
is the Laplacian, in Cartesian coordinates.
The potential of finding the particle in the infinitesimal
volume d3r = dx dy dx is |Ψ(r,t)|2 d3r, and the
normalization condition reads
173

With the integral taken, over all space. If the potential is
independent of time. there will be complete set of
stationary states,
where the spatial wave function ψn satisfies the time-
independent Schrodinger equation:
The general solution to the (time-independent) Schrodinger
equation is
with the complex constants cn determined by initial wave
function, Ψ(r,0), in the usual way. (If the potential admits
continuum states, then the sum in above equation becomes
an integral).
n=1,2,3,….
 
 


r
d
r
r
c n
n
3
0
,
)
(

2
,..
2
,
1
3

 




n
n
c
r
d
174

Separation of Variables
Typically, the potential is function only of the distance
from the origin. In that case it is natural to adopt spherical
coordinates, (r,θ,) (see Fig.1). In spherical coordinates the
Laplacian takes the form,
In spherical coordinates, then, the time-independent
Schrodinger equation reads
We begin by looking for solutions that are separable into
products:
ψ (r,θ,) = R(r) Y (θ,)
175

Then
Fig.1: Spherical coordinates:
Radius r, polar angle θ, and
azimuthal angle 
z
x
y
r
p
ȓ

θ
176


x
y
z

r
r
cos

r sinsin
Note that
 = 0   = 2
In math notation, 
and  are swapped
Switch from Cartesian to Spherical
Coordinates
•r is the distance from the origin to
the point
• is the angle compared to the z-axis
• is the angle of the projected
“shadow”
compared to the x-axis
sin cos
sin sin
cos
x r
y r
z r
 
 




0
0
0 2
r
 
 
  
 
  177

dividing by Y R and multiplying by -2mr 2/ћ2:
178

The Angular Equation
Multiplying by Y sin2θ
Separation variables Y (θ,) = (θ) Ф()
179

Since Φ(ϕ) must be periodic with period 2π, m must be an
integer m = 0, ±1, ± 2, …
The significance of m is essentially the same as in two
dimensions: If we fix r and θ and let ϕ vary, then we move
around a circle about the z-axis
180

z


r
0
radius
181

182

The solution is
Where is the associated Legendre function
183

The radial Equation
The angular part of the wave function, Y(θ,), is the same
for all spherically symmetric potentials; the actual shape of
the potential, V(r), affects only the radial part of the wave
function, R(r), which is determined by
Let u (r )  r R(r ) ,
So that R= u/r , dR/dr = {r(du/dr) – u}/r2,
(d/dr)[r2(dR/dr)] = rd2u/dr2 , and hence:
184

This is called the redial equation; it is identical in form to
the one-dimensional Schrodinger equation.
Example: Consider the infinite spherical well,
Outside the well the wave function is zero; inside the
well the radial equation
185

Where
The boundary condition u (a) = 0. The case l = 0 is easy,
The actual radial wave function is R(r) = u(r) / r, and
[cos(kr)] / r blows up as r  0. So we must choose B = 0.
The boundary condition then requires sin(ka) = 0, and
hence ka = n, for some integer n. The allowed energies
are evidently
186

The Box Potential
The Rectangular Box Potential:
The potential energy of the particle is considered to be
equal zero inside the box and it is infinity at the
boundaries(surface) and in the remaining space.
Consider first the case of a spinless particle of mass m
confined in a rectangular box of sides (a, b, c)
which can be written as V(x, y, z) = Vx(x) + Vy(y) + Vz(z),
with
187

the potentials Vy(y) and Vz(z) have similar forms. The
wave function ψ(x, y, z) must vanish at the walls of the
box, x = 0, x = a, y = 0, y = b, z = 0 and z = c .
a
b
c
y
x
z
V(x,y,z)
=∞
V(x,y,z)=0
188

S.E in three dimension is

 E
z
y
x
m

















 2
2
2
2
2
2
2
2


 E
m


 2
2
2

The boundary condition
0
)
,
,
(
)
0
,
,
(
0
)
,
,
(
)
,
0
,
(
0
)
,
,
(
)
,
,
0
(






c
y
x
y
x
z
b
x
z
x
z
y
a
z
y






Using the separation variable )
(
)
(
)
(
)
,
,
( z
Z
y
Y
x
X
z
y
x 

The total Hamiltonian is break down in three cases
l
dimensiona
one
for
case
same
the
)
(
)
(
2 2
2
2


 x
X
E
dx
x
X
d
m
x







 x
a
n
A
x
X x
x

sin
)
(
is
solution
the
189

The same for Y and Z








 y
b
n
A
y
Y y
y

sin
)
( 





 z
c
n
A
z
Z z
z

sin
)
(
 
 

b c
a
z
y
x
z
y
x
dz
dy
dx
0 0
0
1
)
,
,
(
)
,
,
( 

abc
A
A
A z
y
x
8






















c
z
n
b
y
n
a
x
n
abc
z
y
x z
y
x
n
n
n z
y
x



 sin
sin
sin
8
)
,
,
(








 2
2
2
2
2
2
2
8 c
n
b
n
a
n
m
h
E z
y
x
n
n
n z
y
x
190

Two particles problems
The Hydrogen Atom
The hydrogen atom consists of an electron and a proton.
The H atom is a bound state of a proton and an electron.
The approximation of the potential energy of the electron-
proton system is electrostatic.
From Coulomb’s law,
the potential energy:
r
e
r
V
0
2
4
)
(



-e
(electron)
r
+e
(proton)
The hydrogen atom 191

Spherical polar coordinate system































 2
2
2
2
2
2
2
sin
1
sin
sin
1
1
1










r
r
r
r
r
Angular momentum operator

 )
( 2
2
2
2
z
y
x L
L
L
L 






















 2
2
2
2
2
sin
1
sin
sin
1








 
L
2
2



L
2
2
2
2
2
2 1

r
L
r
r
r
r


 












192

The hydrogen atom potential
energy is given by: e-
P+
r

V(r) 
e2
r
V
(potential)
0 r
The time-independent Schrödinger equation in three dimensions is
then:






E
r
e
z
y
x
m



















0
2
2
2
2
2
2
2
2
4
2

The radial equation:
It describes a particle with angular momentum L = √l(l+1). which behaves like a
particle in a one-dimensional effective potential of the form
193

The problem is to solve this equation for u (r) and
determine the allowed electron energies E.
Let
(For bound states, E < 0 , so  is real.)
The radial equation becomes
We define a new variable ρ so that
194

As    , the constant term in the brackets dominates, so
approximately
The general solution is u () = A e- + B e
but e blows up (as    ), so B = 0. Evidently,
u ()  A e- ,
as   0 the centrifugal term dominates; approximately,
then
The general solution is u () = C  l+1 + D  -l ,
but  -l blows up (as   0 ), so D = 0. Thus
u ()  C  l+1
195

For small .
The next step is to peel off the asymptotic behavior,
introducing the new function v ():
u () =  l+1 e - v () ,
And
the radial equation
196

Finally, we assume the solution, v (), ca be expressed as a
power series in  :
Differentiating term by term,
Differentiating again,
197

This recursion formula determines the coefficients. Now
let's see what the coefficient look like for large j (this
corresponds to large  , where the higher powers
dominate).
198

and hence u () = A  l+1 e 
To be complete
So the allowed energies are:
This is the famous Bohr formula.
199

Where
is the so-called Bohr radius.
Evidently the spatial wave functions for hydrogen are
labeled by three quantum numbers (n , l , and m):
nlm (r,θ,) = Rnl(r) Yl
m (θ,) ,
200

CHAPTER SEX
Approximation Methods for Stationary States
• perturbation theory
• The variational method
• WKB method
• Time-Independent Perturbation Theory
•Nondegenerate Perturbation Theory
201

Most problems encountered in quantum mechanics cannot
be solved exactly. Exact solutions of the Schrodinger
equation exist only for a few idealized system. To solve
general problems, one must restore to approximation
methods. In this chapter we consider approximation
methods that deal with stationary states corresponding to
time-independent Hamiltonians. To study problems of
stationary states, we focus on three approximation methods:
perturbation theory, the variational method, and WKB
method.
Perturbation theory is designed to predict the eigenvalues
and eigenfunctions of a perturbed Hamiltonian H that
differs only slightly from an unperturbed Hamiltonian H0
whose eigenvalues and eigenfunctions are known. The
difference between H and H0 is called the perturbation. 202

The variational method was initially derived from a
mathematical theorem that formulates a relation between
differential equations with boundary conditions, integral
equations, and matrices of infinite order. In situations
where exact solutions to the problem could not found, the
matrix representations lead to convenient and fairly reliable
approximate solutions.
The WKB method is useful for finding the energy
eigenvalues and wave functions of systems for which the
classical limit is valid. Unlike perturbation theory, the
variational and WKB methods do not require the existance
of a closely related Hamiltonian that can be solved exactly.
203

Perturbation Theory
Time-Independent Perturbation Theory
Suppose that we solved the time-independent Schrodinger equation
for some potential and obtained a complete set of orthonormal
eigenfunctions 0
n and corresponding eigenvalues E0
n.
nm
m
n
n
n
n E
H







0
0
0
0
0
0
This is the problem that we completely
understand and know solutions for.
Now we slightly perturb the potential. For example, we raise a little
bit the bottom of the infinite square well or put a little bump there:
x
a
V(
x)
Infinite square well with small perturbation
little
bump
204

The problem of the perturbation theory is to find eigenvalues and
eigenfunctions of the perturbed potential, i.e. to solve approximately
the following equation:
H
H
H
E
H n
n



 0
,


This method is most suitable when Ĥ is very close to a Hamiltonian
Ĥ0 that can be solved exactly. In this case Ĥ can be split into two
time-independent parts
using the known solutions of the problem
perturbation
0
0
0
0
n
n
n E
H 
 
Suppose we change this potential only slightly; e.g., we could add a
slight ‘bump’ in the bottom of the well. It is not likely that we can
solve for the S.E of this new Hamiltonian H exactly, but let’s try to
find an approximate solution.
0
0
0
0
n
n
n E
H 
 
205

H = H0 + H΄
rewrite the Hamiltonian introducing explicitly a parameter ¸ that
determines how small is a particular term
Where 1 ≥ λ ≥ 0
Ĥ ΄ = λŴ
The application of the approximation methods to the
study of stationary states consists of finding the energy
eigenvalues En and the eigenfunctions |n of a time-
independent Hamiltonian Ĥ that does not have exact
solutions: Ĥ |n = En |n . or
The eigenvalue becomes:
(Ĥ0 + λŴ ) |n
We consider two separate cases depending on whether
the exact solutions of Ĥ0 are nondegenerate or
degenerate. Each of these two cases requires its own
approximation scheme. 206

Nondegenerate Perturbation Theory
non-degeneracy occurs when each eigenstate corresponds
to a unique energy. If Ĥ0 has no degenerate eigenvalues:
that is, for every energy En
(0) there corresponds only one
eigenstate |n :
n
n
n E
H 
 )
0
(
0
ˆ 
Where the exact eigenvalues En
(0) and exact eigenfunction
n are known.
The main idea of perturbation theory consists in assuming
that the perturbed eigenvalues and eigenstates can both be
expanded in power series in the parameter λ.
207

When λ = 0 the two above expression yield the unperturbed solutions:
En = En
(0) and |n = |n . or n = n
Here En
(1) is the first order correction to the nth eigenvalue, and n
(1) is
the first order correction to the nth eigenfunction; En
(2) and n
(2) are the
second order corrections, and so on.
The lowest order (λ0) this yields H0 n
0 = En
0 n
0 .
......]
)[
( 2
2
1
0
0




 n
n
n
H
H 




]
.....)[
( 2
2
1
0
2
2
1
0
n
n
n
n
n
n E
E
E 




 





......
)
(
)
( 1
2
0
2
0
1
0
0
0






 n
n
n
n
n H
H
H
H
H 






...
]
(
)
( 0
2
1
1
2
0
2
0
1
1
0
0
0






 n
n
n
n
n
n
n
n
n
n
n
n E
E
E
E
E
E 







,
.
).........
( 0
0
0
0
0
n
n
n E
H 

 
,
).......
( 0
1
1
0
0
1
0
1
n
n
n
n
n
n E
E
H
H 



 



,
)...
( 0
2
1
1
2
0
1
2
0
2
n
n
n
n
n
n
n
n E
E
E
H
H 




 




208

The first order (λ1)
The second order (λ2)
multiplying by (n
0) and integrating.
But H 0 is Hermitian, so
209

and this cancels the first term on the right.
Moreover, , so
This is the fundamental result of first order perturbation
theory;
210

So if m = n, we are in trouble (also 2nd order energy
correction).
Having m=n means we have two (or more) states with the
same energy = Degenerate states
211

Example: The unperturbed wave functions for the infinite
square well are the equation
Suppose first that we perturb the system by simply raising
the "floor" of the well by a constant amount V0 as shown
Fig. In that case H' = V0, and the first order correction to the
energy of the nth state is:
0
:
n
Hamiltonia
on
Perturbati V
H 

212

V(x)
x
a
V0
Constant perturbation over the whole well.
213

The corrected energy levels, then, are En  En
0 + V0 ; they
are simply lifted by the amount V0 . The only surprising
thing is that in this case the first order theory yields the
exact answer. Evidently, for a constant perturbation all the
higher corrections vanish. If, on the other hand, the
perturbation extends only halfway across the well (as in
Fig.)
V(x)
x
a
V0
a/2 214

then,
In this case every energy level is lifted by V0 /2. That’s not
the exact result, presumably, but it does seem reasonable as
a first order approximation.
Q)- Find the first order correction to the wave function.
Second-Order Energies
Now, cut the perturbation to only a half-way across the well
215

216

Typically, the perturbation (H) will “break” the
degeneracy: As we increase λ (from 0 to 1), The common
unperturbed energy E 0 splits into two(as in Fig.).
The essential problem is this: when we turn off the
perturbation, the upper state reduces down to one linear
combination of a
0 and b
0 , and the lower state reduce to
some other linear combination,
but we don’t know a priori
what these good linear
Combination will be.
For this reason we can’t even
Calculate the first-order energy
λ
1
E0
E
Lifting of a degeneracy by a perturbation.
217

because we don’t know what unperturbed states to use. For the
moment, therefore, let’s just write “good” unperturbed states in the
general form
Keeping α and β adjustable. We want to solve Schrodinger equation,
H  = E  with H = H 0 + λ H  and
E = E0 + λE1 + λ2E2 + … ,  = 0 + λ 1 + λ2 2 + …
0
0
0
b
a 

 

218

Degenerate Perturbation Theory
If the unperturbed states are degenerate that is, if two (or
more) distinct states (a
0 and b
0) share the same energy
then ordinary perturbation theory fails. However the
perturbation usually removes the degeneracy, so the
levels split.
Twofold Degeneracy
We have two states a
0 and b
0 that are degenerate, i.e
they have the same energy E0, Spilt states will have
orthogonal.
b
b
a
a E
H
E
H 


 0
0
0
0 ˆ
ˆ 

ˆ
ˆ 0
0
0
0
0
0
0
0
b
b
a
a E
H
E
H 


 

1
,
0 0
0
0
0
0
0


 b
b
a
a
b
a 





0
1
1
0
0
1
0



 E
E
H
H 


 219

Linear combination of these states
0
0
0
b
a 

 

is also an eigenstate of H0 with eigenvalue E0 .
We want to solve
H
H
H
E
H 


 0
,


.........
....., 1
0
1
0





 


E
E
E
0
1
1
0
0
0



 E
E
H
H 




This time we multiply this equation from the left by a
0 and
integrate, i.e take inner product with a
0.
0
0
1
1
0
0
0
0
1
0
0







 a
a
a
a E
E
H
H 



1
0
0
1
0
0
n
a
n
a E
E
H 



 


  1
0
0
0
n
b
a
a E
H 


 


0
0
0
b
a 

 

220

0
0
1
0
0
1
0
0
0
0
b
a
a
a
b
a
a
a E
E
H
H 










 




1 0
ab
aa
b
a
a
a
w
w
H
H
E








 0
0
0
0
1




ab
aa w
w
E 

 

1
)
,
,
(
,
0
0
b
a
j
i
H
w j
i
ij 

 

Wij are known since we know i
0  we can calculate them
Similarly, the inner product with b
0 yields
βE 1 = αWba + βWbb
now solve this system of equations for E1
where
ab
aa w
w
E 

 

1
bb
ba w
w
E 

 

1
Notice that the W’s are (in principle) known “matrix elements” of H,
with respect to the unperturbed wave functions a
0 and b
0
221

 
bb
ab
ba
ab
ab
ab
bb
ba
aa
ab
ab
aa
w
w
w
w
w
E
w
w
w
E
w
E
w
w
w
E






















1
1
1
1





 





















2
2
1
1
2
1
1
1
1
1
1
4
)
(
2
1
0
)
(
)
(
)
(
0
if
)
)(
(
)
(
)
(
ab
ab
aa
ab
aa
ab
ba
ba
ab
bb
aa
bb
aa
ba
ab
bb
aa
aa
bb
ba
ab
aa
w
w
w
w
w
E
w
w
w
w
w
w
w
w
E
E
w
w
w
E
w
E
w
E
w
w
w
w
E
E








Fundamental result of degenerate perturbation theory: two roots
correspond to two perturbed energies (degeneracy is lifted). 222

aa
bb
w
E
w
E






1
1
0
If
0
If












 0
0
0
0
1
b
b
bb
a
a
aa
H
w
H
w
E




if we could guess some good linear combinations a
0 and b
0, then
we can just use nondegenerate perturbation theory.
This is the fundamental result of degenerate perturbation
theory; the two roots correspond to the two perturbed
energies.
223

But what if α is zero? In that case β = 1, Wab = 0, and E1 = Wbb.
This is actually include in the general result with plus sign (the
minus sign corresponds to α = 1 , β = 0).
Then using
E
E0
1
E
1
E
d=2
 
,
a b
 
0 0
'
bb b b
W H
 

0 0
'
aa a a
W H
 

0 0
'
ab a b
W H
 

0
ab ba
W W
 
If
aa
a W
E 
 
1
0

 bb
b W
E 
 
1
0

 224

Problem: Suppose we put a delta-function bump in the center of the
infinite square well











2
a
x
H  where  is a constant
(a) Find the first-order correction to the allowed energies. Explain
why energies are not perturbed for even n.
(b) Find the first three nonzero terms in the expansion below of the
correction to the ground state,
0
0
0
0
0
m
n
m m
n
n
m
n
E
E
H



 
 



225

The Variational Principle
There are many problems of wave mechanics which
cannot be conveniently treated either by direct solution of
the wave equation or by use of perturbation theory. The
Helium atom is such a system. No direct method of
solving the wave equation has been found for this atom,
and the application of perturbation theory is unsatisfactory
because the first approximation is not accurate enough.
One of the approximation methods that is suitable for
solving such problems is the virational method, which is
also called the Rayleigh-Ritz method. The variational
method is one way of finding approximations to the
lowest energy eigenstate or ground state, and some excited
states. But not very useful for the study of excited states.
226