Quantum mechanics
Review of classical mechanics (Newtonian mechanics).
Formulated by Sir Isaac Newton (1642-1727 ~ 300 years ago).
Newtons laws
1. Every object must be at rest or moving at a constant speed unless a net force is acting
on it.
2. The rate of change of momentum of an object is equal to the force acting on it.
3. Every action has an equal and opposite reaction.
Classical mechanics describes the motion of objects based on two conditions. They are
Newton’s second law and the condition that the total energy of the object is constant (KE
+ V) = constant.
Force applied to an object = mass of the object  acceleration of the object.
F= ma ,
a = acceleration
Force = rate of change of momentum.
dp
F=
dt
d(mv)
m = mass of the object.
F=
dt
m dv
d  dx  m d 2 x
F=
= m  =
dt
dt  dt 
dt 2
At constant acceleration,
dv
m = ma
dt
dv = a dt
v
t
u
0
 dv = a  dt
where, v = final velocity, u = initial velocity
vuv = at t0
v  u = at
v = u + at
dx
Using v = dt where x = displacement
1
𝑑𝑥
= 𝑢 + 𝑎𝑡
𝑑𝑡
𝑠
𝑡
𝑡
∫ 𝑑𝑥 = ∫ 𝑢 𝑑𝑡 + ∫ 𝑎𝑡 𝑑𝑡
0
0
0
2 𝑡
𝑡
[𝑥]0𝑠 = 𝑢[𝑡]𝑡0 + 𝑎 [ ]
2 0
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Removing t from the above two equations yields, v 2 = u 2 + 2as
Theorem of equipartition of energy.
For a collection of particles at the temperature T, at thermal equilibrium, the average
contribution to the total energy from each quadratic term is 1/2kT. Where k = 1.381x10-23
J K-1 (Boltzmann constant).
Eg. Total energy of a particle having three velocity components,
E = ½ mvx2 + ½ mvy2 + ½ mvz2
= ½ kT + ½ kT + ½ kT
= 3/2 kT
Eg. Energy of a particle attached by a spring (force constant k) and free to move in 1-D,
E = ½ mvx2 + ½ kx2
= kT
Newtonian mechanics is successful in describing motion of large objects. Evidence began
to appear after 1900 that Newtonian mechanics is unsuccessful in describing the motion
of elementary particles (electrons, protons etc.), atoms and molecules.
Paradoxes in classical physics
Black body radiation
(quantization of energy)
The intensity and wavelength of electromagnetic radiation emitted by a black body
depends on temperature. A black body can absorb all the wavelengths of electromagnetic
radiation falling on it. The closest example for a black body is a pinhole on a hollow dark
8kT
sphere. Classical prediction of the intensity of emitted radiation is given by,   4
λ
(Rayleigh-Jeans law).
 = energy emitted per unit volume per unit wavelength, k = Boltzmann constant, T =
temperature,  = wavelength of emitted radiation. According to this equation, intensity
2
should be infinite at very small wavelengths. This is known as ultraviolet catastrophe.
Eg. Coal (red color) temp 1200 0C.
Tungsten filament (white) 2500 0C
Surface of the Sun (Bright white) 5700 0C
Blackbody radiation is explained by considering, that the energy is emitted by oscillators,
where the energy is quantized. This idea was first put forward by Max Planck (Nobel
Prize 1918) is known as Quantization of energy.
8𝜋ℎ𝑐 2
1
𝐸 = 𝜆5 𝑒 ℎ𝑐/(𝑘𝑇𝜆)−1 where h= Plancks constant, k = Boltzmann constant, c = speed of
light, T= temperature,
Photoelectric effect
(quantization of energy and particle nature of radiation)
Evidence to particle nature of radiation and quantization of energy.
First observed by Heinrich Rudolf Hertz in 1887 (Nobel prize did not exist at this
time). A beam of light falling on a metal surface in an evacuated space ejects electrons
from the surface. This causes a current known as the photocurrent. Electrons cannot be
ejected when the frequency of light is below a certain value called the threshold
frequency (0).
3
Incident light ejects electrons causing a photocurrent. When the potential (bias) is
reversed (opposite to the direction shown in the figure), the current decreases. The
current can be completely stopped by increasing the reverse bias above a certain value.
This is called the stopping potential. The photocurrent depends on the intensity of light
and the stopping potential depends on the frequency of incident light. When the
frequency of light is below a certain value (0 ) no electrons can be ejected.
Albert Einstein in 1905 hypothesized (Nobel prize-1921) that light consists of
“quanta” each having an amount of energy, E = h. where h = Planck’s constant,
6.626 x 10 -34 Js, and  = frequency of light.
Kinetic energy of an electron = KE = ½ mv2 (v=velocity of electron).
KE = h-w0 where, w0 = work function of the metal (amount of energy needed to bring
an electron to the surface).
When light having a frequency 0 is used, the kinetic energy of electrons ejected becomes
zero.
Therefore, h0 -w0 = 0
w0 = h0
Electrons cannot be ejected when the frequency of light is below 0.
A light quantum is considered as a particle. This experiment illustrates that the
electromagnetic radiation, considered as waves also has a particle nature. This is known
as corpuscular nature (particle nature) of radiation.
Compton effect
(quantization of energy and particle nature of radiation)
Evidence to particle nature of radiation.
Observed by Arthur Compton in 1923 (Nobel prize-1927). X-ray is scattered by
electrons in carbon atoms in graphite. The wave length of scattered X-ray is larger than
4
the wave-length of incident X-ray and increases with the scattering angle. According to
classical wave theory, the wave length of scattered radiation should be the same as that of
incident radiation. Compton pointed out that X-ray is a collection of photons, each having
energy E = h and the wave length increases (c = ) when energy is transferred to the
electrons in graphite.
This is another example that shows the particle like nature of radiation.
Wavelength of scattered radiation is large due to the transfer of energy to electrons in the
graphite sheet (λ/ > λ).
Electron diffraction
(wave nature of particles)
When a beam of electrons is incident on a single crystal of Ni, a diffraction pattern is
obtained. Diffraction is a property of waves. When waves overlap in the same phase, they
undergo constructive interference. When they overlap in opposite phases, they undergo
destructive interference. Diagram below shows constructive and destructive interference
of waves.
5
This experiment confirmed that the electrons have wave properties.
6
Diagram shows the scattering of electrons (behaving as waves) by atoms in two layers of
Ni. The wave scattered by the lower layer travels an extra distance of AB+BC. This extra
distance is equal to 2d sin θ. When this extra distance is an integer multiple of the wave
length, the two waves meet on the screen in the same phase, giving rise to constructive
interference. When the extra distance is half of a wavelength, the waves meet in opposite
phases annihilating each other (destructive interference).
Criteria for constructive interference is,
2d sin θ = n 
where n is an integer and d is the spacing between planes containing Ni atoms..
Electron was discovered by J.J. Thompson in 1897 and characterized as a particle. He
won the Nobel prize for this in 1906.
Electron diffraction, which shows the wave nature of electrons was discovered by G.P.
Thompson in 1927 (son of J.J. Thompson). He won the Nobel prize with Davisson and
Germer in 1937.
Other examples for wave nature of particles:
Neutron diffraction, diffraction of molecular beams of H2 or He on LiF crystals.
Wave – particle duality
For photons,
E=hυ (h=Planck’s constant , υ = frequency of photon)
According to Einstein’s equation, E=mc2
Combining the two equations,
hυ = mc2
Using, c = λν
hc
h
mc 2 
, canceling c yields; mc 


Momentum = mass x velocity, Therefore, the momentum of a photon, p = mc and,
h

p
Going along with the momentum-wavelength relation for photons, Louis deBroglie
(1892-1987) (Nobel prize-1929) proposed that matter also has particle and wave nature.
He proposed the idea of matter waves (Wave and particle like nature of matter is known
as wave particle duality). The associated wavelength of a matter wave is given by,
h
de Broglie relation
p
p = momentum of the particle, h = Planck’s constant,  = associated wavelength.

Waves
7
A wave can be represented by a sine or cosine function.
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sin 𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑜𝑢𝑠𝑒
E  E 0 sin [
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
cos 𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑜𝑢𝑠𝑒
2
( x  vt)] gives the amplitude of the electric field of a propagating

electromagnetic wave, where x = position, v = velocity, t = time. The quantity, 2π/λ is
2x
] gives the amplitude of the
called the wave vector (k). When time = 0, E  E 0 sin [

wave frozen in time. A wave frozen in time is called a stationary wave. The graph below
shows a stationary wave (Amplitude on vertical axis and x-on horizontal axis).
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
0.5
1
1.5
2
2.5
-0.5
-1
-1.5
Y  sin [
2x

]
1.5
1
0.5
0
0
-0.5
-1
-1.5
8
Y  cos [
2x

]
A matter wave associated with a particle can also be written as a sine function.
(psi) is used to represent matter waves.
 2

  A sin  ( x  vt) 



2
 = probability of finding the particle at a given location and given time. This is equal
to the square of the amplitude of the matter wave (probability density).
Functions
Functions describe the relationship between two or more variables. y = mx + c gives the
relationship between x and y. x = independent variable, y = dependent variable.
f(x) = mx + c
f(x) = 3x + 5
f(x) = 3x2 + 5x +2
Functions may contain more than one independent variables.
f(x,y) = 3x2 + xy + 4
f(x,y) = exy
Operators
An operator instructs to do something to the entity that follows it.
d
instructs us to differentiate the function that follows.
dx
The operator acts on the entity written to the right of it.
d 2
x  2x
dx
d
3x 2  2 x  5  6 x  2
dx
A cap on a symbol ( eg. D̂ ) says that the symbol represents an operator.
Operator algebra
Operators can be added or subtracted to generate new operators.
d
Eg. Suppose that letters A and B represent two operators as, Aˆ 
and Bˆ  x . A new
dx
operator Ĉ can be constructed by adding  and B̂ .
Cˆ  Aˆ  Bˆ
d
Cˆ 
x
dx
Operators can be multiplied to generate new operators. However, the sequence of
operators must be preserved. Change of sequence may change the result of the action of
the new operator on a function.
9
Let Mˆ  Aˆ Bˆ
ˆ  d x
M
dx
Consider the action of M̂ on the function f(x) = x + 1.
d
Mˆ f ( x) 
x ( x  1)
dx
The action of several operators on a function is evaluated by executing the instructions in
the right most operator in the sequence of operators, on the function, getting the result
and then executing the instructions in the next rightmost operator on the result. When
there are many operators, this process is repeated.
The right most operator in the above sequence is x. Therefore, we need to multiply the
function by x. The result is,
x( x  1)  x 2  x
d
The next operator in the sequence is
. Now we must differentiate the result with
dx
respect to x.
d 2
( x  x)  2 x  1 .
dx
Therefore, Mˆ f ( x)  2 x  1
Aˆ Bˆ f ( x) may not be equal to Bˆ Aˆ f ( x) all the time.
Linear operators
An operator is said to be a linear operator, if it satisfies the following.
Aˆ [ f ( x)  g ( x)]  Aˆ f ( x)  Aˆ g ( x)
and
Aˆ cf ( x)  cAˆ f ( x)
where f(x) and g(x) are two functions and c is a constant.
Eigenfunctions and eigenvalues
The function f(x) is said to be an eigenfunction of the operator  if it satisfies the
relationship, Aˆ f ( x)  k f ( x) , where k is a constant.
d
Eg. Let Aˆ 
and f(x) = e2x
dx
d
Aˆ f ( x)  e 2 x
dx
2x
2e
d
f(x) is an eigenfunction of the operator
with the eigenvalue 2.
dx
10
An equation to get the energy of a particle (Schrödinger equation, 1920) (Nobel
prize-1933)
h
Associated wavelength of a particle according to the deBroglie relation,  
p
mv 2
p2

2
2m
Combining the above two equations, the following relation is obtained.
h2
 2 4 2  2 k 2
h
2
where  
and k 
kinetic energy  2 

2
2

2m
2 m 2m 
To combine the particle nature and wave nature of matter we start with the wave
 2

equation,   A sin 
( x  vt)  .
 

 2 x 
At t = 0 (stationary or time independent) the wave equation becomes,   A sin 

  
2
or  A sin( kx) . Here k 
is called the wave vector.

Can we obtain the kinetic energy of the particle associated with this matter wave? We
h2
need to extract kinetic energy  2 from the wave equation by operating on it with an
2 m
d
operator. Try the operator
,
dx
d
 2 x  2
 2 x 
A sin 
A cos


dx
   
  
operate on the result with the same operator one more time.
2
d 2
 2 x   4
 2x 
A cos
A sin 


2
dx 

  
  
This double operation can be written as,
2
d d
 2 x   4
 2x 
A sin 
A sin 


2
dx dx

  
  
or
2
d2
 2 x   4
 2x 
A sin 
A sin 

 ---(1)
2
2
dx

  
  
 2 x 
We have gotten the same function back after the operation. Therefore, A sin 
 is an
  
d2
 4 2
eigenfunction of the operator,
with
the
eigenvalue
. We are partially
dx 2
2
1
successful, we have obtained a result containing 2 , which is present in the expression

kinetic energy of a particle having mass m and speed v, =
11
for the kinetic energy. What if we multiply the equation (1) by
 2
. Then we get,
2m
2
2
 2 d 2
 2 x   4
 2x 
A
sin

A sin 


.
2
2
2m dx
  
   2m
 2 d 2
  K .E. 
2m dx 2
In effect we have obtained the same function multiplied by kinetic energy of the particle.
 2 d 2
we can recognize that
is the operator for kinetic energy. We modify this
2m dx 2
equation by including the potential energy operator V as follows.
  2 d 2


  E 

V
2
2
m
dx


Action of kinetic energy plus potential energy operator on the wavefunction yields total
energy and the same function again. The operator for total energy is called the
Hamiltonian operator (H).
H  E
This equation is known as the Schrödinger equation.
Postulates of quantum mechanics
(postulate = assume as necessary condition, basis for reasoning)
1. The state of a system is described by a function , of the coordinates and time. This
function called the state function (wave function) contains all the information that can be
determined about the system.
2. To every physical observable there corresponds a linear Hermitian operator.
3. The only possible values that can result from measurements of the physical observable G
are the eigenvalues gi of the equation, G i = gi i . Where G is the operator corresponding
to the physical observable G.
4. If G is any linear Hermitian operator that represents a physical observable, then
eigenfunctions i of the eigenvalue equation, G i = gi i form a complete set.
12
5. If (q,t) is the wave function of a system at time t, then the average value of a physical
 G d
observable G at time t is  G  
  d
*
*
6. The time development of the state of an undisturbed system is given by the Schrödinger
time-dependent equation, i

 H
t
Born interpretation of the wavefunction
1.  contains all the information we know about the particle.
2. * is proportional to the probability of finding the particle at each point in space
(probability density).
* dx is proportional to the probability of finding the particle between x and x + dx.
3.  must be single valued,  must be continuous, first and second derivatives of  must
be finite, * must be finite (quadratically integrable).
Under these conditions, solutions for the Schrodinger equation exist only for certain
energies. This is the quantization of energy.
Normalization of the wavefunction
b
 * dx  1
This is the probability of finding the particle between a and b.
a
Orthogonality of the wavefunction
b

i
* j dx  0
This condition is satisfied if i and j are two different wave
a
functions where the overlap between the two functions is zero.
Heisenberg’s uncertainty principle
It is impossible to specify simultaneously, with arbitrary precision, both momentum and
the position of a particle.
The above statement can be presented in the form of a mathematical expression as
follows.

x p x 
2
x = error in position, px = error in momentum.
Application of quantum mechanics to simple systems
Particle in a 1-D box.
13
The diagram above shows a one dimensional (1-D) box. The potential energy of a
particle within the region x = 0 through x = l is zero. To put the particle outside of this
region, an infinite amount of potential energy must be supplied to it (which is
impossible). Therefore, the particle will be found only inside the box. Now set up the
Schrӧdinger equation for a particle in the box.
ℏ2 𝑑 2 𝜓
−
= 𝐸𝜓
2𝑚 𝑑𝑥 2
General solution to this equation could be written as,
𝜓 = 𝐴 𝑠𝑖𝑛(𝑘𝑥) + 𝐵 𝑐𝑜𝑠(𝑘𝑥) where A, B and k are to be determined.
Boundary conditions and normalization condition could be used to evaluate these
constants.
Boundary conditions: The wave function must go to zero at x = 0 and x = l
Applying the boundary condition x = 0,
𝜓 = 𝐴 𝑠𝑖𝑛(0) + 𝐵 𝑐𝑜𝑠(0)
Since sin(0) = 0 and con (0) = l,
𝜓 = 𝐵, but the wave function must be zero when x = 0. Therefore, B = 0
Then the wave function is,
𝜓 = 𝐴 𝑠𝑖𝑛(𝑘𝑥)
Applying the boundary condition, x = l,
𝜓 = 𝐴 𝑠𝑖𝑛(𝑘𝑙)
But 𝜓 = 0 at x = l.
The function sin(θ) goes to zero when θ = nπ, where n is an integer
n = 0 is not acceptable as the wave function become zero at any x.
Therefore, 𝑘 =
𝑛𝜋
𝑙
and n = 1,2,3,4……..
𝑛𝜋𝑥
)
𝑙
To determine the constant A, the wave function is normalized.
𝜓 = 𝐴 𝑠𝑖𝑛 (
Using the condition that the probability of finding the particle within the box = 1.
𝑙
∫0 𝜓 ∗ 𝜓 𝑑𝑥 = 1
14
𝑙
∫ 𝐴 sin(
0
𝑛𝜋𝑥
𝑛𝜋𝑥
) 𝐴 sin(
) 𝑑𝑥 = 1
𝑙
𝑙
𝑙
2
𝐴 ∫ (sin(
0
𝑛𝜋𝑥 2
)) 𝑑𝑥 = 1
𝑙
cos(2θ) = 1-2 sin2(θ)
Using
𝑙
1
2𝑛𝜋𝑥
(1 − 𝑐𝑜𝑠 (
) ) 𝑑𝑥 = 1
2
𝑙
𝐴2 ∫
0
𝑙
𝐴2 [∫
0
𝑙
1
2𝑛𝜋𝑥
𝑑𝑥 − ∫ 𝑐𝑜𝑠 (
) 𝑑𝑥 ] = 1
2
𝑙
0
𝑙
1
𝑙
2𝑛𝜋𝑥
𝐴 [ [𝑥]𝑙0 −
[𝑠𝑖𝑛 [
]] ] = 1
2
2𝑛𝜋
𝑙
0
2
1
𝐴2 [2 (𝑙 − 0) −
𝑙
2𝑛𝜋
[𝑠𝑖𝑛 [
2𝑛𝜋𝑙
𝑙
] − sin(0)] ] = 1
Since sin (2nπ) = 0,
𝐴2
𝑙
=1
2
2
𝐴=√
𝑙
Therefore, the wave function is,
2
𝑛𝜋𝑥
𝜓 = √ 𝑙 𝑠𝑖𝑛 (
𝑙
) and n = 1, 2, 3 …
15
𝜓 ∗ 𝜓 = probability of finding the particle at any location (probability density)
𝜓 ∗ = complex conjugate of the wave function
Complex conjugate is btained by replacing i in the wave function by -i where 𝑖 = √−1 )
If the wave function does not contain i then 𝜓 ∗ = 𝜓
Different values of n correspond to different energy states.
n=1 lowest energy state (ground state)
n=2 first excited state
n=3 second excited state etc.
Energy of the particle can be found by using the wave function in the Schrӧdinger
equation.
ℏ2 𝑑 2
2
𝑛𝜋𝑥
2
𝑛𝜋𝑥
−
(√ 𝑠𝑖𝑛 (
)) = 𝐸 √ 𝑠𝑖𝑛 (
)
2
2𝑚 𝑑𝑥
𝑙
𝑙
𝑙
𝑙
2
On the left-hand side of the equation, the second derivative of √ 𝑙 𝑠𝑖𝑛 (
𝑛𝜋𝑥
𝑙
) must be
evaluated.
ℏ2 (− 𝑛2 𝜋 2 )
𝑛𝜋𝑥
𝑛𝜋𝑥
−
𝑠𝑖𝑛
(
)
=
𝐸
𝑠𝑖𝑛
(
)
√2/𝑙
√2/𝑙
2𝑚
𝑙2
𝑙
𝑙
Now cancelling the similar terms on the left and right sides of the equation,
𝐸=
ℏ2 𝑛 2 𝜋 2 𝑛 2 ℎ 2
=
2𝑚 𝑙 2
8𝑚𝑙 2
The particle has its lowest energy when n = 1 (the ground state)
Expectation values.
When the wavefunction in not an eigen function of the operator that represents a physical
observable, the expectation value of the physical observable (average value of the physical
observable) can be calculated.
Wave function of a particle in a one-dimensional box, is not an eigen function of the
position operator. Also, the wavefunction is not an eigen function of the momentum
operator Therefore, the expectation values for position and momentum are calculated.
16
Expectation value of position.
𝑙
∫0 𝜓 ∗ 𝑥 𝜓 𝑑𝑥
<𝑥 >= 𝑙
∫0 𝜓 ∗ 𝜓 𝑑𝑥
<𝑥 >=
𝑙 2
𝑛𝜋𝑥
2
𝑛𝜋𝑥
∫0 √ 𝑙 sin( 𝑙 ) 𝑥 √ 𝑙 sin( 𝑙 ) 𝑑𝑥
𝑙 2
𝑛𝜋𝑥 2
𝑛𝜋𝑥
∫0 √ 𝑙 sin( 𝑙 ) √ 𝑙 sin( 𝑙 ) 𝑑𝑥
𝑙
2
𝑛𝜋𝑥
𝑙
𝑙
< 𝑥 > = ∫0 √ sin(
2
𝑛𝜋𝑥
𝑙
𝑙
) 𝑥 √ sin(
) 𝑑𝑥
𝑙
2
Since the wavefunction is normalized, ∫0 √ 𝑙 sin(
𝑛𝜋𝑥
𝑙
2
) √ 𝑙 sin(
𝑛𝜋𝑥
𝑙
) 𝑑𝑥 =1
2 𝑙
𝑛𝜋𝑥
< 𝑥 > = ∫ 𝑥 𝑠𝑖𝑛2 (
) 𝑑𝑥
𝑙 0
𝑙
2 𝑙 1
2𝑛𝜋𝑥
< 𝑥 > = ∫ 𝑥 (1 − cos (
) 𝑑𝑥
𝑙 0 2
𝑙
1 𝑙
1 𝑙
2𝑛𝜋𝑥
< 𝑥 > = ∫ 𝑥 𝑑𝑥 − ∫ 𝑥 cos (
) 𝑑𝑥
𝑙 0
𝑙 0
𝑙
1 𝑥2
< 𝑥 > = [ ]𝑙0
𝑙 2
1 𝑙 𝑙
2𝑛𝜋𝑥
− ∫
𝑥 𝑑sin(
)
𝑙 0 2𝑛𝜋
𝑙
𝑙
1 𝑙2 0
1
2𝑛𝜋𝑥 𝑙
1
2𝑛𝜋𝑥
<𝑥 >= [ − ] −
[𝑥 sin (
)] +
∫ sin (
) 𝑑𝑥
𝑙 2 2
2𝑛𝜋
𝑙
2𝑛𝜋 0
𝑙
0
<𝑥 >=
𝑙
1
2𝑛𝜋𝑙
1 𝑙
2𝑛𝜋𝑥 𝑙
−
[𝑙 sin (
) − 0] +
[− cos(
)]0
2
2𝑛𝜋
𝑙
2𝑛𝜋 2𝑛𝜋
𝑙
<𝑥 >=
𝑙
1
1 𝑙
2𝑛𝜋𝑙
[𝑙 sin(𝑛𝜋) − 0] −
−
[cos (
) − cos(0)]
2
4𝑛𝜋
2𝑛𝜋 2𝑛𝜋
𝑙
<𝑥 >=
𝑙
1
1 𝑙
[0 − 0] −
−
[1 − 1]
2
4𝑛𝜋
2𝑛𝜋 2𝑛𝜋
<𝑥 >=
𝑙
2
17
Expectation value of momentum.
𝑙
∫0 𝜓 ∗ 𝑝𝑥 𝜓 𝑑𝑥
< 𝑝𝑥 > =
𝑙
∫0 𝜓 ∗ 𝜓 𝑑𝑥
< 𝑝𝑥 > =
𝑙 2
𝑛𝜋𝑥
𝑑
2
𝑛𝜋𝑥
∫0 √ 𝑙 sin( 𝑙 ) (−𝑖ℏ 𝑑𝑥 ) √ 𝑙 sin( 𝑙 ) 𝑑𝑥
𝑙 2
𝑛𝜋𝑥 2
𝑛𝜋𝑥
∫0 √ 𝑙 sin( 𝑙 ) √ 𝑙 sin( 𝑙 ) 𝑑𝑥
𝑙
2
Since the wavefunction is normalized, ∫0 √ 𝑙 sin(
𝑛𝜋𝑥
𝑙
2
) √ 𝑙 sin(
𝑛𝜋𝑥
𝑙
) 𝑑𝑥 = 1
𝑙
2
𝑛𝜋𝑥
𝑑
2
𝑛𝜋𝑥
< 𝑝𝑥 > = ∫ √ sin(
) (−𝑖ℏ ) √ sin(
) 𝑑𝑥
𝑙
𝑙
𝑑𝑥
𝑙
𝑙
0
𝑙
2
𝑛𝜋𝑥
𝑛𝜋
2
𝑛𝜋𝑥
< 𝑝𝑥 > = ∫ √ sin(
) (−𝑖ℏ ) √ cos (
) 𝑑𝑥
𝑙
𝑙
𝑙
𝑙
𝑙
0
< 𝑝𝑥 > =
−𝑖ℏ𝑛𝜋 𝑙
𝑛𝜋𝑥
𝑛𝜋𝑥
∫ 2 sin(
) cos (
) 𝑑𝑥
𝑙
𝑙
𝑙
0
−𝑖ℏ𝑛𝜋 𝑙
2𝑛𝜋𝑥
< 𝑝𝑥 > =
∫ sin(
) 𝑑𝑥
𝑙
𝑙
0
−𝑖ℏ𝑛𝜋 𝑙
2𝑛𝜋𝑥 𝑙
[𝑐𝑜𝑠 (
)]
𝑙 2𝑛𝜋
𝑙
0
−𝑖ℏ𝑛𝜋 𝑙
2𝑛𝜋𝑙
2𝑛𝜋 (0)
< 𝑝𝑥 > =
[𝑐𝑜𝑠 (
) − 𝑐𝑜𝑠 (
)]
𝑙 2𝑛𝜋
𝑙
𝑙
−𝑖ℏ𝑛𝜋 𝑙
[𝑐𝑜𝑠(2𝑛𝜋) − 𝑐𝑜𝑠(0)]
< 𝑝𝑥 > =
𝑙 2𝑛𝜋
< 𝑝𝑥 > =
𝑐𝑜𝑠(2𝑛𝜋) = 1 𝑎𝑛𝑑 𝑐𝑜𝑠(0) = 1
Therefore, < 𝑝𝑥 > = 0
Quantum Chemistry
1. Which of the following functions are eigen functions of d2/dx2.
(a). ex (b). x2 (c). sin x (d). 3 cos x (e). sin x + cos x , give the eigen values.
18
2. For a particle in a 1-D box, find the following values for the state with quantum number
n.
(a). <x> (b). <x2> (c). <px> (d). <p2x>
Show that Heisenberg's uncertainty principle is valid for any value of n, (i.e. show that x

px  ).
2
Where x=|<x2> - <x>2|1/2 and px=|<px2> - <px>2|1/2.
3. If Y is an un normalized wave function and N is a constant such that NY is normalized,
express |N| in terms of Y.
4. Benzene has its carbon atoms arranged is a regular hexagon with side length 1.39A0. It
has six pi electrons. Adopt the solutions from particle in a 2-D box problem assuming that
benzene molecule can be approximated to a square with the side length 3.92A0. Calculate
the wave length of the radiation for the excitation of an electron from the highest occupied
level to the lowest unoccupied level.
5. Construct the Schrödinger equation for a particle (mass = m) under a potential v = kx2/2.
Show that y = A exp(-bx2/2) is a solution to the Schrödinger equation when k 
 2b 2
.
m
What is the eigen value for the energy of the system?
(hint: construct the Schrödinger equation and then show that the given wave function is a
solution by application).
6. A particle is having mass m is in a state described by the wave function  = A eikx + B
e-ikx if the particle is in a constant potential field of magnitude V show that the total energy of
the particle is given by
k 2 2
V .
2m
19
If the particle is having a linear momentum k , show that the constant B must be zero.
Derive an expression for * (probability density of finding the particle) when B=0 and
sketch as a function of x in the interval x=0 through x=L.
7. Identify the functions that are acceptable as wave functions in the limit x = 0 through x =
. An acceptable wave function is a function which is finite, continuous, single valued,
quadratically integrable and its first derivative continuous, and second derivatives finite. (a)
Y = sinx, (b) Y = x2, (c) Y = ax2+b, (d) Y = exp(-x2), (e) Y = exp(-ax). Where a and b are
constants.
8. (a) State whether the following statements are true or false. If false explain why.
(i) If properly designed, the position and momentum of a particle can be experimentally
determined without any uncertainty.
(ii) Classical mechanics is always better because it can be used to predict things like
position, momentum energy etc. of a particle simultaneously without uncertainty.
(iii) Quantum mechanics can be used to predict the x-position and y-momentum of a
particle simultaneously with no uncertainty.
(iv) Quantum mechanics incorporates particle and wave nature of matter. Wave nature
begins to predominate as the mass of a particle become small.
(b) Write down expressions for expectation value of the x-position and y-momentum of a
particle constrained (restricted) to a 2-D box with a side length L. Potential energy
outside the box is infinity.
(c) Write down the Schrodinger equation for a particle (mass = m) confined to the 2-D
box (side lengths L1 and L2).
 n x 
 n y 
2
Show that  
sin  1  sin  2  is a solution to the Schrodinger equation
L1 L2
 L1 
 L2 
(by substitution). Give the expression for energy.
(d) A planer molecule containing four conjugated electrons can be approximated by a 2D box. Derive an expression to calculate the wavelength of radiation absorbed by the
molecule.
9. A particle restricted to move in x-direction is described by the wave function

  B cos kx where B is the normalisation constant and k  .
L
L
L
(a) Normalise the function in the range x 
through x  and find B.
2
2
L
L
(b) Sketch the wave function in the range x 
through x  .
2
2
20
Some useful integrations
𝑚
𝑥𝑚
1
[𝑠𝑖𝑛(2 𝑎𝑥)]𝑚
∫ 𝑐𝑜𝑠 2 (𝑎 𝑥) 𝑑𝑥 = [ ] +
𝑛
2
4𝑎
𝑛
𝑛
∫ 𝑥 𝑠𝑖𝑛 (𝑎𝑥) 𝑑𝑥 =
∫ 𝑥 𝑐𝑜𝑠 (𝑎𝑥) 𝑑𝑥 =
1
𝑥
𝑠𝑖𝑛( 𝑎𝑥) − cos (ax)
2
𝑎
𝑎
1
𝑥
𝑐𝑜𝑠 (𝑎𝑥) + 𝑠𝑖𝑛 (𝑎𝑥)
2
𝑎
𝑎
∫ 𝑥 2 𝑠𝑖𝑛 (𝑎𝑥) 𝑑𝑥 =
2𝑥
𝑎2 𝑥 2 − 2
𝑠𝑖𝑛
(𝑎𝑥)
−
𝑐𝑜𝑠 (𝑎𝑥)
𝑎2
𝑎3
∫ 𝑥 2 𝑐𝑜𝑠 (𝑎𝑥) 𝑑𝑥 =
2𝑥
𝑎2 𝑥 2 − 2
𝑐𝑜𝑠
(
𝑎𝑥)
+
𝑠𝑖𝑛 (𝑎𝑥)
𝑎2
𝑎3
∫ 𝑥 𝑠𝑖𝑛
2
𝑥 2 𝑥 𝑠𝑖𝑛 (2𝑎𝑥) 𝑐𝑜𝑠 (2𝑎𝑥)
(𝑎𝑥) 𝑑𝑥 =
−
−
4
4𝑎
8𝑎2
∫ 𝑥 2 𝑠𝑖𝑛2 (𝑎𝑥) 𝑑𝑥 =
𝑥3
𝑥2
1
𝑥 𝑐𝑜𝑠 (2𝑎𝑥)
− ( − 3 ) 𝑠𝑖𝑛 (2𝑎𝑥) −
6
4𝑎 8𝑎
4𝑎2
Intended learning outcomes.
After following this course, you should be able to do the following.
Describe the experimental observations led to the formulation of quantum mechanics.
Construction of operators to calculate physical states/properties of non-interacting
particles.
Apply operators to functions.
Calculate expectation values of physical observables.
Calculate probability from wave function.
Apply and solve Schrodinger equation to noninteracting particles restricted to 1-D, 2-D
and 3-D boxes.
Use boundary conditions and normalization condition to obtain wave functions.
21