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Quantum Mechanics
Lecture-8
Concept of Modern Physics
by “A. Beiser”
Page 1
Exercise: A Particle limited to the x-axis has the wave
function  =ax between x=0 and x=1; =0 elsewhere.
(a)Find the probability that the particle can be found x=0.45
and x=0.55.
(b)Find the expectation value < x > of the particle’s position.
(a) 0.0251 a2
(b) a2/4
Page 2
Particle in a one dimensional Box (infinite square well potential)
• A particle is confined to a one-dimensional
region of space between two impenetrable
walls separated by distance L
– This is a one- dimensional “box”
• The particle is bouncing elastically back
and forth between the walls
– As long as the particle is inside the box, the
potential energy does not depend on its
location. We can choose this energy value to be
zero
• V= 0, 0 < x < L, V  , x ≤ 0 and x ≥ L
Page 5
Particle in a one dimensional Box (infinite square well potential)
• Since the walls are impenetrable, there is zero
probability of finding the particle outside the box.
Zero probability means that
ψ(x) = 0, for x < 0 and x > L
• The wave function must also be 0 at the walls (x = 0
and x = L), since the wavefunction must be
continuous
– Mathematically, ψ(0) = 0 and ψ(L) = 0
2
 2  x 

 V  x   E  x 
2m x 2
• In the region 0 < x < L, where V = 0, the
Schrödinger equation can be expressed in the form
 2  x 

 E  x 
2m x 2
2
Page 6
Particle in a one dimensional Box (infinite square well potential)
• We can re-write it as
 2  x 
2mE
  2 x
2
x
 2  x 
2


k
x
2
x
k 
2
2mE
2
The most general solution to this differential equation is
ψ(x) = A sin kx + B cos kx
– A and B are constants determined by the properties of the
wavefunction as well as boundary and normalization conditions
Page 7
Particle in a one dimensional Box (infinite square well potential)
Since the particle cannot have infinite energy, it cannot exist outside the box.
Therefore, the wave function  must be zero outside the box.  must be also zero
at the walls, that is, at x = 0 and x = L, for otherwise there would be discontinuities
at the walls.
1.
Sin(x) and Cos(x) are finite and single-valued functions
2.
Continuity: ψ(0) = ψ(L) = 0
• ψ(0) = A sin(k0) + B cos(k0) = 0  B = 0 Not admissible
 ψ(x) = A sin(kx)
• ψ(L) = A sin(kL) = 0  sin(kL) = 0  kL = nπ, n =0, ±1, ±2…
( n = 0 is not admissible because it yields  zero everywhere which means that the
particle is no where).
kn 
2

L
n
 2 kn
En 
2m
2

 ( n) 2  2 2 
2

 2


h
2
L
n  
n
En 
 
2 
2 
2m
 2mL 
 8mL 
Page 8
Particle in a one dimensional Box (infinite square well potential)
The particle can not have arbitrary energy , but
can have certain discrete energy corresponding
to n=1,2,3,…
n2h2
En 
8mL2
Each permitted energy is called ‘eigen value’ of the
particle and constitute the energy level of the
system, and the integer ‘n’ that specifies an energy
level En is called its principal quantum number.
The wave function  corresponding to each eigen
value is called eigen functions.
Lowest level n = 1, energy not
zero why ?
Page 11
Conclusions
n 2 h 2 n 2 2 2
En 

2
8mL
2mL2
Energy of the particle bounded in a box is quantized.
Particle can not have zero energy but has minimum energy and called as
zero point energy. The state corresponding to this energy is called ground
state.
The particle can not have zero kinetic energy, because from Heisenberg
uncertainty principle, the uncertainty in position of the trapped particle in
a box is Δx = L, hence Δp as well as velocity of the particle and their kinetic
energy can not be zero.
According to classical mechanics, when a particle is placed in a box, it can
have zero energy or continuous kinetic energy. Thus the quantization of
energy is a specific result derived from quantum mechanics.
If one assumes v = 0 for a particle in a box, the de-Broglie wave associated
with it will be λ =(h/mv) = , which is an absurd result because there should
be node at the boundary for the bounded particle.
Page 12
An e- is in a box 0.10 nm across, which is the order of magnitude of
atomic dimensions. Find its permitted energies.
n 2h 2
En 
8mL2
n 2 (6.63 1034 J.s) 2
En 
 6.0  1018 n 2 J
31
10
2
8(9.1 10 kg)(1.0  10 m)
 38n 2eV
The minimum energy of the e- can have 38eV, corresponding to n=1.the sequence of
energy levels continues with E2=152 eV, E3=342 eV, and so on. If such box existed, the
quantization of a trapped e-’s energy would be a prominent feature of the system. ( and
indeed energy quatization is prominent in the case of an atomic election.)
A 10 g marble is in a box 10 cm across. Find its permitted energies.
n 2 (6.63  1034 J.s) 2
En 
 8.28  1065 n 2 J
2
1
2
8(1.0  10 kg)(1.0  10 m)
The minimum energy of the marble can have 8.2810-65 J, corresponding to n=1. A
marble with this K.E. has a speed of only 1.310-31 m/s and therefore can not be
experimentally distinguished from a stationary marble.
Hence in the domain of everyday experience, quantum effects are unnoticeable, which
accounts for the success of Newtonian Mechanics in this domain. (mL2>>h2)
Page 13
Particle in a one dimensional Box (infinite square well potential)
 2  x 
2mE
  2 x
2
x
 2  x 
2


k
x
2
x
 (x)  Asin(kx)  Bcos(kx)
 (0)  Bcos(0)  0
 (L)  Asin(kL)  0
k n L  n

kn  n
L
2mE n
 n 
  
2
 L 
2
kn2
n 2h 2
En 
8mL2
n
 (x)  Asin( x)
L
Page 14
Particle in a one dimensional Box (infinite square well potential)
Let us now consider the eigenfunctions of the particle. Substituting B=0 and k=n/L
in the equation of general solution.
Solutions of the Schrödinger equation are
E4
E4 = L
nx
2
 n (x)  Asin
L
To find the value of A, we use the normalization

condition
2
|
n
E3
E3 =2L
3
E2
E2 =L
E1
E1 =2L
(x) | dx  1

L
A 2  sin 2
0
nx
dx  1
L
A2  L / 2  1
A  2/L
The normalized eignfunctions of the particle
 n (x) 
2
nx
sin
L
L
2L
n=1,2,3,...
n
Initial 2wavefunctions
 and Energy
2
n 
n 2h 2
En 
for the first four states in
8mL2
a one-dimensional particle in a box
Page 15
Particle in a one dimensional Box (infinite square well potential)
Wavefunctions
2
nx
 n (x) 
sin
L
L
n=1,2,3,...
Probability: *n  n
n=1,2,3,...
Note particle most likely
to be found in the middle
for n=1
Energy:
(a) The first two wavefunctions, (b) the corresponding
probability distributions, and (c) a representation of the
probability distribution in terms of the darkness of
shading.
n 2h 2
E=
8mL2
n=1,2,3,...
Page 16
No. of antinodes in eigenfunction = n, No. of nodes =No. of antinodes+1
Page 17
Electron in the 10nm Wide Well with Infinite Barriers
2 2

2
En  n E1 , where E1 
2mL2
• Calculate E1=?
3.142 (1.05 1034 )2
E1 
2  9.11031  (10 109 )2
E1  6 1022 J  0.00375 eV  3.75 meV
• Assume that a photon is absorbed, and
the electron is transferred from the
ground state (n = 1) to the second
excited state (n = 3)
What was the wavelengths of the photon?
Page 18
Electron in the 10nm Wide Well with Infinite Barriers
• E1 = 3.75 meV
Eground  E1  0.00375 eV
Third excited state is E3
E3  3  E1  9  0.00375 eV  0.0338 eV
2
(h )  E3  E1  0.0338  0.00375  0.030 eV
34
c
6.63  10  3  10
h  0.030eV,  

0.030  1.6  1019
1240
λ
 41333 nm  41 μm
0.03
8
Page 19
Particle in the Infinite Potential Well
For the n th state
2
x 
n 
sin 
n
L
 L 
2 2 2
En 
n
2
2mL
En  n 2E1
Probability to Find particle in the Right Half of the Well
L
L
2
2
|

(x)
|
dx

[
sin
kx]
dx
L/2
L/2 L
2
L
2
2 L1 1
  sin 2 (kx)dx  [
]
L L/2
L 22 2
Page 20
Some trajectories of a particle in a box according to
Newton's laws of classical mechanics (A), and according
to the Schrödinger equation of quantum mechanics (BD). In (B-D), the horizontal axis is position, and the
vertical axis is the real part (blue) or imaginary part (red)
of the wave function. The states (B,C,D) are energy
eigenstates.
Page 21
Page 22
Problems: Normalization.
1.
Determine normalization constant
i ) ( x)  N sin(nx / L), 0  x  L Ans : N 
2
L
ii) ( x)  N exp kx , 0  x   Ans : N  2k
 x2

iii) ( x)  N exp  2  ix  ,    x   Ans : N 
 2a



iv) ( x)  N cos x ,   x 
Ans : N 
2
2
2
1
a 
8
3


2
1
8
i
)
Ans
:
N

,
ii
)
Ans
:
N

2
k
iii
)
Ans
:
N

,
iv
)
An
:
N

L
3
a
23
Page 23
Hint for solution of (iii) problem


 * ( x ) ( x )dx  1


N2


 x2 
exp  2 dx  1
 a 

2aN 2

Say
x
 t , dx  adt if x  , t  
a
 
exp  t 2 dt  1

 
2aN 
 1
 2 
1
N
a 
2
Page 24
Problems: Particle in a box.
1. An particle is confined to an onedimensional infinity potential well of width
0.2 nm. It is found that when the energy of
the particle is 230 eV, its eigenfunction has 5
antinodes. Find the mass of the particle and
show that it can never have energy equal to
1 keV.
Answer
m =9.1 x10E-31 Kg
n = 10.4
Page 25
Problems: Particle in a box.
1.
Find the probability that a particle trapped in a box L wide can
be found between 0.45L and 0.55L for the ground and first
excited state.
2.
Write down Schrödinger equation for a one dimensional box,
obtain the expression for eigen function and eigen values. If
length of the box ix 25Å, calculate the probability of finding
the particle with in an interval of 5Å at the centre of the box
when it is in the state of least energy.
3.
A particle is in a cubic box with infinitely hard walls whose
edges are L long. The wave function of the particle is given by
by
8
 n x   n yy   n zz 
 ( x, y, z )  A sin  x  sin 
sin
Ans
:
A


 
3
 L 
 L 
 L 
L
Find the normalization constant A.
26
Page 26