Transcript Chapter (6) Introduction to Quantum Mechanics  is a single valued function , continuous, and finite every where Example(6.1) Normalizing the Wavefunction • The initial wavefunction.

Chapter (6)
Introduction
to
Quantum Mechanics
 is a single valued function , continuous, and finite every where
Example(6.1) Normalizing the Wavefunction
• The initial wavefunction of a particle is given as (x,o)= C exp(x /xo), where C
and xo are constants.
1- Sketch  function.
2-Find C in terms of xo such that (x, 0) is normalized.
• Solution The given wavefunction is symmetric, decaying
exponentially from the origin in either direction, as shown in
Figure ,The decay length xo represents the distance over which
the wave amplitude is diminished by the factor 1/e from its
maximum value  (0, 0) = C.
• The normalization requirement is
Because the integrand is unchanged when x •
changes sign we may evaluate the integral
over the whole axis as twice that over the
half-axis x > 0, where x = x. Then
EXAMPLE 6.2
Calculate the probability that the particle in the
preceding example will be found in the interval
 x0  x  x0
solution
or about 86.5%, independent of xo.
Wavefunction For A Free Particle
• A free particle is one subject to no force
The wave number k and frequency  of free particle matter waves are
given by the de Broglie relations
For non relativistic particle  is related to k as
From
p2
k 2 2 k 2
  (k ) 


2m 2m 2m
(1)
The wave function for a free particle can be
represented by a plane wave
(2)
This is an oscillation with wavenumber k, frequency  , and amplitude A
(travelling wave). If each plane wave constituting the packet is assumed to
propagate independently of the others according to equation (2), the
packet at any time is
Example
Find the wavefunction (x, 0) that results from taking the function
a(k ) 
Solution
Write
c

e
 2 k 2
, , where C and  are constants
or
For a free particle U(x)=0, ,
Schrödinger equation :
2

(ik )(ik )eikx  Eeikx
2m
 2k 2
Which is the total energy
E
2m
The Particle In A Box ( Infinite Square Well )
The particle can never be found outside, i.e.(x)=0 outside.
Inside the box U(x)=0
Schrödinger equation is
 2 d 2 ( x)

 E ( x)
2
2m dx
d 2 ( x)
2E
2m E
2
2
or



(
x
)


k

(
x
),
k

dx2
2m
2

U
(1)
0
k is the wavenumber of oscillation, the solution to this equation is .
Boundary condition: from the boundary conditions one obtains A and B
n=1,2,….

L
n
 n ( x)  A sin
x
L
Using
Relation (1) becomes
Which shows that the particle energies are
quantized and gives the energy levels
n  1  E1 
 2 2
Ground state or zero point energy
2m L2
2 2  2
n  2  E2 
 4 E1
2
mL
9 2  2
n  3  E3 
 9 E1
2
2m L
n  4  E4  16E1
Exited state
Notice that E=0 is not allowed, that is the particle can never be at rest.
E1 is zero-point energy > 0
Wave function:Return to

 1 ( x)  A sin
x
L
2
 2 ( x)  A sin
x
L
3
x
L
Probability:-
 3 ( x)  A sin
n  2,  2
2
dx  A2 sin
L
 2
2
2
L
at x    2
4
2
at x 
3L
at x 
 2
4
2
x
L
2
2
 A2 sin  0  0 i.e theparticel is not found at x 
 A2 sin
2

2
2
3
 A2 sin
2
 A
2
maxim um
2
 A
2
maxim um
L
2
Normalization:-
1 2
L
2n 1 1 2
L
2n
1 2
1  A [x 
sin
x]0  A [ L 
sin
L]  A L
2
2n
L
2
2n
L
2
2
2
2
A   A
L
L
 n ( x) 
2
2n
sin
x
L
L
The Quantum Oscillator
Consider the problem of a particle subject to a linear restoring force
F = - kx. Here x is the displacement of the particle from equilibrium (x
=0) and k is the spring constant.
*The potential energy is
The angular frequency ,

k
m
The total energy ,
The quantum oscillator is described by the potential energy
1 2 1
U ( x)  kx  m 2 x 2
2
2
In the Schrödinger equation
 2 d 2

 U ( x) ( x)  E ( x)
2
2m dx
 2 d 2 1
2 2


m

x  ( x)  E ( x)
2
2m dx
2
d 2 2m 1
2 2

(
m

x  E )
2
2
dx
 2
The solution may be in the form
Where C0 and  are constants, using this solution, we get
d
x 2
 c0 ( )(2 x)e
dx
d 2
2
2 x 2
x 2

c
(


)
(
2
x
)
e

c
(
2

)
e
0
0
dx2
 c0 [4 2 x 2  2 ]e x
2
• Comparing with (1), we get
2m 1
4 x  2 . m 2 x 2
 2
2 2
m

m
4 2 



2
2
2m
  2 m  2  
2  2 E  E 


2

m
m
2
2
2
This is the ground state energy
With Eigen function:
1
E 0  
2
m 2 x 2
 0 ( x)  c0 exp[
]
2
Normalization of the ground state wavefunction
Exited state
 ( x)  xe
d
 x 2
 x(2 x)e
 x 2
e
 x 2
dx
d 2
2  x 2
 x 2
 x 2
 x(2 x) e
 4 x e
 2 x e
2
dx
 [4 x  6 x]e
2
2
 x 2
 [4 x  6 ] x e
2
2
 x 2
Comparing with (1)
m 2 2
m
4 



2
2
2m
6m 2m
6  2 E 
 2 E

2

3 2
3
This is the first excited state Eigen energy So

 E  E1  
the energy levels for the harmonic
1 
2
oscillator is
1
.
n=0,1,2,…
E n  ( n  ) 
2
2
The separation between any two levels is

E0 is the ground state energy level
Expectation values
•A particle described by the wavefunction  may occupy various places
x with probability given by 2
•We have several values of x, we need the average value of x
•The average value of x, written x , is called the expectation value and
defined as
Which gives the average positionx of a particle. The average value
of any function f(x) is
In quantum mechanics the standered derivation, x, is called the uncertainty
in position, and given by
The degree to which particle position is fuzzy is given by the magnitude
of x. the position is sharp only if x=0
EXAMPLE: Location of a Particle in a Box
Compute the average position x and the quantum uncertainty
in this value,  x, for the particle in a box, assuming it is in the
ground state.
Solution
The ground state wavefunction is
with n =1 for the ground state. The average position is calculated
as