chapter (6) introduction to quantum mechanics. is a single valued function, continuous, and finite...
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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
00 xxx 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
The wave function for a free particle can be represented by a plane wave
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
(2)
From m
k
m
k
m
pk
222)(
2222
(1)
Example Find the wavefunction )x, 0( that results from taking the function
,)(22ke
cka
Solution
Write
, where C and are constants
or
For a free particle U)x(=0 , ,
Schrödinger equation :
Em
k
Eeeikikm
ikxikx
2
))((222
2
Which is the total energy
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(=0Schrödinger equation is
222
2
2
2
2
22
2),()(
2
)(
)()(
2
mEkxkx
m
E
dx
xdor
xEdx
xd
m
0L
U
(1)
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,….
xL
nAxn
sin)(
Using Relation )1( becomes
Which shows that the particle energies are quantized and gives the energy levels
14
12
22
3
12
22
2
2
22
1
164
92
93
42
2
21
EEn
EmL
En
EmL
En
mLEn
Ground state or zero point energy
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
xL
Ax sin)(1
xL
Ax 2
sin)(2
xL
Ax 3
sin)(3
Probability-:
imumAAL
xat
imumAAL
xat
LxatfoundnotisparticeltheeiA
Lxat
xL
Adxn
max2
3sin
4
3
max2
sin4
2.00sin
2
2sin,2
22
22
2
22
22
2
222
2
2
22
2
Normalization:-
xL
n
Lx
LA
LA
LALL
n
n
LLAx
L
n
n
LxA
n
2sin
2)(
22
2
1]
2sin
2[
2
1]
2sin
2[
2
11
2
2210
2
The Quantum Oscillator Consider the problem of a particle subject to a linear restoring forceF = - 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, m
k
The total energy,
The quantum oscillator is described by the potential energy
222
2
1
2
1)( xmkxxU In the Schrödinger equation
)2
1(
2
)()(2
1
2
)()()(2
2222
2
222
22
2
22
Exmm
dx
d
xExxmdx
d
m
xExxUdx
d
m
The solution may be in the form
Where C0 and are constants, using this solution, we get
2
22
2
]24[
)2()2()(
)2)((
220
022
02
2
0
x
xx
x
exc
ecexcdx
d
excdx
d
• Comparing with (1), we get
2
22
24
2
1.
24
2
22
2
2
222
222
22
m
m
mEE
m
mm
xmm
x
This is the ground state energy
2
10 E
With Eigen function:
]2
exp[)(22
00 xm
cx
Normalization of the ground state wavefunction
Exited state
2
32
1
3
2
2
626
24
1
22
2
222
EE
Emm
Em
mm
22
222
22
2
]64[]64[
24)2(
)2(
)(
2222
22
2
xx
xxx
xx
x
exxexx
exexexxdx
d
eexxdx
d
xex
Comparing with (1)
This is the first excited state Eigen energy So the energy levels for the harmonic oscillator is . )
2
1( nEn
The separation between any two levels is
n=0,1,2,…
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 position 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
x
EXAMPLE: Location of a Particle in a BoxCompute the average position x and the quantum uncertaintyin this value, x, for the particle in a box, assuming it is in the
ground state.
SolutionThe ground state wavefunction is
with n =1 for the ground state. The average position is calculatedas