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Chapter 8 Quantum Theory:
Techniques and Applications
Using Quantum Mechanics on Simple Systems
, Phys. Chem. 2nd Ed
T. Engel, P. Reid (Ch15)
Objectives
• Using the postulates to understand the
particle in the box (1-D, 2-D and 3-D)
Outline
1.
2.
3.
4.
The Free Particle
The Particle in a One-Dimensional Box
Two- and Three-Dimensional Boxes
Using the Postulates to Understand the
Particle in the Box and Vice Versa
8.1 The Free Particle
•
For a free particle in a one-dimensional space
on which no forces are acting, the Schrödinger
equation is
ℏ 2 d 2ψ ( x )
−
= − Eψ ( x )
2
2 m dx
ψ (x )
•
is a function that can be differentiated
twice to return to the same function
(
2 mE / ℏ )x
+ ikx
2π
2 mE
(
x
)
=
A
e
=
A
e
ψ
+
+
where k =
=
2
λ
ℏ
ψ − (x ) = A e −i (2 mE / ℏ )x = A e −ikx
+i
+
2
2
−
−
8.1 The Free Particle
•
If x is restricted to the interval − L ≤ x ≤ L then
the probability of finding the particle in an
interval of length dx can be calculated.
P(x )dx =
ψ * ( x )ψ ( x )dx
L
∫ψ * (x)ψ (x )dx
−L
=
A+ A+ e −ikx e + ikx dx
L
A+ A+ ∫ e −ikx e + ikx dx
−L
=
dx
2L
8.2 The Particle in a OneOne-Dimensional Box
•
When consider particle confined to a box in 1D, the potential is
V(x) =0,
(for 0<x <a)
V(x) =∞, (for x <0, or x ≥a)
8.2 The Particle in a OneOne-Dimensional Box
•
Consider the boundary condition satisfying 1-D,
ψ (0) = ψ (a ) = 0
•
The acceptable wave functions must have the
form of
 nπx 
ψ n = A sin 
, for n = 1, 2,3, 4...
 a 
•
Thus the normalized eigenfunctions are
ψ n (x ) =
2  nπx 
sin

a  a 
Physical Chemistry Fundamentals: Figure 8.3
Fig. 8.3 The first five
normalized
wavefunctions of a
particle in a box. Each
wavefunction is a
standing wave, and
successive functions
possess one more half
wave and a
correspondingly
shorter wavelength.
8.2 The Particle in a OneOne-Dimensional Box
•
Energies for the Particle in a Box
k 2ℏ 2
Ek =
2m
ka = nπ (n = 1,2,...)
n 2π 2 ℏ 2 n 2 h 2
En =
=
(n = 1,2,...)
2
2
2 ma
8ma
Zero - point energy of a particle in a box :
h2
E1 =
8ma 2
- the lowest, irremovable energy
Physical Chemistry Fundamentals: Figure 8.2
Fig. 8.2The allowed
energy levels for a particle
in a box. Note that the
energy levels increase as
n2, and that their separation
increases as the quantum
number increases.
Example 1
From the formula given for the energy levels for
the particle in the box, En = h 2 n 2 / 8ma 2 for n = 1, 2, 3,
4… , we can see that the spacing between adjacent
levels increases with n. This appears to indicate
that the energy spectrum does not become
continuous for large n, which must be the case for
the quantum mechanical result to be identical to
the classical result in the high-energy limit.
Example 1
A better way to look at the spacing between levels
is to form the ratio (En+1 − En ) / En . By forming this
ratio, we see that ∆E / E becomes a smaller
fraction of the energy as n → ∞ .
This shows that the energy spectrum becomes
continuous for large n.
Solution
We have,
En+1 − En
=
En
(
[
]
h 2 ( n +1) 2 − n 2 / 8 ma 2
h 2 n 2 / 8 ma 2
)
2n + 1
=
n2
which approaches zero as n → ∞ . Both the level spacing
and the energy increase with n, but the energy increases
faster (as n2), making the energy spectrum appear to be
continuous as n→∞
•The correspondence principle states that classical
mechanics emerges from quantum mechanics as high
quantum numbers are reached.
Physical Chemistry Fundamentals: Figure 8.4
The probability density for a
particle in a box with length a (or
L) is
2
a
 nπx 

 a 
ψ 2 ( x) = sin 2 
(8.8)
Fig. 8.4 (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.
Example 8.1 Using the particle in a box solutions (p.292)
• What is the probability, P, of locating the electron between x = 0 (the
left-hand end of a box) and x = 0.2 nm in its lowest energy state in a
cbox of length 1.0 nm?
• Method The value of Ψ2dx is the probability of finding the particle in the
small region dx located at x; therefore, the total probability of finding the
particle in the specified region is the integral of Ψ2dx over that region.
The wavefunction of the particle is given in eqn 8.4b with n = 1.
• Answer The probability of finding the particle in a region between x = 0
and x = l is
2l
nπ x
l
1
2π nl
P = ∫ψ dx = ∫ sin 2
dx = −
sin
a0
a
a 2nπ
a
0
l
2
n
ψ n ( x) =
2
 nπ x 
sin 

a
 a 
(8.4b)
• Set n = 1 and l = 0.2 nm, which gives P = 0.05. The result corresponds
to a chance of 1 in 20 of finding the particle in the region. As n becomes
infinite, the sine term, which is multiplied by 1/n, makes no contribution
to P and the classical result, P = l/a, is obtained.
• Test 8.4 Calculate the probability that a particle in the state with n = 1
will be found between x = 0.25a and x = 0.75a in a box of length a (with
x = 0 at the left-hand end of the box).
• Correct Answer: 0.82
A particle in a box – Key points
• (a) The energies of a particle constrained to move in
a finite region of space are quantized;
• (b) The energies and wavefunctions for a particle
moving in a box are labelled by quantum numbers.
The wavefunctions of a particle constrained to move
in a one-dimensional box are mutually orthogonal
sine functions with the same amplitude but different
wavelengths.
• The zero point energy is the lowest, irremovable
energy of a particle in a box.
• The correspondence principle states that classical
mechanics emerges from quantum mechanics as high
quantum numbers are reached.
Physical Chemistry Fundamentals: Figure 8.5
Fig. 8.5 A twodimensional square well.
The particle is confined
to the plane bounded by
impenetrable walls. As
soon as it touches the
walls, its potential
energy rises to infinity.
Motion in two dimensions
- Separation of variables
Physical Chemistry Fundamentals: Figure 8.6
Fig. 8.6 The wavefunctions for a particle confined to a rectangular
surface depicted as contours of equal amplitude.
(a) n1 = 1,n2 = 1, the state of lowest energy, (b) n1 = 1,n2 = 2,
(c) n1 = 2,n2 = 1, and (d) n1 = 2, n2 = 2.
Figure 8.7 Degeneracy L1 = L2 = L
Fig. 8.7 The wavefunctions
for a particle confined to a
square surface. Note that one
wavefunction can be
converted into the other by a
rotation of the box by 90°.
The two functions correspond
to the same energy.
Degeneracy and symmetry
are closely related.
Motion in two and more dimensions - Key points
• (a) The separation of variables technique
can be used to solve the Schrodinger equation
in multiple dimensions. The energies of a
particle constrained to move in two or three
dimensions are quantized.
• (b) Degeneracy occurs when different
wavefunctions correspond to the same
energy. Many of the states of a particle in a
square or cubic box are degenerate.
8.3 TwoTwo- and ThreeThree-Dimensional Boxes
•
•
1-D box is useful model system as it allows
focus to be on quantum mechanics instead of
mathematics.
For 3-D box, the potential energy is
V ( x, y , z ) = 0 for 0 < x < a, 0 < y < b, 0 < z < c;
V ( x, y , z ) = ∞ otherwise
•
Inside the box, the Schrödinger equation can
be written as
h2  ∂ 2
∂2
∂2 
 2 + 2 + 2 ψ (x , y , z ) = Eψ ( x, y , z )
−
2m  ∂ x ∂y
∂z 
8.3 TwoTwo- and ThreeThree-Dimensional Boxes
•
The total energy eigenfunctions have the form
nxπx nyπy nzπz
ψnxnynz ( x, y, z) = Nsin
sin
sin
a
b
c
•
And the total energy has the form
2
2
2

n
h  nx
nz 
y
E=  2 + 2 + 2 
8m  a b c 
2
8.4 Using the Postulates to Understand the Particle in the
Box and Vice Versa
Postulate 1
The state of a quantum mechanical system is
completely specified by a wave function ψ ( x , t ) .
The probability that a particle will be found at time
t in a spatial interval of width dx centered at x0 is
given by ψ * ( x0 , t )ψ (x0 , t )dx .
• This postulate states that all information
obtained about the system is contained in the wave
function.
Example 88-2
Consider the function ψ ( x ) = c sin (πx / a ) + d sin (2πx / a)
a. Is ψ (x ) an acceptable wave function for the
particle in the box?
b. Is ψ (x ) an eigenfunction of the total energy
operator?
c. Is ψ (x ) normalized?
Solution
a. If ψ (x ) is to be an acceptable wave function, it
must satisfy the boundary conditions ψ (x ) =0 at
x=0 and x=a. Its first and second derivatives must
ψ (x
)
also be
well-behaved
functions between x=0 and
x=a. This is the case for ψ (x ) . We conclude that
ψ (xψ)
(x) = c sin (πx / a ) + d sin (2πx / a )
is an acceptable wave function for the particle in
the box.
Solution
b. Although ψ ( x ) = c sin (πx / a ) + d sin (2πx / a )
may be an
acceptable wave function, it need not be an
eigenfunction of a given operator. To see if ψ (x ) is
ψ (x )
an eigenfunction
of the total energy operator, the
operator is applied to the function:
2 (x )
2 2
ℏ2 d ψ

2
π
x

ℏ
π 


 πx 
 2πx  
πx
−
 c sin ( a ) + d sin 
 c sin   + 4d sin 
  =
 
2 
2 
2 m dx 
 a   2ma 
a
 a 
The result of this operation is not ψ (x ) multiplied
by a constant. Therefore, ψ (x) is not an
eigenfunction of the total energy operator.
Solution
c. To see if ψ (x ) is normalized, the following
integral is evaluated:
2
 πxψ (x )  2πx 
c
sin
∫0  a  + d sin a  dx
a

x
 πx   2πx 
2 π
2  2πx 
ψ (x
)
= ∫ c * c sin
+
d
*
d
sin
+
(
cd
*
+
c
*
d
)
sin
 


  sin 
dx
 a
 a 
 a   a 
0 
a
2
2  πx 
2
 πx   2πx 
2  2πx 
= ∫ c sin  dx + ∫ d sin 
 dx + ∫ (cd * +c * d ) sin  sin 
dx
 a
 a 
 a  a 
0
0
0
a
a
a
Solution
Using the standard integral ∫ sin 2bydy = y / 2 − (1/ 4b ) sin by
and recognizing that the third integral is zero
2
 2 ψ2(x
2
 π)x 
2  2πx  
c
sin
+
d
sin
 

 dx
∫ 
a
 a 
0
2 a
a (sin 2π − sin 0 ) 
2a
a (sin 4π − sin 0 )  a 2
2
= c  ψ−(x )
+
d
−
=
c
+
d

 2
 2
4π
8π
2
a
(
)
Solution
Therefore, ψ (x ) is not normalized, but the function
ψ (x )
2
 πx 
 2π
c
sin
+
d
sin
 


a
a 
 a



is normalized for the condition that c 2 + d 2 = 1
ψ (x )
Note that a superposition wave function has a more
complicated dependence on time than does an
eigenfunction of the total energy operator.
Solution
For instance, ψ (x ) for the wave function under
consideration is given by
2  −iE / h  πx 
−iE
ψ
(x
)
ψ ( x, t ) =
ce
sin
+
ce


a 
a 
1
ψ (x )
2
/h
 2πx 
sin 
 ≠ ψ ( x ) f (t )
 a 
This wave function cannot be written as a product of a
function of x and a function of t. Therefore, it is not a
standing wave and does not describe a state
whose properties are, in general, independent of time.
8.4 Using the Postulates to Understand the Particle in the
Box and Vice Versa
• Acceptable Wave Functions for the Particle
in a Box
Example 88-3
What is the probability, P, of finding the particle in
the central third of the box if it is in its ground
state?
Solution
For the ground state, ψ 1 (x ) = 2 / a sin (πx / a ) . From
the postulate, P is the sum of all the probabilities of
finding the particle in intervals of width dx within
the central third of the box. This probability is
given by the integral
2
P=
a
 πx 
∫a / 3sin  a dx
2a/ 3
2
Solution
Solving this integral,
2 a a
P=  −
a  6 4π
2π
 4π
− sin
 sin
3
3


  = 0.609

Although we cannot predict the outcome of a single
measurement, we can predict that for 60.9% of a
large number of individual measurements, the
particle is found in the central third of the box.
8.4 Using the Postulates to Understand the Particle in the
Box and Vice Versa
Postulate 3
In any single measurement of the observable that
corresponds to the operator  , the only values
that will ever be measured are the eigenvalues of
that operator.
8.4 Using the Postulates to Understand the Particle in the
Box and Vice Versa
Postulate 4
If the system is in a state described by the wave
function ψ (x, t ) , and the value of the observable a is
measured once each on many identically prepared
systems, the average value of all of these
measurements is given by
∞
a =
ˆ ψ (x , t )dx
*
(
x
,
t
)
A
ψ
∫
−∞
∞
∫ψ * (x, t )ψ (x, t )dx
−∞
8.4 Using the Postulates to Understand the Particle in the
Box and Vice Versa
• Expectation Values for E, p, and x for a
Superposition Wave Function
Example 88-4
Assume that a particle is confined to a box of
length a, and that the system wave function is
ψ ( x ) = 2 / a sin (πx / a )
a. Is this state an eigenfunction of the position
operator?
b. Calculate the average value of the position that
would be obtained for a large number of
measurements. Explain your result.
Example 88-4
a. The position operator
xˆ = x . Because ,
xψ (x ) = x 2 / a sin (πx / a ) ≠ cψ (x )
where c is a constant, the wave function is not an
eigenfunction of the position operator.
Example 88-4
b. The expectation value is calculated using the
fourth postulate:
x =
2
2   πx  
2   πx 
 πx 
sin
x
sin
dx
=
xsin   dx
 
  
∫
∫
a 0   a 
a 0   a 
a
a
a
Using the standard integral
a
2
x 2 cos 2bx x sin 2bx
∫0 x (sin bx ) dx = 4 − 8b 2 − 4b
Example 88-4
We have
a

 2πx 
 2πx  
 x sin 

 2 cos
 a2  a
2 x
2  a 2 a 2
a 
a 


x =
−
−
=  − 2 − 0  + 2  =
2
a4
π  
a  4 8π
π 
 8π  2
4


8 


a


a

0
The average position is midway in the box. This is
exactly what we would expect, because the particle
is equally likely to be in each half of the box.