Download Techniques and Applications - Angelo Raymond Rossi

The Classical Free Particle
Advanced Physical Chemistry
Chemistry 5350
The simplest classical system imaginable is a particle with no forces
acting on it:
Q UANTUM T HEORY: T ECHNIQUES
AND A PPLICATIONS
In one dimension, the motion of a free particle can be described as:
F = ma = m
d2 x
=0
dt2
This differential equation can be solved to obtain
Professor Angelo R. Rossi
x = x0 + v 0 t
http://homepages.uconn.edu/rossi
Department of Chemistry, Room CHMT215
The University of Connecticut
The initial position x0 and the initial velocity v0 arise from constants of
integration.
Fall Semester 2013
To give them explicit values, the boundary conditions of the problem,
the initial position and velocity must be known.
[email protected]
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
The Quantum Mechanical Free Particle
Free Particle Approach to the Wave Function
The wave nature of the electron has been clearly shown in experiments
like the photoelelectric effect
The connection to the Schrödinger equation can be made by examining
wave and particle expressions for energy:
particle
⇐=
p2
2m
E
wave
=⇒
2
hν
Asserting the equivalence of these two expressions for energy
p̂2 → −~2
What is the nature of the wave for an electron?
The wave is the wave function for the electron.
Starting with the expression for a traveling wave in one dimension, the
connection can be made to the Schrodinger equation.
This process makes use of the de Broglie relationship between
wavelength and momentum and the Planck relationship between
frequency and energy.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
∂2
∂x2
Ê → i~
∂
∂t
and putting in the quantum mechanical operators for both brings us to
the Shrodinger equation.
The time-dependent Schrödinger equation takes the form
−
3
Fall Semester 2013
~2 ∂ 2 Ψ(x, t)
∂Ψ(x, t)
= i~
2m ∂x2
∂t
Last Updated: October 20, 2013 at 8:16pm
4
Free Particle Approach to the Wave Function
The Time-Independent Wave Function for a Free Particle
This gives a plane wave solution of the form
The time-independent solutions for a free-particle wave function are
Ψ(x, t) = Aei
2π −iωt
λ
The solutions are plane waves with one moving to the right (positive x direction), and the other moving to
the left (negative x direction)
Using the de Broglie relation gives
2πp
p
p
2π
=
= h = = k;
λ
h
~
2π
The general solutions are
p (electron momentum) = ~k
ψk (x) = A+ eikx + A− e−ikx
The energy is given by
Using the Planck relationship
Ek =
ω×
ψ − (x) = A− e−ikx
ψ + (x) = A+ e+ikx
= Aeikx−iωt
~
~ω
E
=
= ;
~
~
~
E (electron energy) = ~ω
p2
k 2 ~2
=
2m
2m
The wave functions and energies (eigenfunctions and eigenvalues of Ĥ) are labeled with the index k.
The wave function Ψ(x, t) = Aeikx−iωt is a complex function which can be expanded in the form
k=
Ψ(x, t) = A cos(kx − ωt) + iA sin(kx − ωt)
2π
=
λ
r
2mEk
~2
Either the real or imaginary part of this function could be appropriate for a given application.
In general, one is interested in particles which are free within some kind of boundary but have the boundary
conditions set by a potential.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
5
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
Probability of Determining the Position of a Free Particle
The Particle in a One-Dimensional Box
A plane wave is not localized, and it is not possible to speak of its position as in the
same way one can for a free classical particle.
The particle in a box consists of a particle of mass m confined between two walls at
x = 0 and x = a.
6
However, the probability of finding the particle in an interval of length dx can be
calculated
The free-particle wave functions cannot be normalized over the interval
−∞ < x < +∞, but if x is restricted to the interval −L < x < +L
•
•
A− A+ e−ikx eikx dx
dx
ψ ∗ (x)ψ(x)dx
=
=
P (x)dx = R L
RL
∗ (x)ψ(x)dx
−ikx eikx dx
2L
ψ
A
A
e
− + −L
−L
P (x)dx is independent of x.
This result states that the particle is equally likely to be anywhere in the interval,
which is equivalent to saying that nothing is known about the position of the
particle.
The wave function must be zero at the walls, and the solution for the wave function
yields just sine waves.
The idealized situation of a particle in a box with infinitely high walls is an application
of the Schrödinger equation which yields some insights into particle confinement.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
7
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
8
The Particle in a One-Dimensional Box:
Potential and Schrödinger Equation
The Particle in a 1-D Box: Boundary Conditions
The impenetrable walls of the particle in the box are modeled by making
the potential energy infinite outside of a region of width a:
Outside of the box, where the potential is infinite, the second derivative
would be infinite if ψ(x) were not zero for all values of x outside the box.
V (x) = 0 for a > x > 0
V (x) = ∞ for x ≥ a and x ≤ 0
ψ(0) = ψ(a) = 0
Moreover, inside the box where V (x) = 0, the wave function must be
continuous.
How does this change in potential affect the eigenfunctions that were
obtained for the free particle?
The above statements are boundary conditions that any well-behaved
wave function for a one-dimensional box must satisfy.
The Schrödinger equation is written in the following form:
−
Fall Semester 2013
The solution to Schrödinger equation for a particle inside the box is
identical to that for a free particle.
~2 d2 ψ(x)
+ V (x)ψ(x) = Eψ(x)
2m dx2
Last Updated: October 20, 2013 at 8:16pm
9
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
The Particle in a 1-D Box: Boundary Conditions
The Particle in a 1-D Box: Acceptable Wave Functions
For ease in applying the boundary conditions, the solution of the Schrödinger
equation in the box is written as
Therefore, we conclude that
ψn (x) = A sin
ψ(x) = A sin kx + B cos kx
Now apply the boundary conditions
nπx
a
10
The requirement that ka = nπ with n being an integer will turn out to
have important consequences for the energy spectrum of a particle in a
box.
Acceptable wave functions for the particle in a box must have the form:
ψ(0) = 0 + B = 0
ψ(a) = A sin ka = 0
The first condition can only be satisfied by the condition that B = 0.
ψn (x) = A sin
The second condition can be satisfied if either A = 0 or if ka = nπ with n being an
integer.
nπx
a
,
for n = 1, 2, 3, 4, . . .
Each different value of n corresponds to a different eigenfunction.
Setting A to zero would mean the wave function will always be zero which is
unacceptable because then there is no particle in the interval.
But there IS a particle in the interval a > x > 0
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
11
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
12
The Particle in a 1-D Box: Normalized Wave Functions
The Particle in a 1-D Box: Eigenvalues and Eigenfunctions
The constant A can be determined by normalization by realizing that
ψ ∗ (x)ψ(x)dx represents the probability of finding the particle in the
interval dx centered at x
Ra
Ra ∗
dx = 1
ψ (x)ψ(x)dx = A∗ A 0 sin2 nπx
a
0
What are the eigenvalues that go with these eigenfunctions?
~ d
− 2m
2
A=
2
;
a
ψn (x) =
q
2
a
sin
~ d
− 2m
dx2
2
nπx
2
q
2
a
sin
Last Updated: October 20, 2013 at 8:16pm
= En ψn (x)
nπx
a
=
~2
2m
nπ 2
a
q
2
a
sin
nπx
a
the following result for the energy (En ) is obtained
a
En =
Fall Semester 2013
n (x)
dx2
Applying the total energy operator to the eigenfunctions will give back
the eigenfunction multiplied by the energy eigenvalue:
because the probability of finding the particle somewhere in the entire
interval is one.
q
2ψ
13
The Particle in a 1-D Box vs The Free Particle
Fall Semester 2013
~2
2m
nπ 2
a
=
h2 n 2
,
8ma2
= En ψn (x)
for n = 1, 2, 3, . . .
Last Updated: October 20, 2013 at 8:16pm
14
The Particle in a 1-D Box: Energies and Wave Functions
An important difference between the energies of a particle in a box and
the energies of a free particle is that the energy for a particle in a box
can take on only discrete values.
The lowest four energy levels for the particle in the box are shown in the adjacent
figure and are superimposed on an energy versus distance diagram.
The energy of a particle in the box is quantized, and n is a quantum
number.
The eigenfunctions are also shown on
this diagram.
Another important result is that the lowest allowed energy is greater
than zero.
The variation of the wave function with
time is a standing wave which turns out
to be a general result for a stationary
state.
The particle in the box has a nonzero minimum energy known as zero
point energy.
A standing wave has nodes that are at
fixed distances as a function of time but
move in time to produce a traveling
wave.
Energies and Wave Functions
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
15
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
16
The Particle in a 1-D Box: Probability Density
The Particle in a 1-D Box: Calculating Energies
What is the ground state energy for an electron that is confined to a
potential well with a width of 0.2 nm?
Using the formula for the allowed energies for a particle in a box,
The probability of finding the electron
outside the box is zero.
The probability of finding the electron
within an interval dx in the box depends
both on the position and on the wave
function.
En =
Although | ψ(x) |2 can be zero at nodal
positions, | ψ(x) |2 dx is never zero for a
finite interval dx inside the box.
E1 =
Last Updated: October 20, 2013 at 8:16pm
for n = 1, 2, 3, . . .
(6.626 x 10−34 J s)2 (1)2
= 1.506 x 10−18 J/e−
8 (9.109 x 10−31 kg)(0.2 x 10−9 m)2
E1 = (1.506 x 10−18 J/e− )(6.022 x 1023 mol−1 ) ×
Probability Density
Fall Semester 2013
h2 n 2
,
8ma2
17
Fall Semester 2013
1 kJ
= 907 kJ mol−1
103 J
Last Updated: October 20, 2013 at 8:16pm
The Particle in a 1-D Box: Calculating Probabilities
The Particle in a 1-D Box: Calculating Expectation Values
What is the probability of finding the particle in the box in a small region of space?
The average value of x (< x >) for a particle in any of the eigenstates of
a one-dimensional box is
Z
Z
nπx
a
2 a
∗
x sin2
dx = =< x >
< x >n = ψn (x)xψn (x)dx =
a 0
a
2
For a particle in a one-dimensional box, the ground-state wave function is given as
ψ(x) =
r
2
πx
sin(
)
a
a
What is the probability that the particle is in the middle third of the box?
Thus, the particle is in the middle of box for all values of the quantum
number n which is reasonable.
The average value of x2 in the nth eigenstate ( < x2 >n ) is
Z
Z
2 a 2 2 nπx
< x2 >n = ψn∗ (x)x2 ψn (x)dx =
x sin
dx
a 0
a
a π 2 a2
< x2 > n =
−2
4
2πn
3
The probability density at point x in the interval dx is given by
πx
2
sin2
dx
a
a
P (x)dx = ψ ∗ (x)ψ(x)dx =
P
P
Fall Semester 2013
a
2a
≤x≤
3
3
a
2a
≤x≤
3
3
=
2
a
=
2
a
Z
2a
3
a
3
1
a
x−
sin
2
4π
sin2
πx
dx
a
2πx
a
2a
3
a
3
= 0.609
Last Updated: October 20, 2013 at 8:16pm
18
19
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
20
The Particle in a 1-D Box:
Classical Limit and the Correspondence Principle
The Particle in a 1-D Box:
Classical Limit and the Correspondence Principle
The Correspondence Principle states that quantum mechanical
predictions approach the predictions of classical mechanics as the
quantum number approaches infinity.
Examine the spacing between levels by forming a ratio:
Looking at the equation below shows the dependence of the total
energy eigenvalues on the quantum number n:
En =
En+1 − En
=
En
!
=
2n + 1
n2
which approaches zero as n → ∞.
h2 n2
, for n = 1, 2, 3, . . .
8ma2
Both the level spacing and energy increase with n, but the energy
increases faster (≈ n2 ) making the energy spectrum appear to be
continuous as n → ∞.
However, it is not immediately obvious that the energy spectrum will
become continuous in the classical limit of very large n because the
spacing between adjacent energy levels increases with increasing n.
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
h2 [(n+1)2 −n2 ]
8ma2
h2 n 2
8ma2
21
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
The Particle in a Box: 2-D and 3-D Systems
The Particle in a Box: 2-D and 3-D Systems
The potential energy of a three-dimensional box is given by:
This differential is solved by assuming that ψ(x, y, z) has the form
V (x, y, z) = 0 for 0 < x < a; 0 < y < b;
V (x, y, z) = ∞ otherwise
0<z<c
ψ(x, y, z) = X(x)Y (y)Z(z)
As for the one-dimensional box, the amplitude of the eigenfunctions of
the total energy is zero outside the box.
which is the product of three functions, each of which depends on only
one of the variables.
This assumption is valid
Inside the three-dimensional box, the Schrödinger equation can be
written
−
Fall Semester 2013
~2
2m
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
22
•
•
because V (x, y, z) is independent of x, y, and z inside the box.
because the potential is of the form
ψ(x, y, z) = Eψ(x, y, z)
Last Updated: October 20, 2013 at 8:16pm
V (x, y, z) = Vx (x)+Vy (y)+Vz (z)
23
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
24
The Particle in a Box: 2-D and 3-D Systems
The Particle in a Box: 2-D and 3-D Systems
Substituting ψ(x, y, z) = X(x)Y (y)Z(z) into the Schrödinger equation for
three-dimensional systems gives
Each of the equations has the same form as the equation that was
solved for the one-dimensional problem:
1 d2 X(x)
1 d2 Y (y)
1 d2 Z(z)
~2
+
+
= EX(x)Y (y)Z(z)
−
2m X(x) dx2
Y (y) dy 2
Z(z) dz 2
ψnx ,ny ,nz (x, y, z) = N sin
The total energy has the form:
The energy E has independent contributions from the three
coordinates: E = Ex + Ey + Ez which allows the original differential
equation in three variables to be reduced to three differential equations,
each with one variable:
−
nx πx
ny πy
nz πz
sin
sin
a
b
c
E=
h2
8m
n2x n2y n2z
+ 2 + 2
a2
b
c
~2 d2 X(x)
~2 d2 Y (y)
~2 d2 Z(z)
= Ex X(x); −
= Ey Y (y); −
= Ez Z(z)
2
2
2m dx
2m dy
2m dz 2
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
25
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
The Particle in a Box: 2-D and 3-D Systems
The Particle in a Box: 2-D and 3-D Systems
Because this is a three-dimensional problem, the eigenfunctions depend on three
quantum numbers.
If a = b = c, the energy can be written as
For a particle in a two-dimensional box, the total energy eigenfunctions
are given by
E=
ψnx ny (x, y) = N sin
h2
n2 + n2y + n2z
8ma2 x
ny
2
1
1
nx πx
ny πy
sin
a
b
The total energy in terms of nx , ny , a, b is given by:
Now, more than one set of the three quantum numbers can have the same energy:
nx
1
2
1
26
nz
1
1
2
E=
h2
8m
n2x n2y
+ 2
a2
b
In this case, the energy level is degenerate, and the number of states that have the
same energy level is the degeneracy of the level.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
27
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
28
The Particle in a Box: 2-D and 3-D Systems
The Classical Harmonic Oscillator
To understand vibrations in molecules, it is important to
understand the quantum mechanical treatment of a harmonic oscillator.
As background, it is necessary to review the classical
treatment of harmonic oscillator.
The simplest example of a harmonic oscillator is a mass
connected to a wall by means of an idealized spring in
the absence of gravity.
Contour plots of several eigenfunctions
are shown here. The x and y directions of
the box lie along the horizontal and vertical directions, respectively. The amplitude has been displayed as a gradation
in colors with regions of positive and negative amplitude noted.
As shown in the adjacent figure, the displacement of the
mass is given by the x coordinate, and the origin of the
coordinate system is taken at the equilibrium postion.
The mass oscillates about its equilibrium postion, and
the motion is said to be harmonic if the force F of the
spring is directly proportional to the displacement, x.
F = −kx
The negative sign comes from the fact that the force F is
opposite to the displacement.
The constant k is referred to as a force constant and is
small for a weak spring but large for a stiff spring.
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
29
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
The Classical Harmonic Oscillator
Simple Harmonic Motion
Since force is expressed as mass times accleeration, the
equation of motion in the x directions is
Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear
elastic restoring force given by Hooke’s Law.
30
The motion is sinusoidal in time and demonstrates a single resonant frequency
d2 x
+ kx = 0
dt2
The general solution of this differential equation is
x(t) = A sin ωt + B cos ωt
q
k
where ω =
is the angular frequency in radians per
m
second. Since
q ω = 2πν, the resonant frequency is given
as ν =
1
2π
A mass on a spring will trace out a sinusoidal pattern as a function of time, as will any object vibrating in
simple harmonic motion.
k
.
m
For example, while walking in a straight line at constant speed while carrying the vibrating mass, the mass
will trace out a sinusoidal path in space as well as time.
If the spring is initially stretched so that the mass is a
position x0 and the velocity is zero, then let go, the time
course of the motion is represented by
x(t) = x0 cos ωt
Since F = dV
= −kx, integration of this equation yields
dx
V = 12 x2 which is the equation of a parabola as shown
in the adjacent figure.
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
31
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
32
Quantum Harmonic Oscillator: Schrödinger Equation
Eigenvalues of the Harmonic Oscillator
The Schrödinger equation for a harmonic oscillator may be obtained by using the
classical spring potential
r
1
k
1
ω=
(angular frequency)
V (x) = kx2 = mω 2 x2 ;
2
2
m
Substituting the function
ψ(x) = Ce−α
into the Schrödinger equation and imposing boundary
conditions leads to the ground state energy for the quantum harmonic oscillator:
As we saw previously, the Schrödinger equation with this form of the potential is
−
~2 d2 ψ(x) 1
+ mω 2 x2 ψ(x) = Eψ(x)
2m dx2
2
E0 =
The imposition of boundary conditions leads to a discrete
energy spectrum.
x2
2
The lowest state accessible to the system has a non-zero
energy, i.e. the zero point energy which is also true for
the particle in the box.
This is a Gaussian function which satisfies the requirement that the wave funcion
approach zero at infinity which allows the wave function to be normalized.
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
1
~ω
= hν
2
2
The general solution to the Schrödinger equation produces a sequence of evenly spaced energy levels characterized by a quantum number, n.
Since the second derivative of the wave function must give back the square of x plus a
constant times the original function, the following form is suggested:
ψ(x) = Ce−α
x2
2
33
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
Eigenfunctions of the Harmonic Oscillator
Eigenfunctions of the Harmonic Oscillator
The wave functions for the quantum harmonic oscillator contain the Gaussian form
which allows them to satisfy the necessary boundary conditions at infinity.
The Hermite polynomials for the first four levels of the harmonic oscillator are as
follows:
The wave functions for the first two energy levels are given by:
ψ0 =
where α =
km
~2
12
α 41
π
e−
αx2
2
;
ψ1 =
4α3
π
1
4
x e−
H0 (ξ) = 1;
An even function is a function that satisfies f (x) = f (−x), and an odd function is a
function that satisfies f (x) = −f (−x)
In general, the wave functions of the harmonic oscillator are given by
αx2
2
;
Nν =
1
1
(2ν ν!) 2
αx2
Since e− 2 is the wave functions for the harmonic oscillator is even, the even-odd
character is determined by the Hermite polynomials.
α 41
Thus, the harmonic oscillator wave functions are even when ν is even and odd when ν
is odd making it easy to evaluate integrals.
π
1
and the Hν (α 2 x) are called Hermite Polynomials
Fall Semester 2013
H3 (ξ) = 8ξ 3 − 12ξ
A property of the Hermite polynomials is that Hν (ξ) is an even function if ν is even
and an odd function if ν is odd.
.
1
H2 (ξ) = 4ξ 2 − 2;
where ξ is a dummy variable.
αx2
2
The wave functions have the shape of a Gaussian probability function.
ψν = Nν Hν (α 2 x)e−
H1 (ξ) = 2ξ;
34
Last Updated: October 20, 2013 at 8:16pm
35
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
36
Quantum Harmonic Oscillator: Eigenfunctions and Probabilities
Probability Distributions for the Quantum Oscillator
The solution of the Schrödinger equation for the
quantum harmonic oscillator gives the wave functions, the energy levels as well as the probability
distributions for the quantum states of the oscillator.
The square of the wavefunction gives the probability
of finding the oscillator at a particular value of x.
There is a finite probability that the oscillator will
be found outside the potential well as indicated by
the smooth curve which is forbidden in classical
physics.
The most probable value of position for the lower
states of a quantum oscillator is very different from
the classical harmonic oscillator where it spends
more time near the end of its motion.
As the quantum number increases, the probabability distribution becomes more like that of the classical oscillator
The normalized wavefunctions for the first four energy states gives are displayed on the left above.
The probability of finding the oscillator at any given value of x is the square of the wavefunction, and these
probabilities are shown at right above.
The wavefunctions for higher n have more ”humps” within the potential well, and this corresponds to a
shorter wavelength, a higher momentum according to the de Broglie equation, and therefore higher energy.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
37
Comparison of Classical and Quantum
Probabilities for the Harmonic Oscillator
Fall Semester 2013
The tendency to approach classical behavior for high quantum
numbers is called the correspondence principle.
For the ground state of the quantum harmonic oscillator, the
correspondence principle seems far-fetched, since the classical and quantum predictions for the most probable location
are in total contradiction.
If the equilibrium position for the oscillator is taken to be x = 0,
then the quantum oscillator predicts that for the ground state,
the oscillator will spend most of its time near that center point.
On the other hand, the classical spends very little time there
because it has the maximum speed as it passes through x =
0.
It will spend most of its time near the end points of its oscillation where the velocity is small.
In going to higher and higher states of the quantum oscillator
by increasing n, the most probable location shifts to the edges
of the well.
There is still the undulation of the probability which is characteristic of the wave solution, but at least the overall trend of
the probability begins to look more like the classical probability
which is shown by the dashed line.
A comparison between the classical and quantum
probability of finding the object which is oscillating
at a given distance, x, from the equilibrium position
can be made.
For the classical case, the probability is greatest
at the ends of the motion since it is moving more
slowly and comes to rest instantaneously at the extremes of the motion.
The relative probability of finding it in any interval,
dx, is just the inverse of its average velocity over
that interval.
For the quantum mechanical case the probability of
finding the oscillator in an interval, dx, is the square
of the wavefunction, and that is very different for the
lower energy states.
In the diagram for the ground state (n = 0), the
maximum probability is at the equilibrium point x =
0.
Last Updated: October 20, 2013 at 8:16pm
38
The Correspondence Principle and the Quantum Oscillator
The harmonic oscillator is a good example of how
different quantum and classical results can be.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
39
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
40
Quantum Harmonic Oscillator: Zero-Point energy
Quantum Harmonic Oscillator and the Uncertainty Principle
If ∆x and ∆p are the uncertainty in position and momentum, respectively, the ground
state energy for the quantum harmonic oscillator
The most surprising difference for the quantum case is the zero-point vibration for the
n = 0 ground state.
E0 =
This implies that molecules are not completely
at rest, even at absolute zero temperature.
can be shown to be the minimum energy allowed by the uncertainty principle.
The quantum harmonic oscillator has implications far beyond the simple model presented
here.
This is a very significant physical result because it tells us that the energy of a system
described by a harmonic oscillator potential cannot have zero energy.
Physical systems such as atoms in a solid lattice or polyatomic molecules in the gas
phase cannot have zero energy even at absolute zero temperature.
It is the foundation for the understanding
of complex modes of vibration in larger
molecules, the motion of atoms in a solid lattice, the theory of heat capacity, and other
properties of molecules and solids.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
~ω
= ∆x∆p
2
The energy of the ground vibrational state is often referred to as zero point vibration.
The zero point energy is sufficient to prevent liquid helium-4 from freezing at
atmospheric pressure, no matter how low the temperature.
41
Characteristics of Vibrational Motion in Molecules
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
42
Two Masses Connected by a Spring: Harmonic Motion
When two masses are connected by a spring, there are two
equations of motion:
The discussion of the free particle and the particle in the box was useful for understanding how
translational motion in various potentials is described in the context of wave-particle duality.
d2 x1
= k(x2 − x1 − l0 )
dt2
d2 x2
= k(x2 − x1 − l0 )
dt2
In applying quantum mechanics to molecules, two other types of motion that molecules can undergo are
vibration and rotation.
m1
The simplest vibrational motion that can be imagined is that which occurs in a diatomic molecule.
where l0 is the equilibrium length of the spring.
The force on m1 is equal and opposite to the force on m2 as
required by Newton’s third law.
The oscillatory motion depends only on the relative coordinate
x = x2 − x1 − l0 which yields
d2 x
= −k
dt2
m2
1
1
+
m1
m2
x
The mass term can be taken to be the reciprocal of the rem2
duced mass µ defined by µ = mm1+m
. and
1
The equation for harmonic motion also applies to mass m1 connected to mass m2 as shown in the figure
above which represents an idealized diatomic molecule.
µ
d2 x
dt2
2
+ kx = 0
This equation is the same form as the previous equation for
the harmonic oscillator with m replaced by the reduced mass
and x is now the relative coordinate.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
43
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
44
Schrödinger Equation for a Vibrating Diatomic Molecule
Vibrational Frequencies in Diatomic Molecules
The following table provides transition frequencies and calculated force
constants for the n = 0 to n = 1 vibrational level for diatomic molecules.
A diatomic molecule vibrates somewhat like two masses on a spring with
a potential energy that depends upon the square of the displacement from
equilibrium.
The wave-particle of mass µ vibrating around its equilibrium position is described by a set of wave functions ψn (x)
Molecule Frequency/(1013 Hz) Force Constant (N/m)
HF
8.72
970
HCl
8.66
480
HBr
7.68
410
HI
6.69
320
CO
6.42
1860
NO
5.63
1530
To find these wave functions and corresponding allowed energies of vibrational motion, the following Schrödinger equation must be solved:
−
~2 d2 ψn (x)
2µ
dx2
kx2
+
2
ψn (x) = En ψn (x)
The energy levels of the quantum harmonic oscillator are quantized at equally
spaced values and given by
En = (n +
1
2
)~ω = (n +
1
2
)hν
For a diatomic molecule, the natural frequency is of the form
ω=
s
k
µ
;
ν =
1
2π
s
k
µ
m m
where the reduced mass (µ) is given by µ = m 1+m2
1
2
The form of the frequency is the same as that for the classical simple harmonic oscillator.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
45
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
Comparison of Harmonic and More Realistic Potentials
Characteristics of Vibrational Motion in Molecules
The adjacent figure shows the potential energy,
V (x), as a function of bond length, x for a diatomic molecule. xe represents the equilibrium
bond length.
The existence of a stable chemical bond implies that a minimum energy
exists at some internuclear distance called the bond length
46
The configuration of atoms is dynamic rather than static, and so the
chemical bond should be thought as having springs rather than a rigid
bar connecting two atoms.
The zero of energy is chosen to be the bottom of
the potential.
The red curve depicts a realistic potential in which
the molecule dissociates for large values of x.
Thermal energy increases the vibrational amplitude of their atom about
their equilibrium positions.
The orange curve shows a harmonic potential,
V (x) = 21 kx2 which is a good approximation to the
realistic potential near the bottom of the well.
However, for T ≈ 300 K, only the lowest one or two vibrational levels
are occupied for most molecules.
In real systems, energy spacings are equal only for
the lowest levels where the potential is a good approximation harmonic potential.
Therefore, it is a good approximation to employ the functional form
V (x) = 12 kx2 near the equilibrium bond length:
The potential becomes steeply repulsive at short
distances and levels out at large distances.
The anharmonic terms which appear in the potential for a diatomic molecule are useful for mapping
the detailed potential of such systems.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
47
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
48
The Classical Rigid Rotor
Diatomic Molecule as a Classical Rigid Rotor
In the adjacent figure, a particle is rotating
around a fixed axis and has angular momentum and kinetic energy.
The kinetic energy of a particle T in circular motion about a fixed point is expressed
in terms of angular velocity or angular momentum
L2
1
T = Iω 2 =
2
2I
A model of a rigid diatomic molecule, i.e. with the bond
distance fixed, provides a good example of a rigid rotor.
The two masses rotate around their center of mass, satisfying the condition r1 m1 = r2 m2 .
Because there is no potential energy, the classical Hamiltonian of a rigid rotor is just the kinetic energy.
V (x, y, z) = 0
The rotational kinetic energy is
where I = mr2 is the moment of inertia, L =
Iω is the angular momentum, and ω is the
angular velocity.
In the expression for the rotational kinetic energy, the moment of inertia is analogous to
mass, and angular velocity ω is analogous to
linear velocity v for linear motion.
T =
where I is the moment of inertia of the diatomic molecule
I = m1 r12 + m2 r22 =
Last Updated: October 20, 2013 at 8:16pm
m1 m2
R2 = µR02
m1 + m2 0
The rotational kinetic energy may also be written in terms
of the angular momentum
T =
Fall Semester 2013
1 2
Iω
2
49
Fall Semester 2013
L2
L2
=
2I
2µR02
Last Updated: October 20, 2013 at 8:16pm
Quantum Mechanical Motion in 2-D: Rotation on a Ring
Quantum Mechanical Rotation in 2-D: Spherical Coordinates
The quantum mechanical operator for a rigid rotor is given as
∂2
∂2
~2 ∂ 2
~2
+
+
Ĥ = − ∇2 = −
2µ
2µ ∂x2 ∂y 2 ∂z 2
~2 ∂ 2
∂2
∂2
−
+
+
ψ = Eψ
2µ ∂x2 ∂y 2 ∂z 2
The relationship between Cartesian coordinates (x, y, z) and spherical coordinates (r, θ, φ) is given below:
50
In discussing rotation, it is more convenient to use spherical coordinates
because they reflect they symmetry of the system being considered.
The differential volume element in this coordinate system is
dτ = r2 sin θdrdθdφ
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
51
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
52
Quantum Mechanical Rotation in 2-D: Schrödinger Equation
Quantum Mechanical Rotation in 2-D: Boundary Conditions
In spherical coordinates, the Laplacian operator ∇2 is given by
∂
1
∂
1 ∂
1
∂
∂2
+
∇2 = 2
R2
+ 2 2
sin
θ
R ∂R
∂R
∂θ
R sin θ ∂φ2 R2 sin θ ∂θ
To obtain the solutions of the Schrödinger equation that describe the physical
problem, it is necessary to introduce the boundary condition
Φ(φ + 2π) = Φ(φ)
This condition states that there is no way to distinguish the particle that has rotated n
times around the circle from one that has rotated n + 1 times.
Applying the single-valued condition
For R fixed at R0 and θ fixed in a plane at 90◦ , the Schrödinger takes
the form:
~2 d2 Φ(φ)
−
= EΦ(φ)
2µR02 dφ2
eiml [φ+2π] = e[iml φ]
Thus,
This equation has the same form as the Schrödinger equation for a free
particle with two linearly independent solutions
Using Euler’s relation
cos 2πml + i sin 2πml = 1
Φ+ (φ) = A(+φ) eiml φ and Φ− (φ) = A(−φ) e−iml φ
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
e2πiml = 1
To satisfy this condition, ml = 0, ±1, ±2, ±3, . . .
53
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
Quantum Mechanical Rotation in 2-D: Eigenfunctions
Quantum Mechanical Rotation in 2-D: Eigenvalues
The motivation for using the subscript l will become clear when rotation
in two- and three-dimensions is considered.
The normalization constant is
Z 2π
Φ∗ml (φ)Φml (φ)dφ = 1
Putting the eigenfunctions back into the Schrödinger equation allows
the corresponding eigenvalues Eml to be calculated
E ml =
54
~2 m2l
~2 m2l
=
for ml = 0, ±1, ±2, ±3, . . .
2
2µR0
2I
0
(A+φ )
2
Z
2π
e
−iml φ iml φ
e
A(+φ)
A(−φ)
Fall Semester 2013
dφ = (A+φ )
0
2
Z
States with +ml and −ml have the same energy even though the wave
functions corresponding to these states are orthogonal to each other.
2π
dφ = 1
Energy levels with ml 6= 0 are twofold degenerate.
0
1
=√
2π
1
=√
2π
Last Updated: October 20, 2013 at 8:16pm
55
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
56
Quantum Mechanical Rotation in 3-D: Motion on a Sphere
Quantum Mechanical Rotation in 3-D: Motion on a Sphere
In the case just considered, the motion was constrained to two dimensions.
The Schrödinger equation with zero potential energy in
three-dimensions is given as
∂2
~2 1 ∂
1
1
∂
∂
2 ∂
−
+
R
+
sin
θ
ψ = Eψ
2µ R2 ∂R
∂R
∂θ
R2 sin2 θ ∂φ2 R2 sin θ ∂θ
Now imagine the more familiar case of a diatomic molecule freely rotating in
three-dimensional space.
Again, we transform to the center of mass coordinate system, and the rotational
motion is transformed to the motion of a particle on the surface of a sphere of radius
R0 .
Since the two masses of the rigid rotor are at fixed distances from the
origin, R is constant, and we can ignore the derivatives with respect to
R:
~2
1 ∂2
1 ∂
∂
−
+
sin
θ
Y (θ, φ) = E Y (θ, φ)
2I sin2 θ ∂φ2 sin θ ∂θ
∂θ
The bond length is assumed to remain constant as the molecule rotates giving rise to
the term rigid rotor.
The kinetic and potential operators are are expressed in spherical coordinates.
The potential energy V (R, φ, θ) is zero because there is no hindrance to rotation.
Y (θ, φ) is the product of two functions
Y (θ, φ) = Θ(θ)Φ(φ)
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
57
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
Quantum Mechanical Rotation in 3-D: Motion on a Sphere
Eigenfunctions and Eigenvalues of a Particle on a Sphere
One obtains two separate differential equations:
Two quantum numbers l and ml which arise in the solution of this eigenvalue equations on the previous
m
slide and are represented by the function Yl l (θ, φ) and are called spherical harmonics.
1 d2 Φ(φ)
= −m2l
Φ(φ) dφ2
1
d
dΘ(θ)
sin θ
sin θ
+ β sin2 θ = m2l
Θ(θ)
dθ
dθ
The first several spherical harmonics are given in the table below:
Y00 =
β=
1
1
(4π) 2
1
3 2
Y10 = 4π
cos θ
1
3 2
1
sin θeiφ
Y1 = 8π
1
3 2
Y1−1 = 8π
sin θe−iφ
2µR02 E
~2
To ensure Y (θ, φ) remains single-values and finite, the following
conditions must be met:
ml
The eigenfunctions Yl
β = l(l + 1), for l = 0, 1, 2, 3, . . .
ml
(θ, φ) =
so that
El =
ml = −l, −(l − 1), −(l − 2), . . . , 0, . . . , (l − 2), (l − 1), +l
Last Updated: October 20, 2013 at 8:16pm
Y20 =
Y21 =
Y2−1 =
Y22 =
5
16π
1
1
15
2
(3 cos2 θ − 1)
2 sin θ cos θeiφ
8π
1
15 2
sin θ cos θe−iφ
8π
1
15 2
2 θe2iφ
sin
32π
(θ, φ) satisfy the eigenvalue equation
ĤYl
and
Fall Semester 2013
58
~2
m
l(l + 1)Yl l (θ, φ) for l = 0, 1, 2, 3, . . .
2I
~2
l(l + 1) for l = 0, 1, 2, 3, . . .
2I
The numerical factor in front of these functions ensures that they are normalized over the intervals
0 ≤ θ ≤ π and 0 ≤ φ2π.
59
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
60
Plots of p and d Linear Combinations of the Spherical Harmonics
Review of Classical Angular Momentum
Angular momentum is a vector and has components in the x, y, and z
directions.
r 3
1 px = √ Y11 + Y1−1 =
sin θ cos φ
4π
2
To develop the quantum mechanical operators for angular momentum
in the x, y, and z directions, we need to review the classical expressions
for angular momentum in three dimensions.
r 3
1 1
Y1 − Y1−1 =
sin θ sin φ
py = √
4π
2i
r
3
pz = Y10 =
cos θ
4π
r
5
3 cos2 θ − 1
dz2 = Y20 =
16π
r 15
1 dxz = √ Y21 + Y2−1
sin θ cos θ cos φ
4π
2
r
15
1 1
Y2 − Y2−1
dyz = √
sin θ cos θ sin φ
4π
2i
r
15
1 dx2 −y2 = √ Y22 + Y2−2
sin2 θ cos 2φ
16π
2
r 15
1 2
Y2 − +Y2−2
dxy = √
sin2 θ sin 2φ
16π
2i
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
A particle rotating around a fixed
axis has an angular momentum
and kinetic energy.
L = Iω = mvr
ω is the angular velocity.
61
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
Review of Classical Angular Momentum
Review of Classical Angular Momentum
In classical mechanics, the angular momentum rotating about a fixed
point is represented by the vector L in the direction perpendicular to the
plane of motion.
If a mass m is rotating about a fixed point with linear velocity v , the
angular momentum L is given by the cross product of the radius r and
the linear momentum p
The vectors r and p may be expressed in terms of their components and unit vectors i ,
j , and k pointing along the x, y, and z axes.
62
r = xii + yjj + zkk
p = px i + py j + pz k
The cross product of r and p may be calculated from a determinant:
i
j
k L = r × p = x y z px py pz L = r × mvv = r × p
The cross product of two vectors r and p is a vector of magnitude
|rr | |pp| sin θ where θ is the angle between r and p .
Thus, the angular momentum vector for the circular motion in the
previous figure points up.
L = (ypz − zpy )ii + (zpx − xpz )jj + (xpy − ypx )kk
Lx = ypz − zpy
Ly = zpx − xpz
Lz = xpy − ypx
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
63
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
64
Quantum Mechanical Angular Momentum
Review of Classical Angular Momentum
The quantum mechanical operators for the angular momentum are
obtained by replacing the components of momentum with their
corresponding quantum mechanical operators. Specifically,
The square of the angular momentum is given by the dot product of L
with itself:
L • L = L2 = L2x + L2y + L2z
pˆx −→ −i~
∂
;
∂x
pˆy −→ −i~
∂
;
∂y
pˆz −→ −i~
The square of the angular momentum is a scalar.
∂
∂z
If no torque acts on a particle, its angular momentum is conserved, i.e.
is a constant.
This substitution yields:
In classical mechanics, all possible values of L and E are permitted.
∂
∂
Lˆx = −i~ y
−z
∂z
∂y
∂
∂
ˆ
Ly = −i~ z
−x
∂x
∂z
∂
∂
−y
L̂z = −i~ x
∂y
∂x
Last Updated: October 20, 2013 at 8:16pm
Fall Semester 2013
65
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
Angular Momentum: Quantum Mechanical Operators
Quantum Mechanical Angular Momentum Operators
The operator for the square of the angular momentum is given by:
It is readily shown that Lˆx and Lˆy , Lˆy and Lˆz , Lˆx and Lˆz do not commute with each
other.
Lˆx , Lˆy , and Lˆz all commute with Lˆ2 .
Lˆ2 = |L̂|2 = L̂ • L̂ = Lˆ2x + Lˆ2y + Lˆ2z
Therefore, we can measure precisely the square of the total angular momentum, and
one, but only one of its components.
It is often more convenient to use the angular momentum operators in
spherical coordinates:
∂
∂
Lˆx = i~ sin φ + cot θ cos φ
∂θ
∂φ
∂
∂
Lˆy = i~ − cos φ + cot θ sin φ
∂θ
∂φ
Lˆ2 = −~2
Fall Semester 2013
L̂z = −~
1 ∂
sin θ ∂θ
∂
∂φ
sin θ
∂
∂θ
+
1 ∂2
sin2 θ ∂φ2
Last Updated: October 20, 2013 at 8:16pm
66
Thus, if the magnitude of the total angular momentum
q
√
L| = L2 = L2x + L2y + L2z
|L
is measured, and Lz is measured, it is not possible to measure Lx or Ly precisely.
The eigenfunction of Lˆ2 is also an eigenfunction of Lˆz but is not an eigenfunction of
Lx or Ly which is an essential difference between classical and quantum mechanical
systems.
67
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
68
Eigenvalues of Angular Momentum Operators
Magnetic Quantum Numbers
Since Lˆ2 and Lˆz commute, it is possible to construct an function that is an
eigenfunction of both operators.
Operating on the spherical harmonics with
Lˆz ,
Lˆz Y ml (θ, φ) = ml ~ Y ml (θ, φ)
For a classical rigid rotor,
l
thus, for the quantum mechanical operator
ml = −l, −l + 1, . . . , 0, . . . , l − 1, +l
Lˆ2 = 2I Ĥ
and
Lˆ2 Ylml (θ, φ) = l(l + 1)~2 Ylml (θ, φ)
yields the following eigenvalues for the z component Lz of the angular momentum
Lz = ml ~;
l = 0, 1, 2, . . .
According to this eigenvalue equation, the total angular momentum for a rigid rotor
can have only the values
L2 = l(l + 1)~2
The possible orientations of the momentum
vector L for l = 1 with respect to a particular
direction are shown in the adjacent figure.
l = 0, 1, 2, . . .
Last Updated: October 20, 2013 at 8:16pm
69
Magnitude of Quantum Mechanical Angular Momentum
Since the x and y components of the angular
momentum are indeterminant, the vector can
be rotated about the z axis to lie anywhere on
the conical surface.
Since the Lx and Ly components are unknown, L can be described only as being on
the surface of a cone as shown in the adjacent figure.
The magnitude of the orbital angular momen1
tum L is [l(l + 1)] 2 ~. and the maximum component of Lz in a particular direction is l~.
Thus, the magnitude of the angular momentum L is greater than its z component so that
the angular momentum vector cannot point in
the direction of an applied magnetic field or
along a unique axis.
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
ml = −l, −l+1, . . . , 0, . . . , l−1, +l
ml is referred to as the magnetic quantum
number.
l is referred to as the angular momentum quantum number.
Fall Semester 2013
l
with
L2 = 2IT
71
Fall Semester 2013
Last Updated: October 20, 2013 at 8:16pm
70