Download Physical Chemistry II

CHEM 32: Physical Chemistry II
Andrew Rosen
May 11, 2013
Contents
0 Math Review
0.1
Complex Numbers
3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 The Dawn of the Quantum Theory
3
3
1.1
Blackbody Radiation Could Not Be Explained by Classical Physics . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Planck Used a Quantum Hypothesis to Derive the Blackbody Radiation Law
. . . . . . . . . . . . . . . . . .
3
1.3
Einstein Explained the Photoelectric Eect with a Quantum Hypothesis . . . . . . . . . . . . . . . . . . . . .
4
1.4
The Hydrogen Atomic Spectrum Consists of Several Series of Lines . . . . . . . . . . . . . . . . . . . . . . . .
4
1.5
The Rydberg Formula Accounts for All the Lines in the Hydrogen Atomic Spectrum . . . . . . . . . . . . . .
4
1.6
Louis de Broglie Postulated That Matter Has Wavelike Properties
. . . . . . . . . . . . . . . . . . . . . . . .
5
1.7
de Broglie Waves Are Observed Experimentally . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.8
The Bohr Theory of the Hydrogen Atom Can Be Used to Derive the Rydberg Formula . . . . . . . . . . . . .
5
1.9
The Heisenberg Uncertainty Principle States That the Position and the Momentum of a Particle Cannot be
Specied Simultaneously with Unlimited Precision
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 The Wave-Particle Duality
6
7
2.1
Wave vs. Particle Facts
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
Probability of an Event
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.3
Alternative Expressions for the Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . .
8
3 The Schrödinger Equation and a Particle In a Box
3.1
The Hamiltonian Operator
3.2
The Schrödinger Wave Equation
3.3
3.4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
8
8
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
3.2.1
Time Dependent SWE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
3.2.2
Derivation of Temporal and Spatial Parts
9
3.2.3
Time Independent SWE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
3.2.4
Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Particle In a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wave Functions Must Be Normalized . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
3.4.1
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
3.4.2
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
3.4.3
Normalize the Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
3.5
The Average Momentum of a Particle in a Box is Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
3.6
Higher-Ordered Particle in a Box
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.7
3.8
Two-Dimensional Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.7.1
Derivation using Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.7.2
Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Higher-Ordered Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1
4 Postulates and Tunneling
14
4.1
Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
4.2
The Potential Step - Example of Tunneling
14
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 The Harmonic Oscillator - Model for Vibration
16
5.1
Hooke's Law
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
5.2
Quantum-Mechanical Oscillator and Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
6 The Rigid Rotor - Model for Rotation
6.1
16
Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Molecular Spectroscopy
16
17
7.1
Vibration-Rotation Interaction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
7.2
Spacing of Lines and Overtones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
8 The Hydrogen Atom
18
9 Partition Functions
18
2
0
Math Review
0.1 Complex Numbers
• For an equation
z = x + iy ,
• The asterisk symbol (e.g.
the operator
z∗)
Re(z) = x
and
Im(z) = y
denotes that the complex conjugate is taken
|Z|2 = Z · Z ∗
• The modulus of a number is the same as the absolute value
|Z| =
p
|Z|2
• For conversion between imaginary and Cartesian coordinates,
x = r cos θ
• To convert a standard complex number of form
1
are used :
a + bi into a complex number in the form of reiθ , the following equations
p
r = a2 + b2
θ = tan−1
• To convert a complex number of the form
reiθ
y = r sin θ
b
a
into one of the form
a + bi,
the following equations, Euler's formula is
used:
reiθ = r (cos θ + i sin θ)
1
The Dawn of the Quantum Theory
1.1 Blackbody Radiation Could Not Be Explained by Classical Physics
• Wavelength and frequency are related by,
λν = c
• A blackbody is an ideal body that can absorb and emit all frequencies of light, known as blackbody radiation
• The Rayleigh-Jeans law states that the radiant energy density between an innitesimal change in frequency is2 ,
dρ
8πν 2
= 3 kB T
dν
c
• From this equation came the ultraviolet catastrophe, which suggested that, at high frequencies, the radiant energy
density diverges as
ν2
1.2 Planck Used a Quantum Hypothesis to Derive the Blackbody Radiation Law
• The Planck distribution law for blackbody radiation solves the ultraviolet catastrophe by xing the average energy
per oscillator using quantized values
• The energy for oscillator number
n
3
was derived as ,
E = nhν
1 When computing the Arc tangent, be careful since the calculator will simplify the result and multiple θ values can have the same Arc tangent
2 k ≡ 1.380658 × 10−23 J · K −1
B
3 h ≡ 6.6260755 × 10−34 J · S
3
• The quantized blackbody formula is thus,
1
dρ
8πhν 3
=
dν
c3 ehν/kB T − 1
• The Wien displacement law states,
2.90 × 10−3
T
λmax =
m
=
hc
4.965kB T
1.3 Einstein Explained the Photoelectric Eect with a Quantum Hypothesis
• The photoelectric eect is the ejection of electrons from the surface of radiated metal
• Unlike theory at the time, experiment showed that the kinetic energy of the ejected electrons is independent of the
intensity of the incident radiation
Additionally, there is a threshold frequency (ν0 ) characteristic for each metal where no
e−
for
ν < ν0
regardless of
intensity (contrasts classical mechanics)
∗
∗
∗
Above
Above
Above
ν0 ,
ν0 ,
ν0 ,
the number of electrons is proportional to intensity
the kinetic energy of electrons is independent of intensity
the kinetic energy of the ejected electrons depends on
ν − ν0
• As such4 ,
Ee− = hν − hν0
• The work function is dened as
φ ≡ hν0
• Therefore,
Ee− = hν − φ
From this, the negated y-intercept of an electron energy vs. frequency plot will yield the work function
Also, Planck's constant can be derived from the slope of the line
1.4 The Hydrogen Atomic Spectrum Consists of Several Series of Lines
• The characteristic emission spectrum of an atom is called its line spectra
Reciprocal wavelength is known as a wavenumber
• The Balmer formula for the hydrogen spectra is,
ν̃ =
1
ν
= = 109680
λ
c
1
1
− 2
22
n
cm−1 (n = 3, 4, ...)
1.5 The Rydberg Formula Accounts for All the Lines in the Hydrogen Atomic Spectrum
• The more general Rydberg formula states5 ,
ν̃ = RH
1
1
− 2
n21
n2
cm−1 (n2 > n1 )
4 φ is the work function of a metal, which is analogous to an ionization energy of an isolated atom. It typically has units of electron volts (eV)
where 1 eV = 1.602 × 10−19 J
5 The Rydberg constant, R , is equal to 109677.581 cm−1
H
4
1.6 Louis de Broglie Postulated That Matter Has Wavelike Properties
• Relating related the momentum of a photon to wavelength as follows,
λ=
h
h
=
p
mv
1.7 de Broglie Waves Are Observed Experimentally
• X-ray diraction occurs because the interatomic spacings in the crystal are about the same as the wavelength of the
x-rays
The distance between rings has to do with the interatomic spacing
• Electron microscopes are devices that used electrons, specically their wave-like properties, to observe matter that a
light microscope simply cannot
1.8 The Bohr Theory of the Hydrogen Atom Can Be Used to Derive the Rydberg Formula
I have included the following derivation. It should not be memorized, but it's helpful to understand it. I have also tried
to make this equation relatively general by including
Z
µ. It still only works for atoms with one electron.
Z = 1. Also, by rearranging constants it is possible to have
a0 are typically easier to use on a calculator since the other
and mentioning
However, for the most frequently used hydrogen Bohr model,
alternate yet equal versions of the equations. The ones with
equations raise very small constants to higher powers (exception: TI-nspire) :
1. Coulomb's law relates the radius of orbit (r ), charge on the electron (e), and the force holding the electron in a circular
6
orbit (f ) ,
f=
qnucleus · qelectron
4π0 r2
(a) Equating this to the centripetal force of the electron orbit (this assumes that the electron circles the nucleus in a
stationary, circular orbit) yields the following, where
f=
Z
is the proton number:
Ze2
me v 2
=
2
4π0 r
r
2. Classical physics assumes that the electron emits electromagnetic radiation, loses energy, and should collapse into the
nucleus
(a) The assumption that the de Broglie waves of the orbiting electron must be in phase and that the electron had
stationary orbits solved this issue. This quantum approach is given by,
2πr = nλ
(b) Since
λ=
h
h
=
,
p
mv
7
substituting this into equation 2a yields the angular momentum ,
mvr =
(c) Plugging equation 2b into the equation 1 and solving for
r
nh
2π
yields the following equation. Physically, it means the
electron has a radii of orbit that must satisfy:
r=
4π0 n2 ~2
me e2 Z
6 is the permitivity of free space and has a value of 8.85419 × 10−12 C2 · N−1 ·m−2
0
7~ ≡ h
2π
5
i. Solving for
v
instead yields,
e2
4π0 n~
v=
ii. Setting
n=1
Z=1
and
gives the Bohr radius for the hydrogen atom,
a0 =
4π0 ~2
= 5.2917725 × 10−11 m
m e e2
iii. Using the Bohr radius is helpful since
rn = a0 n2
3. The potential energy of the electron in the Bohr model separated from the positive nucleus by a distance
r
is,
2
V (r) = −
Ze
4π0 r
4. Consequently,
E = KE + V (r) =
1
Ze2
me v 2 −
2
4π0 r
5. Then, a little algebra is used:
me v 2
(a) Rearrange the equation from step 1a to solve for
(b) Then substitute equation 2c into 5a by substituting for
4
En = −
and substitute it into equation 4
r.
The simplied result is,
me e Z
e2 Z 2
=
−
820 h2 n2
8π0 a0 n2
2
i. The negative sign indicates that the energies are less when the proton and electron are innitely separated
ii.
n=1
n
is the ground state energy and higher
values are excited states
6. Finally, for hydrogen or a hydrogen-like ion with only one electron (and using the eective mass
µ
in place of
me )
the
change in energy can be found as,
m e e4 Z 2
∆E =
820 h2
(a) Using
a0
1
1
− 2
2
n1
n2
= hν =
hc
λ
instead gives:
e2 Z 2
∆E =
8π0 a0
1
1
− 2
2
n1
n2
= hν =
hc
λ
• The Lyman series occurs when excited electrons return to ground-state, and the Balmer series occurs when excited
electrons fall back to
n=2
7. For accuracy, the eective mass,
µ,
can be substituted for
mass of the electron:
µ=
(a) Note that
a0
8. Optional: Since
ν̃
with
me
is simply
me
where
m1
will be the mass of the nucleus, and
m2
is the
m1 m2
m1 + m2
is not a perfect constant since it does not use the reduced mass
λ−1 ,
one can derive the Rydberg Constant by solving for
4
R∞ =
4
ṽ .
Thus,
2
µe
me e
e
≈ 2 3 =
2
3
80 ch
80 ch
8π0 a0
1.9 The Heisenberg Uncertainty Principle States That the Position and the Momentum of
a Particle Cannot be Specied Simultaneously with Unlimited Precision
• The act of locating the electron leads to a change in its momentum
• As such, the Heisenberg Uncertainty Principle states,
∆x∆p ≥
6
~
2
2
The Wave-Particle Duality
2.1 Wave vs. Particle Facts
• In the double slit experiment, light exhibited diraction - a wave property
• Einstein's photoelectric eect proved that light has particle-like properties
• de Broglie was the rst to predict that matter can behave as a wave according to
λ=
h
h
=
p
mv
Note: The de Broglie wavelength does NOT work in
for anything but light waves
ν=
c
λ
since
c
is the speed of
light!
Do not use this equation
• Davisson and Germer's experiment proved the de Broglie wavelength for electron diraction
• When
h
is much smaller when compared to other variables, the quantum limit approaches the classical limit
• For normal (90◦ ) incidence (d is the spacing of scattering centers),
∆l = d sin θ
• Each photons needs to feel both slits simultaneously so it can interfere with itself (diuse and large)
However, the photon is still detected as a tiny localized spot on lm (photon is small)
• Wave nature of light and particles is interpreted as a statistical prediction
• The measurement process can and does aect the outcome of the event
2.2 Probability of an Event
The recipe for predicting what happens:
• The probability of an event happening is given by the square of the absolute magnitude of a probability amplitude (ψ ∗ )
P = ψ ∗ ψ = |ψ|2
• When an event can occur in two or more ways, then
ψ
for the event is the sum of the
ψi
values for each path considered
separately
P = |ψ1 + ψ2 |2
• If we perform an experiment that can identify which way the event occurred, then the probability is the sum of
probabilities for each alternative
P = |ψ1 |2 + |ψ2 |2
• The measurement process forces the event to have a single outcome
If many events are averaged, you get a probability distribution
• Wave Particle Duality states:
The Photoelectric Eect had 1 path, and light behaved as a particle
The Double Slit Experiment had 2 paths, and light was wave-like
• A plane wave has the form eikx where
k=
2π
λ
• A standing wave has the form e−iωt since it becomes a number at any given t
7
2.3 Alternative Expressions for the Heisenberg Uncertainty Principle
• Three forms of the Heisenberg Uncertainty Principle are:
3
∆x∆px ≥
~
2
∆E∆t ≥
~
2
∆ω∆t ≥
~
2
The Schrödinger Equation and a Particle In a Box
3.1 The Hamiltonian Operator
• The mathematical operators8 used in this course are all linear (order matters though). This means that an operator
follows the following rule:
Ω̂
Ω̂ (c · f ) = c Ω̂ (f )
• The commutation relationship states that,
x̂P̂ − P̂ x̂ = i~
• The Hamiltonian operator is dened as follows:
Ĥ = T̂ + V̂
• Kinetic Energy is described as,
T =
p2
1
mv 2 =
2
2m
• The corresponding quantum mechanical operato
T̂ =
−~2 2
∇
2m
To not restrict the discussion to one-dimension,
∇2 ≡
∂2
∂2
∂2
+
+
∂x2
∂y 2
∂z 2
• Finally, the Hamiltonian operator can be expressed as follow:
Ĥ = −
~2 2
−
∇ + V (→
r , t)
2m
3.2 The Schrödinger Wave Equation
3.2.1 Time Dependent SWE
• The time-dependent Schrödinger Wave Equation (SWE) states,
i~
−
∂Ψ (→
r , t)
−
= ĤΨ (→
r , t)
∂t
The solution to a SWE is called a wave function and is the probability amplitude that encodes all information
about the system it describes
• As an example, here is the time-dependent SWE with variables substituted for the operators in one dimension,
i~
∂Ψ (x, t)
~2 ∂ 2 Ψ(x, t)
=−
+ V (x, t)Ψ(x, t)
∂t
2m ∂x2
8 All operators have a carat above the letter
8
(1D Time Dependent SWE)
3.2.2 Derivation of Temporal and Spatial Parts
If we assume
−
−
Ψ (→
r , t) = ψ (→
r ) θ (t):
1. Therefore,
i~
∂
−
−
[ψ (→
r ) θ (t)] = Ĥ [ψ (→
r ) θ (t)]
∂t
∂ θ (t)
−
−
i~ψ (→
r)
= θ (t) Ĥψ (→
r)
∂t
2 2
i~ ∂θ (t)
1
−~ ∇
→
−
→
−
→
−
→
−
ψ ( r ) θ (t) yields:
= →
ψ( r ) + V ( r )ψ( r )
θ (t) ∂t
2m
ψ(−
r)
2. Using properties of partial derivatives,
3. Dividing through by
(a) We have separated time dependence from spatial dependence
(b) To have a solution, each side must be equal to a constant, and that constant is energy
i~ ∂θ(t)
∂θ(t)
= E → i~
= Eθ (t)
θ (t) ∂t
∂t
2
−~ 2
→
−
−
−
∇ + V ( r ) ψ (→
r ) = Eψ (→
r)
2m
4. The temporal piece is:
5. The spatial piece is:
6. Using separation of variables on
−iE
∂θ (t)
=
dt
θ (t)
~
yields
θ (t) = Ae−(iE/~)t
3.2.3 Time Independent SWE
• The Time-Independent SWE states,
−
−
Eψ (→
r ) = Ĥψ(→
r)
• An example of this in one dimension is given as,
Eψ(x) = −
~2 d2 ψ(x)
+ V (x)ψ(x)
2m dx2
(1D Time Independent SWE)
3.2.4 Postulates
1. Quantum mechanical systems are described by a wavefunction,
(a) All information about the system is encoded in
(b) Born Interpretation states
time
i.
−
Ψ (→
r , t)
−
Ψ (→
r , t)
is a probability amplitude for the presence of the particle at position
t
−
|Ψ (→
r , t) |2 = Probability
(c) Additionally, using
ψ
Density
as spatial-dependent and
τ is the generic volume
ˆ ∞
ψ ∗ ψ dτ ≡ 1
−∞
(d)
−
Ψ (→
r , t)
dτ = sin θdθdφr2 dr
is used for spherical coordinates
9
element,
→
−
r
and
3.3 Particle In a Box
• Above is a diagram of the theoretical particle in a box. Outside of the domain,
• Inside the box
Ĥ = −
~2 d2
2m dx2
and outside the box
Ĥ = −
V (x) = ∞
~2 d 2
+∞
2m dx2
• The particle in a box problem is a good model for electrons in a conned geometry, such as molecules with conjugated
π
systems and quantum dots/wells
ψ(x) = 0
• Since it is impossible for the particle to be outside of the box,
Since
ψ(x)
outside of the domain
is a position, it must be a continuous function
• The general solution of the 1D Time Independent SWE where
V (x) = 0
is
ψ(x) = A cos (kx) + B sin (kx)
√
k≡
•
ψ(0) = 0
because
ψ=0
This implies that
• Since
for
x≤0
A=0
and
because
ψ
2mE
~
is continuous
cos(0) = 1.
It does not provide any information for
A = 0,
ψ(L) = B sin (kx) = 0
•
B=0
is rejected because it yields
ψ(x) = 0
for all
x,
which isn't true
Alternatively, we can state the following to make
sin
be zero,
k=
Therefore,
En =
h2 n2
8mL2
The energy of transition for an electron is,
∆En =
nπ
L
h2
n2f − n2i
2
8me L
The corresponding wave function is,
ψn (x) = B sin
10
nπx L
B
0≤x≤L
3.4 Wave Functions Must Be Normalized
• The wave function must be normalized such that the sum of all probabilities is 1
ψnorm = N ψ
∞
ˆ
∗
−1/2
ψ ψ dτ
N=
−∞
ˆ
• It is also true that
∗
ψm
ψn dτ = δm,n
This relationship holds:
δm,n = 1
m = n,
when
and
δm,n = 0
(orthonormal) when
For spherical coordinates and functions that are spherically symmetrical:
ˆ
∞
ˆ
π
ˆ
ˆ
2π
∞
f (r)4πr2 dr
f (r)r2 sin θdrdθdφ =
0
0
0
0
• The ground state energy (lowest energy state,
m 6= n
E1 )
This is known as the zero-point energy and is
is always greater than zero
h2
8mL2
E1 =
3.4.1 Example 1
Normalize
1. If
2
ψ = e−ax
2
ψ = e−ax
2. Then,
:
for instance, it can be normalized by doing
N = q´
∞
−∞
=
ψ ∗ ψdx
3. Finally,
1
ψnorm =
2a
π
1/4
2a
π
´∞
ψ ∗ ψ dx =
−∞
´∞
2
e−2ax dx =
−∞
r
π
2a
1/4
2
e−ax
• For even functions that have symmetry,
ψ(x) = ψ(−x)
• For odd functions that have antisymmetry
ψ(−x) = −ψ(x)
• The following rules apply: odd × even = odd, even × even = even, odd × odd = even
• The integrals of odd functions are zero
3.4.2 Example 2
Normalize
ψ0 = Ae−x
2
1. Set up the integral:
2. Factoring out constants yields:
3.
ψ0
2
ψ1 = Bxe−ax
´ ∗
´∞
2
2
ψ0 ψ1 dτ = −∞ A∗ e−x Bxe−x dx
and
A∗ B =
´∞
−∞
2
xe−2x dx
is an even function due to its symmetry (the Gaussian curve) and
4. Therefore, the integral is 0 since
x
is odd and
e−2x
2
ψ1
is an odd function due to its antisymmetry
is even and is already normalized
11
3.4.3 Normalize the Particle in a Box
1. Set up the integral:
´∞
ψ ∗ ψdx =
−∞
2. Simplify the integral:
´L
0
B ∗ B sin2
´0
0∗ 0dx +
−∞
nπx ´L
0
B ∗ sin
nπx L
B sin
nπx L
dx +
´∞
L
0∗ 0dx
dx
L
(a) There is a nifty integral trick that is as such: For
´H
L
nπx sin2
H
=
H − L ´ nπ 2
H − L nπ H − L
sin z dz =
=
0
nπ
nπ
2
2
i. This also works for cosine
(b) Therefore, the integral is equal to
3. As such,
N =B= p
1
L/2
r
=
(B ∗ B)
L
2
2
L
(a) This can also be found because the wavefunction must be equal to one along the domain, so
(B ∗ B)
L
=1
2
4. Therefore,
r
ψn (x) =
• To nd the probability density, use
nπx 2
sin
L
L
(1D Particle in a Box)
ψ∗ ψ
• To nd the probability a particle is in a certain region,
ˆ
x2
ψ ∗ ψ dx =
x1
2
L
ˆ
x2
sin2
nπx x1
• For a one-dimensionl box, each energy state has a corresponding
L
dx
n-half-cycle
of sine where the numbers of nodes are is
n − 1,
• The average momentum of a particle in a box is zero, indicating it is likely to move in either direction
3.5 The Average Momentum of a Particle in a Box is Zero
ˆ
2
hp i =
ψn∗ (x)ψn (x)P̂x2 dx
• Evaluating this integral along the limits of the box (0 to L) yields the following where
σp2 = hp2 i =
n2 π 2 ~2
nπ~
∴ σP =
2
L
L
12
P̂x2 =
d2
−~2 2 :
dx
• To prove the Heisenberg Uncertainty Principle,
~
σx σP =
2
1/2
~
π 2 n2
>
−2
3
2
This is true because the square-root can never be less than unity
3.6 Higher-Ordered Particle in a Box
• For a three-dimensional particle in a box,
ψnx ,ny ,nz =
8
Lx Ly Lz
1/2
sin
nx πx
Lx
sin
ny πy
Ly
sin
nz πz
Lz
• As a result,
Enx ,ny ,nz
h2
=
8m
n2y
n2z
n2x
+
+
(Lx )2
(Ly )2
(Lz )2
!
3.7 Two-Dimensional Particle in a Box
3.7.1 Derivation using Separation of Variables
The process is similar for a two-dimensional box. For our case, let's assume it's a square 2D box such that
1. If
ψ(x, y) = X(x)Y (y),
2. Inside the box,
3. Dividing by
then
V (x, y) = 0
X(x)Y (y)
6.
so
yields
−
∂2
∂2
X(x)Y (y) + 2 X(x)Y (y) + V (x, y) = EX(x)Y (y)
∂x2
∂y
~2
(Y (y) + X 00 (x) + X(x)Y 00 (y)) = E (X(x)Y (y))
2m
X 00 (x) Y 00 (y)
−2mE
+
=
X(x)
Y (y)
~2
Y 00 (y)
−2mEy
=
Y (y)
~2
r
r
2
2
nx πx
ny πy
Separation of variables then yields X(x) =
sin
sin
and Y (y) =
Lx
Lx
Ly
Ly
!
n2y
h2
n2x
Additionally, Enx ,ny =
+
8m (Lx )2
(Ly )2
4. Solving the above yields
5.
~2
−
2m
Ĥψ(x, y) = Eψ(x, y)
−2mEx
X 00 (x)
=
X(x)
~2
and
3.7.2 Properties
• If the box is square (Lx
= Ly )
then the energies can be written out as (1, 1), (1,2), (2, 1), (2, 2), etc., for
values
The energies will be degenerate for
(1, 2)
and
(2, 1),
• Free Electron Molecular Orbital Model (FEMO)
Treat electron in a MO as a particle in a box
13
for instance
nx
and
ny
3.8 Higher-Ordered Particle in a Box
• For a three-dimensional particle in a box,
ψnx ,ny ,nz =
8
Lx Ly Lz
1/2
sin
nx πx
Lx
sin
ny πy
Ly
sin
nz πz
Lz
• As a result,
Enx ,ny ,nz
• Each energy level can hold 2
4
h2
=
8m
n2y
n2z
n2x
+
+
2
2
(Lx )
(Ly )
(Lz )2
!
π -electrons
Postulates and Tunneling
4.1 Postulates
• Postulate 1:
ψ
completely describes a quantum-mechanical system
• Postulate 2: For every observable in classical mechanics, there is a corresponding quantum mechanical linear operator
(specically, Hermitian Operators) where we choose
x̂
x̂ = x
• Postulate 3: For an operator
Â,
and
p̂
so that
p̂ = −i~
x̂p̂ − p̂x̂ = i~
∂
∂x
the only observables are equal to the eigenvalues of
Case 1: If the system is in a state described by an eigenfunction,
Case 2:
ψ = c cos (kx).
Here,
ψ
is not a function of
p̂.
ψn ,
of
Â
When
◦
◦
p̂
acts on
an
1 ikx 1 −ikx
ce + ce
with
2
2
then we always measure
We can also write
ψ
theorem.
∗
Â
as
Euler's
ψ = c cos (kx):
1) Only returns e-values of
p̂
for answers
2) The probability of measuring a particular e-value is given by the square of the coecient weighting the
corresponding e-function in the superposition sum
◦
3) We can only predict the average outcome of many experiments, or the possible outcomes and their
probabilities for a single measurement
• Postulate 4: Expectation Value =
If
i
h
Â, B̂ = ÂB̂ − B̂ Â = 0,
• Postulate 5: Free particle where
´ ∗
ψ Âψdτ
hai = hÂi = ´ ∗
ψ ψdτ
then we can measure a and b without uncertainty
V (x) = 0
A free particle is not bound and has no quantization
4.2 The Potential Step - Example of Tunneling
14
• When
E ≥ V0 ,
x=0
the particle decelerates at
• In quantum mechanical regions I and II,
x>0
but either stops or continues on to
ψI (x = 0) = ψII (x = 0)
and
0
ψI0 (x = 0) = ψII
(x = 0)
so that the wave
functions are nite and continuous
• As seen in the diagram, an energy below
V0
will have a decaying exponential, termed an evanescent wave, when it hits
the potential step
−h2 d2
ψI = EψI
2m dx2
• Region I:
• Region II:
−h2 d2
ψII + V0 ψII = EψII ,
2m dx2
• Therefore,
ψI = Aeikx + Be−ikx
• The restrictions on
ψ
and
must be that
0
0
ψII = A0 eik x + B 0 e−ik x
ψ(x)
h2 d2
ψII = (E − V0 )ψII
2m dx2
p
√
2m (E − V0 )
2mE
0
where k =
and k =
~
~
which is equivalent to:
−
is continuous, nite, and that
ψ 0 (x)
V =∞
is continuous except if
for all
x
This can give you boundary conditions to solve for the constants. However, two will be missing
One more constraint can be found from the assumption that the particles originate from the
−x
direction as a
right-going wave such that there are no particles coming in from the rightmost region (this makes the rightgoing
constant of the last region zero)
2
The last constraint can be found from the ux, which is dened as Flux=v|A|
N
constant representing probability density when squared,
N,
where
A
is the pre-exponential
is the total number of particles, and
v
is the velocity
in that region
∗
∗
Recall that
p = mv .
Therefore,
v=
k~
m
Reected ux is found based on the leftgoing wave before the boundary (B ), and transmitted ux is found
0
based on the rightgoing wave after the boundary (A )
• To nd the transmittance, simply divide the rightgoing ux after the boundary (region II in this case) by the rightgoing
ux before the boundary (region I in this case). As in, it's the amount that went through
In this example, it would be
This will simplify down to,
vII |A0 |2 N
vI |A|2 N
T =
T =
k 0 4k 2
k (k + k 0 )2
• To nd the fraction reected, simply divide the leftgoing ux before the boundary by the rightgoing ux before the
boundary
In this example, it would be
This will simplify to
vI |B|2
vI |A|2
R=
2
∗
R=
(k − k 0 )
2 , which is always greater than or equal to zero
(k + k 0 )
This goes against classical mechanics because, even when
• Additionally, there should be no particles lost; therefore,
• When
E < V0 , k
E > V0 ,
there will be some particles reected
R+T =1
is imaginary; therefore, the exponential terms of the wavefunction are real and not plane waves
When the wave has a probability for transmittance even when
• The larger total energy of the particle (lower
V0
and/or higher
E)
E < V0 ,
the wave is called an evanescent wave
means larger
p
and thus shorter
amplitude
• The smaller total energy of the particle means smaller
p
and thus longer
15
λ.
Also, larger amplitude
λ.
Also, smaller
5
The Harmonic Oscillator - Model for Vibration
5.1 Hooke's Law
• The classical harmonic oscillator is
• The total energy is conserved as
1
V = kx2 , F = −kx,
2
E =V +T =
1 2
kx
2
and
1/2
k
ω = 2πν =
µ
for a spring oscillating between
−x
and
+x
5.2 Quantum-Mechanical Oscillator and Hermite Polynomials
• The fundamental vibrational frequency is given by
• Note that the value of
k
1
2πc
ṽ =
1/2
k
µ
is the same regardless of the isotope (even though
µ
is dierent).
k
is only a function of bond
order
~2 ∂ 2 ψ 1 2
+ kx ψ = Eψ
2m ∂x2
2
√
kµ
2
, Hn
α1/2 x e−αx /2 where α =
~
1
α 1/4
−
• The SWE can be set up as follows (note that there is potential energy now):
ψn (x) = Nn Hn
• The wavefunction for the harmonic oscillator is given by
Hermite Polynomial, and
Nn
is the normalization constant given by
Nn =
1/2
(2n n!)
is the
π
Hermite polynomials are either even or odd. The Hermite polynomial has the same parity as
n
2
The symmetric integral of an odd function is zero, and because ψn (x) is always even, hxi must be odd. Therefore,
the average displacement of a harmonic oscillator is zero for all states. This is the same for hpi
n ≥ 0,
• There is quantum tunneling with the harmonic oscillator,
and there are
n nodes,
and a zero-point energy of
1
hν
2
(for a 1-D case), which is dierent from the PIB
• Allowable energies are
E=
n+
1
2
hν =
n+
~ω
where
n
is the quantum number
There is thus equal spacing between the energies of the vibrational states
For the harmonic oscillator,
∆n = ±1
(typically
confusing for me to use since it looks like
6
1
2
v
is used for the harmonic oscillator energy state, but it's too
ν)
The Rigid Rotor - Model for Rotation
6.1 Energy Levels
• To describe a rotating body, we use the moment of inertia,
• The possible energies9 for a rigid rotor are
It is also true that
As such,
EJ =
I,
~2
J (J + 1),
2I
such that,
where
I=
J ≥0
P
(integers)
∆J = ±1
~2
∆EJ =
(J + 1)
I
• A molecule must posses a permanent dipole moment to have rotational energy
• Frequently, the constant
B=
h
8π 2 I
is used such that
B̃ =
h
8π 2 cI
9 Note that the units for energy here are in Hz and must be multiplied by h to yield joules
16
mi ri2 = µR2
• Additionally10 ,
ν = 2B (J + 1),
This is useful because a frequency (or wavenumber) can be read o of a vibrational spectrometer. This could then
I
I = µR2 ,
be used to calculate
nd
7
∆ν = 2B
and thus for an energy transition
R
from
using the denition of
B
shown above and the fact that
∆ν = 2B .
Finally, with
I,
you can
and thus you have the bond length
Molecular Spectroscopy
7.1 Vibration-Rotation Interaction
• Since a molecule can rotate and vibrate at same time, we need to combine the rigid rotor and SHO models to nd a
better approximation of the internuclear energies
• The vibrational energy of a molecule in wavenumbers is
• The rotational energy of a molecule in wavenumbers if
• As such,
Ẽ = G(v) + F (J)
and
v,
where
∆J = +1
it is called an R-Branch (high frequency side)
If
∆J = −1
it is called a P-Branch (low frequency side)
There are no lines at
ṽ
because the transition of
ν̃obs
are inaccurate, as
ν̃R
To derive the equations of
and
ν̃P ,
B̃
consider a
∆J = 0
as described earlier
v
B̃
for a
v=0→1
is dependent on
∆ER = ν̃ + 2B̃ (J 00 + 1)
v
transition and apply
ν̃R = Ẽ1,J+1 − Ẽ0,J and ν̃P = Ẽ1,J−1 − Ẽ0,J
∗ ν̃R = ν̃ + 2B̃1 + 3B̃1 − B̃0 J + B̃1 − B̃0 J 2
∗ ν̃P = ν̃ − B̃1 + B̃0 J + B̃1 − B̃0 J 2
The vibration rotation interaction states that
is forbidden
is a function of
Example:
• The energies of each branch are
F (J) = B̃J (J + 1)
so is the change in energy
If
• Technically, the equations of
J
F (J)
where
∆E = E 0 − E 00
where we dene
• Since energy is a function of
G(v)
1
ν̃
G(v) = v +
2
and
v
∆E
transition
such that
1
B̃v = B̃e − αe v +
2
∆EP = ν̃ − 2B̃J 00
7.2 Spacing of Lines and Overtones
• The spacing of spectral lines are not even because the centrifugal force causes the bond to stretch slightly
The corrected equation is
• To account for overtones,
F (J) = B̃J (J + 1) − D̃J 2 (J + 1)
2
1
1
G(v) = ν̃e v +
− x̃e ν̃e v +
,
2
2
2
where
where
x̃e
D̃
is the centrifugal distortion constant
is the anharmonicity constant
• The terminology is that the 0 to 1 transition is the fundamental transition, 0 to 2 is the rst overtone, 0 to 3 is the
second overtone, etc.
• The selection rule for anharmonic oscillator is any integral value for
• The electronic transition energy is given by
∆v
with higher integral values being weaker
Ẽ = T̃e + G0 (v 0 ) − G00 (v 00 ) + F 0 (J 0 ) − F 00 (J 00 )
• The Franck-Condon Principle states that electron motion does not cause appreciable nuclei motion
10 Note: This can be rewritten using ν̃ and B̃
17
8
The Hydrogen Atom
• For the hydrogen atom,
ψ(r, θ, φ) = Rn,` (r)Y`m` (θ, φ)
• The angular solution is based on Legendre Polynomials (m
= 0)
and Associated Legendre Functions (any
m)
The shape of the orbitals
• The radial solution is based on the Laguerre Polynomials
Tells how the probability density dies o
• The wavefunctions of a rigid rotor are spherical harmonics given by
•
L̂2 Y`m` (θ, φ) = ` (` + 1) ~2 Y`m` (θ, φ)
•
L̂z Y`m` (θ, φ) = m` ~Y`m` (θ, φ),
•
Ylm 's
Note that
m = 0, ±1, ±2, ..., ±`
1/2 Rn,` = −
so
(n − ` − 1)!
2n (n + `)!
Such that
En =
2
na0
where
` (` + 1) ~2
`+3/2
is the eigenvalue of
L̂2
with an eigenvalue of
m` ~
` = 0, 1, 2, 3, ...
r` e−r/(na0 ) L2`+1
n+`
2r
na0
where
L
is the Laguerre Polynomial and
a0 =
0 h2
πµe2
e2
8π0 a0 n2
• The variation principles states that we can pick any guess at
L̂z
are the eigenfunctions of
and that
Y
This means that a trial wavefunction,
φ,
ψ,
we'll call it
φ,
and then
´ ∗
φ Ĥφdτ
´
= Eφ ≥ E0
φ∗ φdτ
can be created. To improve the accuracy of this model the energy can
be minimized by changing the parameter(s). Adding more parameters to alter can also improve the accuracy of
this trial wavefunction
9
Partition Functions
• For partition functions,
a = g exp (−E/kT ),
where
g
is the number of degeneracies
• Furthermore, the fraction of molecules in this state is
fi = ai /Q
18
where
Q=
P
a