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CHEM 3010
Introduction to
Quantum Chemistry,
Spectroscopy,
& Statistical Thermodynamics
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Rene Fournier: Petrie 303, [email protected],
ext 30687
prerequisite: CHEM 2011
Required Text: Physical Chemistry by Thomas
Engel and Philip Reid (Pearson Prentice Hall) 3rd
edition (2013), ISBN-10: 0-321-81200-X ISBN-13:
978-0-321-81200-1
Course web site (soon): go to www.yorku.ca/renef
and follow the CHEM 3010 link.
Lectures: LSB 105, MWF 10:30–11:20
Office hours: Petrie 303, MWF 13:00-14:00
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Chapter 12
From Classical to
Quantum Mechanics
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In early 1900’s, classical physics failed to explain
• why atoms, with their electrons orbiting around
a nucleus, are stable
• why the spectrum of the H atom is discrete
• the blackbody radiation
• the photoelectric effect
• electron diffraction
A “new physics” was needed.
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Two key concepts in quantum mechanics:
• energy quantization:
the amount of energy exchanged between molecules
and radiation (light) is a discrete variable, ie, it is
not continuous
• wave-particle duality:
light (a wave) sometimes shows particle-like pro-
perties; and “particles” (e−, p+, atoms, . . . ) so-
metimes show wave-like properties.
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Rutherford, 1910: atoms have a positive charge
concentrated at the center (nucleus) and negative
charge (electrons, e−) spread out around it.
Classically, orbiting e− should radiate energy and
quickly spiral down to the nucleus.
How can atoms be stable ? ? ?
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H atoms only emit light with specific (discrete)
wavelengths λ, see Fig. 12.11, p 304.
A fit to data shows that
1
λ= RH
(1
n21
− 1
n22
)
with RH = 109678 cm−1
n1 = 1, 2, 3, . . .
and n2 = n1 + 1, n1 + 2, n1 + 3, . . .
But why ?
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Blackbody radiation
Fig. 12.1, p295
Light is emitted, re-absorbed, and re-emitted many
times: it is in thermodynamic equilibrium with
the solid.
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Classical physics: the radiation energy density for
frequency between ν and ν + dν:
ρ(ν, T ) =8πν2
c3Eosc =
8πν2kBT
c3
kB = 1.381× 10−23 J K−1
c = 2.998× 108 m s−1
ρ(ν, T ): J m−3 s
ρ(ν, T ) ∝ ν2, so
∫ ∞0
ρ(ν, T ) dν = C ×∫ ∞
0ν2dν →∞
The “UV catastrophe” predicted by classical physics
makes no sense!
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Planck (1900) hypothesized that oscillating dipoles
in the solid could only contain energy hν, 2hν,
3hν, . . . , and nothing in between: their energy is
“quantized”. That led to . . .
Eosc =hν
ehν/kBT − 1
and a formula for ρ(ν, T ) in perfect agreement
with expt when
h = 6.6261× 10−34 J s
Note that at high T (kBT � hν)
Eosc =hν
(1 + hν/kBT + . . .) − 1= kBT
as in the classical case.
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Photoelectric effect (Fig. 12.3, p297)
Light of intensity I and frequency ν causes N e−
to be emitted with kinetic energy Ee.
1. N = 0 when ν < ν0
2. N ∝ I
3. Ee is independent of I
4. Ee ∝ (ν − ν0)
5. an e− is emitted even when I is very small, as
if all light energy was concentrated in one
spot, like a particle!
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Einstein, 1905: light is made of “grains” (photons)
each with an energy proportional to frequency:
Ephoton ∝ ν
Ephoton = C × ν
e− are bound to the metal by an energy φ, the
work function. If one photon causes the emission
of one e−,
Ee = Cν − φ = C(ν − ν0)
A fit to data gives C = h !?
Ephoton = hν
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Bohr’s H atom
1. the e− goes around the proton, with orbit cir-
cumference 2πr
2. for stable orbits Coulombic and centrifugal forces
must cancel out: e2/(4πε0r2) = mev
2/r
This gives v2 = e2/(4πε0mer)
3. the e− can jump between orbits 1 and 2 by ab-
sorbing light energy E2 − E1 = ∆E = hν
4. the angular momentum is mevr = n(h/2π)
with n = 1, 2, 3,. . .
This is not a trivial assumption!
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m2e ×
(e2
4πε0mer
)× r2 = n2(h/2π)2
Rearranging gives the size of stable orbits, rn.
rn = n2 h2
4π2
4πε0mee2
= n2 0.5291772× 10−10m
The energy in a stable orbit, En, is the sum of
potential and kinetic energies
En =−e2
4πε0r+mev
2
2
=−e2
4πε0r+
mee2
8πε0mer
=−e2
8πε0rn
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En = − 1
n2
mee4
8ε20h2
= − 1
n22.17987× 10−18 J
in perfect agreement with the H atom spectrum!
Furthermore,
2.17987× 10−18 J = hcRH
Same h again!?
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DeBroglie, 1924: every object has an associated
wavelength λ
λ =h
p=
h
mv
Davisson and Germer, 1927, observed diffraction
of an e− beam off a NiO crystal.
If the beam has very few e−, we see individual e−
impacts, but the diffraction pattern is preserved.
He atoms bouncing off a Ni surface also form a
diffraction pattern.
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e− diffraction by a double slit
A low intensity e− beam shows particle-like be-
havior.
But accumulating signals over a long time reveals
the diffraction pattern ⇒ wave-like behavior.
⇒ each e− goes through both slits at once and
“interferes with itself”. Each e− is wave-like . . .
until it hits the photographic plate.
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Chapter 13
The Schrodinger Equation
(S equation)
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• Quantum mechanics (QM) is strictly “better”
(more accurate) than classical mechanics (CM)
• QM is much more complicated
• QM and CM give nearly identical results when
1. masses are big enough ( ' 100 a.m.u.)
2. dimensions are big enough (' 1000 A)
3. T is high enough
• Better: CM may be used when
(1) λ = h/p is small compared to a charac-
teristic length for the phenomenon
(2) ∆E is small compared to kBT
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Diffraction of e− and Xe atoms
through a 1 A slit
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Vibrations of H2 and Ar2 at T = 300 K
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Diffraction of e−, He+, and Xe+ off a surface
Take λ = 1 A:
v =h
mλ= 6.63× 10−24/m
KE = mv2/2 =h2
2mλ2
=1
m× 2.20× 10−47 J
=1
m× 1.37× 10−28 eV
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v (m/s) E (eV)
e− 7.3 ×106 150
He+ 998 0.021
Xe+ 30 0.00063
At rt kBT = 0.026 eV
c = 2.998× 108 m/s
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Traveling wave:
velocity v = λν
vertical displacement ψ:
ψ(x, t) = A sin[2π(x/λ− νt)]note that
ψ(x± λ, t) = ψ(x, t)
ψ(x, t± 1/ν) = ψ(x, t)
because that adds ±2π inside [. . .] in both cases.
Let
k = 2π/λ
ω = 2πν
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then
ψ(x, t) = A sin(kx− ωt)
but
sinx = cos(x− π/2)
and the choices for x = 0 and t = 0 are arbitrary.
So
ψ(x, t) = A cos(kx− ωt + φ)
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Complex numbers
The imaginary number i is defined by
i2 = −1
A complex number z has a real part a and ima-
ginary part ib: z = a + ib, with a and b real.
For z = a + ib, define
z∗ = a− ib
Then
zz∗ = a2 − aib + aib− i2b2 = a2 + b2
so zz∗ is always real. We also define
Re(z) = a
Im(z) = b
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Physical quantities (energy, position, mass, etc.)
are always real of course. But complex numbers
and functions “z” can be used as intermediate
variables to describe physical systems, and the
physical quantities can then be expressed in terms
of zz∗, Re(z), or Im(z).
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Complex plane representation of z = a + ib:
a = r cos θ ⇒ θ = cos−1(a/r)
b = r sin θ ⇒ θ = sin−1(b/r)
r =√a2 + b2
Euler’s formula:
eiθ = cos θ + i sin θ
So z can be written as z = reiθ and
ψ(x, t) = A Re[ei(kx−ωt+φ)
]
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Standing wave.
If you pluck a string, a mixture of λ’s coexist for a
short time, but they all interfere and destruct ex-
cept λ = 2L, λ = 2L/2, λ = 2L/3, . . .λ = 2L/n,
n = 1, 2, 3,. . . .
The “quantization” of standing waves, E = hν,
and λ = h/p, suggested that the classical diffe-
rential equation for waves may apply to particles,
with λclassical changed to h/p.
. . .
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Time-dependent S equation, with ~ = h/2π:
−~2
2m
∂2Ψ(x, t)
∂x2+ V (x, t)Ψ(x, t) = i~
∂Ψ(x, t)
∂t
When V (x, t) = V (x), it reduces to the time-
independent S equation
[−~2
2m
∂2
∂x2+ V (x)
]ψ(x) = Eψ(x)
where Ψ(x, t) = ψ(x) e−iEt/~
The quantity in [brackets] is an operator, the
Hamiltonian operator (or energy operator).
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Numbers, functions, and operators
If x and y are real numbers, a function f is a rule
(or algorithm) that inputs a number x and out-
puts another number y: y = f (x).
An operator O is an algorithm that inputs a func-
tion f and outputs another function g:
g(x) = O[f (x)]
Examples of operators: “multiply by 3”, “take the
derivative w.r.t. x”, “take the square of”, . . .
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Note: a functional is an algorithm that inputs a
function f and outputs a number y
y = F [f (x)]
for example, “integrate f between 0 and infinity”,
or “take the minimum of f over the interval from
0 to 5”.
A transform is an algorithm that inputs a func-
tion f of variable x, and outputs a different func-
tion g of a different variable t:
g(t) = F [f (x)]
for ex., the Fourier transform (often used for time-
to-frequency mapping).
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From the time-dependent S equation to
the time-independent S equation
When the potential is time-independent, V = V (x) (e.g., for the proton-
electron pair of a H atom), we can write the total wavefunction Ψ(x, t)
as a product
Ψ(x, t) = ψ(x)φ(t) (1)
To show that it is correct, substitute the right-hand side (r.h.s.) in the
time-dependent S equation with V (x, t) changed to V (x).
φ(t)
[−~2
2m
∂2ψ(x)
∂x2+ V (x)ψ(x)
]= i~ψ(x)
∂φ(t)
∂t(2)
φ(t) Hψ(x) = i~ψ(x)∂φ(t)
∂t(3)
The last equation is just a definition of H , the hamiltonian operator.
Next, divide boths sides by φ(t)ψ(x).
(Hψ(x)
ψ(x)
)= i~
1
φ(t)
∂φ(t)
∂t(4)
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(1) the r.h.s. has units J s s−1 or J (energy). Therefore, Hψ(x)/ψ(x)
also has units of energy.
(2) the r.h.s. does not depend on x, so the l.h.s. does not depend on
x either. In other words, Hψ(x)/ψ(x) represents a constant energy.
Call it E, and multiply both sides by φ(t).
Eφ(t) = i~dφ(t)
dt(5)
(−iE/~)φ(t) = −i2 dφ(t)
dt=dφ(t)
dt(6)
so
φ(t) = e−iEt/~ (7)
We also have Hψ(x)/ψ(x) = E, or
Hψ(x) = Eψ(x) (8)
with
H =−~2
2m
∂2
∂x2+ V (x, t)
Equation (8) is the time-independent S equation.
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Now let’s repeat the derivation with V = V (x, t) instead. Equation (2)
becomes
φ(t)−~2
2m
∂2ψ(x)
∂x2+ φ(t)V (x, t)ψ(x) = i~ψ(x)
∂φ(t)
∂t(9)
Divide both sides by ψ(x)φ(t):
1
ψ(x)
−~2
2m
∂2ψ(x)
∂x2+ V (x, t) =
1
φ(t)i~∂φ(t)
∂t(10)
The r.h.s. does not depend on x so we can write more simply
1
ψ(x)
−~2
2m
∂2ψ(x)
∂x2+ V (x, t) = f (t)
If we write the last equation twice, for times t2 and t1, and subtract one
from the other, the terms 1ψ(x)
−~2
2m∂2ψ(x)∂x2 cancel out and we’re left with
only
V (x, t2)− V (x, t1) = f (t2)− f (t1) (11)
The r.h.s. does not depend on x, therefore the l.h.s. does not depend on
x: the potential is a function of time only, V = V (t). This shows that
we can decompose Ψ(x, t) as a product ψ(x)φ(t) only if V (x, t) = V (x)
(first case) or V (x, t) = V (t) (second case). If the potential depends on
both x and t, we can not write Ψ(x, t) = ψ(x)φ(t).
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We have the time-independent S equation.
Now what ??
What are the meanings of E, H , and ψ(x)?
• E is the energy of the system in a stationary
state, a constant
• H is a mathematical operation that gives us
E if we know ψ
• with H , we can set up a differential equation
whose solutions are En and ψn(x), n = 1, 2, 3, . . .
• ψ(x) is a fictitious wave associated with a par-
ticle (or a system). It has no direct meaning. But
ψ(x)2dx has meaning.
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Eigenvalue Equations
The time-independent S equation is
Hψ(x) = Eψ(x)
H is an operator. The S equation is an “eigen-
value equation” (EE) where E is the eigenvalue
and ψ(x) is the eigenfunction. Here’s a simpler
example of EE:
O =d
dx
Of (x) = αf (x)
The solutions to this EE: f (x) = Ceax, α = a.
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Note that
(1) once O is known, the EE solutions (f (x), α)
can be found;
(2) for a given O, there are infinitely many
solutions;
(3) in this example, the eigenvalues a form
a continuum. However, if we force f (x) to satisfy
some boundary conditions, the eigenvalues will
form a discrete set (which is still infinite).
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For the time-independent S equation:
1. from V (x)⇒ H
2. solve Hψ = Eψ ⇒ (En, ψn(x)), n = 1, 2, 3. . .
3. ψ(x)⇒ physical properties of stationary states
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The eigenfunctions of a QM operator, including
H , form a complete and orthogonal set of func-
tions.
Here’s a simple example of orthogonal functions.
R1(x) = 1 if 0.5 ≤ x < 1.5
= 0 otherwise
R2(x) = 1 if 1.5 ≤ x < 2.5
= 0 otherwise
. . .
Rk(x) = 1 if k − 1/2 ≤ x < k + 1/2
= 0 otherwise
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These functions are orthogonal because
∫ ∞−∞
R∗k(x)Rl(x)dx = 0 if k 6= l
They are normalized in the sense that
∫ ∞−∞
Rk(x)dx = 1
The Rk’s are orthonormal.
The Rk’s can be used to represent, approximately,
any function. For ex., take g(x) = x2, for x > 0.
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g(x) ≈∑k
ckRk(x)
g(x) ≈ bR0(x) + (1 + b)R1(x)
+ (4 + b)R2(x) + (9 + b)R3(x) + . . . (A)
where b = 1/12. Why the “1/12”? The best ck’s
are obtained by taking
ck =
∫ ∞−∞
g∗(x)Rk(x)dx
That’s the average value of g(x) on the interval
k − 12 to k + 1
2:
ck =
∫ k+0.5
k−0.5x2dx
=1
3x3|k+0.5
k−0.5 = . . . = k2 +1
12
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Equation (A) is approximate because the Rk’s are
too wide, they give only a “coarse grained” repre-
sentation of g(x). If we make the rectangular Rk’s
infinitely thin, infinitely high, and normalized, we
get a complete set. Specifically, let
δ(x− xk) = 1/w if |x− xk| < w/2
= 0 otherwise∫ ∞−∞
δ(x− xk)dx = 1
and take the limit for w → 0. δ(x− xk) is called
a “Dirac delta function”.
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As it turns out, for any complete set of orthonor-
mal functions ψk, and any function g(x), we have
g(x) =
∞∑k=1
ck ψk(x)
ck =
∫ ∞−∞
g∗(x)ψk(x) dx
However, for the eigenfunctions φj of QM ope-
rators, the definition of normalization is slightly
different. It is
∫ ∞−∞
φ∗j(x)φj(x) dx = 1
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The function-vector analogy
The concepts of orthogonality, normalization, and completeness are easier to
grasp for vectors. Functions can be viewed as a natural generalization of vectors.
Viewed that way, the concepts of orthogonality, normalization, and completeness
for sets of functions do not seem so strange.
First, consider vectors in two-dimensional (2D) space. The position of any point
in 2D can be given with two coordinates x and y. The arrow that goes from the
origin to that point is a vector which we represent by ~r = |x, y >. Vectors can
also be represented by columns of numbers:
~r =
(x
y
)(12)
The transpose of a vector is that same vector written as a row of numbers. The
notation for the transpose of a vector is like this:
~rT = < x, y| =(x y
)(13)
Instead of ~r, “x” and “y”, it is preferable to use “~x”, “x1”, and “x2”, because
later on we will generalize definitions to the n-dimensional case. So
~x = |x1, x2 > =
(x1
x2
)(14)
~xT = < x1, x2| =(x1 x2
)(15)
The length of a vector is |~x| = (x21 + x2
2)1/2. When |~x| = 1, we say that ~x is
normalized.
By definition, the multiplication of a vector ~x = |x1, x2 > by a number c gives
c~x = |cx1, cx2 >. Geometrically speaking, multiplying a vector by c leaves its
orientation unchanged but changes its length by a factor c.
By definition, addition of two vectors gives ~x + ~y = |x1, x2 > +|y1, y2 >=
|x1 + y1, x2 + y2 >.
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The inner product ~xT~y of two vectors ~x and ~y is defined as the sum x1y1+x2y2.
The notation is
~xT~y = < x1, x2|y1, y2 > =(x1 x2
) (y1
y2
)= x1y1 + x2y2 (16)
Note that
~xT~y = x1y1 + x2y2 = y1x1 + y2x2 = ~yT~x
~xT~y = ~yT~x
The two basis vectors ~b1 = |1, 0 > and ~b2 = |0, 1 > are normalized and
orthogonal:
< 1, 0|1, 0 > = 1 + 0 = 1
< 0, 1|0, 1 > = 0 + 1 = 1
< 1, 0|0, 1 > = 0 + 0 = 0
Furthermore, ~b1 and ~b2 form a complete set for the representation of 2D vectors
because any vector ~v = |v1, v2 > can be written as a linear combination of ~b1 and~b2.
~v = |v1, v2 >= v1~b1 + v2~b2
Vectors ~b1 and ~b2 are just one of infinitely many possible choices of orthonormal
basis. Geometrically speaking, any two unit length vectors that make a right
angle are orthonormal. For instance, we could take s =√
2/2, and define a new
basis ~c1 = |s, s > and ~c2 = |s,−s >.
< s, s|s,−s > = s2 − s2 = 0
< s, s|s, s > = s2 + s2 = 1
< s,−s|s,−s > = s2 + s2 = 1
Any vector ~v = |v1, v2 > can be written as a linear combination in that new basis
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with coefficients a1 and a2.
|v1, v2 > = a1|s, s > +a2|s,−s >
For an orthonormal basis there is a simple formula to find the coefficients a1 and
a2. We illustrate it with |v1, v2 > and basis { ~c1,~c2 }.
a1 = ~cT1 ~v =< s, s|v1, v2| >= sv1 + sv2
a2 = ~cT2 ~v =< s,−s|v1, v2| >= sv1 − sv2
|v1, v2 >?= a1~c1 + a2~c2
= (sv1 + sv2)|s, s > +(sv1 − sv2)|s,−s >= |s2v1, s
2v1 > +|s2v2, s2v2 > +|s2v1,−s2v1 > +| − s2v2, s
2v2 >
= |2s2v1, 2s2v2 >
= |v1, v2 >
The definitions and formulas carry over to the general case of vectors in a n-
dimensional space.
|~x| = (~xT~x)1/2 = (< x|x >)1/2
= (< x1, x2, x3, . . . , xn|x1, x2, x3, . . . , xn >)1/2
= (x21 + x2
2 + x23 + . . .+ x2
n)1/2
c~x = c|x1, x2, x3, . . . , xn >
= |cx1, cx2, cx3, . . . , cxn >
~x+ ~y = |x1 + y1, x2 + y2, x3 + y3, . . . , xn + yn >
~xT~y = < x|y >= x1y1 + x2y2 + x3y3 + . . .+ xnyn
48
and one can define a set of n orthonormal basis vectors, for instance,
~b1 = |1, 0, 0, . . . , 0 >~b2 = |0, 1, 0, . . . , 0 >~b3 = |0, 0, 1, . . . , 0 >
. . .
~bn = |0, 0, 0, . . . , 1 >
~bTi~bj = δij
where δij, Kronecker’s delta, is 1 if i = j and 0 otherwise. As before, there are
infinitely many possible choices of orthonormal basis sets. For instance, if the
“. . . ” represent a series of 0’s, we could have:
~a1 = |0.48797,−0.31553, 0.81383, . . . >
~a2 = | − 0.63085,−0.77187, 0.07900, . . . >
~a3 = |0.60324,−0.55196,−0.57571, . . . >
~a4 = |0, 0, 0, 1, . . . , 0 >. . .
~an = |0, 0, 0, . . . , 1 >
As before, we can use an orthonormal basis to represent any vector ~v.
~v =∑j
cj~bj =∑j
(~bTj ~v)~bj =∑j
(~vT~bj)~bj
We can show that the formula for cj is correct:
~v =∑j
cj~bj or ~vT =∑j
cj~bTj
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If we multiply both sides by ~bk on the right, we get
~vT~bk =∑j
cj (~bTj~bk)
Since the vectors ~bj are orthonormal, ~bTj~bk = δjk and
~vT~bk =∑j
cjδjk
~vT~bk = ck = ~bTk~v
Now, consider a function f(x) defined over the range 0 ≤ x ≤ 1, for example
f(x) = 2x2 − x3. Suppose we don’t know the formula “2x2 − x3”; instead, we
only know the value of f(x) for certain values of x:
x: 0.000 0.200 0.400 0.600 0.800 1.000
f : 0.000 0.072 0.256 0.504 0.768 1.000
With these values, we can represent f(x) as a 6D vector
|0.000, 0.072, 0.256, 0.504, 0.768, 1.000 > .
Even without knowing “2x2 − x3”, we could use the table for all kinds of appro-
ximate calculations. For instance we could calculate∫ 1
0 f(x)dx:
∫ 1
0f(x)dx ≈ 0.1(0.000) + 0.2(0.072) + 0.2(0.256)
+0.2(0.504) + 0.2(0.768) + 0.1(1.000)
= 0.42000
The true value of the integral is (2 · 13x
3 − 14x
4|10 = 0.4166666. . . . Likewise, if
we had two functions f and g tabulated for the same 6 values of x, we could
approximate the integral of their product:
∫ 1
0f(x)g(x)dx ≈ 1
6(f1g1/2 + f2g2 + f3g3 + f4g4 + f5g5 + f6g6/2)
50
With a more complete table (a “finer grain” description of f) we could do more
accurate calculations. For example, with
x: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
f : 0.000 0.019 0.072 0.153 0.256 0.375 0.504 0.637 0.768 0.891 1.000
we would get∫ 1
0 f(x)dx ≈ 0.41750:
∫ 1
0f(x)dx ≈ 0.1×
(1
20 + 0.019 + 0.072 + 0.153 + . . .+ 0.891 +
1
21.000
)= 0.41750
and with n = 101 values of f for x = 0.00, 0.01, 0.02, . . . 0.99, 1.00, we get∫ 10 f(x)dx ≈ 0.4166750000. It’s clear that, in the limit of large n, we have a
perfect representation of f(x) and we can make exact calculations. If we also
have infinitely many values of g(x), we can calculate the overlap of f and g using
the inner product formula of vectors.∫ 1
0f(x)g(x)dx =
1
n
(1
2f1g1 + f2g2 + f3g3 + . . .+
1
2fngn
)When n→∞, multiplying the first and last term by 2 makes no difference, and
dividing by n− 1 instead of n makes no difference, so:
∫ 1
0f(x)g(x)dx =
1
n− 1(f1g1 + f2g2 + f3g3 + . . .+ fngn)
=1
n− 1
∑j
fjgj
∫ 1
0f(x)g(x)dx =
j=n∑j=1
fjgjdx
In the limit n→∞, the r.h.s. in the last equation is an integral. We could apply
51
the same reasoning for functions defined over x = 0 to x =∞, or over x = −∞to x = ∞. The conclusion is that, conceptually, a function f(x) can be viewed
as an infinite-dimensional vector. So all the results obtained for vectors carry
over to functions. In particular, if we have a complete basis of (infinitely many)
orthonormal functions φj(x)
< φj|φk > =
∫ ∞−∞
φ∗j(x)φk(x)dx = δjk
then we can write any function f(x) as a linear combination of the φj’s
f(x) =∑j
cjφj(x)
cj =
∫ ∞−∞
f ∗(x)φj(x)dx =
∫ ∞−∞
φ∗j(x)f(x)dx
52
Chapter 14
The Postulates of QM
53
We will write the postulates for a single particle in
1D, with spatial coordinate “x” and time “t”, but
they are easily generalized to systems of n parti-
cles in 3D by changing
x⇒ x, y, z
dx⇒ dxdydz
x⇒ x1, y1, z1, x2, . . . , yn, zn
etc.
54
Postulate 1
The wavefunction (wf) Ψ(x, t) completely speci-
fies the state of a system. Ψ∗(x0, t)Ψ(x0, t)dx =
|Ψ(x, t)|2dx is the probability of finding the par-
ticle between x0 and x0 + dx at time t.
In CM, positions x and velocities dx/dt completely
specify the state of a system. (I assume we know
the mass and charge of every particle.)
Note 1.1. Multiplying Ψ(x, t) by eiθ does not
change Ψ∗(x0, t)Ψ(x0, t)dx. So the phase factor
(θ) is immaterial.
55
Note 1.2. Since |Ψ(x, t)|2dx is a probability, we
must have
|Ψ(x, t)|2 ≥ 0∫ ∞−∞|Ψ(x, t)|2 dx = 1
Note 1.3. Ψ(x, t) must be single-valued and smooth:
∂2Ψ/∂x2 is not infinite, or, ∂Ψ/∂x has no discon-
tinuity.
56
Postulate 2
Every measurable property P has a corresponding
QM operator P , and the eigenvalues of that ope-
rator are real.
P P
kinetic energy T T = (−~2/2m)∂2/∂x2
potential energy V V = V (x)·energy E H = T + V
position x x = x·momentum px px = −i~ ∂/∂x
angular momentum Lx = −i~(y∂/∂z − z∂/∂y)
Ly, Lz : 4
57
Postulate 3
If A is a QM operator and Aφj(x) = ajφj(x),
then a measurement of property “A” must yield
one of the aj’s.
Note 3.1. The aj’s may form a discrete set, or
a continuous set.
58
Postulate 4
If we prepare a very large number N of identical
systems in the same state Ψ(x, 0) at time t = 0
(Ψ is normalized) and we then measure property
“A” for each system, the average of those mea-
surements will be
Avg(a) = < a >=
∫ ∞−∞
Ψ∗(x, t) (AΨ(x, t)) dx
59
Note 4.1. If Ψ(x, t) is one of the eigenfunctions
φj(x, t) of A, then
< a > =
∫ ∞−∞
φ∗j(x, t)(Aφj(x, t)) dx
=
∫ ∞−∞
φ∗j(x, t)ajφj(x, t)) dx
= aj
∫ ∞−∞
φ∗j(x, t)φj(x, t)) dx
= aj
60
Note 4.2. The eigenfunctions of A form a com-
plete set so we can write
Ψ =∑j
cjφj
< a > =
∫ ∞−∞
∑j
c∗jφ∗j
A
∑k
ckφk
dx
=∑j
∑k
c∗jck
∫ ∞−∞
φ∗j Aφk dx
=∑j
∑k
c∗jck ak
∫ ∞−∞
φ∗jφk dx
The φj’s are orthonormal, so the integral is 0 if
j 6= k and 1 if j = k. We represent this by the
“Kronecker delta” δjk:
δjk = 1 if j = k
= 0 if j 6= k
61
So
< a > =∑j
∑k
c∗jckakδjk
< a > =∑k
|ck|2ak
|ck|2 ≥ 0. We assumed that Ψ and the φj’s are
normalized. In that case
∑k
|ck|2 = 1
|ck|2: probability that the wf Ψ will change to φkas a consequence of measuring property “A”.
62
Note 4.3. Suppose that
Ψ = 0.632φ1 + 0.707φ2 + 0.316φ3
and Hφj = Ejφj with E1 = −10, E2 = −5,
E3 = −1.
< E > = 0.6322(−10) + 0.7072(−5) + 0.3162(−1)
= 0.40(−10) + 0.50(−5) + 0.10(−1)
= −6.60
< E > (and, in general, any property average) is
a weighted average of eigenvalues.
63
Suppose we prepare many identical systems in
that state Ψ and measure E for each system (#1
in the table below). We leave the systems un-
perturbed, and then measure E again (#2 in the
table). We would get something like this:
system 1 2 3 4 5 6 7 8 . . .
#1 -10 -1 -10 -10 -5 -10 -5 -5 . . .
#2 -10 -1 -10 -10 -5 -10 -5 -5 . . .
3 important things to notice:
1) the outcome of measurements performed on
identically prepared systems is probabilistic.
64
2) measuring property “A” of a system changes
its wf to an eigenfunction of A with eigenvalue aj(−10,−5, or −1 in our example), and aj is the
measured value. In general, measuring a pro-
perty changes the state of the system.
3) If, after the first measurement of “A”, the
system is left undisturbed, every subsequent mea-
surement of “A” will give the same aj.
65
Postulate 5
The time evolution of a system is deterministic
and given by
HΨ(x, t) = i~∂Ψ(x, t)
∂t
66
Note 5.1. Take an operator A, with eigenfunc-
tions Φj(x, t), that does not depend on time.
AΦj(x, t) = ajΦj(x, t)
We can write Φj(x, t) = φj(x)f (t) and preserve
the eigenvalue equation. To show that, substitute:
A(φj(x)f (t))?= ajφj(x)f (t)
f (t)Aφj(x) = f (t)ajφj(x)
Aφj(x) = ajφj(x)
So if A is time-independent it has time-independent
eigenfunctions φj(x).
67
Chapter 15
QM of simple systems
68
the free particle (FP)
By definition, no force acts on a “free particle”:
F = 0. But F = −dV/dx, so V is a constant.
We are free to choose any value for that constant:
take V (x) = 0.
In CM F = md2x/dt2 = 0, so x = x0 + v0t.
The FP moves at constant speed v0 from left to
right if v0 > 0, and right to left if v0 < 0.
69
In QM we note that V is independent of t and
substitute V (x) by 0 in the time-independent S
equation.
−~2
2m
d2ψ
dx2= Eψ(x)
d2ψ
dx2=
(−2mE
~2
)ψ(x)
ψ(x) equals its own second derivative times a cons-
tant, and it is finite everywhere. So ψ(x) can be
written either as: (i) a combination of sine and
cosine; or (ii) an exponential with imaginary ar-
gument. The latter is easier, so
ψ+(x) = A+ eikx
ψ−(x) = A− e−ikx
k =√
2mE/~
70
Euler’s formula tells us that kx is the argument
of cosine and sine functions, so kx = 2πx/λ.
2π/λ = k =√
2mE/~
E =~2k2
2m
k can be any real value, so E is continuous. Since
V = 0, E is just the kinetic energy of the FP. The
CM expression is E = mv2/2. So
2π/λ =
(2mmv2
2
)1/2
/(h/2π)
1/λ = mv/h
λ = h/mv = h/p
just like the DeBroglie formula.
71
When we get ψ(x) by solving the time-independent
S equation, we can get the full wf by multiplying
ψ(x) by e−iEt/~. Define ω = E/~,
Ψ+(x, t) = A+ eikx e−iωt = A+ e
i(kx−ωt)
The r.h.s. is the equation of a traveling wave going
from left to right. Likewise, Ψ−(x, t) is a wave
traveling from right to left. Ψ+(x, t) and Ψ−(x, t)
are called plane waves. They are constant energy,
constant velocity, waves.
Ψ+(x, t) and Ψ−(x, t) can not be normalized be-
cause they do not go to zero as x → ±∞. Their
square describes a density made of infinitely many
particles: |Ψ+(x, t)|2 = A2+e−ikxeikx = A2
+. If
we setA2+ = 1/2L, then |Ψ+(x, t)|2dx = A2
+dx =
(1/2L)dx, which means that there is one particle
for every 2L units of length.
72
the particle in a box (PIB)
VII(x) = 0 when 0 ≤ x ≤ a and
VI(x) = VIII(x) =∞The S equation is
d2ψ
dx2=
2m
~2(V (x)− E)ψ
d2ψdx2 can not be infinite, so ψ(x) = 0 in regions I
and III. ψ(x) is continuous, so
ψ(0) = ψ(a) = 0
73
The solutions are like for the FP, but now it’s eas-
ier to work with sine and cosine, so
ψ(x) = A sin kx + B cos kx
What are A,B, k?
0 = ψ(0) = A sin 0 + B cos 0 = B (1)
B = 0
0 = ψ(a) = A sin ka (2)
A 6= 0 or else ψ(x) = 0 everywhere. So ka = nπ
and k = nπ/a.
74
Next we use normalization to calculate A.
1 =
∫ a
0|ψ(x)|2 dx
=
∫ a
0A2 sin2(nπx/a) dx (3)
1 = A2[x
2− sin(2nπx/a)
(4nπ)/a
∣∣∣∣a0
= A2(a/2)
A = (2/a)1/2
so
ψn(x) =
√2
asin(nπx/a) n = 1, 2, 3, . . .
75
To find the energies (eigenvalues of H) we calcu-
late Hψ(x)/ψ(x).
−~2
2m
d2ψ
dx2=−~2
2m
(−n2π2
a2
)(2
a
)1/2
sin(nπxa
)
=h2n2
8ma2ψ(x)
So
En =h2n2
8ma2n = 1, 2, 3, . . .
76
Notes:
1. the only difference between the FP and the PIB
are the boundary conditions ψ(0) = ψ(a) = 0.
They give rise to wf localization and energy quan-
tization.
2. the lowest energy (zero point energy, ZPE) for
the PIB is E1 = h2/8ma2, it is not zero.
3. When a → ∞ the PIB becomes equivalent
to the FP
4. See Fig. 15.2 p346: as n↗, E ↗, the number
of nodes ↗ and λ↘.
5. Ψn(x, t) = (2/a)1/2 sin(nπx/a) e−iEt/~ is a
standing wave. In general
V = V (x)⇒ standing wave
V = V (x, t)⇒ traveling wave
77
6. The probability density |ψ(x)|2 is not uniform
(see Fig. 15.3 p347)
7. QM becomes like CM when T,m, or a increase.
78
Take a = 1 mm, m = 9.1× 10−31 kg, T = 300 K.
kBT = 4.14× 10−21 J
n =
(8ma2Enh2
)1/2
≈
(8ma2kBT
h2
)1/2
= 2.62× 105
P1: probability of finding the e− between
x = 0 and ∆x = 10−6 m;
P2: probability of finding the e− between
∆x = 10−6 m and 2∆x = 2× 10−6 m.
sin2(nπx/a) has roughly 262 maxima between x =
0 and ∆x = 10−6; but it might have 261 or 263.
The same goes for the interval ∆x to 2∆x.
79
So
δP =|P1 − P2|
12(P1 + P2)
/1
262= 0.004
So P1 and P2 (and P3, P4, . . . ) are equal to within
0.4%. δP will get even smaller if any one of these
things happen:
1. the box gets bigger, a↗
2. the temperature gets higher, T ↗
3. the particle gets heavier, m↗
4. the precision on measurements go down, ∆x↗
QM and CM become indistinguishable when a, T ,
m, or ∆x become large: this is the
correspondence principle
80
the PIB in 3D
V (x, y, z) = 0 when 0 ≤ x ≤ a
0 ≤ y ≤ b
0 ≤ z ≤ c
−~2
2m
(∂2
∂x2+∂2
∂y2+∂2
∂z2
)ψ = Eψ
Try
ψ(x, y, z) = X(x)Y (y)Z(z)
ψ = XY Z
This is called “separation of variables”.
81
−~2
2m
(Y Z
∂2X
∂x2+ XZ
∂2Y
∂y2+ XY
∂2Z
∂z2
)= EXY Z
divide by XY Z:
−~2
2m
(1
X
∂2X
∂x2+
1
Y
∂2Y
∂y2+
1
Z
∂2Z
∂z2
)= E (1)
−~2
2m
(X ′′
X
)=
~2
2m
(Y ′′
Y+Z ′′
Z
)+ E
The l.h.s. does not depend on y or z: call it Ex.
But the r.h.s. does not depend on x. Therefore
Ex is also independent of x: Ex is a constant.
Likewise,
−~2
2mY ′′/Y = Ey
−~2
2mZ ′′/Z = Ez
82
Ey and Ez are constants. Equation (1) becomes
Ex + Ey + Ez = E
and it can be split into 3 equations:
−~2
2m
∂2X
∂x2= ExX
4
These 3 equations are just like the S equation for
the PIB in 1D, they have the same solutions.
83
X(x) =
(2
a
)1/2
sin(nxπx/a)
Y (y) =
(2
b
)1/2
sin(nyπy/b)
Z(z) =
(2
c
)1/2
sin(nzπz/c)
and
Ex =n2xh
2
8ma2; Ey =
n2yh
2
8mb2; Ez =
n2zh
2
8mc2
E = Ex + Ey + Ez
The lowest energy levels have (nx, ny, nz) = (1,1,1),
(2,1,1), (1,2,1), (1,1,2), . . .
84
• we have 3 quantum numbers (QN) because it
is a 3D system
• ZPE= E111 = (h2/8m)(1/a2 + 1/b2 + 1/c2)
• If a = b = c, E211 = E121 = E112 and we
say that the second energy level is triply degene-
rate (g = 3).
• If a = b > c, E211 = E121: the second energy
level is doubly degenerate (g = 2).
• high symmetry boxes (cubes) show more de-
generacies than low symmetry ones (a 6= b 6= c).
85
the PIB and the postulates
If we know the energy E, we don’t know the posi-
tion x. For ex., if E = E3, we have this |ψ3(x)|2:
The particle could be anywhere (except x = 0,
a/3, 2a/3, a). If we measure x precisely it means
that |ψ(x)|2 now looks like this:
The new ψ(x) is not an eigenfunction of H , so we
don’t know E.
86
When n = 1, the probability of finding the parti-
cle in the middle third of the box is
Prob. =
∫ 2a/3
a/3
(2
a
)sin2(πx/a) dx
. . .
= 0.609
In CM it would be 0.333. With n = 1000, that
probability would also be very close to 0.333.
87
Eigenfunctions:
Ψn(x, t) = ψn(x) e−iEnt/~
The wf of a PIB is not necessarily one of the eigen-
functions. For ex., it could be
Ψ(x, t) = c1ψ1(x)e−iE1t/~ + c2ψ2(x)e−iE2t/~
Note that Ψ(x, t) 6= φ(x)f (t). Since ψ1 and ψ2
are orthonormal, we must have
|c1|2 + |c2|2 = 1
If we have many identical PIB with the same Ψ(x, t)
and measure E, we will get E1 |c1|2 of the times
and E2 |c2|2 of the times. The average of all these
measurements will be
< E > = |c1|2E1 + |c2|2E2
88
< E > =
∫ a
0
(c∗1ψ∗1 + c∗2ψ
∗2) H (c1ψ1 + c2ψ2) dx
= c∗1c1
∫ a
0
ψ∗1 H ψ1 dx
+ c∗2c2
∫ a
0
ψ∗2 H ψ2 dx
+ c∗1c2
∫ a
0
ψ∗1 H ψ2 dx
+ c∗2c1
∫ a
0
ψ∗2 H ψ1 dx
= |c1|2E1
∫ a
0
ψ∗1ψ1 dx
+ |c2|2E2
∫ a
0
ψ∗2ψ2 dx
+ c∗1c2E2
∫ a
0
ψ∗1ψ2 dx → 0
+ c∗2c1E1
∫ a
0
ψ∗2ψ1 dx → 0
= |c1|2E1 + |c2|2E2
89
For eigenstate “n”:
< p > =
∫ a
0ψ∗n(x)
(−i~ d
dx
)ψn(x) dx
. . .
= 0
The particle has equal probabilities of moving from
left to right at speed “v” and moving from right
to left at speed “v”.
Likewise, we can use a symmetry argument to
show that < x >= a/2.
90
Chapter 16
The PIB and chemistry
91
A particle in a box of finite depth has
V (x) = 0 − a/2 ≤ x ≤ a/2
= V0 |x| > a/2
In region II ψ(x) must resemble(
2a
)1/2cos(nπx/a).
In regions I and III we have
d2ψ
dx2=
2m(V0 − E)
~2ψ(x)
92
and ψ(x)→ 0 when x→ ±∞ (otherwise we can
not normalize ψ). So, in I and III, ψ has the form
ψ(x) = Ae−κx when x > a/2
= Beκx when x < −a/2
For a full solution, see p381, P16.7.
Eigenfunctions ψn(x): see p362, Fig 16.1.
In CM, E = V + T . In regions I and III, V > E,
therefore T < 0.
How can the kinetic energy be negative?
93
It isn’t. In QM we can not get the kinetic energy
at a point in space, x, by taking E − V (x). We
can calculate
< T > =
∫ ∞−∞
ψ∗n(x)
(−~2
2m
d2
dx2
)ψn(x) dx
The integrand is negative for x < −a/2 or x >
a/2, but the integral as a whole is positive.
When E < V0, regions I and III are classically
forbidden.
The wf “leaking” out of the box is called tun-
nelling.
94
Overlap of valence atomic orbitals (AOs)
Put two boxes in close proximity. The two wf
ψ1(x) die out rapidly outside their respective boxes.
But the two wf ψ5(x) extend far outside their
boxes and overlap.
Likewise, in Br2, the core AOs (1s, 2s, 2p, 3s, 3p,
3d) do not extend very far. But the valence AOs
(4s, 4p) are big, they overlap, and they are respon-
sible for the Br–Br chemical bond.
95
π-bonded networks
Molecules with alternating single and double bonds
have π molecular orbitals (MOs). The e− in π
MOs are loosely bound and can delocalize.
96
The PIB model of π bonds:
1. the e− move in 1D along the chain
2. e− do not interact
3. n e− fill the n/2 lowest π MOs
Take n = 6 for instance. The length of the box is
a ≈ (n + 1)× 1.4 A = 2.65(n + 1) a0
97
Atomic units (a.u.):
charge : 1 a.u. = |e| = 1.602177× 10−19 C
mass : 1 a.u. = me = 9.109383× 10−31 kg
length : 1 a.u. = a0 = 0.5291772 A
action : 1 a.u. = ~ = 1.054572× 10−34 J s
energy : 1 a.u. = 2Ry = 4.359748× 10−18 J
time : 1 a.u. = 2.418882× 10−17 s
98
With n = 6 e−, the frontier orbitals are n/2 = 3
(HOMO) and n/2 + 1 = 4 (LUMO).
∆E = E4 − E3 = (42 − 32)h2
8ma2
=
(42 − 32
72
)0.702713 a.u.
=
(1
7
)0.702713 a.u.
= 0.10039 a.u.
99
Likewise for n = 4 and n = 8 we find
∆E =
(1
5
)0.702713 a.u. = 0.14054 a.u.
∆E =
(1
9
)0.702713 a.u. = 0.07808 a.u.
The absorption wavelengths are
∆E = hν = hc/λ
λ = hc/∆E
We get these values:
n λcalc. (nm) λexp. (nm)
4 324 345
6 454 375
8 584 390
100
Note that
1. λcalc. are about right, in the UV-vis
2. λcalc.↗ as n↗, like λexp.
3. λcalc. can be in error by 50% or more
101
Metals
In a metal like Cu, the outermost valence e− are
loosely bound and delocalized over the entire solid.
Take a 1 cm long chain of Cu atoms.
dCu−Cu ≈ 2.5 A (4.72a0) so there are
10−2m
2.5× 10−10m= 4× 107
Cu atoms in the chain. Using the PIB model as
before, the frontier orbitals are nHOMO = 2×107
and nLUMO = 2×107 + 1. The energy difference
between them is
102
∆E = Eg =1
2× 107 + 1
(4π2
8× 4.722
)= 1.11× 10−8 a.u.
= 3.01× 10−7 eV
= 0.0000291 kJ/mol
At rt kBT = 2.5 kJ/mol: ∆E ≈ 1.2× 10−5 kBT .
Many e− are thermally excited, and metals are
good conductors of heat.
The energy levels are so close that they form a
nearly continuous energy band.
103
Apply a potential difference ∆V :
e− flow ⇒ electric current. Insulators have a
large Eg, they do not conduct electricity.
Eg (eV)
metals, semimetals ∼ 0
semiconductors <4
insulators ≥4
104
Chapter 17
Commutation of operators
105
Take two QM operators A and B:
Aψn(x) = αnψn(x)
Bφj(x) = βjφj(x)
Measuring “A” changes the wf to one of the ψn’s
and gives A = αn;
measuring “B” changes the wf to one of the φj’s
and gives B = βj.
We can measure A and B simultaneously and
get precise values A = αn and B = βj only if
φj(x) = ψn(x). In other words,
Bψn(x) = βnψn(x)
106
A and B have the same eigenfunctions, so
A Bψn(x) = A (βnψn(x))
= βn (Aψn(x))
= βnαnψn(x)
B Aψn(x) = B (αnψn(x))
= αn (Bψn(x))
= αn βnψn(x)
A Bψn(x) = B Aψn(x)
This is true for any ψn(x), so it’s also true for any
f (x) =∑k ckψk(x). Therefore
A B = B A
107
By definition, the commutator of A and B is
[A, B] ≡ AB − BA
Properties A and B can be measured simulta-
neously with perfect precision iff [A, B] = 0.
108
The kinetic energy T = −~2
2md2
dx2 and momentum
px = −i~ ddx commute: [T , px] = 0.
For the PIB in 1D, V (x) = 0 in the box, so it
would seem that [V , px] = 0, [H, px] = 0, and
precise values of energy and momentum can be
obtained simultaneously.
However, V (x) =∞ outside the box. V changes
abruptly at x = 0 and x = a, V and px do not
commute there, so they do not commute in gene-
ral. Precise values of energy and momentum can
not be obtained simultaneously for the PIB.
109
The Stern-Gerlach experiment
Magnet: S pole (z > 0) and N pole (z < 0).
A beam of Ag atoms going between the S and N
poles is split in 2.
So the operator Sz for “measure the z compo-
nent of the magnetic moment of Ag” has only
2 eigenvalues.
Szα = mz1α
Szβ = mz2β
and α and β form a complete set.
110
Next, we pass the Ag beam between the poles of
another magnet, with S-N along the x axis. It
splits in 2 again, so
Sxδ = mx1δ
Sxγ = mx2γ
Splitting the beam with a 3rd magnet having S-N
along z produces 2 beams again.
Together, these results show that Sz and Sx do
not have the same eigenfunctions: we can not
measure mz and mx simultaneously.
111
Heisenberg’s Uncertainty Principle (HUP)
[x, px]f (x) = x
(−i~ d
dx
)f −
(−i~ d
dx
)(xf )
= −i~xf ′ + i~(f + xf ′)
= i~f
[x, px] = i~·
We can not measure x and px exactly simultane-
ously. The HUP:
∆px∆x ≥ ~2
This “uncertainty” can not be detected for macro-
scopic objects because ~ is so small (∼ 10−34).
112
Note 1: a more general way to state the HUP is
the generalized uncertainty principle (GUP):
∆A∆B ≥ 1
2|i < [A, B] > |
Note 2: the GUP does not apply to situations
where the wavefunction ψ, or Aψ, or Bψ, or ABψ,
or BAψ, fails to satisfy the usual requirements of
a wavefunction (continuous, single-valued, norma-
lizable).
see http://arxiv.org/pdf/quant-ph/0011115.pdf
113
Chapter 18
Vibration and rotation of molecules
114
A N -atom molecule with Ne electrons has diffe-
rent types of motion.
• 3Ne electronic degrees of freedom (d.o.f.)
• 3 translations (along x, y, z)
• 3 (or 2) rotations (around the x, y, and z axes)
• 3N − 6 (or 3N − 5) vibrations
115
vibrations in formaldehyde
116
• electrons: electronic spectroscopy (UV-vis), pho-
toelectron spectra, electronic structure theory
(molecular orbitals, electrons’ correlation,
etc.)
• translations: FP and PIB models
• rotations: microwave spectroscopy,
rigid rotor model
• vibrations: infra-red (IR) and Raman spec-
troscopy, electron energy loss (EELS) spectra,
harmonic oscillator model and
normal mode analysis
117
Interaction potential in diatomics
V (r) ≈ De
(1− e−a(r−Re)
)2Morse
V (r) ≈ 1
2k(r −Re)2 harmonic
The Morse potential is more realistic, but the har-
monic potential is much simpler and it is OK near
the minimum (r ≈ Re).
118
The harmonic oscillator
2 atoms of mass m1 and m2 lie on the x axis at
x1 and x2.
x = (x2 − x1) − Re (Re is the equilibrium dis-
tance). For the interaction potential, we make the
harmonic approximation:
V (x) ≈ 1
2kx2
Then, the forces on atoms are
−F1 = F2 = −dVdx
= −kx
119
Newton’s law:
m1a1 = −m2a2
a1 = −m2
m1a2
d2x
dt2=
d2
dt2((x2 − x1)−Re)
= a2 − a1
= a2 +m2
m1a2
=
(m1 + m2
m1
)a2
(m2
m2
)=
(m1 + m2
m1m2
)m2a2
=F2
µwhere µ =
m1m2
m1 + m2
120
So we have
F2 = −kx = µd2x
dt2
The relative motion of the 2 atoms under the in-
fluence of forces F1 and F2 is equivalent to the
motion of a single object, of mass µ and coordi-
nate x, under the influence of force F2 = −kx.
121
the harmonic oscillator: CM
We need to solve
−kx = µd2x
dt2
d2x
dt2= −
(k
µ
)x(t)
Let ω =√k/µ.
x is real so it is a combination of sine and cosine:
x(t) = A cos(ωt) + B sin(ωt)
For simplicity, take x = 0 at t = 0. A = 0 and
x(t) = B sin(ωt)
The velocity v is
122
v(t) =dx
dt= Bω cos(ωt)
At t = 0 the velocity v0 = Bω cos(0) = Bω.
B = v0
õ/k
So
x(t) = v0
√µ/k sin(ωt)
If ν is the frequency, 1/ν = T is the period: the
argument of the sine function must be 2πνt so
that
2πν(t + 1/ν)− 2πνt = 2π
So we have
2πν = ω =
√k
µ
ν =1
2π
√k
µ
123
1. the frequency ν = 12π
√kµ is independent of the
energy stored in the oscillator.
2. by definition v = 0 at the turning points ±xt.So E = 1
2kx2t
3. when x = 0, V (x) = 0 and E = µv20/2.
Comparing with above gives xt = v0
õ/k
4. The average kinetic energy Tavg and average
potential energy Vavg can be calculated by
Tavg = 1T
∫ T0 (µv2/2)dt
and
Vavg = 1T
∫ T0 (kx2/2)dt
The result is Tavg = Vavg = 14µv
20 = 1
4kx2t
5. In CM v0 can be any real number, so E can be
any real number. In particular, E can be zero.
124
The average kinetic energy in the oscillator is
Tavg =1
T
∫ T
0
(µv2/2)dt
=1
T
1
2µv2
0
∫ T
0
cos2(√k/µ t)dt
=1
2Tµv2
0
(t/2 +
1
4
õ/k sin(2
√k/µ t)
)T0
=1
2Tµv2
0
(T/2 +
1
4
õ/k[sin(2
√k/µ 2π
√µ/k)− sin(0)]
)=
1
2Tµv2
0
(T/2 +
1
4
√µ/k[sin(4π)− sin 0]
)=
1
2Tµv2
0 (T/2) =µv2
0
4
The average potential energy in the oscillator is
Vavg =1
T
∫ T
0
(kx2/2)dt
=k
2T
∫ T
0
B2 sin2(ωt)dt
=k
2T
v20µ
k
(t/2− 1
4ωsin(2ωt)
)T0
=µv2
0
2T
(T/2− 1
4ω[sin(2
√k/µ 2π
√µ/k)− sin(0)]
)=
µv20
2T(T/2− [sin(4π)− sin(0)]) =
µv20
4
125
The rigid rotor in CM
The rigid rotor model of a diatomic molecule has
atoms of mass m1 and m2 separated by a fixed
distance r.
The rotating diatomic is equivalent to a single ob-
ject of mass µ on a circular orbit a distance r from
the origin.
µ =m1m2
m1 + m2
126
v = |~v| is constant, and ω = dθdt is constant.
v = rdθ
dt= rω
dθ =v
rdt
v is constant, but the direction changes:
|d~v| = vdθ =v2
rdt
|d~v|dt
= ac =v2
r
127
ac is the centripetal acceleration.
Velocity, position and angular velocity ω have mag-
nitude and orientation ⇒ ~v, ~r and ~ω:
~ω is perpendicular to the ~r-~v plane.
Notes:
(1) the last equation in (18.20) is wrong (~v and ~r
are perpendicular, not parallel);
(2) α = 0 in our case.
128
T =1
2µv2 =
1
2µr2ω2
=1
2Iω2 (∗)
where I = µr2. The linear momentum is ~p = m~v.
Define the angular momentum ~
~ = ~r × ~p
~ is perpendicular to the ~r-~v plane and its magni-
tude ` = |~| = rp = |~r||~p|. The kinetic energy T
of the rotating object is
T =p2
2µ=r2p2
2µr2=`2
2I(∗∗)
(*) and (**) show that ω, I , and ` are defined in
such a way that the expressions for T for linear
and circular motions are analogous.
129
linear rotation
motion
µ I = µr2
v = dxdt ω = dθ
dt
p = mv ` = rmv
T = µv2
2 T = Iω2
2
T = p2
2µ T = `2
2I
In CM ω, ` and T are continuous variables.
130
The QM harmonic oscillator
The S equation with V (x) = 12kx
2 is
− ~2
2µ
d2ψn(x)
dx+kx2
2ψn(x) = Enψn(x)
Solving that is tricky
. . .
. . .
. . .
The eigenvalues are
En =
(n +
1
2
)hν n = 0, 1, 2, 3, . . .
ν =1
2π
√k
µsame as in CM
131
The eigenfunctions are
ψn(x) = AnHn(√αx) e−αx
2/2
• An is a normalization factor (units: m−1/2)
• Hn(√αx) is the n’th Hermite polynomial: it
gives its nodal structure to ψn(x)
• e−αx2/2 makes ψn go to zero as x→ ±∞
• α =√kµ/~ (units: m−2)
132
Take y =√αx: the first 6 Hermite polynomials
are
H0(y) = 1
H1(y) = 2y
H2(y) = 4y2 − 2
H3(y) = 8y3 − 12y
H4(y) = 16y4 − 48y2 + 12
H5(y) = 32y5 − 160y3 + 120y
Hn(y) is even when n is even:
e.g., H4(−y) = H4(y)
Hn(y) is odd when n is odd:
e.g., H5(−y) = −H5(y)
An =1√
2nn!
(απ
)1/4
133
The energy is discrete, and the ZPE is hν/2.
ZPE= h4π
√kµ: the ZPE ↗, and quantum effects
become more obvious, as . . .
• µ↘: the atoms get lighter
• k ↗: the bond gets stiffer (the oscillator is con-
fined to a smaller range of x)
Note also that
• in QM, as in CM, Tavg = Vavg = E/2.
• ψn(x) resemble the wf of the PIB of finite depth.
In fact . . .
• . . . in the limit k → 0, we get a FP
134
e−αx2/2 is an even function (f(−x) = f(x))
An is a constant, it is even
Hn is even if n is even, odd if n is odd
odd × odd = even
even × even = even
even × odd = odd
So ψn is even if n is even, odd if n is odd.
If f(x) is odd ∫ ∞−∞
f(x) dx = 0
for ex.: ∫ ∞−∞
x dx = 0∫ ∞−∞
x3 dx = 0
. . .
If n is odd ∫ ∞−∞
ψm(x)ψm+n(x) dx = 0
More generally, one can show that∫ ∞−∞
ψm(x)ψn(x) dx = δmn
The ψn(x) form an orthonormal set.
If n is even ∫ ∞−∞
ψm(x) x ψm+n(x) dx = 0
One implication for IR spectra is that n = 0 → n = 2 transitions (“overtones”)
are forbidden. In practice, those peaks are very weak in IR spectra.
135
We verify that ψ1(x) is a solution to the S equation.
dψ1
dx=
d
dx
[1
2
(απ
)1/4(2√αx) e−αx
2/2]
= C × d
dx
(x e−αx
2/2)
= C ×(e−αx
2/2 + xe−αx2/2(−αx)
)= C ×
(e−αx
2/2(1− αx2))
d2ψ1
dx2 = C ×(e−αx
2/2(−αx)(1− αx2) + e−αx2/2(−2αx)
)= Ce−αx
2/2 ×(−αx+ α2x3 − 2αx
)= Ce−αx
2/2 ×(α2x3 − 3αx
)We sub that into the S equation
−~2
2µCe−αx
2/2(α2x3 − 3αx) +kx2
2C x e−αx
2/2 ?= E1C x e
−αx2/2
Divide both sides by Cxe−αx2/2:
−~2
2µ(α2x3 − 3αx)
1
x+kx2
2?= E1
−~2
2µ(α2x2 − 3α) +
kx2
2?= E1
−~2
2µ
(kµ
~2 x2 − 3α
)+kx2
2?= E1
−kx2
2+
3α~2
2µ+kx2
2?= E1
136
E1?=
3~2
2µ
√kµ
~
E1?=
3~2
√k
µ
E1?=
3h
2
(1
2π
√k
µ
)
E1?=
3h
2hν yes!
The r.h.s. is a constant which shows that Cxe−αx2/2 is indeed an eigenfunction.
And E1 = (1 + 12)hν, which agrees with the general formula En = (n+ 1
2)hν.
137
Rigid rotor in QM
The hamiltonian for N nuclei is approximately
Hnuclei ≈ Htrans. + Hrot. + Hvib.
Htrans. = Htrans.(X, Y, Z)
X =∑j
mjxj ÷∑j
mj
Hvib. = Hvib.(q1, q2, q3, . . .)
Hrot. = Hrot.(θ, φ)
Separation of variables gives
Enuclei = Etrans. + Evib. + Erot.
ψnuclei = ψtrans. × ψvib. × ψrot.
138
Consider the rotation of mass µ in a plane, a fixed
distance r from the origin, with variable angle φ.
We have V (φ) = 0, which is similar to the FP
and PIB, but with a periodic boundary condi-
tion:
Φ(φ + 2πn) = Φ(φ)
As a result, the wf and energies are similar to
those of the FP and PIB.
Φ(φ) = Aeimlφ ml = 0,±1,±2,±3, . . .
E =m2lh
2
2µ(2πr)2=m2l~
2
2I
ml = 0: the rotor is not moving and E = 0;
ml > 0: clockwise rotation;
ml < 0: anticlockwise rotation.
139
With rotation in the xy plane, the QM operator
for the z component of angular momentum is
ˆz = −i~ d
dφ
The wf that are eigenfunctions of the S equations
are also eigenfunctions of ˆz:
ˆz(Aeimlφ) = −i~A d
dφ(eimlφ)
= −i~ (iml) (Aeimlφ)
= ~ml Φml(φ)
So the eigenvalues of ˆz are ml~.
Note: the SI units for angular momentum are like
those of rp: mkgms−1 = kg m2 s−2 s = J s,
same units as h and ~.
140
Rotations in 3D, QM
Angles θ and φ are a natural choice of coordi-
nates. V (θ, φ) = 0. Unfortunately, T (θ, φ) is
complicated (see Eq. (18.44)) and solving the S
equation is difficult.
We will look only at the eigenfunctions Ymll (θ, φ)
and eigenvalues El and try to get insight from
them.
141
Spherical Harmonics
Ymll (θ, φ) = Θl(θ)Aeimlφ
Y 00 = N0
0
Y 01 = N0
1 cos θ
Y ±11 = N1
1 sin θ e±iφ
Y 02 = N0
2 (3 cos2 θ − 1)
Y ±12 = N1
2 sin θ cos θ e±iφ
Y ±22 = N2
2 sin2 θ e±2iφ
The same s-type, p-type, and d-type functions
that go into the definition of atomic orbitals.
142
Energy of the 3D rigid rotor
E ≡ El =~2
2Il(l + 1) l = 0, 1, 2, 3, . . .
Level l has degeneracy (2l + 1)
l = 0 1 2 3 4 5 6 . . .
S P D F G H I . . .
143
The spherical harmonics Ymll (θ, φ) are eigenfunc-
tions of H and of ˆ2 and ˆz:
ˆ2 Ymll (θ, φ) = ~2 l(l + 1) Y
mll (θ, φ)
ˆz Ymll (θ, φ) = ml ~ Y
mll (θ, φ)
So |~| = ~√l(l + 1).
ˆx, ˆy, and ˆz do not commute:
[ ˆx, ˆy] = i~ ˆz 4
H commutes with ˆz, but not with ˆx or ˆy.
[ ˆx, H ] 6= 0 ; [ ˆy, H ] 6= 0
[ˆz, H ] = 0
144
Why is z different from x or y? It is only a matter
of convention.
One can simultaneously measure precise values
of the energy, the square of the angular mo-
mentum, and one component of the angular
momentum.
In CM one could measure |~|, `x, `y, and `z si-
multaenously.
p426, Fig. 18.16, 18.17, and 18.18 give a semi-
classical depiction of ~.
In the classical limit, l→∞, all orientations of ~
are possible.
145
Chapter 19
Vibrational and rotational spectroscopy:
diatomic molecules
146
In absorption spectroscopy, the frequency ν of ab-
sorbed light matches an energy difference in the
molecule.
Ef − Ei = ∆Emolecule = Ephoton = hν
147
• ∆E vary over orders of magnitude.
• Line intensities I vary over orders of magnitude.
They can be zero: forbidden transitions.
• Natural linewidth: ∆(∆E) ∼ ~2∆t.
•Molecular interactions further broadens spectral
lines: inhomogeneous broadening.
• Hot bands originate from excited states.
148
type of technique spectral typical
transition range ν(cm−1)
nuclear NMR radio 0.05
spin states
rotational microwave 0.2–1
states
vibrational IR, EELS infrared 50–4000
states Raman
valence electronic UV-vis 12000–50000
electrons
core X-ray
electrons EXAFS X-ray ∼ 105–108
Auger
149
For CH3Cl we have . . .
• maybe 3 electronic states of interest?
• 9 vibrational QNs, each could be 0, 1, 2, . . .
• 3 rotational QNs, 2 of which determine energy
levels:
J = 0, 1, 2, . . .
K = −J,−J + 1, . . . , J − 1, J
If we assume 10 possible J , we get 1 + 3 + 5 +
. . . + 19 = 100 combinations of J and K.
If each vibrational QN can be 0 or 1, we get
29 = 512 vibrational states.
Total: 3× 100× 512 = 153,600 energy levels !
150
Light absorption (qualitative)
Light carries an oscillating electric field.
Take ν = 1000 cm−1, λ = 10−5m = 105 A.
At a given t, the electric field felt by a molecule of
dimension < 10 A is linear.
151
When νlight = νvibration the vibration is always
in sync with the oscillating field and energy can
be transferred from the light to the molecule
⇒ photon absorption.
If νlight 6= νvibration: no absorption.
If there is no static dipole: no absorption in IR.
In fact, there is no absorption when there is no
change of dipole associated with the vibration(s).
152
Consider levels 1 and 2 with populations N1 and
N2. The probability of . . .
. . . absorption ∝ N1 and B12,
spontaneous emission ∝ N2 and A21,
stimulated emission ∝ N2 and B21
LASER — Light Amplification by Stimulated Emis-
sion of Radiation
153
At equilibrium
N1B12ρ(ν) = N2 (B21ρ(ν) + A21)
ρ(ν): radiation density at frequency ν ;
the limit ρ(ν)→∞ gives B12 = B21
A21/B21 = 8πhν3/c3
154
Vibrational spectra have relatively few lines
and
are determined to a large degree by local modes
of vibration — CH stretch, CO stretch, CH2 wag
and scissor, CCO bend, HCCH torsion, etc.
155
At rt most diatomic molecules are in their ground
vibrational state. Take Br2, ν = 323 cm−1:
λ =1
323cm = 3.1× 10−5 m
∆E = hc/λ = 6.4× 10−21 J
N1
N0= e−6.4×10−21/kBT = e−1.5487 = 0.21
Take F2, ν = 892 cm−1:
N1
N0= e−1.5487×(892/323) = 0.014
For N2, ν = 2331 cm−1:
N1
N0= e−1.5487×(2331/323) = 0.000014
156
For H2, ν = 4160 cm−1:
N1
N0= e−1.5487×(4160/323) = 2.2× 10−9
n = 1 → n = 2, n = 2 → n = 3, . . . transitions
are called overtones. Overtones are forbidden.
The fundamental transition, n = 0 → n = 1,
is often the only one observed.
In a N–atom molecule, we expect at most 3N −6
intense peaks.
157
We can calculate the stiffness of a bond, k, from
ν01.
∆E = hcν01 =h
2π
√k
µ
k = (2πc)2 ν201 µ
= 5.891830× 10−5 × ν201 µ
where k is in SI (N m−1), ν01 is in cm−1, and µ
is in g/mol. For 1H35Cl ν01 = 2991 cm−1 and
k = 5.891830× 10−5 × 29912 ×(
1.008 · 34.969
1.008 + 34.969
)= 516 N m−1 = 516 kg s−2
158
DCl has the same k as HCl, butmD = 2.014 g/mol.
ν01 ∝ 1/√µ. For DCl, ν01 = 2145 m−1.
H/D isotopic substitutions produce large shifts.
12C/13C isotopic substitutions produce much smaller
shifts.
159
For a harmonic potential, V (r) = 12k(r−re)2 and
En =
(n +
1
2
)hν
For a Morse potential
V (r) = De
(1− e−α(r−re)
)2
α =
√k
2De
En = hν
(n +
1
2
)+
(hν)2
4De
(n +
1
2
)2
D0 = De −
(hν
2− (hν)2
16De
)
160
For N2, k = 2295 N m−1, re = 1.098 A, and
De ≈ D0 +hν
2= 1.593× 10−18 J
α =
√k
2De= 2.684 A
−1
How much stretch is needed to bring the potential
energy of N2 1/4 of the way up toward dissocia-
tion?
V (r) =De4
= De(1− e−y)2
y = − ln(1−√
1/4) = 0.6931
r − re = y/α = y
(2× 1.593× 10−18
2295
)1/2
= 0.26 A
161
N2 stretch for a potential energy of . . .
0.25De : 0.26 A
0.50De : 0.46 A
0.75De : 0.75 A
0.90De : 1.11 A
0.95De : 2.47 A
0.99De : 3.07 A
162
Selection rules: approximate rules for transitions
having nonzero probability.
For absorption mediated by the electric dipole,
intensity ∝ transition dipole squared
(µij)2 = (µijx )2 + (µ
ijy )2 + (µ
ijz )2
µijx =
∫ψ∗i µxψj
For a diatomic on the x axis, µx is a function of
interatomic distance r = re + x:
µx ≈ µ0 + x(t)dµ
dx|x=0 + . . .
µ0: static dipole
x(t)dµdx|x=0: dynamic dipole
163
For a 0 to m transition
µm0x = µ0
∫ψ∗mψ0
+dµ
dx|x=0 ×
∫ψ∗m xψ0
The first term is zero by orthogonality.
In the harmonic approximation, the 2nd term is
AmA0dµ
dx|x=0
∫Hm(α1/2x)xH0(α1/2x) e−αx
2dx
For even m:∫
even× odd× even× even = 0
If m is odd the integral may be nonzero. As it
turns out, it is zero for all m except m = 1. So
the only allowed transition from n = 0 is to n = 1.
164
More generally transitions due to electric dipole
interaction obey the selection rule
∆n = +1 for absorption
∆n = −1 for emission
165
Note:
• µ0 is irrelevant for intensity except that . . .
• . . . molecules with µ0 = 0 by symmetry also
have dµdx|x=0 = 0 and are IR inactive
• 99.57% of Earth’s atmosphere is made of N2,
O2 and Ar which are IR inactive
• H2O (0.25%), CO2 (0.040%), NOx, CH4, . . .
are IR active “greenhouse gases”
166
Beer-Lambert’s law
I/I0 = e−εM`
I0: incident light on the sample
I : transmitted light
`: optical path length
M : concentration of absorbers
ε = ε(λ) (L mol−1 m−1): molar absorption coef-
ficient
• ε(λ) ≈ 0 unless hc/λ matches an energy dif-
ference with nf = ni ± 1.
• high-symmetry molecules have fewer than (3N−6) allowed fundamental transitions. For ex., CH4
has only 2 (not 9) in IR spectra. See chapter 27.
167
• every line is broadened, mostly due to inter-
molecular interactions.
• ε(λ) ∝ the square of |~µij|: difficult to calcu-
late!
168
In the CO spectrum, Fig. 19.10, p442:
Why is the peak so broad?
Why does it have 2 asymetric parts on either side
of a dip?
⇒ rotational structure
169
Recall
EJ =~2
2IJ(J + 1) J = 0, 1, 2, . . .
EJ = hcB J(J + 1)
B = h/(8π2cµr2e)
The selection rule for rotational transitions (dipole
mechanism):
∆J = ±1
µ0 6= 0
(in Raman the rule is ∆J = 0,±2)
170
J J′ ∆E/hcB
0 1 +2
1 2 +4
2 3 +6
3 4 +8
1 0 −2
2 1 −4
3 2 −6
Transitions with ∆J = J ′−J = +1 are at higher
energy than hνvib.: R branch.
∆J = −1 transitions are at a lower energy than
hνvib.: P branch.
171
Rovibrational transition energies are
∆E = ∆Evib. + ∆Erot.
= hcνvib + 2hcB(J + 1) (R branch)
= hcνvib − 2hcBJ (P branch)
172
On p448, Fig. 19.18, J = 7 seems the most po-
pulated initial state for CO. At what temperature
was the spectrum recorded?
Boltzmann:
NJ ∝ (2J + 1)e−EJ/kBT
= (2J + 1)e−hcBJ(J+1)/kBT = (2J + 1)ey
When NJ is maximum we have
dNJdJ
= 0
0 = 2ey + (2J + 1)ey(−hcBkBT
)(2J + 1)
0 = 2−(hcB
kBT
)(2J + 1)2
173
So
T =
(hcB
kB
)(2J + 1)2
2
=
(hc× 193.13 m−1
kB
)152
2
= 313 K
The spectrum in Fig. 19.18 was recorded near rt.
174
How many rotational levels
are there between the n = 0 and n = 1
vibrational levels, roughly?
For H2
νvib. = 4401 cm−1
B = 60.853 cm−1
hcνvib.hcB
=4401
60.853= 72 = J(J + 1)
so J = 8: H2 has 9 rotational levels between n = 0
and n = 1.
I2 has roughly 75 rotational levels between n = 0
and n = 1.
175
Equilibrium bond lengths of diatomics
can be obtained
from microwave spectra
Spacings between lines in Fig. 19.18 show that
B ≈ 2.0 cm−1. Actually, B = 1.9313 cm−1.
µ =
(12.011× 16.00
12.011 + 16.00
)1
1000 gkg NA
= 1.13925× 10−26 kg
193.13 m−1 =
(h
8π2cµr2e
)
re =
(h
8π2cµ× 193.13 m−1
)1/2
re = 1.128 A
176
Selection Rules
pure rotational spectra
absorption:
µ0 6= 0
and
∆J = ±1
and
∆MJ = 0,±1
Raman scattering:
∆J = 0,±2
177
rovibrational spectra
absorption:
∆n = ± 1
and
∆J = ± 1
Raman scattering:
∆n = ± 1
and
∆J = 0,± 2
178
Chapter 20: the H atom
179
In atomic units (e = 1, me = 1, a0 = 1, ~ = 1,
and 4πε0 = 1) the potential felt by the electron is
V (r, θ, φ) = V (r) = −1/r.
Reduced mass: mpme/(mp +me) = 0.9994557
V does not depend on θ or φ. Therefore, sepa-
ration of variables leads to
ψ = R(r)Ymll (θ, φ)
Ymll (θ, φ) are the familiar s, p, d, f , . . . functions:
spherical harmonics.
180
Unlike the rigid rotor, r is variable
⇒ 3rd QN “n”
n = 1, 2, 3, . . .
l = 0, 1, 2, . . . , n− 1
ml = 0,±1,±2, . . . ,±l
181
The R(r) = Rn,l(r) have the form
Rn,l(r) = N Pn,l(r) e−Zr/n
N : normalization factor
Z: 1 for H, 2 for He+, 3 for Li2+, etc.
e−Zr/n: ψn,l,ml(r, θ, φ)→ 0 when r →∞
Pn,l(r) has (n− 1) radial nodes, it makes ψn,l,ml
with same l and ml but different n mutually or-
thogonal.
see p468, bottom
see 474, Fig. 20.6
182
Energies:
En =−Z2
2n2in a.u.
=−Z2 × 27.211383 eV
2n2
=−Z2 × 4.35974382× 10−18 J
2n2
Note: in H the energy does not depend on l or
ml, and En has degeneracy 2n2.
In many-electron atoms, electron removal ener-
gies (“orbital energies”) depend on n and on l.
183
Electron density
ρ(r, θ, φ) = |ψn,l,ml(r, θ, φ)|2
Probability of finding the e− within a shell of
thickness dr a distance r from the proton:
=
∫θ
∫φ|ψn,l,ml
(r, θ, φ)|2 r2 sin θ drdθ dφ
= R2n,l(r) r
2dr
Spherically-averaged density: 14π R
2n,l(r)
Note: atoms have a spherical e− density.
See p476 Fig. 20.9 / p479 Fig. 20.10 /
p480 Fig. 20.12
184
Chapter 21:
Many-electron atoms
185
The hamiltonian for a N -electron atom, in a.u.,
with the nucleus at the origin:
H = Te + Ven + Vee
Te =
N∑i=1
−1
2
(∂2
∂x2i
+∂2
∂y2i
+∂2
∂z2i
)
Ven =
N∑i=1
−Z|~ri|
Vee =
N∑i=2
i−1∑j=1
1
|~ri − ~rj|
Te and Ven are separable — they are sums of N
one-electron terms — but the e−–e− repulsion po-
tential Vee is not. Therefore . . .
186
• the exact wf is not a product of one-electron
functions
• the exact energy is not a sum of one-electron
energies
187
Orbital approximation
ψ(~r1, ~r2, . . . , ~rn) ≈ φ1(~r1)φ2(~r2) . . . φn(~rn)
~r1 : x1, y1, z1
φi(~r) = Rn,l(r)Ymll (θ, φ)
φi: an atomic orbital (AO) with n, l,ml and an
effective nuclear charge Zeff ≤ Z.
188
Li 1s22s1: Zeff (2s) ≈ 1.3 , Zeff (1s) ≈ 2.5
Ionization energies (eV):
IE1 ≈1.32
2× 22× 27.211 = 5.7
IE2 ≈2.52
2× 12× 27.211 = 85
IE3 =32
2× 12× 27.211 = 122.4
Expt (NIST): 5.3917, 75.640, and 122.454
189
e− with l 6= 0 have nonzero angular momentum
and magnetic moment.
e− with l = 0 have a zero angular momentum
. . . and zero magnetic moment?
Why is Ag [Kr]4d105s1 deflected by a magnet?
e− have intrinsic angular momentum, or spin,
s = 12. In analogy to ~l and l, there must be
2s + 1 = 2
possible states. Call those states “α” and “β”.
190
The z component of spin
sz = ms~ = ±1
2~
so
sz α = (~/2)α
sz β = −(~/2) β
and
s2α = ~2 s(s + 1)α =3
4~2α
s2 β =3
4~2 β∫
α∗α dσ =
∫β∗ β dσ = 1∫
α∗ β dσ =
∫β∗α dσ = 0
spin-orbital: AO × (α or β)
191
e− are indistinguishable: the probability of finding
an e− at ~r1 and another e− at ~r2 can be written
either
|ψ(~r1, ~r2)|2 dVor
|ψ(~r2, ~r1)|2 dV
where dV is a small volume element. In short-
hand notation,
|ψ(1, 2)|2 = |ψ(2, 1)|2
so
ψ(1, 2) = ψ(2, 1) (a)
or ψ(1, 2) = −ψ(2, 1) (b)
There are 2 kinds of particles in the world: bosons
obey equation (a), and fermions obey (b).
192
fermions bosons
electron photon
muon Higgs boson
quarks gluons
proton W, Z
neutron
Nuclei made of p protons and n neutrons are . . .
• fermions if n + p is odd
• bosons if n + p is even
193
Pauli:
The wavefunction of a n-electron system
changes sign under the exchange of
any two of its electrons
194
For He in its g.s. 1s2:
ψ(1, 2) = 1s(1)α(1)1s(2)β(2)− 1s(2)α(2)1s(1)β(1)
= 1s(1)1s(2) [α(1)β(2)− β(1)α(2)]
With the convention fg . . . h = f (1)g(2) . . . h(n),
we have
ψ(1, 2) = 1s1s [αβ − βα]
For He in the 1s12s1 configuration, we can either
make a symmetric (s) combination of AOs, or an
antisymmetric (a) combination:
φs = 1s2s + 2s1s
φa = 1s2s− 2s1s
195
4 combinations of spin functions:
fs1 = αα
fs2 = ββ
αβ
βα
The last 2 are unacceptable because they are nei-
ther symmetric (s) nor antisymmetric (a). But we
combine them to get
fs3 = αβ + βα
fa = αβ − βα
To get wf that are antisymmetric overall, we com-
bine φs with fa, and φa with fs1, fs2, fs3.
196
ψS = (1s2s + 2s1s) [αβ − βα]
ψT1 = (1s2s− 2s1s) [αα]
ψT2 = (1s2s− 2s1s) [ββ]
ψT3 = (1s2s− 2s1s) [αβ + βα]
Energy depends only on the spatial part of the wf.
We have a singly degenerate level ES (singlet) and
a triply degenerate level ET (triplet).
ET < ES
197
Slater determinants generate antisymmetric func-
tions from any number n of spin-orbitals. For ex.
for Li 1s22s1:
1s(j)α(j) ⇒ 1s(j)
1s(j)β(j) ⇒ 1s(j)
2s(j)α(j) ⇒ 2s(j)
ψ(1, 2, 3) =1√3!
∣∣∣∣∣∣∣1s(1) 1s(1) 2s(1)
1s(2) 1s(2) 2s(2)
1s(3) 1s(3) 2s(3)
∣∣∣∣∣∣∣=
1√6
(1s(1)
∣∣∣∣∣1s(2) 2s(2)
1s(3) 2s(3)
∣∣∣∣∣ − 1s(1)
∣∣∣∣∣1s(2) 2s(2)
1s(3) 2s(3)
∣∣∣∣∣+ 2s(1)
∣∣∣∣∣1s(2) 1s(2)
1s(3) 1s(3)
∣∣∣∣∣)
198
=1√6
(1s1s2s− 1s2s1s− 1s1s2s
+1s2s1s + 2s1s1s− 2s1s1s)
We get all n! permutations of n electrons in n spin-
orbitals, each with the right sign for antisymmetry.
199
Variational Theorem
Suppose hamiltonian H has eigenfunctions ψ0, ψ1,
ψ2, . . . and eigenvalues E0 < E1 < E2 < . . .
Any trial function φt can be written as
φt = c0ψ0 + c1ψ1 + c2ψ2 + . . .
200
The mean energy for state φt is
Et =
∫φ∗t (Hφt)
=
∫(c∗0ψ
∗0 + c∗1ψ
∗1 + c∗2ψ
∗2 + . . .)
×(H(c0ψ0 + c1ψ1 + c2ψ2 + . . .)
)
=
∫(c∗0ψ
∗0 + c∗1ψ
∗1 + c∗2ψ
∗2 + . . .)
× (c0E0ψ0 + c1E1ψ1 + c2E2ψ2 + . . . )
201
Et = c∗0c0E0
∫ψ∗0ψ0
+ c∗0c1E1
∫ψ∗0ψ1
+ c∗0c2E2
∫ψ∗0ψ2
+ . . .
+ c∗1c0E0
∫ψ∗1ψ0
+ c∗1c1E1
∫ψ∗1ψ1
+ c∗1c2E2
∫ψ∗1ψ2
+ . . .
+ c∗2c0E0
∫ψ∗2ψ0
+ c∗2c1E1
∫ψ∗2ψ1
+ c∗2c2E2
∫ψ∗2ψ2
+ . . .
202
Et = |c0|2E0 + |c1|2E1 + |c2|2E2 + . . .
1 = |c0|2 + |c1|2 + |c2|2 + . . .
So
Et = E[φt] ≥ E0
If we calculate Et as we vary φt, we can never
go below the true ground state energy E0.
203
Hartree-Fock
Self-Consistent Field Method
• make a Slater determinant of spin-orbitals
• the n-electron S equation breaks into n 1-electron
equations
• vary the Zeff of each AO, and other parameters,
so as to minimize Et
⇒ energy, electron density, dipole, an-
gular momentum, etc.
→ orbitals, atomic charges, bond orders, etc.
204
Shell structure of atoms: see p494, Fig. 21.6
Zeff : see p496, Table 21.2
IE ≈ −εHOMO: see p 497, Fig. 21.9
205
Aufbau principle: to get the g.s. configuration, fill
orbitals in order of increasing energy. For atoms:
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f 5g
6s 6p 6d . . . . . .
7s 7p . . . . . . . . .
then, transfer one e− from a s or f AO to the next
d AO for the elements
Cr, Cu / Nb, Mo, Ru, Rh, Pd∗, Ag /
La, Ce, Gd / Pt, Au / Ac, Th, Pa, U
∗ for Pd, transfer two e−.
206
Ionization energy (IE), energy required for
M →M+ + e−
Electron affinity (EA), energy required for
M−→M + e−
Mulliken’s electronegativity χM :
χM = (IE + EA)/2
Pearson-Parr hardness η:
η ≈ (IE − EA)/2
207
AOs with Zeff and QN “n”:
• IE and EA ↗ as Zeff ↗ or n↘,
left to right in the periodic table (PT) and
bottom to top in the PT
• radius of AO ↘ as Zeff ↗ and n↘
• range of IEs: 4.2—24.6 eV
• range of EAs: 0—3.6 eV
• superhalogens have EA > 3.6 eV (BH4)
IE, EA, and atomic radius: see p500 Fig. 21.12
208
Chapter 22:
Quantum States of
Many-electron atoms
209
The QNs for the H atom are n, l, ml and ms.
In N -electron atoms, e− are indistinguishable and
we do not have 4N QNs, we have only 5:
• angular momentum L
• spin S
• z component of ~L, ML
• z component of ~S, MS
• total angular momentum J
210
To a good approximation, the energy of an atom
depends on its
(1) electronic configuration (e.g., 1s22s22p2)
(2) L
(3) S
(4) J
Rules for ML and MS:
ML = −L,−L + 1,−L + 2, . . . ,+L
MS = −S,−S + 1,−S + 2, . . . ,+S
211
A set of degenerate atomic states with same L and
S is called a term.
L = 0, 1, 2, 3, 4, . . . ⇒ S, P, D, F, G . . .
The multiplicity is (2S + 1) by definition
Terms are represented by symbols 2S+1L, e.g.,
3P , 2D, 1S, . . .
If we take J into account, the notation becomes2S+1LJ , e.g.,
3P2, 3P1, 3P0, 1S0, . . .
212
But how do we get L, S and J?
We can not get L and S simply by adding up
the individual l and s of N electrons. We have to
follow a multi-step procedure.
213
Rules for getting L and S
from an electronic configuration
(1) Closed subshells contribute 0 to L and S.
Ne 1s22s22p6: L = 0 and S = 0, so 2S + 1 = 1,
the term is 1S and the only possible values of ML
and MS are 0.
He (1s2), Be (1s22s2), and Mg (1s22s22p63s2)
also have a singly degenerate 1S ground state.
214
(2) If only one e− is unpaired, the possible ml
and ms for that e− are the possible ML and MS
for the atom. Take
Sc 1s22s22p63s23p64s23d1
l = 2, ml = −2,−1, 0,+1,+2
s = 1/2, ms = −1/2,+1/2
SoML = −2,−1, 0,+1,+2 andMS = −1/2,+1/2.
Since ML varies between −L and +L, we con-
clude that L = 2.
Since MS varies between −S and +S, we con-
clude that S = 1/2, and 2S + 1 = 2.
Term 2D, 10-fold degenerate (2× 5 = 10)
215
(3) If there are two unpaired e− in different sub-
shells with l1, s1 = 12, and l2, s2 = 1
2, the possible
L and S are
L = l1 + l2, l1 + l2 − 1, . . . , |l1 − l2|
S = s1 + s2, . . . , |s1 − s2|
=1
2+
1
2, . . . , |1
2− 1
2|
= 1, 0
216
Example, C 1s22s22p13d1
Lmax = 1 + 2 ; Lmin = |1− 2|
L = 3, 2, 1
S = 1, 0
All 6 combinations are possible, giving terms and
degeneracies
3F : g = 211F : g = 73D : g = 151D : g = 53P : g = 91P : g = 3
a total of 21 + 7 + 15 + 5 + 9 + 3 = 60 states
grouped into 6 energy levels.
217
Note: with 1s22s22p13d1, there are 6 2p spin-
orbitals to choose from, and 10 3d spin-orbitals to
choose from, for a total of 60 states, which agrees
with the sum of term degeneracies.
218
(4) If there are many unpaired e−, all of which
are in different subshells, combine the li and siof 2 e−, then combine the resulting L and S with
the next e−, etc., until we’re done.
Example, Li 1s12p13d1. We can tell that there
are 2 × 6 × 10 = 120 different states. What are
the terms?
1s12p1: Lmax = 1 + 0, Lmin = |1− 0|.
So L = 1 and S = 1 or 0
(a) combine (1s12p1) (L = 1, S = 1) with 3d1:
Lmax = 1 + 2, Lmin = |2− 1| so L = 3, 2, 1
Smax = 1+1/2, Smin = |1−1/2| so S = 3/2, 1/2
219
That gives us
4F : g = 282F : g = 144D : g = 202D : g = 104P : g = 122P : g = 6
(b) combine (1s12p1) (L = 1, S = 0) with 3d1:
L = 3, 2, 1 as before; S = 1/2. We get
2F : g = 142D : g = 102P : g = 6
28 + 14 + 20 + 10 + 12 + 6 + 14 + 10 + 6 = 120
as it should. There are 9 energy levels and 120
quantum states derived from 1s12p13d1.
220
The excited electronic configuration
C 1s22s12p13p13d1
generates 2 × 6 × 6 × 10 = 720 quantum states
and the terms
5G, 3G, 3G, 3G, 1G, 1G, 5F, 3F, 3F,
3F, 1F, 1F , 5D, 3D, 3D, 3D, 1D, and 1D
221
(5) If there are two unpaired e− in the same sub-
shell, generate all possible pairs (ML,MS) and
work your way back to the possible L and S.
Take C 1s22s22p2: there are C62 ways of putting
two e− in 6 spin-orbitals:
C62 =
6!
2!(6− 2)!
=6× 5× 4× 3× 2
2× 4× 3× 2
=6× 5
2= 15
ms = −12
ms = +12
l = -1 0 +1
222
It yields 15 (ML, MS) pairs:
ML = -2 -1 -1 0 0 -1 -1 0 0 0 1 1 1 1 2
MS = 0 0 1 0 1 -1 0 -1 0 0 0 1 -1 0 0
(a) x x x x x
(b) X X X X X X X X X
(c) o
(a) Find the largest ML: +2. −L ≤ ML ≤ L,
so one of the terms must have L = 2. Among
ML = 2 cases, find the largest MS: it is 0. That
shows L = 2 is paired up with S = 0 ⇒ 1D.
Generate all (ML, MS) pairs of 1D
ML = −2,−1, 0,+1,+2
MS = 0
and cross them out from the table (see the x’s in
row (a)).
223
(b) Find the largest remaining ML (in columns
without “x”): +1. −L ≤ ML ≤ L, so we must
have a term with L = 1. Among ML = 1 cases,
find the largest MS: +1. So L = 1 is matched
with S = 1 ⇒ 3P . Generate all (ML, MS) pairs
of 3P
ML = −1, 0,+1
MS = −1, 0,+1
Cross those 9 combinations from the table (see X’s
in row (b)).
(c) Find the largest remaining ML: there is only
one, ML = 0 matched with MS = 0 (see “o” in
row (c)). That can only come from a 1S term.
224
All combinations are now accounted for. We have
3 terms — 1D, 3P , 1S. The degeneracies (5, 9,
and 1) add up to 15 as they should. We conclude
that the 1s22s22p2 configuration gives rise to 15
quantum states and 3 energy levels.
For d3, we would start by generating the C103 =
120 pairs (ML,MS) and follow the same steps as
in (5).
This can get tedious . . . But it has already been
done for all common configurations, see Table 22.3
on page 514.
225
the QN “J”
Given L and S, the possible J values are
J = L + S, L + S − 1, . . . , |L− S|
For instance we found that C 1s22s22p2 gives rise
to 3 terms — 1D, 3P , 1S. Rewriting these terms
with all possible J ’s gives
1D : L = 2, S = 0, so J = 2 ⇒ 1D2
3P : L = 1, S = 1, so J = 2, 1, 0 ⇒ 3P2,3P1,
3P0
1S : L = 0, S = 0, so J = 0 ⇒ 1S0
226
The degeneracy of term (2S+1)LJ is (2J + 1).
Hence, for a p2 configuration, we have 5+(5+3+
1) + 1 = 15 states, as before.
Once we have a configuration, and all of its terms,
we wish to find the most stable term. For that,
we use . . .
227
. . . Hund’s rules!
1/ The terms with highest multiplicity (2S + 1)
are lowest in energy.
2/ For a given multiplicity, the term with largest
L has the lowest energy.
3a/ If a subshell holds fewer than (2l + 1) e−
(it is less than half-filled) the lowest J has the
lowest energy.
3b/ If a subshell holds more than 2l e−, the
highest J has the lowest energy.
228
Examples
atom config. terms ground
state
C p2 1S0,1D2,
3P2,3P1,
3P03P0
O p4 1S0,1D2,
3P2,3P1,
3P03P2
F p5 2P3/2,2P1/2
2P3/2
V d3 4F, 4P, 2H, . . . 4F3/2
Co d7 4F, 4P, 2H, . . . 4F9/2
see NIST Atomic Spectra Database
229
Selection rules for
atomic absorption and emission
(dipole approximation)
• ∆l = ±1
• ∆L = 0,±1
• ∆J = 0,±1
• ∆S = 0