1

CHEM 3010

Introduction to

Quantum Chemistry,

Spectroscopy,

& Statistical Thermodynamics

2

Rene Fournier: Petrie 303, renef@yorku.ca,

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

3

Chapter 12

From Classical to

Quantum Mechanics

4

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.

5

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.

6

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 ? ? ?

7

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 ?

8

Blackbody radiation

Fig. 12.1, p295

Light is emitted, re-absorbed, and re-emitted many

times: it is in thermodynamic equilibrium with

the solid.

9

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!

10

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.

11

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!

12

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

13

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!

14

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

15

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!?

16

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.

17

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.

18

Chapter 13

The Schrodinger Equation

(S equation)

19

• 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

20

Diffraction of e− and Xe atoms

through a 1 A slit

21

Vibrations of H2 and Ar2 at T = 300 K

22

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

23

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

24

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

25

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)

26

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

27

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).

28

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)

]

29

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.

. . .

30

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).

31

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”, . . .

32

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).

33

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)

34

(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.

35

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).

36

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.

37

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.

38

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).

39

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

40

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

41

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.

42

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

43

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”.

44

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

45

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 >.

46

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

47

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

49

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