CHAPTER 8 1

CHAPTER 8 The Quantum Theory of Motion

I. Translational motion.

A. Single particle in free space, 1-D.

1. Schrodinger eqn

€

H ψ = Eψ

−!2

2md2

dx2ψ = Eψ ; no boundary conditions

2. General solution:

ψ = Aeikx +Be−ikx 3. Energy eigenvalues

E = k2!2

2m

Note: all values of k are permitted, not just integers. Therefore the

energy of a free particle is not quantized. E can take on any (+) value.

4. Note: ψ can also be written as: ψ = Aʹ′ sin(kx) + Bʹ′ cos(kx) Since:

€

eiθ = cos θ +i sin θ

€

e−iθ = cos θ – i sin θ

CHAPTER 8 2

B. A particle in a one-dimensional box. 1. The potential function V(x). V(x) = 0 for 0 < x < L V(x) =

€

∞ elsewhere

2. Schrodinger equation:

€

ˆ H ψ = Eψ where

€

H = kinetic energy operator + V(x)

=−!2

2md2

dx2+ V(x)

So we can write:

−!2

2md2

dx2ψ(x)+ V(x)ψ(x) = Eψ(x)

V(x) = 0 in box = ∞ outside box Above is equivalent to equation:

−!2

2md2ψ(x)dx2

= Eψ(x) plus boundary condition.

Boundary condition: x restricted to interval from 0 to L (in which case

V(x) = 0) and ψ is required to vanish at x=0 and x=L. Boundary condition allows us to drop V(x) term.

Now, need to find ψ(x) which is a solution of this differential eqn. In other words, find a function ψ such that, if we differentiate with

kinetic energy operator we obtain Eψ; that is, a constant times ψ.

CHAPTER 8 3

Solution:

ψn= Bsin nπx

L

#

$%%

&

'(( n=1, 2, 3, …

whole series normaliz. quantum number of solutions coefficient (state number) ψ1, ψ2, ψ3,…& associated energy eigenvalues

En=n2h2

8mL2

Problem Let’s check to see if this is a solution:

−!2

2md2

dx2ψn=−!2

2md2

dx2Bsin nπx

L

$

%&&

'

())

=−!2

2mddxBcos nπx

L

$

%&&

'

())nπL

=−!2

2mBnπL

−( )sin nπxL

$

%&&

'

())nπL

=+!2

2mB nπL

$

%&&

'

())

2

sin nπxL

$

%&&

'

())

=!2

2mn2π2

L2*

+,,

-

.//Bsin nπx

L

$

%&&

'

())

n2h2

8mL2 ψn

= Enψn

or

Hψn= E

nψn

CHAPTER 8 4

Interpretation: Lowest energy quantum state: n=1

€

ψ1 =2Lsin πx

L; E1 =

h2

8mL2

First excited state n=2

€

ψ2 =2Lsin2πx

L; E2 =

4h2

8mL2

Energy level diagram: Wavefunctions for n=1 through 5 Limit of large n: ψ∞*ψ∞ ≈ uniform probability (just like classical result)

CHAPTER 8 5

Correspondence principle: limit (Quantum Results) = Classical results

€

n → ∞ Orthogonality Condition: If ψn is a correct, well-constructed solution to the Schrodinger

equation, then the wavefunctions are mutually orthogonal:

dx ψi*

0

L

∫ ψj= 1 if i = j

and

€

dx ψi*

0

L

∫ ψj = 0 if i ≠ j

ψi is said to be orthogonal to ψj. Illustration of orthogonality. Here multiply n=1 and n=3 wave

functions together and then integrate the product from 0 to L. Should get 0.

CHAPTER 8 6

Practical Applications (particle in 1-D box): Energy levels of delocalized π electrons in long conjugated molecules: hexatriene

€

En =n2h2

8mL2

where: En is the energy levels for π electrons m is electron mass L ≈ 6 times C-C bond length Spectroscopic transitions in the hexatriene system: 6 electrons fill 3 levels: remember, Pauli Exclusion principle

excludes more than 2 e- from occupying a given quantum state.

C= C -- C = C -- C = C

C -- C -- C -- C -- C -- C

Delocalized π-bond system Approximately 1-D

6 electrons

L Length

ground state

n=4 n=3

n=2

n=1

xx

xx

xx

1st excited state

n=4 n=3

n=2

n=1

x

x ΔE = hν, where ν is the

frequency of light promoting transition xx

xx

CHAPTER 8 7

So can use: Particle-in-box to predict ν-spectrum Pauli Excl princ ΔE=hν

C. Particle in a 2-D Box (i.e. a particle confined to a surface)

1. Picture a = length in x-direction b = length in y-direction

2.

€

ˆ H ψ = Eψ

€

ˆ H =−2

2m∂2

∂x2 +∂2

∂y2

$

% &

'

( ) partial differential equation now

Solved by separation of variables technique (see text).

3.

€

Enxny=

h2

8mnx

2

a2 +ny

2

b2

"

# $ $

%

& ' ' nx,ny =1,2,3...∞

€

ψnx ,ny=

4ab

sin nxπxa

$

% &

'

( ) sin

nyπy

b

$

% &

'

( )

4. Special case: Energy level diagram when a=b=L (square box)

Energy (3,1) (1,3) 10h2/8mL2 (2,2) 8h2/8mL2

(2,1) (1,2) 5h2/8mL2 degeneracy

(1,1) 2h2/8mL2 (nx,ny)

CHAPTER 8 8

5. Attempt to draw pictures of ψ2 . (1,1) (1,2) (2,1) (2,2) The above is first example of degeneracy = two or more different wave

functions (quantum states) having the same energy. e.g., states (1,2) and (2,1) states (1,3) and (3,1)

D. Particle in 3-D Box: 1. c a b

€

ψnx ,ny ,nz=

8abc

sin nxπxa

$

% &

'

( ) sin

nyπy

b

$

% &

'

( ) sin

nzπzc

$

% &

'

( )

€

Enx ,ny ,nz=

h2

8mnx

2

a2 +ny

2

b2 +nz

2

c2

"

# $ $

%

& ' '

CHAPTER 8 9

2. Energy level diagram for a=b=c (CUBE)

II. Vibrational Motion.

A. The 1-D Harmonic Oscillator.

1. An important model for bond vibration.

V(x) =

€

12 kx2

where k= bond force constant (stiffness)

x = displacement of bond length b from equilibrium length bo

x = b - bo

x = 0 represents bond at its equilibrium distance

F=-kx (Force)

CHAPTER 8 10

2. Classical trajectory.

x(t) = A sin(2πνt) (if vibration starts at x=0 moving in (+) direction)

A = amplitude of vibration (can be any value)

ν = frequency of oscillation

€

ν =12π

kµ

let’s also define

€

ω ≡2πν =kµ

µ =

€

m1m2

m1 + m2 “reduced mass”

3. Quantum mechanical Schrodinger Eqn.

Hψ = Eψ where H = EK+ V

EK=−!2

2µd2

dx2as usual

V =k2

x2

S.E. is then:

−2

2µd2

dx2ψ(x)+ k

2x2ψ(x) = Eψ

4. Stationary Solutions: ψv(x) vibrational quantum number = v = 0,1,2,3…∞ Energies

€

Ev = v + 12( )hν

Ev = v + 12( )ω

= 12ω,

32ω,

52ω

(evenly spacedlevels)

m1 m2

CHAPTER 8 11

where:

€

ν =12π

kµ

OR

€

ω =kµ

Wavefunctions:

€

ψν x( ) = NvHv(y)e−y2 /2;where y ≡

xα

and α ≡ ( 2

µk)1 /4

€

Nv = normalization constant

Hv(y)=the Hermite polynomials

The Hermite polynomials Ground state

€

ψo x( ) = Noe−x2 /2α2

= Gaussian function symmetric function around x=0 (said to be an even function)

CHAPTER 8 12

€

ψo(x) =ψo(−x) property of even function 1st excited state

€

ψ1 x( ) = N12xα

e−x2 /2a2

x times Gaussian

an odd function

€

ψ1 x( ) = −ψ1 −x( ) ψ1 is odd because:

€

even function

e−x2 /2α2

$

% & &

'

( ) ) ×

odd functionx

$

% &

'

( )

= odd function( )

€

ψ2 x( ) = N24x2

α2 −2%

& '

(

) * e−x

2 /2α2

even power even power even of x of x (even function) (x0=1) ψ2 is an even function

CHAPTER 8 13

Here are the first 5 vibrational state wavefunctions:

Here are the probability density functions ψ*ψ

CHAPTER 8 14

5. Notes on Parity (symmetry of functions). ψ0 has even parity ψ1 has odd parity ψ2 has even parity ψ3 has odd parity etc.

(even function) x (even function) = even function

(even function) x (odd function) = odd function

(odd function) x (odd function) = even function

Integrals:

€

dx f x( )−∞

∞

∫ = 2 dx f x( )0

∞

∫ if f(x) is even

€

dx f x( )−∞

∞

∫ = 0 if f(x) is odd

Very useful properties in solving certain integrals. Some functions, however, have neither even or odd parity.

Examples:

Knowing the wavefunctions we can calculate a variety of properties

of the harmonic oscillator in any given quantum state v, such as average position <x>, average <x2>, the average potential energy <kx2/2>. Here < > denotes “expectation value”.

See text for examples.

0 0

CHAPTER 8 15

III. Rotational Motion and Angular Momentum.

A. Rotation in Two Dimensions (particle on a ring)

1. Physical picture: particle of mass m constrained to move on a circular path of radius r in the x-y plane.

2. Treat classically first. Energy is all kinetic, so classically E = p2/2m Classical angular momentum around the z-axis is Jz =+ or -pr So we can write:

€

E =Jz2

2mr2 =Jz2

2I; where I=moment of inertia

3. Apply quantization condition simply using de Broglie equation

€

since Jz = ±pr

and

λ =hp

then

Jz = ±hrλ

opposite signs denote opposite directions of travel

However, not all wavelengths λ are allowed but only those that exactly repeat themselves around the ring (they satisfy periodic b.c.).

CHAPTER 8 16

Thus, only wavelengths obeying the following condition will work:

λ =2πrm

wherem= 0,±1,±2,.... (here is the angular

momentum quantum #)J

z= m

So we have the quantization of angular momentum.

4. Now what about the energies:

E =Jz2

2I=m22

2I

5. The wavefunctions:

ψm(φ) = eimφ

(2π)1/2

CHAPTER 8 17

B. The 2-Particle Rigid Rotor (rotation in 3-D)

1. Physical Picture.

masses m1 and m2 moment of inertia = I = µd2 3-dimensional problem central force problem (that is, the 2 particles are held at fixed distance by inwardly

directed forces which counteract centrifugal forces.) Importance: good quantum mechanical model for rotation of

molecules. 2. Full Coordinate System:

3. Simplifying Features:

Since both particles rotate about a fixed center of mass (C.M.), knowledge of the position of 1 particle automatically implies position of 2nd particle is known.

Thus, really a 1-particle problem with an effective mass µ (called the reduced mass) moving on surface of sphere of radius d.

€

µ =m1m2

m1 + m2

CHAPTER 8 18

So: • translational motion of system

2 particle separates as whole (motion of the C.M.) problem into

• internal (rotational) motion characterized by a single effective particle of mass µ confined to motion on sphere of radius d.

this is the part of interest here The rigid rotor problem is mathematically homologous with a single particle of mass m=µ on a sphere of radius r=d. Still 3-D problem! 4. S.E. Hψ = Eψ

€

ˆ H = −2

2µ∇2 + ˆ V

∇2 is Laplacian operator:

€

∂2

∂x2+

∂2

∂y2+

∂2

∂z2 (in Cart. Coord.)

V = 0, if the particle remains constrained to the spherical surface. 5. For convenience in central force problems, convert to spherical polar

coordinates. (See “The Chemist’s toolkit 7B.1, page 295, Atkins 10th) x = r sin θ cos φ y = r sin θ sin φ z = r cos θ r fixed (= d) in rigid rotor

CHAPTER 8 19

∇2ψ =∂2ψ

∂r2+2r∂ψ

∂r+1r2

1sin2 θ

∂2ψ

∂φ2+1sinθ

∂

∂θsinθ ∂ψ

∂θ

&

'((

)

*++

&

'((

)

*++

=0 radial portion (kinetic energy of radial motion = 0 since r=d rigid)

angular portion

=1d2

Λ2ψ( ) where r=d

€

Λ ≈ orbital angular momentum operator

€

Λ2 = the legendrian operator

€

ˆ H ψ = Eψ −!2

2IΛ2ψ = Eψ

where I = µd2 = moment of inertia of rotor 6. Solution ψ(θ, φ) (function of the 2 polar angles)

ψ(θ, φ) =

€

Y,m θ, φ( ) where Y is special family of functions called

the “spherical harmonics”

Quantum state can be specified by 2 quantum numbers:

, m

= 0, 1, 2… orbital angular momentum quantum number

m = -, …0…+ magnetic quantum number

7. Now refer to Table 8C.1 in Atkins. (Note: H atom wave functions have these very same functions times a radial function.)

8. Energies of the rotor.

€

E = + 1( ) 2

2I depend only on , not m.

(m=-1 0 +1 =1

€

E = 2 3( ) 2

2I

€

E = 1 2( ) 2

2I

CHAPTER 8 20

9. Degeneracy = (2 + 1) 10. Angular momentum J and Jz.

€

J = + 1( ) magnitude of the angular momentum

€

Jz = m angular momentum about the z-axis

Table 8C.1 The spherical harmonics

11. Artist’s rendering of spherical harmonics using ml=0

(m=-2) -1 0 1 2 =2

=0, m=0

€

E = 0

CHAPTER 8 21

CHAPTER 8 22

Conversion of notation to the rigid diatomic rotor:

Use EJ = J J+1( ) !2

2I

where J= rotational energy level of rotating diatomic

CHAPTER 8 23

IV. Intrinsic Spin of Microscopic Particles

A. Properties

1. Spin emerges when special relativity is applied to quantum mechanics.

2. Microscopic particles possess an intrinsic angular momentum about their axis called spin.

3. This is treated differently than orbital angular momentum because it has different boundary conditions.

4. The magnitude of the spin angular momentum of a particle is determined by the spin quantum number s, and is given by:

€

s(s +1){ }1 /2

5. The value of s for a given type of particle is fixed. It cannot change. For an electron or proton, s=1/2.

6. The component of spin angular momentum along an axis z defined by an external magnetic field is dictated by the spin magnetic quantum number ms.

ms can take on values from -s, -s+1, … s-1, s

For an electron or proton where s=1/2 ms can take on values -1/2 and +1/2 Properties of angular momentum

J=2

M=-1 0 1 J=1

J=0

CHAPTER 8 24

V. Tunneling. A. Definition: passage of a particle through an energy barrier that exceeds

the energy of the particle. i.e. passage through a classically forbidden zone.

B. Example.

Classically, the particle of energy E trapped in the well will remain

trapped forever. In Q.M., it has finite probability of escape. In fact, ψ of the particle has a small contribution through and outside

barrier.

CHAPTER 8 25

κ = spatial decay parameter =

€

2m Vb −E( )2

# $ %

& %

' ( %

) %

1 /2

C. Transmission probability ∝

€

e−κL

Note that the larger κ is, the more ψ is damped as it passes through

barrier.

As κ ↑ Transmission Probability ↓

κ depends on Vb – barrier height Vb ↑ κ ↑ Prob ↓ E – energy of particle E ↑ κ ↓ Prob ↑ m – mass of particle m ↑ κ ↑ Prob ↓ D. Exact calculation of transmission probability through square barriers can

be obtained by properly piecing together solutions to S.E. in each region.

Matching boundary conditions provide us simultaneous eqns to solve to

obtain weighting coefficient for each region.

CHAPTER 8 26

E. Result:

Transmission Probability

€

T = 1+eκL −e−κL( )216ε 1− ε( )

%

& '

( '

)

* '

+ '

−1

€

ε =EV

where E is the energy of incoming particle, V is barrier height

F. Approx. result in limit of high and long barrier.

€

κL >> 1

eκL >> e−κL

Simplest results:

T ≈ 16ε 1− ε( )e−2κL T depends on L, m, E, V

CHAPTER 8 27

VI. Approximation Techniques.

A. Most interesting systems cannot be solved exactly in terms of known analytic functions. Two main approximation approaches are used:

1. Variation theory (will introduce with molecular orbital theory)

2. Perturbation theory (here)

B. Time-independent perturbation theory

1. Hamiltonian split into easy part (0-th order part) with known eigenfunctions and eigenvalues, and the complicating part (the perturbation):

€

ˆ H = ˆ H (0) + ˆ H (1)

2. Energy eigenvalues are then represented by a series of corrections, with the 0-th term being the eigenvalue of the non-perturbed Hamiltonian.

€

E = E(0) +E(1) +E(2) + ... 3. Wavefunctions work the same way, with the 0-th order term being

the eigenfunctions of the o-th order Hamiltonian.

€

ψ =ψ(0) +ψ(1) +ψ(2) + ... 4. It can be shown that the 1st order correction to the energy (of the

ground state) can be written as:

€

Eo(1) = ψo

(o)* ˆ H (1)∫ ψo(o)dτ

Notice that the 1st order energy correction requires only a knowledge

of the 0-th order wavefunction (the unperturbed solution). It is the perturbation Hamiltonian averaged over the unperturbed

probability distribution. 5. The 2nd-order energy correction is a bit more complicated:

€

Eo(2) =

Hno(1) 2

Eo(o) −En

(o)n≠0∑

where we have introduced the notation often used in QM:

Ωnm

= ψn*∫ Ωψ

mdτ

CHAPTER 8 28

C. Time-dependent perturbation theory

1. Extremely useful in spectroscopy for calculating transition probability between quantum states due to electromagnetic wave (light).

2. Perturbation Hamiltonian is now time-dependent:

€

ˆ H = ˆ H (0) + ˆ H (1)(t) 3. Let’s specifically treat the perturbing effect of an oscillating electric

field of frequency ω and amplitude E interacting with a molecule’s dipole moment µz:

€

ˆ H (1)(t) = −µzEcosωt 4. It can be shown that the rate of transition between two quantum

states i and f can be written as:

€

wf←i ∝E2 ψf*µzψi∫

2

The term in absolute value brackets is called the transition dipole

moment, which is central to the understanding of spectroscopy. It is a measure of the charge redistribution that accompanies a transition.