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Sixth Edition

+ Lab Manual

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Lecture 1: Classical Mechanics and the Schrödinger Equation

This lecture covers the following parts of Atkins 1. Further information 4. Classical mechanics (pp 911- 914 )2. 11.3 The Schrödinger Equation (pp 294)

Lecture-on-line Introduction to Classical mechanics and the Schrödinger equation (PowerPoint) Introduction to Classical mechanics and the Schrödinger equation (PDF)

Handout.Lecture1 (PDF) Taylor Expansion (MS-WORD)

Tutorials on-line The postulates of quantum mechanics (This is the writeup for Dry-lab-II)( This lecture has covered (briefly) postulates 1-2)(You are not expected to understand even postulates 1 and 2 fully after this lecture) The Development of Classical Mechanics Experimental Background for Quantum mecahnics Early Development of Quantum mechanics The Schrödinger Equation The Time Independent Schrödinger Equation

Audio-Visuals on-line Quantum mechanics as the foundation of Chemistry (quick time movie ****, 6 MB)Why Quantum Mechanics (quick time movie from the Wilson page ****, 16 MB) Why Quantum Mechanics (PowerPoint version without animations) Slides from the text book (From the CD included in Atkins ,**)

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Classical Mechanics

A particle in 3-D has the following attributes

X

Y

Z

1. Mass m

m

mass

r r

Posit ion 2. Position

r r

r v = d

r r /dt

velocity

3. Velocity r v

Rate of change of position with time

Expression for total energy

ET =Ekin+Epot(r r )

The total energy of a particle with position

r r ,

mass m and velocity r v also has energy

Kinetic energy dueto motion

Potential energy due to forces

v p

v v small mass large velocity

v v

large mass small velocity

or

Linear Momentum and Kinetic Energy

Ek =12

mv2

The kinetic energy can be written as :

r p =m v v

Or alternatively in terms of the linear momentum:

as:

Ek =p2

2m

A particle moving in a potential energy field V is subject to a force

V(x)

X

F=-dV/dx

Force in one dimension

Force in direction of decreasing potential energy

The potential energy and force

v F =−dV

dxw e x−

dVdy

w e yPotential energy V

The force has the direction of steepest descend

Force F

v F =−(dV/dx)

r ex −(dV/dy) v ey−(dV/dz) v ez

v F =−

v∇ V =−gradV

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The expression for the total energy in terms of the potential energy and the kinetic energy given in terms of the linear momentum

The Hamiltonian will take on a special importance in the transformation from classical physics to quantum mechanics

E =Ekin + Epot=

p2

2m+V(

r r )

is called the Hamiltonian

H =

p2

2m+V(

r r )

The Classical Hamiltonian

Quantum Mechanics

The particle is moving in the potential V(x,y,z)

Classical HamiltonianWe consider a particle of mass m,

Linear momentum

r p =m

r v

and positionr r

r r

r p =

X

Y

Z m

Positionmass

mv r

Linear Momentum

Classical Hamiltonian

r r

r p =

X

Y

Z m

Positionmass

mv r

Linear Momentum

The classical Hamiltonian is given by

H =12m

px2 + py

2 + pz2( ) +V(x,y,z)

H =

12m

r p⋅ r p +V( v r ) = 1

2mp2 +V( v r)

Quantum Mechanical Hamiltonian

The quantum mechanical Hamiltonian ˆ H is constructed by thefollowing transformations :

HClass → ˆ H =12m

ˆ px2 + ˆ py

2 + ˆ pz2( ) +V( ˆ x, ˆ y, ˆ z)

Classical Mechanics Quantum Mechanics

x px ˆ x −> x ; ̂ px−>hiδδx

y py ˆ y−> y ; ̂ py−>hiδδy

z pz ˆ z−> z ; ̂ pz−>hiδδz

Here h ' h - bar'=

h2π

is a modification of Plancks constanth

h=1.05457 × 10−34 Js

ˆ H =12m

( ˆ px2 + ˆ py

2 + ˆ pz2) +V( ˆ x, ˆ y, ˆ z)

= 12m

[(hiδδx

×hiδδx

) + (hiδδy

×hiδδy

) + (hiδδz

×hiδδz)] +V(x,y, z)

We have

hiδδy

×hiδδy

=h2

i2δδy

×δδy

=−h2δ2

δy2

Thus

ˆ H =−

h2

2m[δ2

δx2 +δ2

δy2 +δ2

δz2]+V(x,y,z)

By introducing the Laplacian : ∇2 =δ2

δx2+

δ2

δy2+

δ2

δz2 we have

ˆ H =−h2m

∇2 +V(x, y, z)

It is now a postulate of quantum mechanics that :

the solutions Ψ(x, y, z) to the Schrödinger equation

ˆ H Ψ(x, y, z)=EΨ(x, y, z)

−h2

2m∇2Ψ(

r r ) +V( r r )Ψ(

r r) =EΨ( r r)

−h2

2m[δ2Ψδx2

+δ2Ψδy2

+δ2Ψδy2

] +V(x,y,z)Ψ =EΨ

Contains all kinetic information about a particle moving in the Potential V(x,y,z)

ˆ H =−

h2

2m[δ2

δx2 +δ2

δy2 +δ2

δz2]+V(x,y,z)

What you should learn from this lecture

Definition of :

Linear momentum (pm),

kinetic energy(p2

2m);

Potential Energy

Relation between force F

and potential energy V (r F =-

r ∇ V)

The definition of the Hamiltonian (H)

as the sum of kinetic and potential energy,

with the potential energy written in terms

of the linear momentum

For single particle: H=

p2

2m+V(

r r )

You must know that :The quantum mechanical Hamiltonian ˆ H is constructed from the classical Hamiltonian H by the transformation

HClass → ˆ H =12m

ˆ px2 + ˆ py

2 + ˆ pz2( ) +V( ˆ x, ˆ y, ˆ z)

Classical Mechanics Quantum Mechanics

x px ˆ x −> x ; ̂ px−>hiδδx

y py ˆ y−> y ; ̂ py−>hiδδy

z pz ˆ z−> z ; ̂ pz−>hiδδz

The position of the particle is a function of time.

Let us assume that the particle at t =tohas the position

r r (to )

and the velocity r v (to ) =(d

r r /dt)t=to

What is v r (to +Δ )t = v r(t1)= ?

v r (to +Δt)=

v r (to)+(d

v r /dt)t=to Δt+

12

(d2v r /dt2)t=to Δt2

v r (to +Δt)=

v r (to)+

v v (to)Δt+

12

(d2v r /dt2)t=to Δt2

By Taylor expansion around r r (to )

or

Newton's Equation and determination of position..cont

v r (t o )

v r (to

+Δt)

(d2 v r / dt2 )t=to

Δt2 (d

v r / dt )t=toΔt

Appendix A

v r (t o )

v r (to

+Δt)

v v (t o )Δ t (d

2v r / dt2 )t=toΔt2

v r (to +Δt)=

v r (to)+

v v (to)Δt+

12

(d2v r /dt2)t=to Δt2

v F (to ) =−

v∇ V =−gradV=m(d2 v r /dt2 )t=to

However from Newtons law:

v r (to +Δt)=

v r (to)+

v v (to)Δt-

12m

(gradV)t=t0Δt2Thus :

Newton's Equation and determination of position..contAppendix A

v r (t o )

v r (to

+Δt) v v (to)Δt

- 1m(gradV)

t=toΔt

v r (to +Δt)=

v r (to)+

v v (to)Δt-

12m

(gradV)t=t0Δt2

Newton's Equation and determination of position..cont

At the later time : t1 =to +Δt we have

v r (t1+Δt)=

v r (t1)+(d

v r /dt)t=t1Δt+

12(d2v

r /dt2)t=t1Δt2(1)

The last term on the right hand side of eq(1) can again be determined from Newtons equation

v F (t1 ) =−

v∇ V =−gradV=m(d2 v r /dt2 )t=t1

as

(d2v r / dt2 )t=t1 =−

1m(gradV)t=t1

Newton's Equation and determination of position..cont

Appendix A

We can determine the first term on the right side of eq(1) By a Taylor expansion of the velocity

v r (t1+Δt)=

v r (t1)+(d

v r /dt)t=t1Δt+

12m

(gradV)t=t1Δt2(1)

(d

v r /dt)t=t1 =(d

v r /dt)t=t0 +

12(d2v

r /dt2)t=t0Δtor

(d

v r /dt)t=t1 =v v (to)−

12m

(gradV)t=toΔt

Where both: v v (to ) and

1m

(gradV)t=to are known

Newton's Equation and determination of position..contAppendix A

The position of a particle is determined at all times from the position and velocity at to

v v (t2)=(d

v r /dt)t=t2 =v v (t1)−

1m

(gradV)t=t1Δt

Newton's Equation and determination of position..cont

v r (t2+Δt)=

v r (t2)+

v v (t2)Δt+12(d2v

r /dt2)t=t2Δt2

(d2v

r /dt2)t=t2 =−1m

(gradV)t=t2

At t2 =t0 +2Δt what about v r (t2+Δt) ?

v r (t2 )

v r (t2 +Δt)

v v (t2 )Δt

- 1m(gradV)

t= t 2

Δ t

Appendix A