Chapter (6)Introduction

to

Quantum Mechanics

is a single valued function , continuous, and finite every where

Example(6.1) Normalizing the Wavefunction

• The initial wavefunction of a particle is given as (x,o)= C exp(x /xo), where C and xo are constants.

1- Sketch function. 2-Find C in terms of xo such that (x, 0) is normalized.• Solution The given wavefunction is symmetric, decaying

exponentially from the origin in either direction, as shown in Figure ,The decay length xo represents the distance over which the wave amplitude is diminished by the factor 1/e from its maximum value (0, 0) = C.

• The normalization requirement is

•Because the integrand is unchanged when x changes sign we may evaluate the integral

over the whole axis as twice that over the half-axis x > 0, where x = x. Then

EXAMPLE 6.2 Calculate the probability that the particle in the preceding example will be found in the interval

00 xxx solution

or about 86.5%, independent of xo.

Wavefunction For A Free Particle

• A free particle is one subject to no force

The wave number k and frequency of free particle matter waves are given by the de Broglie relations

For non relativistic particle is related to k as

The wave function for a free particle can be represented by a plane wave

This is an oscillation with wavenumber k, frequency , and amplitude A (travelling wave). If each plane wave constituting the packet is assumed to propagate independently of the others according to equation (2), the packet at any time is

(2)

From m

k

m

k

m

pk

222)(

2222

(1)

Example Find the wavefunction )x, 0( that results from taking the function

,)(22ke

cka

Solution

Write

, where C and are constants

or

For a free particle U)x(=0 , ,

Schrödinger equation :

Em

k

Eeeikikm

ikxikx

2

))((222

2

Which is the total energy

The Particle In A Box ( Infinite Square Well )

The particle can never be found outside, i.e.)x(=0 outside. Inside the box U)x(=0Schrödinger equation is

222

2

2

2

2

22

2),()(

2

)(

)()(

2

mEkxkx

m

E

dx

xdor

xEdx

xd

m

0L

U

(1)

k is the wavenumber of oscillation, the solution to this equation is.

Boundary condition: from the boundary conditions one obtains A and B

n=1,2,….

xL

nAxn

sin)(

Using Relation )1( becomes

Which shows that the particle energies are quantized and gives the energy levels

14

12

22

3

12

22

2

2

22

1

164

92

93

42

2

21

EEn

EmL

En

EmL

En

mLEn

Ground state or zero point energy

Exited state

Notice that E=0 is not allowed, that is the particle can never be at rest. E1 is zero-point energy > 0

Wave function:-Return to

xL

Ax sin)(1

xL

Ax 2

sin)(2

xL

Ax 3

sin)(3

Probability-:

imumAAL

xat

imumAAL

xat

LxatfoundnotisparticeltheeiA

Lxat

xL

Adxn

max2

3sin

4

3

max2

sin4

2.00sin

2

2sin,2

22

22

2

22

22

2

222

2

2

22

2

Normalization:-

xL

n

Lx

LA

LA

LALL

n

n

LLAx

L

n

n

LxA

n

2sin

2)(

22

2

1]

2sin

2[

2

1]

2sin

2[

2

11

2

2210

2

The Quantum Oscillator Consider the problem of a particle subject to a linear restoring forceF = - kx. Here x is the displacement of the particle from equilibrium )x =0( and k is the spring constant.

*The potential energy is

The angular frequency, m

k

The total energy,

The quantum oscillator is described by the potential energy

222

2

1

2

1)( xmkxxU In the Schrödinger equation

)2

1(

2

)()(2

1

2

)()()(2

2222

2

222

22

2

22

Exmm

dx

d

xExxmdx

d

m

xExxUdx

d

m

The solution may be in the form

Where C0 and are constants, using this solution, we get

2

22

2

]24[

)2()2()(

)2)((

220

022

02

2

0

x

xx

x

exc

ecexcdx

d

excdx

d

• Comparing with (1), we get

2

22

24

2

1.

24

2

22

2

2

222

222

22

m

m

mEE

m

mm

xmm

x

This is the ground state energy

2

10 E

With Eigen function:

]2

exp[)(22

00 xm

cx

Normalization of the ground state wavefunction

Exited state

2

32

1

3

2

2

626

24

1

22

2

222

EE

Emm

Em

mm

22

222

22

2

]64[]64[

24)2(

)2(

)(

2222

22

2

xx

xxx

xx

x

exxexx

exexexxdx

d

eexxdx

d

xex

Comparing with (1)

This is the first excited state Eigen energy So the energy levels for the harmonic oscillator is . )

2

1( nEn

The separation between any two levels is

n=0,1,2,…

E0 is the ground state energy level

Expectation values • A particle described by the wavefunction may occupy various

places x with probability given by 2

• We have several values of x, we need the average value of x• The average value of x, written x , is called the expectation value

and defined as

Which gives the average position of a particle. The average value of any function f(x) is

In quantum mechanics the standered derivation, x, is called the uncertainty in position, and given by

The degree to which particle position is fuzzy is given by the magnitude of x. the position is sharp only if x=0

x

EXAMPLE: Location of a Particle in a BoxCompute the average position x and the quantum uncertaintyin this value, x, for the particle in a box, assuming it is in the

ground state.

SolutionThe ground state wavefunction is

with n =1 for the ground state. The average position is calculatedas