Wave nature of particles

We have thus far developed a model of atomic structure basedon the particle nature of matter:Atoms have a dense nucleus of positive charge with electronsorbiting the nucleus in rather peculiar quantum orbits.Electrons are thus negatively charged particles.Or are they ?….

Wave nature of particles

De Broglie wavelength: De Broglie postulated that in analogy to light, matter could also have particle and wave characteristics.

Recall the energy, momentum relations for photons:

E = hf p =h

!

Similarly a wavelength and frequency for matter waves can be definedas

f =E

h! =

h

p

Wave nature of particles

De Broglie wavelength: De Broglie postulated that in analogy to light, matter could also have particle and wave characteristics.

Through what potential difference must an electron be accelerated tohave a de Broglie wavelength of 500nm ?

p = mv =h

!" v =

h

m!

v =h

m!=

6.63"10#34 J s

9.11"10#31 kg( ) 500 "10#9 m( )= 1456 m/s

e!V =1

2mv

2" !V =

mv2

2e=

9.11#10$31 kg( ) 1456 m/s( )2

2 1.6 #10$19 C( )= 6.0 #10$6 V

Wave nature of particles

De Broglie wavelength: De Broglie postulated that in analogy to light, matter could also have particle and wave characteristics.

A proton is confined to a nucleus of diameter 10fm. What is its kineticenergy?

p = mv =h

!" v =

h

m!

v =h

m!=

6.63"10#34 J s

1.67 "10#27 kg( ) 10 "10#15 m( )= 3.970 "107 m/s

K =1

2mv

2=

1

21.67 !10"27 kg( ) 3.970 !107 m/s( )

2

!1 eV

1.6 !10"19 J= 8.2 MeV

De Broglie wavelength and the Bohr atomDe Broglie’s matter wave provided an explanation of the quantization ofangular momentum in the Bohr atom:

2!r = n"

Waves travelling in opposite directions in a confined space can set up astanding wave due to constructive interference.

A standing wave in a circle is formed when an integer number ofwavelengths fits around the circumference:

Using the expression for the de Broglie wavelength we obtain thecondition for quantized angular momentum:

! =h

p=

h

mv

" mvr = nh

2#= n!

Davisson-Germer ExperimentDavisson and Germer showed that electron do behave like matter waves and can be diffracted.

Electrons are reflected from the surface of a nickel target.The surface layer of atoms in nickel acts as a diffraction grating.

Davisson-Germer ExperimentDavisson and Germer showed that electron do behave like matter waves and can be diffracted.

Kinetic energy of incoming electrons:

Substituting into the expression for the de Broglie wavelength,

1

2mv

2= eV ! v = 2Ve m

! =h

mv=

h

2Vem

V = 54V !" =h

2Vem= 0.167nm

Davisson-Germer ExperimentDavisson and Germer showed that electron do behave like matter waves and can be diffracted.

Now consider the electrons hitting thesurface of nickel:

After reflection from the surface, the electrons are observed to makea pattern of bright and dark spots. Can you explain why?

Davisson-Germer ExperimentDavisson and Germer showed that electron do behave like matter waves and can be diffracted.

Constructive interference occurs when thepath difference between adjacent reflectedelectron waves is an integer number ofwavelengths:

D-G experimental value for the electron wavelength agreed well withthe de Broglie prediction

Lattice spacing d in nickel is 0.215nm.1 st diffraction maximum was observed at50 degrees:

d sin! = n"

! =d sin"

1= 0.167nm

Matter waves (but waves matter too)

• Electron diffraction was demonstrated by Davisson and Germer and G. P.Thomson (son of J. J. Thomson).

• Subsequently, diffraction for other matter waves such as atoms, neutronsand even molecules has been demonstrated.

• Atom interferometers can be constructed using standing waves of lightas atomic mirrors and beam splitters

Matter waves (but waves matter too)

• All matter can have wavelike properties.

• The electron microscope is based on the wave properties of electrons.

Check outhttp://www.pbrc.hawaii.edu/microangela/for more cool pictures

Matter wavesWhy do we not observe matter wave effects for macroscopic objects?

De Broglie of a 1g object moving at 10m/s:

De Broglie wavelength of an electron moving at 0.02c:

Macroscopic objects have wavelengths that are far too small to interactwith slits etc to show interference or diffraction.

! =h

p=

h

mv=6.63"10

#34Js

1g( )(10m / s)= 6.63"10

#32m

! =h

p=

h

mv=

h

0.02mec= 0.36nm

Matter wavesAn electron (or any other microscopic particle) has a particular finiteextent (size).

How do we describe a wave of a particular finite extent?

Wave packetsSingle wave:Consider a single wave which is a solution to the wave equation:

!2 y

dx2=

1

v2

!y2

!t2or

k 2

"2

!y2

!t2

Such a wave has infinite extent and does not represent physical waves.All real waves are limited to finite regions: wave packets.

y = Acos kx !"t( )

k =2!

" ! = 2" f =

2"

T

v

p= ! f =

"

k

Wave number: Angular frequency:

Phase velocity: Amplitude: APeriod: T

Wave packetsSuperposition of waves:Consider the superposition (addition) of two waves of equal amplitudebut slightly different wavelengths and frequencies:

y = y1+ y

2= Acos k

1x !"

1t( ) + Acos k2x !"2

t( )

y x,t( ) = 2Acos!kx

2"!#t

2

$%&

'()

cos kx "#t( )

k =k

1+ k

2

2, ! =

!1+!

2

2 ,

"k = k2# k

1, "! =!

2#!

1

The new wave has a wave envelope modulating a high frequency wave.

Wave packetsSuperposition of waves:

y x,t( ) = 2Acos!kx

2"!#t

2

$%&

'()

cos kx "#t( )

k =

k1+ k

2

2, ! =

!1+!

2

2 , "k = k

2# k

1, "! =!

2#!

1

The velocity of the envelope is different from the velocity of the wavewithin the envelope:

vp=!

k=

!1+!

2( ) 2

k1+ k

2( ) 2

vg="!

"k

Wave packetsSuperposition of waves:

Adding two waves is not enough to create a wavepacket that is localized.One must add many waves to obtain a wave packet:

Wave packetsSuperposition of waves:Reciprocity:If the spatial extent of the envelope decreases then the range ofwavelengths required increases.

If the temporal extent of the envelope decreases, then the range offrequencies required increases:

!x!k " 1

!t!# " 1

Wave packetsSuperposition of waves:Reciprocity:A radio station broadcasts at 10.0 MHz. What range of frequencies isrequired to broadcast a 0.80µs pulse?

!t!f " 1

# !f "1

!t=

1

0.8µs= 1.250MHz

Central frequency :10.0MHz

Frequency range :10.0MHz ± 0.6250MHz

# 9.375MHz < f < 10.625MHz

Wave packetsSuperposition of waves:

Phase velocity of the wavepacket:

Group velocity:

Taking the derivative, weget a dispersion relation:

v

g=

d

dkkv

p( ) = vp+ k

dvp

dk

dvp

dk= 0 ! v

g= v

p (non dispersive medium)

dvp

dk" 0 ! v

g" v

p (wave packet spreading)

v

g=

d!

dk=

d

dkkv

p( )

v

p=!

k

Wave packetsMatter wave packets:Individual matter waves have frequency and wavelength

v

g=

d

dkkv

p( ) = vp+ k

dvp

dk= u

f =E

h! =

h

p=2"

k

vp= f ! =

E

p=!

2k 2c2+ m2c4

!k= c 1+

mc2

!2k 2

" c

The phase velocity of matter waves is greater than than or equal to thespeed of light.A wave packet made up of a range of wave numbers will showdispersion (spreading) even in free space.

The group velocity of the matter wave packet is equal to the particle’sspeed u.

Heisenberg Uncertainty Principle

GaussianGaussian

p = !k , E = !!

!p!x "

!

2

x

k

Minimum uncertainty:

• Our knowledge of conjugate quantities is inherently uncertain.

• No matter how good our measuring instruments are we cannot

simultaneously know x and p or E and t with complete precision.

!x!k =

1

2, !"!t =

1

2

!E!t "!

2

For Gaussian functions:

!p!x =

!

2

!E!t =!

2

Heisenberg Uncertainty Principle

Scientists can use lasers to trap sodium atoms in a box of length 1mm.

What is the minimum temperature that the atoms can have?

!p!x = m!v!x "

!

2#!v "

!

2m!x

Range of possible velocities centred around 0:

!!

4m"x# v #

!

4m"x

T =mvrms

2

3kB

=m

3kB

1

2vmax

!

"#$

%&

2

=1

3kB

!2

64m'x2

!

"##

$

%&&

=1

3 1.38(10)23 J/K( )

1.05(10)34Js( )

2

64 23(1.67 (10)27 kg( ) 1(10)3m( )

2

!

"

####

$

%

&&&&

= 0.1(10)15 K

Currently scientist have achieved cooling of about 1nK !