chapter 5 the wavelike properties of particles. de broglie wavesde broglie waves electron...

Chapter 5Chapter 5

The Wavelike Properties of The Wavelike Properties of ParticlesParticles

• De Broglie WavesDe Broglie Waves• Electron ScatteringElectron Scattering• Wave MotionWave Motion• Waves or Particles?Waves or Particles?• Uncertainty PrincipleUncertainty Principle• Probability, Wave Functions, Probability, Wave Functions,

and the Copenhagen and the Copenhagen InterpretationInterpretation

• Particle in a BoxParticle in a Box

Wave Properties of Matter and Quantum Mechanics Wave Properties of Matter and Quantum Mechanics

I thus arrived at the overall concept which guided my studies: for both I thus arrived at the overall concept which guided my studies: for both matter and radiations, light in particular, it is necessary to introduce the matter and radiations, light in particular, it is necessary to introduce the corpuscle concept and the wave concept at the same time.corpuscle concept and the wave concept at the same time.

- Louis de Broglie, 1929- Louis de Broglie, 1929

Louis de Broglie (1892-1987)

CHAPTER 5CHAPTER 5

The Wavelike Properties of ParticlesThe Wavelike Properties of Particles

• The de Broglie HypothesisThe de Broglie Hypothesis• Measurements of Particles WavelengthsMeasurements of Particles Wavelengths• Wave PacketsWave Packets• The Probabilistic Interpretation of the Wave The Probabilistic Interpretation of the Wave

FunctionFunction• The Uncertainty PrincipleThe Uncertainty Principle• Some Consequences of Uncertainty PrincipleSome Consequences of Uncertainty Principle• Wave-Particle DualityWave-Particle Duality

De Broglie WavesDe Broglie Waves• In his thesis in 1923, Prince Louis V. In his thesis in 1923, Prince Louis V.

de Broglie suggested that mass de Broglie suggested that mass particles should have wave particles should have wave properties similar to electromagnetic properties similar to electromagnetic radiation.radiation.

• The energy can be written as:The energy can be written as:

hf = pc = phf = pc = pff • Thus the wavelength of a matter Thus the wavelength of a matter

wave is called wave is called the De Broglie the De Broglie wavelengthwavelength::

Louis V. de Broglie(1892-1987)

If a light-wave could If a light-wave could also act like a particle, also act like a particle, why shouldn’t matter-why shouldn’t matter-particles also act like particles also act like waves? waves?

The de Broglie HypothesisThe de Broglie Hypothesis

Since light seems to have both wave and particle Since light seems to have both wave and particle properties, it is natural to ask whether matter (electrons, properties, it is natural to ask whether matter (electrons, protons) might also have both wave and particle protons) might also have both wave and particle characteristics. characteristics.

For the wavelength of electron, de Broglie chose:For the wavelength of electron, de Broglie chose:

λλ = h/p = h/p

f = E/hf = E/h

wherewhere EE is the total energy,is the total energy, p p is the momentum, andis the momentum, and

λλ is called the is called the de Broglie wavelengthde Broglie wavelength of the particle. of the particle.

For photons these same equations results directly For photons these same equations results directly from Einstein’s quantization of radiationfrom Einstein’s quantization of radiation E = hfE = hf and and equation for an energy of a photon with zero rest energyequation for an energy of a photon with zero rest energy E = pcE = pc ::

Using relativistic mechanics de Broglie Using relativistic mechanics de Broglie demonstrated, that this equation can also be applied to demonstrated, that this equation can also be applied to particles with mass and used them to physical particles with mass and used them to physical interpretation of Bohr’s hydrogen-like atom. interpretation of Bohr’s hydrogen-like atom.

The de Broglie HypothesisThe de Broglie Hypothesis

hc

hfpcE

The de Broglie WavelengthThe de Broglie Wavelength

Using Using de Brogliede Broglie relation let’s find the relation let’s find the wavelength of awavelength of a 1010-6-6gg particle moving with a particle moving with a speedspeed 1010-6-6m/sm/s::

msmkg

sJ

mv

h

p

h 1969

34

1063.6)/10)(10(

1063.6

Since the wavelength found in this example is so Since the wavelength found in this example is so small, much smaller than any possible apertures, small, much smaller than any possible apertures, diffraction or interference of such waves can not diffraction or interference of such waves can not be observed. be observed.


The situation are different for low energy electrons The situation are different for low energy electrons and other microscopic particles.and other microscopic particles.

Consider a particle with kinetic energyConsider a particle with kinetic energy KK. .


The situation are different for low energy electrons The situation are different for low energy electrons and other microscopic particles.and other microscopic particles.

Consider a particle with kinetic energyConsider a particle with kinetic energy KK. Its . Its momentum is found frommomentum is found from

mKporm

pK 2

2

2

Its wavelength is thenIts wavelength is then

mK

h

p

h

2


mK

h

p

h

2

If we multiply the numerator and denominator byIf we multiply the numerator and denominator by cc we obtain:we obtain:

nmKKeV

nmeV

Kmc

hc 226.1

)10511.0(2

1240

2 62

Where Where mcmc22=0.511MeV=0.511MeV for electrons, andfor electrons, and KK in in electron-volts.electron-volts.


eVinKnmK

,226.1

We obtained the electron wavelength:We obtained the electron wavelength:

Similarly, for proton (Similarly, for proton (mcmc22 = 938 MeV = 938 MeV for protons)for protons)

nmK

p

0286.0


m

Tkv B32

For the molecules of a stationary gas at the absolute For the molecules of a stationary gas at the absolute temperaturetemperature TT, the square average speed of the , the square average speed of the moleculemolecule vv22 is determined by is determined by Maxwell’s LawMaxwell’s Law

Then the momentum of the molecule is:Then the momentum of the molecule is:

Tmkp B3

Knowing that the mass ofKnowing that the mass of HeHe atom, for instance, isatom, for instance, is 6.7x106.7x10-24-24gg, (, (kkBB=1.38x10=1.38x10-23-23J/KJ/K) we obtain for ) we obtain for HeHe wavelength: wavelength:

nmT

He

26.1


.

Similarly, for the molecule of hydrogenSimilarly, for the molecule of hydrogen

nmT

H

78.12

and for the thermal neutronsand for the thermal neutrons

nmT

n

52.2

This calculations shows, that for the acceleratedThis calculations shows, that for the accelerated electronselectrons, for, for atoms of heliumatoms of helium, , hydrogen moleculeshydrogen molecules under the room under the room temperature, fortemperature, for thermal neutronsthermal neutrons and other “slow” light and other “slow” light particles particles de Brogliede Broglie wavelength is on the same order as for soft wavelength is on the same order as for soft X-raysX-rays. .

(a)(a) Show that the wavelength of a nonrelativistic Show that the wavelength of a nonrelativistic neutron is neutron is

where where KKnn is the kinetic energy of the neutron in is the kinetic energy of the neutron in

electron-volts. (b) What is the wavelength of a electron-volts. (b) What is the wavelength of a 1.00-keV1.00-keV neutron? neutron?

m1086.2

λ11

nK




m1086.2

λ11

nK

2

h hp mK

Kinetic energy, Kinetic energy, KK, in this , in this equation is in Joulesequation is in Joules

34 11

27 19

6.626 10 J s 2.87 10 m

2 2 1.67 10 kg 1.60 10 J eV nn

h

mK KK

(a)(a)




m1086.2

λ11

nK

1.00 keV 1000 eVnK (b)(b)

11132.87 10

m 9.07 10 m 907 fm1000

The nucleus of an atom is on the order of The nucleus of an atom is on the order of 1010–14–14 m m in in diameter. For an electron to be confined to a nucleus, its diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be on this order of de Broglie wavelength would have to be on this order of magnitude or smaller. (a) What would be the kinetic energy magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) Given that typical of an electron confined to this region? (b) Given that typical binding energies of electrons in atoms are measured to be binding energies of electrons in atoms are measured to be on the order of a on the order of a few eVfew eV, would you expect to find an , would you expect to find an electron in a nucleus? electron in a nucleus?


14~10 m 34

1914

6.6 10 J s~ 10 kg m s

10 m

hp

2 2 2 42 2 2 4 19 8 31 8~ 10 3 10 9 10 3 10eE p c m c

11 8~10 J ~10 eVE

2 8 6 8~10 eV 0.5 10 eV ~10 eVeK E mc

(a)(a)


(b)(b) 9 2 2 1951 2

14

9 10 N m C 10 C~ ~ 10 eV

10 me

e

ek qqU

r

With itsWith its 0eK U

,

the electron would immediately escape the nucleus

The calculations show, that for the acceleratedThe calculations show, that for the accelerated electronselectrons, for, for atoms of heliumatoms of helium, , hydrogen moleculeshydrogen molecules under the room temperature, forunder the room temperature, for thermal neutronsthermal neutrons and and other “slow” light particles other “slow” light particles de Brogliede Broglie wavelength is on wavelength is on the same order as for softthe same order as for soft X-raysX-rays. So, we can expect, . So, we can expect, that diffraction can be observed for this particlesthat diffraction can be observed for this particles

Electron Interference and DiffractionElectron Interference and Diffraction The electron wave interference was discovered in The electron wave interference was discovered in

1927 by 1927 by C.J. DavissonC.J. Davisson and and L.H.GermerL.H.Germer as they were as they were studying electron scattering from a nickel target at the Bell studying electron scattering from a nickel target at the Bell Telephone Laboratories. Telephone Laboratories.

After heating the target to remove an oxide coating After heating the target to remove an oxide coating that had accumulate after accidental break in the vacuum that had accumulate after accidental break in the vacuum system, they found that the scattered electron intensity is a system, they found that the scattered electron intensity is a function of the scattered angle and show maxima and function of the scattered angle and show maxima and minima. Their target had crystallized during the heating, minima. Their target had crystallized during the heating, and by accident they had observed electron diffraction.and by accident they had observed electron diffraction.

Then Davisson and Germer prepared a target from a Then Davisson and Germer prepared a target from a single crystal of nickel and investigated this phenomenon.single crystal of nickel and investigated this phenomenon.

The Davisson-The Davisson-Germer experimentGermer experiment..

Low energy Low energy electrons scattered at electrons scattered at angleangle ΦΦ from a nickel from a nickel crystal are detected in an crystal are detected in an ionization chamber. The ionization chamber. The kinetic energy of kinetic energy of electrons could be varied electrons could be varied by changing the by changing the accelerating voltage on accelerating voltage on the electron gun. the electron gun.

Scattered intensity vs detector angle for Scattered intensity vs detector angle for 54-ev 54-ev electrons. electrons. Polar plot of the data. The intensity at each angle is indicated by Polar plot of the data. The intensity at each angle is indicated by the distance of the point from the origin. Scattered angle the distance of the point from the origin. Scattered angle ΦΦ is is plotted clockwise started at the vertical axis.plotted clockwise started at the vertical axis.

The same data plotted on a Cartesian graph. The The same data plotted on a Cartesian graph. The intensity scale are the same on the both graphs. In each plot intensity scale are the same on the both graphs. In each plot there is maximum intensity atthere is maximum intensity at ΦΦ=50º=50º, as predicted for Bragg , as predicted for Bragg scattering of waves having wavelengthscattering of waves having wavelength λλ = h/p = h/p..

Scattering of electron by crystal. Electron waves are Scattering of electron by crystal. Electron waves are strongly scattered if the Bragg conditionstrongly scattered if the Bragg condition nnλλ = D Sin = D SinΦΦ is met. is met.

George P. Thomson (1892–1975), George P. Thomson (1892–1975), son of J. J. Thomson, reported son of J. J. Thomson, reported seeing electron diffraction in seeing electron diffraction in transmission experiments on transmission experiments on celluloid, gold, aluminum, and celluloid, gold, aluminum, and platinum. platinum.

In 1925, Davisson and Germer In 1925, Davisson and Germer experimentally observed that experimentally observed that electronselectrons were diffracted (much like were diffracted (much like x-rays) in nickel crystals.x-rays) in nickel crystals.

A randomly oriented polycrystalline A randomly oriented polycrystalline sample of SnOsample of SnO22 produces rings. produces rings.

Test of the de Broglie formulaTest of the de Broglie formula λλ = h/p = h/p. The wavelength is . The wavelength is computed from a plot of the diffraction data for electrons computed from a plot of the diffraction data for electrons plotted against plotted against VV00

-1/2-1/2, where, where VV00 is the accelerating voltage. is the accelerating voltage. The straight line isThe straight line is 1.226V1.226V00

-1/2-1/2 nmnm as predicted from as predicted from λλ = h/(2mE)= h/(2mE)-1/2-1/2

Test of the de Broglie formulaTest of the de Broglie formula λλ = h/p = h/p. The wavelength is . The wavelength is computed from a plot of the diffraction data plotted against computed from a plot of the diffraction data plotted against VV00

-1/2-1/2, where, where VV00 is the accelerating voltage. The straight line is the accelerating voltage. The straight line isis 1.226V1.226V00

-1/2-1/2 nmnm as predicted from as predicted from λλ = h/(2mE) = h/(2mE)-1/2-1/2

001931

34

0

1226.1

1

/106.11011.92

106.61

2 EEeVJkg

sJ

Em

h

A series of a polar graphs of Davisson and Germer’s A series of a polar graphs of Davisson and Germer’s data at electron accelerating potential fromdata at electron accelerating potential from 36 V36 V to to 68 V68 V. . Note the development of the peak atNote the development of the peak at ΦΦ = 50º = 50º to a to a maximum whenmaximum when VV00 = 54 V = 54 V. .

Variation of the scattered electron intensity with wavelength Variation of the scattered electron intensity with wavelength for constantfor constant ΦΦ. . The incident beam in this case wasThe incident beam in this case was 10º 10º from from the normal, the resulting diffraction causing the measured the normal, the resulting diffraction causing the measured peaks to be slightly shifted from the positions computed frompeaks to be slightly shifted from the positions computed from nnλλ = D Sin = D Sin ΦΦ. .

Schematic arrangement used for producing a diffraction Schematic arrangement used for producing a diffraction pattern from a polycrystalline aluminum target. pattern from a polycrystalline aluminum target.

Diffraction pattern produced byDiffraction pattern produced by xx-rays-rays of wavelengthof wavelength 0.071 nm0.071 nm and an aluminum foil target. and an aluminum foil target.

Diffraction pattern produced byDiffraction pattern produced by 600-600-eVeV electrons and an electrons and an aluminum foil target ( de Broigle wavelength of aboutaluminum foil target ( de Broigle wavelength of about 0.05 nm0.05 nm: ): )

Diffraction pattern produced byDiffraction pattern produced by 600-600-eVeV electrons and an electrons and an aluminum foil target ( de Broigle wavelength of aboutaluminum foil target ( de Broigle wavelength of about 0.05 nm0.05 nm: ): ) ..

nmeVeV

nmeV

Vmc

hc

Em

h

p

h

kee 05.0

60010511.02

1240

22 60

2

Diffraction pattern produced byDiffraction pattern produced by 0.0568-eV0.0568-eV neutrons (de Broglie neutrons (de Broglie wavelength ofwavelength of 0.120 nm0.120 nm) and a target of polycrystalline copper. ) and a target of polycrystalline copper. Note the similarity in the pattern produced byNote the similarity in the pattern produced by x-raysx-rays,, electronselectrons, , and and neutronsneutrons. .

Diffraction pattern produced byDiffraction pattern produced by 0.0568-eV0.0568-eV neutrons (de Broglie neutrons (de Broglie wavelength ofwavelength of 0.120 nm0.120 nm) and a target of polycrystalline copper. ) and a target of polycrystalline copper. Note the similarity in the pattern produced byNote the similarity in the pattern produced by x-raysx-rays,, electronselectrons, , and and neutronsneutrons. .

nmeVeV

nmeV

Em

h

knn 120.0

0568.01057.9392

1240

2 6

In the Davisson–Germer experiment, In the Davisson–Germer experiment, 54.0-eV54.0-eV electrons were diffracted from a nickel lattice. If the electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed first maximum in the diffraction pattern was observed at at φφ = 50.0° = 50.0°, what was the lattice spacing , what was the lattice spacing a a between between the vertical rows of atoms in the figure? (It is not the the vertical rows of atoms in the figure? (It is not the same as the spacing between the horizontal rows of same as the spacing between the horizontal rows of atoms.) atoms.)


2 sin 2 cos2

m d d

sin2

d a

1m

2 sin cos sin2 2

a a


2 sin cos sin2 2

a a

2 e

h hp mK

34

10

31 19

6.626 10 J s1.67 10 m

2 9.11 10 kg 54.0 1.60 10 J

10101.67 10 m

2.18 10 0.218 nmsin sin50.0

a

A photon has an energy equal to the kinetic energy of a A photon has an energy equal to the kinetic energy of a particle moving with a speed of particle moving with a speed of 0.9000.900cc. (a) Calculate the . (a) Calculate the ratio of the wavelength of the photon to the wavelength of ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having the particle. (b) What would this ratio be for a particle having a speed ofa speed of 0.00100 0.00100cc ? (c) What value does the ratio of the ? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low two wavelengths approach at high particle speeds?(d) At low particle speeds? particle speeds?

A photon has an energy equal to the kinetic energy of a A photon has an energy equal to the kinetic energy of a particle moving with a speed of particle moving with a speed of 0.9000.900cc. (a) Calculate the . (a) Calculate the ratio of the wavelength of the photon to the wavelength of ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having the particle. (b) What would this ratio be for a particle having a speed ofa speed of 0.00100 0.00100cc ? (c) What value does the ratio of the ? (c) What value does the ratio of the two wavelengths approach at high particle speeds? (d) At two wavelengths approach at high particle speeds? (d) At low particle speeds? low particle speeds?

21K mc mh hp mv

For a particle:For a particle:

For a photon:For a photon:

E K 21

c ch ch chf E K mc

2 11m

ch mv vcmc h

A photon has an energy equal to the kinetic energy of a particle moving with A photon has an energy equal to the kinetic energy of a particle moving with a speed of a speed of 0.9000.900cc. (a) Calculate the ratio of the wavelength of the photon to . (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle the wavelength of the particle. (b) What would this ratio be for a particle having a speed ofhaving a speed of 0.00100 0.00100cc ? (c) What value does the ratio of the two ? (c) What value does the ratio of the two wavelengths approach at high particle speeds?(d) At low particle speeds? wavelengths approach at high particle speeds?(d) At low particle speeds?

2 11m

ch mv vcmc h

(a)(a)

2 2

1 0.91.60

1 0.9 1 1 0.9 1m

(b)(b)

3

2 2

1 0.0012.00 10

1 0.001 1 1 0.001 1m

294.2

A photon has an energy equal to the kinetic energy of a particle moving with a A photon has an energy equal to the kinetic energy of a particle moving with a speed of speed of 0.9000.900cc. (a) Calculate the ratio of the wavelength of the photon to the . (a) Calculate the ratio of the wavelength of the photon to the wavelength of the particle. (b) What would this ratio be for a particle having a wavelength of the particle. (b) What would this ratio be for a particle having a speed ofspeed of 0.00100 0.00100cc ? (c) What value does the ratio of the two wavelengths ? (c) What value does the ratio of the two wavelengths approach at high particle speed? (d) At low particle speed? approach at high particle speed? (d) At low particle speed?

(c)(c)

(d)(d)

As As 1vc

,

and 1 becomes nearly equal to γ. 1 1m

0vc

1 22 2 2

2 2 2

1 11 1 1 1

2 2v v v

c c c

2 2

21

1 2m

v c cvv c

What is “waving”? For matter it is the What is “waving”? For matter it is the probability of finding the probability of finding the particleparticle that waves. that waves. Classical waves are the solution of the classical wave equationClassical waves are the solution of the classical wave equation

Harmonic waves of amplitudeHarmonic waves of amplitude yy00, frequency, frequency f f and periodand period TT::

where the angular frequencywhere the angular frequency ωω and the wave numberand the wave number kk are are defined by defined by

and the and the wave wave or or phase velocityphase velocity vvpp is given byis given by

2

2

22

2 1

dt

yd

fdx

yd

T

txytkxyy

2cos)cos( 00

22

2 kandT

f

fvp

If the film were to be observed at various stages, such If the film were to be observed at various stages, such as after being struck byas after being struck by 28 electrons28 electrons the pattern of the pattern of individually exposed grains will be similar to shown here. individually exposed grains will be similar to shown here.

After exposure by aboutAfter exposure by about 1000 electrons1000 electrons the the pattern will be similar to this.pattern will be similar to this.

And again for exposure of aboutAnd again for exposure of about 10,000 10,000 electronselectrons we will obtained a pattern like this.we will obtained a pattern like this.

Two source interference pattern. If the sources Two source interference pattern. If the sources are coherent and in phase, the waves from the sources are coherent and in phase, the waves from the sources interfere constructively at points for which the path interfere constructively at points for which the path difference difference dsindsinθθ is an integer number of wavelength.is an integer number of wavelength.

Grows of two-slits interference pattern. The photo Grows of two-slits interference pattern. The photo is an actual two-slit electron interference pattern in which is an actual two-slit electron interference pattern in which the film was exposed to millions of electrons. The pattern the film was exposed to millions of electrons. The pattern is identical to that usually obtained with photons. is identical to that usually obtained with photons.

Using relativistic mechanics, Using relativistic mechanics, de Brogliede Broglie was able to was able to derive the physical interpretation of derive the physical interpretation of Bohr’sBohr’s quantization of the quantization of the angular momentum of electron. angular momentum of electron.

He demonstrate that quantization of angular momentum He demonstrate that quantization of angular momentum of the electron in hydrogenlike atoms is equivalent to a standing of the electron in hydrogenlike atoms is equivalent to a standing wave condition:wave condition:

forfor n n = integer= integer2nh

nmvr

Using relativistic mechanics, Using relativistic mechanics, de Brogliede Broglie was able to was able to derive the physical interpretation of derive the physical interpretation of Bohr’sBohr’s quantization of the quantization of the angular momentum of electron. angular momentum of electron.

He demonstrate that quantization of angular momentum He demonstrate that quantization of angular momentum of the electron in hydrogenlike atoms is equivalent to a standing of the electron in hydrogenlike atoms is equivalent to a standing wave condition:wave condition:

forfor n n = integer= integer

The idea of explaining discrete energy states in matter The idea of explaining discrete energy states in matter by standing waves thus seems quite promising. by standing waves thus seems quite promising.

2nh

nmvr

orbitofncecircumferenp

nh

mv

nhr 2

Standing waves around the circumference of a Standing waves around the circumference of a circle. In this case the circle iscircle. In this case the circle is 33λλ in circumference. For in circumference. For example, if a steel ring had been suitable tapped with a example, if a steel ring had been suitable tapped with a hammer, the shape of the ring would oscillate between hammer, the shape of the ring would oscillate between the extreme positions represented by the solid and the extreme positions represented by the solid and broken lines. broken lines.

Wave pulse moving along a string. A pulse have a Wave pulse moving along a string. A pulse have a beginning and an end; i.e. it is localized, unlike a pure beginning and an end; i.e. it is localized, unlike a pure harmonic wave, which goes on forever in space and time. harmonic wave, which goes on forever in space and time.

Two waves of slightly different wavelength and frequency Two waves of slightly different wavelength and frequency produced beats. produced beats.

(a) Shows(a) Shows y(x)y(x) at given instant for each of the two waves. at given instant for each of the two waves. The waves are in phase at the origin but because of the The waves are in phase at the origin but because of the difference in wavelength, they become out of phase and difference in wavelength, they become out of phase and then in phase again. then in phase again.

((b) The sum of these waves. The spatial extend of the groupb) The sum of these waves. The spatial extend of the group ΔΔxx is inversely proportional to the difference in wave numbersis inversely proportional to the difference in wave numbers ΔΔkk, where, where k k is related to the wavelength byis related to the wavelength by k = 2k = 2ππ//λλ. .

BEATSBEATSConsider two waves of equal amplitude and nearly equal Consider two waves of equal amplitude and nearly equal frequencies and wavelengths.frequencies and wavelengths.

The sum of the two waves is (superposition):The sum of the two waves is (superposition):

F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)]

)sin()(

)sin()(

222

111

txkFxF

txkFxF

BEATSBEATS

F(x) = F1(x) + F2(x) = F[sin(k1x – ω1t)+sin(k2x – ω2t)]using the trigonometric relationusing the trigonometric relation

Sinα + Sinβ = 2Cos[(α-β)/2] Sin[(α+β)/2]

withwith α =(k1x – ω1t) β = (k2x – ω2t), we we get:get:

tx

kktx

kkFxF

22sin

22cos2)( 21212121

BEATSBEATS

withwith

222121

2121

kkk

kkk

)sin(22

cos2)( txktxk

FxF

tx

kktx

kkFxF

22sin

22cos2)( 21212121

BEATSBEATS

This is an equivalent of an harmonic wave This is an equivalent of an harmonic wave

whose amplitude is modulated bywhose amplitude is modulated by

We have formed wave packets of extendWe have formed wave packets of extend ΔΔxx and can and can imagine each wave packet representing a particle. imagine each wave packet representing a particle.

)sin(22

cos2)( txktxk

FxF

)sin(0 txkyF

txk

22cos2

NowNow

The particle is in the regionThe particle is in the region ΔΔxx, the momentum of the , the momentum of the particle in the rangeparticle in the range ΔΔkk: :

p = p = ћћk → k → ΔΔp = p = ћћΔΔkk

ΔΔx x ΔΔp ≈ h - Uncertainty Principlep ≈ h - Uncertainty Principle

In order to localize the particle within a region In order to localize the particle within a region ΔΔxx, we , we need to relax the precision on the value of the need to relax the precision on the value of the momentum, momentum, ΔΔpp . .

2

2

22

2

22

kx

kx

kandx

The position of the electron can not be resolved better The position of the electron can not be resolved better than the width of the central maximum of the diffraction patternthan the width of the central maximum of the diffraction pattern ΔΔx ≈ x ≈ λλ/sin/sinθθ. The product of the uncertainties. The product of the uncertainties ΔΔppxx ΔΔxx is is

therefore of the order of Planck’s constanttherefore of the order of Planck’s constant hh. .

Uncertainty PrincipleUncertainty Principle

Consider a wave packet Consider a wave packet ΨΨ(x,t)(x,t) representing an representing an electron. The most probable position of the electron is the electron. The most probable position of the electron is the value of value of xx for which for which IIΨΨ(x,t)I(x,t)I22 is a maximum. is a maximum.

Since Since IIΨΨ(x,t)I(x,t)I22 is proportional to the probability that is proportional to the probability that the electron is at the electron is at xx, and , and IIΨΨ(x,t)I(x,t)I22 is nonzero for a range of is nonzero for a range of value value xx, there is an uncertainty in the value of position of , there is an uncertainty in the value of position of the electron.the electron.

This means that if we make a number of position This means that if we make a number of position measurements on identical electrons – electrons with same measurements on identical electrons – electrons with same wave function – we shall not always obtain the same result.wave function – we shall not always obtain the same result.

In fact, the distribution function for the results of In fact, the distribution function for the results of such measurements will be given by such measurements will be given by IIΨΨ(x,t)I(x,t)I22. .


If the wave packet is very narrow, the uncertainty in the If the wave packet is very narrow, the uncertainty in the position will be small.position will be small.

xxk

2

22

xxk

2

22

kp

However, a narrow However, a narrow wave packet must wave packet must contain a wide range contain a wide range of wave numberof wave number k k::

The momentum is The momentum is related to the wave related to the wave number bynumber by

So, a wide range of So, a wide range of kk values means a wide range of values means a wide range of momentum values.momentum values.


We can see that for all wave packets the ranges We can see that for all wave packets the ranges ΔΔxx and and ΔΔk k are related byare related by

2

12

2

kx

2

1 t

Similarly, a packet that is localized in time Similarly, a packet that is localized in time ΔΔtt must contain must contain a range of frequencies a range of frequencies ΔωΔω, where the ranges are related , where the ranges are related byby

If we multiply these equations by If we multiply these equations by ћћ and use and use p = p = ћћkk and and E = E = ћћωω we obtain we obtain

22

tEandpx

If an excited state of an atom is known to have a lifetime If an excited state of an atom is known to have a lifetime of of 66∙∙1010-7-7 s s, what is the uncertainty in the energy of , what is the uncertainty in the energy of photons emitted by such atoms in the spontaneous photons emitted by such atoms in the spontaneous decay to the ground state?decay to the ground state?

An electron (An electron (mmee = 9.11 × 10= 9.11 × 10–31–31 kg kg) and a bullet () and a bullet (m m = 0.0200 kg= 0.0200 kg) )

each have a velocity of magnitude of each have a velocity of magnitude of 500 m/s500 m/s, accurate to , accurate to within within 0.0100%0.0100%. Within what limits could we determine the . Within what limits could we determine the position of the objects along the direction of the velocity? position of the objects along the direction of the velocity?

An electron (An electron (mmee = 9.11 × 10= 9.11 × 10–31–31 kg kg) and a bullet () and a bullet (m m = 0.0200 kg= 0.0200 kg) )

each have a velocity of magnitude of each have a velocity of magnitude of 500 m/s500 m/s, accurate to , accurate to within within 0.0100%0.0100%. Within what limits could we determine the . Within what limits could we determine the position of the objects along the direction of the velocity? position of the objects along the direction of the velocity? .

31 4 329.11 10 kg 500 m s 1.00 10 4.56 10 kg m sep m v

34

32

6.626 10 J s1.16 mm

4 4 4.56 10 kg m s

hx

p

For the electron,

.

4 30.0200 kg 500 m s 1.00 10 1.00 10 kg m sp m v

325.28 10 m4h

xp

For the bullet,

.

Show that the kinetic energy of a nonrelativistic particle can Show that the kinetic energy of a nonrelativistic particle can be written in terms of its momentum as be written in terms of its momentum as K K = = pp22/2/2mm. (b) Use . (b) Use the results of (a) to find the minimum kinetic energy of a the results of (a) to find the minimum kinetic energy of a proton confined within a nucleus having a diameter of proton confined within a nucleus having a diameter of 1.00 × 1.00 × 1010–15–15 m m. .

Show that the kinetic energy of a nonrelativistic particle can Show that the kinetic energy of a nonrelativistic particle can be written in terms of its momentum as be written in terms of its momentum as K K = = pp22/2/2mm. (b) Use . (b) Use the results of (a) to find the minimum kinetic energy of a the results of (a) to find the minimum kinetic energy of a proton confined within a nucleus having a diameter of proton confined within a nucleus having a diameter of 1.00 × 1.00 × 1010–15–15 m m. .

(a) 2 221

2 2 2

mv pK mv

m m

(b) To find the minimum kinetic energy, think of the minimum momentum uncertainty, and maximum position uncertainty of 1510 m x

. We model the proton as moving along a straight line with

2

xp

xp

2

The average momentum is zero. The average squared momentum is equal to the squared uncertainty:

MeVJ

kgm

sJ

mx

h

mxm

p

m

pK

21.51033.8

)1067.1()10(32

)1063.6(

)(322)(42

)(

213

272152

234

22

2

2

222

The Interpretation of the Wave FunctionThe Interpretation of the Wave Function

Given that electrons have wave-like properties, it Given that electrons have wave-like properties, it should be possible to produce standing electron waves. should be possible to produce standing electron waves. The energy is associated with the frequency of the standing The energy is associated with the frequency of the standing wave, aswave, as E = hfE = hf, so standing waves imply quantized , so standing waves imply quantized energies. energies.

The idea that discrete energy states in atom can be The idea that discrete energy states in atom can be explained by standing waves led to the development by explained by standing waves led to the development by Erwin SchrErwin Schrödingerödinger in 1926 mathematical theory known as in 1926 mathematical theory known as quantum theoryquantum theory, , quantum mechanicsquantum mechanics, or, or wave mechanicswave mechanics. .

In this theory a single electron is described by a In this theory a single electron is described by a wave functionwave function ΨΨ that obeys a wave equation called the that obeys a wave equation called the SchrSchrödinger equationödinger equation. .


The form of the The form of the SchrSchrödingerödinger equation of a equation of a particular system depends on the forces acting on the particular system depends on the forces acting on the particle, which are described by the potential energy particle, which are described by the potential energy functions associated with this forces. functions associated with this forces.

SchrSchrödingerödinger solved the standing wave problem for solved the standing wave problem for hydrogen atomhydrogen atom, the , the simple harmonic oscillatorsimple harmonic oscillator, and other , and other system of interest. He found that the allowed system of interest. He found that the allowed frequencies, combined withfrequencies, combined with E=hfE=hf, resulted in the set of , resulted in the set of energy levels, found experimentally for the energy levels, found experimentally for the hydrogen hydrogen atomatom..

Quantum theoryQuantum theory is the basis for our is the basis for our understanding of the modern world, from the inner understanding of the modern world, from the inner working of the atomic nucleus to the radiation spectra of working of the atomic nucleus to the radiation spectra of distant galaxies. distant galaxies.


The The wave functionwave function for waves in a string is the for waves in a string is the string displacementstring displacement yy. .

The wave function for sound waves can be either The wave function for sound waves can be either the displacement of the air molecules, or the pressurethe displacement of the air molecules, or the pressure PP. .

The wave function of the electromagnetic waves The wave function of the electromagnetic waves is the electric fieldis the electric field EE and the magnetic fieldand the magnetic field BB. .

What is the wave function for the electronWhat is the wave function for the electron ΨΨ? The ? The SchrSchrödinger equationödinger equation describes a single particle. The describes a single particle. The square of the wave function for a particle describes the square of the wave function for a particle describes the probability densityprobability density, which is the probability per unit , which is the probability per unit

volume, of finding the particle at a location.volume, of finding the particle at a location.

The Interpretation of the Wave FunctionThe Interpretation of the Wave Function The probability of finding the particle in some The probability of finding the particle in some

volume element must also be proportional to the size of volume element must also be proportional to the size of volume elementvolume element dVdV. .

Thus, in one dimension, the probability of finding Thus, in one dimension, the probability of finding a particle in a regiona particle in a region dxdx at the positionat the position xx is is ΨΨ22(x)dx(x)dx. If we . If we call this probabilitycall this probability P(x)dxP(x)dx, where, where P(x)P(x) is the probability is the probability density, we have density, we have

P(x) = P(x) = ΨΨ22(x)(x)

The probability of finding the particle inThe probability of finding the particle in dxdx at pointat point xx11

or pointor point xx22 is the sum of separate probabilitiesis the sum of separate probabilities

P(xP(x11)dx + P(x)dx + P(x22)dx)dx. .

If we have a particle at all the probability of finding a If we have a particle at all the probability of finding a particle somewhere must be particle somewhere must be 11..


Then, the sum of the probabilities over all Then, the sum of the probabilities over all the possible values of the possible values of x x must equal must equal 11. That is,. That is,

This equation is called the This equation is called the normalization normalization conditioncondition. If. If ΨΨ is to satisfy the normalization is to satisfy the normalization condition, it must approach condition, it must approach zerozero as as xx is is approaching approaching infinityinfinity..

12

dx

Probability Calculation for a Classical ParticleProbability Calculation for a Classical Particle

It is known that a classical point particle moves back and forth It is known that a classical point particle moves back and forth with constant speed between two walls atwith constant speed between two walls at x = 0x = 0 andand x = 8cmx = 8cm. . No additional information about of location of the particle is No additional information about of location of the particle is known. known.

(a) What is the probability density(a) What is the probability density P(xP(x))? ?

(b) What is the probability of finding the particle at(b) What is the probability of finding the particle at x=2cmx=2cm? ?

(c) What is the probability of finding the particle between(c) What is the probability of finding the particle between x=3.0 x=3.0 cmcm and and x=3.4 cmx=3.4 cm??

A Particle in a BoxA Particle in a Box We can illustrate many of important features of We can illustrate many of important features of

quantum physics by considering of simple problem of particle quantum physics by considering of simple problem of particle of massof mass mm confined to a one-dimensional box of lengthconfined to a one-dimensional box of length L L. .

This can be considered as a crude description of an This can be considered as a crude description of an electron confined within an atom, or a proton confined within electron confined within an atom, or a proton confined within a nucleus. a nucleus.

According to the quantum theory, the particle is According to the quantum theory, the particle is described by the wave functiondescribed by the wave function ΨΨ, whose square describes , whose square describes the probability of finding the particle in some region. Since the probability of finding the particle in some region. Since we are assumingwe are assuming that the particle is indeed inside the box, that the particle is indeed inside the box, the wave function must be zero everywhere outside the box:the wave function must be zero everywhere outside the box: ΨΨ =0 =0 for for x≤0x≤0 and forand for x≥Lx≥L. .

A Particle in a BoxA Particle in a Box The allowed wavelength for a particle in the box The allowed wavelength for a particle in the box

are those where the lengthare those where the length LL equals an integral number equals an integral number of half wavelengths.of half wavelengths.

L = n( L = n( λλnn/2) n = 1,2,3,……./2) n = 1,2,3,…….

This is a standing wave condition for a particle in the This is a standing wave condition for a particle in the box of lengthbox of length LL..

The total energy of the particle is its kinetic energyThe total energy of the particle is its kinetic energyE = (1/2)mvE = (1/2)mv22 = p = p22/2m/2m

Substituting the de Broglie relationSubstituting the de Broglie relation ppnn = h/ = h/λλnn,,

2

2

2

2

222 n

nnn

m

h

m

h

m

pE

A Particle in a BoxA Particle in a Box

Then the standing wave conditionThen the standing wave condition λλnn= 2L/n= 2L/n gives the gives the

allowed energies:allowed energies:

wherewhere

2

2

2

2

222 n

nnn

m

h

m

h

m

pE

12

2

22

8En

mL

hnEn

2

2

18mL

hE

A Particle in a BoxA Particle in a BoxThe equationThe equation

gives the allowed energies for a particle in the gives the allowed energies for a particle in the box.box.

This is the ground state energy for a particle in This is the ground state energy for a particle in the box, which is the energy of the lowest state.the box, which is the energy of the lowest state.

12

2

22

8En

mL

hnEn

2

2

18mL

hE

A Particle in a BoxA Particle in a BoxThe condition that we used for the wave function in the boxThe condition that we used for the wave function in the box

ΨΨ = 0 = 0 atat x = 0 x = 0 andand x = Lx = Lis called theis called the boundary conditionboundary condition..

The The boundary conditionsboundary conditions in quantum theory lead to in quantum theory lead to energy quantization. energy quantization.

NoteNote, that the lowest energy for a particle in the box is , that the lowest energy for a particle in the box is not zeronot zero. . The result is a general feature of quantum theory.The result is a general feature of quantum theory.

If a particle is confined to some region of space, the If a particle is confined to some region of space, the particle has a minimum kinetic energy, which is called particle has a minimum kinetic energy, which is called zero-point energy. The smaller the region of space the zero-point energy. The smaller the region of space the particle is confined to, the greater its zero-point energy.particle is confined to, the greater its zero-point energy.

A Particle in a BoxA Particle in a Box

If an electron is confined (i.e., bond to an atom) If an electron is confined (i.e., bond to an atom) in some energy statein some energy state EEii, the electron can make a , the electron can make a

transition to another energy statetransition to another energy state EEf f with the with the

emission of photon. The frequency of the emission of photon. The frequency of the emitted photon is found from the conservation of emitted photon is found from the conservation of the energythe energy

hf = Ehf = Eii – E – Eff

The wavelength of the photon is thenThe wavelength of the photon is then

λλ = c/f = hc/( = c/f = hc/(EEii – E – Eff))

Standing Wave FunctionStanding Wave Function

The amplitude of a vibrating string fixed atThe amplitude of a vibrating string fixed at x=0x=0 andand x=Lx=L is given is given asas

wherewhere AAnn is a constant andis a constant and is the wave number.is the wave number.

The wave function for a particle in a box are the same:The wave function for a particle in a box are the same:

UsingUsing , we have, we have

xkAy nnn sin

nnk

2

xkAy nnn sin

n

Ln

2 L

n

nL

kn

n

222

Standing Wave FunctionStanding Wave Function

The wave function can thus be writtenThe wave function can thus be written

L

xnAx nn

sin)(

The constantThe constant A Ann is determined by normalization conditionis determined by normalization condition

The result of evaluating the integral and solving forThe result of evaluating the integral and solving for AAnn

is independent fromis independent from nn..

The normalized wave function for a particle in a box are thusThe normalized wave function for a particle in a box are thus

1sin222

dxL

xnAdx nn

LAn

2

L

xn

Lxn

sin

2)(

Graph of Graph of energyenergy vs. vs. xx for a particle in the box, that we also call for a particle in the box, that we also call an infinitely deep well. The set of allowed values for the an infinitely deep well. The set of allowed values for the particle’s total energyparticle’s total energy EEnn isis EE11(n=1)(n=1),, 4E 4E11(n=2)(n=2), , 9E9E11(n=3)(n=3) ….. …..

Wave functionsWave functions ΨΨnn(x)(x) and probability densitiesand probability densities PPnn(x)= Ψ(x)= Ψnn22(x)(x)

forfor n=1n=1, , 22, and, and 33 for the infinity square well potential.for the infinity square well potential.

Probability distribution forProbability distribution for n=10n=10 for the infinity square well for the infinity square well potential. The dashed line is the classical probability densitypotential. The dashed line is the classical probability density P=1/LP=1/L, which is equal to the quantum mechanical distribution , which is equal to the quantum mechanical distribution averaged over a regionaveraged over a region ΔxΔx containing several oscillations. A containing several oscillations. A physical measurement with resolutionphysical measurement with resolution ΔxΔx will yield the classical will yield the classical result ifresult if nn is so large thatis so large that ΨΨ22(x)(x) has many oscillations inhas many oscillations in ΔxΔx..

Photon Emission by Particle in a BoxPhoton Emission by Particle in a Box

An electron is in one dimensional box of lengthAn electron is in one dimensional box of length 0.1nm0.1nm. . (a) Find the (a) Find the ground state energyground state energy. (b) Find the energy . (b) Find the energy in electron-volts of the in electron-volts of the five lowest statesfive lowest states, and then , and then sketch an energy level diagram. (c) Find the sketch an energy level diagram. (c) Find the wavelength of the photon emitted for each transition wavelength of the photon emitted for each transition from the statefrom the state n=3n=3 to a lower-energy state. to a lower-energy state.

The probability of a particle being found in a The probability of a particle being found in a specified region of a box.specified region of a box.

The particle in one-dimensional box of lengthThe particle in one-dimensional box of length LL is in is in the ground state. Find the probability of finding the the ground state. Find the probability of finding the particle (a) anywhere in a region of lengthparticle (a) anywhere in a region of length ΔΔx = 0.01Lx = 0.01L centered atcentered at x = ½Lx = ½L; ; (b) in the region(b) in the region 0<x<(1/4)L0<x<(1/4)L..

Expectation ValuesExpectation ValuesThe most that we can know about the position The most that we can know about the position

of the particle is the probability of measuring a of the particle is the probability of measuring a certain value of this positioncertain value of this position xx. If we measure the . If we measure the position for a large number of identical systems, we position for a large number of identical systems, we get a range of values corresponding to the get a range of values corresponding to the probability distribution. probability distribution.

The average value ofThe average value of xx obtained from such obtained from such measurements is called themeasurements is called the expectation valueexpectation value and and writtenwritten ‹x›‹x›.. The expectation value of The expectation value of xx is the same is the same as the average value ofas the average value of xx that we would expect to that we would expect to obtain from a measurement of the position of a obtain from a measurement of the position of a large number of particles with the same wave large number of particles with the same wave functionfunction ΨΨ(x)(x)..

Expectation ValuesExpectation Values

SinceSince ΨΨ22(x)dx(x)dx is the probability of finding a is the probability of finding a particle in the regionparticle in the region dxdx, the expectation value of , the expectation value of xx is:is:

The expectation value of any functionThe expectation value of any function f(x)f(x) is is given by:given by:

dxxxx )(2

dxxxfxf )()()( 2

Calculating expectation values

Find (a)Find (a) ‹ x ›‹ x › andand (b)(b) ‹ x‹ x22›› for a particle in for a particle in its ground state in a box of lengthits ground state in a box of length LL. .

Complex NumbersComplex Numbers

A complex number has the formA complex number has the form a+iba+ib, ,

withwith ii22=-1=-1 oror i=i=√-1√-1 – – imaginary unitimaginary unit..

aa - - real partreal part; ; bb – – imaginary partimaginary part; ; i i – – imaginary unitimaginary unit

(a +ib) + (c +id) = (a+c) + i(b+d)(a +ib) + (c +id) = (a+c) + i(b+d)m(a +ib) = ma + imbm(a +ib) = ma + imb

(a +ib) (c +id) = (ac - bd) + i(ad + bc)(a +ib) (c +id) = (ac - bd) + i(ad + bc)

The absolute value ofThe absolute value of a + iba + ib is denoted byis denoted by │a+ib││a+ib│ and is given byand is given by │a+ib││a+ib│= = √ a√ a22 + b + b22

Complex NumbersComplex Numbers

The complex conjugate of The complex conjugate of a+iba+ib is denoted by is denoted by (a+ib)(a+ib)** and is given and is given

(a+ib)(a+ib)* * = (a-ib)= (a-ib)

ThenThen

(a+ib)(a+ib)**∙ ∙ (a+ib) = (a-ib) (a+ib)=a(a+ib) = (a-ib) (a+ib)=a22 - b - b22

Polar Form of Complex Numbers

p = p = √ a√ a22 + b + b22 = │a + = │a + ib│ib│

We can represent the We can represent the number number (a + ib)(a + ib) in the in the complex complex xyxy plane. plane.

Then the polar coordinatesThen the polar coordinates

a + ib a + ib ≡≡ p(cos p(cosφφ +isin +isinφφ))

Remembering the Euler Remembering the Euler formula:formula:

eeiiφφ = (cos = (cosφφ+isin+isinφφ))

a + ib = p ea + ib = p eiiφφ

Real axisReal axis

Imag

inar

y ax

isIm

agin

ary

axis

φ

aa

bbp

Euler Euler Identities:Identities:

eeiiφφ = cos = cosφφ + isin + isinφφ

ee-i-iφφ = cos = cosφφ - isin - isinφφ

where i = √-1where i = √-1

Fourier TransformFourier TransformIn quantum mechanics, our basic function is the plane In quantum mechanics, our basic function is the plane

wave describing a free particle, given in equation:wave describing a free particle, given in equation:

We are not interested here in how things behave in time, We are not interested here in how things behave in time, so we chose a convenient time of zero. Thus, our “building so we chose a convenient time of zero. Thus, our “building block” isblock” is

Now we claim that any general, nonperiodic wave function Now we claim that any general, nonperiodic wave function ψψ(x)(x) can be expressed as a sum/integral can be expressed as a sum/integral of this building of this building blocks over the continuum of wave numbers:blocks over the continuum of wave numbers:

)(),( wtkxiAetx

ikxe

dkekAx ikx)()(

Fourier TransformFourier Transform

dkekAx ikx)()(

The amplitude The amplitude A(k)A(k) of the plane wave is naturally a function of the plane wave is naturally a function ofof k k, it tell us how much of each different wave number goes , it tell us how much of each different wave number goes into the sum. Although we can’t pull it out of the integral, the into the sum. Although we can’t pull it out of the integral, the equation can be solved for equation can be solved for A(k)A(k).. The result is: The result is:

dxexkA ikx)(2

1)(

The proper name of for The proper name of for A(k)A(k) is the is the Fourier transformFourier transform of the of the function function ψψ(x)(x)..

11 │x│< a│x│< a00 │x│> a│x│> a

Let use Let use EulerEuler identities: identities: __eeikaika = cos = cosφφ+isin+isinφφ ee-ika-ika = cos = cosφφ-isin-isinφφ eeikaika- e- e-ika -ika = 2isinka= 2isinka

And we can overwrite the equation for And we can overwrite the equation for A(k):A(k):

ψψ(x) =(x) =

ik

eex

ik

edkex

dxexkA

ikaika

a

a

ikxa

a

ikx

ikx

)(

2

1)(

2

11

2

1)(

)(2

1)(

kakk

kakA sin

1

2

2sin

2

2)(

General Wave PacketsGeneral Wave Packets

Any point in space can be described as a linear Any point in space can be described as a linear

combination of unit vectors. The three unit vectors combination of unit vectors. The three unit vectors îî,, ĵ ĵ, , andand constitute a base that can generate any points in constitute a base that can generate any points in space.space.

In similar way: given a In similar way: given a periodic functionperiodic function, any value , any value that the function can take, can be produced by the linear that the function can take, can be produced by the linear combination of a set of basic functions. The basic combination of a set of basic functions. The basic functions are the functions are the harmonic functions (harmonic functions (sin sin or or coscos)).. The set The set of basic functions is actually infinite. of basic functions is actually infinite.

k̂

The General Wave PacketThe General Wave Packet

A periodic function A periodic function f(x)f(x) can be represented by can be represented by the sum of harmonic waves:the sum of harmonic waves:

y(x,t) = y(x,t) = ΣΣ [A [Aiicos(kcos(kiix – x – ωωiit) + Bt) + Biisin(ksin(kiix – x – ωωiit)]t)]

AAii and and BBii ≡ amplitudes of the waves with wave ≡ amplitudes of the waves with wave

number number kkii and angular frequency and angular frequency ωωii..

For a function that is not periodic there is an For a function that is not periodic there is an equivalent approach called equivalent approach called Fourier TransformationFourier Transformation..

Fourier TransformationFourier TransformationA function A function F(x)F(x) that is not periodic can be represented by that is not periodic can be represented by a sum (integral) of functions of the typea sum (integral) of functions of the type

ee±ika±ika = Cos = Cosφφ±iSin±iSinφφ

In math terms it called In math terms it called Fourier TransformationFourier Transformation. Given a . Given a function function F(x)F(x)

wherewhere

f(kf(kjj)) represents the amplitude of base function represents the amplitude of base function ee-ikx-ikx used to used to

represent represent F(x).F(x).

dkekfxF ikx )(2

1)(

dxexFkf ikx )(2

1)(

The SchrThe Schrödinger Equationödinger Equation

The wave equation governing the The wave equation governing the motion of electron and other particles with motion of electron and other particles with mass mass mm, which is analogous to the classical , which is analogous to the classical wave equationwave equation

was found by Schrödinger in 1925 and is now was found by Schrödinger in 1925 and is now known as the known as the Schrödinger equationSchrödinger equation..

2

2

22

2 1

t

y

vx

y

The SchrThe Schrödinger Equationödinger Equation

Like the classical wave equation, the Like the classical wave equation, the SchrSchrödinger equationödinger equation is a partial differential is a partial differential equation in space and time. equation in space and time.

Like Like Newton’s lawsNewton’s laws of motion, the of motion, the SchrSchrödinger equationödinger equation cannot be derived. cannot be derived. It’s validity, like that of It’s validity, like that of Newton’s lawsNewton’s laws, lies in , lies in its agreement with experiment. its agreement with experiment.

We will start from classical description of the We will start from classical description of the total energy of a particle:total energy of a particle:

SchrSchröödinger converted this equation into a wave dinger converted this equation into a wave equation by defining a wavefunction, equation by defining a wavefunction, ΨΨ.. He He multiplied each factor in energy equation with multiplied each factor in energy equation with that wave function:that wave function:

)(2

2

xUm

pUKEE t o t

)(2

2

xUm

pE

To incorporate the de Broglie wavelength of the particle

he introduced the operator, ,which provides the

square of the momentum when applied to a plane wave: 2

22

x

)( tk xie If we apply the operator to that wavefunction: If we apply the operator to that wavefunction:

2222

22 )(

pkdx

x

wherewhere k k is the wavenumber, which equals is the wavenumber, which equals 22ππ//λλ..

We now simple replace the We now simple replace the pp22 in equation forin equation for energy: energy:

)()()(

2 2

22

xExUdx

xd

m

Time Independent SchrTime Independent Schrödinger Equationödinger Equation

This equation is called time-independent Schrödinger This equation is called time-independent Schrödinger equation.equation.

EE is the total energy of the particle.is the total energy of the particle.

The normalization condition now becomesThe normalization condition now becomes

∫ ∫ ΨΨ*(x)*(x)ΨΨ(x)dx = 1(x)dx = 1

)()()()(

2 2

22

xExxUdx

xd

m

A Solution to the SrA Solution to the Srödinger Equationödinger Equation

Show that for a free particle of mass Show that for a free particle of mass mm moving in moving in one dimension the function one dimension the function

is a solution of the time independent Sris a solution of the time independent Srödinger ödinger Equation for any values of the constantsEquation for any values of the constants A A andand BB..

kxBkxAx cossin)(

Energy Quantization in Different SystemsEnergy Quantization in Different Systems

The quantized energies of a The quantized energies of a system are generally system are generally determined by solving thedetermined by solving the SchrSchrödinger equationödinger equation for that for that system.system. The form of the The form of the SchrSchrödingerödinger equation equation depends on the depends on the potential potential energyenergy of the particle. of the particle.

The potential energy for a one-dimensional box fromThe potential energy for a one-dimensional box from x = 0x = 0 toto x = Lx = L is shown in is shown in FigureFigure. This potential energy function is . This potential energy function is called an called an infinity square-well potentialinfinity square-well potential, and is described by:, and is described by:

U(x) = 0,U(x) = 0, 0<x<L0<x<L

U(x) = U(x) = ∞∞, , x<0 or x>Lx<0 or x>L

A Particle in Infinity Square Well PotentialA Particle in Infinity Square Well Potential

Inside the boxInside the box U(x) = 0U(x) = 0, so the , so the SchrSchrödinger equationödinger equation is written:is written:

wherewhere E = ħE = ħωω is the energy of the particle, oris the energy of the particle, or

wherewhere kk22 = 2mE/ħ = 2mE/ħ22

The general solution of this equation can be written The general solution of this equation can be written asas

ψψ(x) = A sin kx + B cos kx(x) = A sin kx + B cos kx

where where AA and and BB are constants. Atare constants. At x=0x=0,, we have we have

ψψ(0) = A sin (k0) + B cos (0x) = 0 + B(0) = A sin (k0) + B cos (0x) = 0 + B

)()(

2 2

22

xEdx

xd

m

0)()( 2

2

2

xkdx

xd


The boundary conditionThe boundary condition ψψ(x)=0(x)=0 at at x=0x=0 thus givesthus gives B=0B=0 and equation becomesand equation becomes

ψψ(x) = A sin kx(x) = A sin kx

We received a We received a sinsin wave with the wavelength wave with the wavelength λλ related to related to wave numberwave number kk in a usual way,in a usual way, λλ = 2 = 2ππ/k/k.. The boundary The boundary conditioncondition ψψ(x) =0(x) =0 atat x=Lx=L givesgives

ψψ(L) = A sin kL = 0(L) = A sin kL = 0

This condition is satisfied if This condition is satisfied if kL kL is any integer times is any integer times ππ, or, or

kknn = n = nππ / L / L

If we will write the wave numberIf we will write the wave number k k in terms of wavelength in terms of wavelength λλ = 2 = 2ππ/k/k, we will receive the standing wave condition for , we will receive the standing wave condition for particle in the box:particle in the box:

nnλλ / 2 = L n = 1,2,3,…… / 2 = L n = 1,2,3,……

A Particle in Infinity Square Well PotentialA Particle in Infinity Square Well PotentialSolvingSolving kk22 = 2mE/ħ = 2mE/ħ2 2 forfor EE and using the standing wave and using the standing wave conditioncondition k = nk = nππ / L / L gives us the allowed energy valuesgives us the allowed energy values::

wherewhere

For each valueFor each value nn, there is a wave function, there is a wave function ψψnn(x)(x) given bygiven by

12

2

22

2222

822En

mL

hn

L

n

mm

kE nn

2

2

1 8mL

hE

L

xnAx nn

sin)(


Compare with the equation we received for Compare with the equation we received for

particle in the box, using the standing wave fitting with particle in the box, using the standing wave fitting with

the constantthe constant AAnn = √2/L = √2/L determined by normalizationdetermined by normalization::

L

xn

Lxn

sin

2)(

Although this problem seems artificial, actually it is Although this problem seems artificial, actually it is

useful for some physical problems, such as a useful for some physical problems, such as a

neutron inside the nucleus.neutron inside the nucleus.

A Particle in a Finite Square WellA Particle in a Finite Square WellThis potential energy function This potential energy function is described mathematically is described mathematically by:by: U(x)=VU(x)=V00, x<0, x<0

U(x)=0, 0<x<LU(x)=0, 0<x<L

U(x)=VU(x)=V00, x>L, x>L

Here we assume thatHere we assume that 0 0 ≤E≤V≤E≤V00. .

Inside the well,Inside the well, U(x)=0U(x)=0, and , and the time independent the time independent Schrödinger equationSchrödinger equation is the is the same as for the infinite wellsame as for the infinite well

)()(

2 2

22

xEdx

xd

m

A Particle in a Finite Square WellA Particle in a Finite Square Well

oror

wherewhere kk22 = 2mE/ħ = 2mE/ħ22.. The general solution is The general solution is

ψψ(x) = A sin kx + B cos kx(x) = A sin kx + B cos kx

but in this case,but in this case, ψψ(x)(x) is not required to be zero atis not required to be zero at x=0x=0, , soso BB is not zero. is not zero.

)()(

2 2

22

xEdx

xd

m

0)()( 2

2

2

xkdx

xd

A Particle in a Finite Square WellA Particle in a Finite Square WellOutside the well, the time independent Outside the well, the time independent SchrSchrödinger equation is ödinger equation is

oror

wherewhere

)()()(

2 02

22

xExUdx

xd

m

0)()( 2

2

2

xdx

xd

0)(2

022 EU

m

The Harmonic OscillatorMore realistic than a particle in a box is the harmonic More realistic than a particle in a box is the harmonic oscillator, which applies to an object of massoscillator, which applies to an object of mass mm on a spring of on a spring of force constantforce constant kk or to any systems undergoing small or to any systems undergoing small oscillations about a stable equilibrium. The potential energy oscillations about a stable equilibrium. The potential energy function for a such oscillator is:function for a such oscillator is:

wherewhere ωω00 = √k/m=2πf = √k/m=2πf is the angular frequency of the is the angular frequency of the oscillator.oscillator. ClassicallyClassically, the object oscillates between, the object oscillates between x = +Ax = +A and and x=-Ax=-A. Its total energy is. Its total energy is

which can have any nonnegative value, including zero.which can have any nonnegative value, including zero.

2202

1221)( xmkxxU

2202

1221 AmmvE

Potential energy function for a simple harmonic Potential energy function for a simple harmonic oscillator. Classically, the particle with energyoscillator. Classically, the particle with energy EE is is confined between the “turning points”confined between the “turning points” –A–A andand +A+A. .

The Harmonic Oscillator

Classically, the probability of finding the particle inClassically, the probability of finding the particle in dx dx is proportional to the time spent inis proportional to the time spent in dxdx, which is, which is dx/vdx/v. . The speed of the particle can be obtained from the The speed of the particle can be obtained from the conservation of energy: conservation of energy:

The classical probability is thusThe classical probability is thus

2202

1221 xmmvE

22

212

)(

xmEm

dx

v

dxdxxPC

The Harmonic OscillatorThe Harmonic Oscillator

The classical probability isThe classical probability is

2202

1221 AmmvE

22

212

)(

xmEm

dx

v

dxdxxPC

Any values of the energy Any values of the energy E E is possible. The lowest energy is is possible. The lowest energy is E=0E=0, in which case the particle is in the rest at the origin. The , in which case the particle is in the rest at the origin. The ShrShröödinger equation for this problem is dinger equation for this problem is

)()(2

1)(

222

2

22

xExxmdx

xd

m


In In quantum theoryquantum theory, the particle is represented by , the particle is represented by

the wave functionthe wave function ψψ(x)(x),, which is determined by solving the which is determined by solving the SchrSchröödingerdinger equation for this potential. equation for this potential.

Only certain values of Only certain values of EE will lead to solution that are will lead to solution that are well behaved, i.e., which approach well behaved, i.e., which approach zerozero as as x x approach approach infinityinfinity. Normalizeable wave function. Normalizeable wave function ψψnn(x)(x) occur only for occur only for

discrete values of the energydiscrete values of the energy EEnn given by given by

wherewhere ff00==ωω00/2/2ππ is the classical frequency of the oscillator.is the classical frequency of the oscillator.

......3,2,1,02

1

2

10

nnhfnEn


wherewhere ff00==ωω00/2/2ππ is the classical frequency of the is the classical frequency of the

oscillator. Thus, the ground-state energy isoscillator. Thus, the ground-state energy is ½ħ½ħωω and the exited energy levels are equally spaced and the exited energy levels are equally spaced byby ħħωω..

......3,2,1,02

1

2

10

nnhfnEn

Energy levels in the simple harmonic oscillator potential. Energy levels in the simple harmonic oscillator potential. Transitions obeying the selection ruleTransitions obeying the selection rule ΔΔn=±1n=±1 are indicated by are indicated by the arrows. Since the levels have equal spacing, the same the arrows. Since the levels have equal spacing, the same energyenergy ħħωω is emitted or absorbed in all allowed transitions. For is emitted or absorbed in all allowed transitions. For this special potential, the frequency of emitted or absorbed this special potential, the frequency of emitted or absorbed photon equals the frequency of oscillation, as predicted by photon equals the frequency of oscillation, as predicted by classical theory. classical theory.


Compare this with uneven spacing of the energy Compare this with uneven spacing of the energy levels for the particle in a box. If a harmonic levels for the particle in a box. If a harmonic oscillator makes a transition from energy leveloscillator makes a transition from energy level nn to to the next lowest energy levelthe next lowest energy level (n-1)(n-1),, the frequencythe frequency ff of the photon emitted is given byof the photon emitted is given by hf = Ehf = Eff – E – Eii. .

Applying this equation gives:Applying this equation gives:

The frequencyThe frequency ff of the emitted photon is therefore of the emitted photon is therefore equal to the classical frequencyequal to the classical frequency ff00 of the oscillator.of the oscillator.

0001 2

1)1(

2

1hfhfnhfnEEhf nn

Wave function for the ground state and the first two excited Wave function for the ground state and the first two excited states of the simple harmonic oscillator potential, the states states of the simple harmonic oscillator potential, the states withwith n=0n=0,, 1 1, and, and 22. .

2n

xm

u2

Probability density Probability density for the simple for the simple harmonic oscillator harmonic oscillator plotted against the plotted against the dimensionless valuedimensionless value

, for , for n=0, 1, n=0, 1,

andand 2 2. The blue . The blue curves are the curves are the classical probability classical probability densities for the densities for the same energy, and same energy, and the vertical lines the vertical lines indicate the classical indicate the classical turning points turning points x = x = ±A±A

2n

Molecules vibrate as harmonic Molecules vibrate as harmonic oscillators. Measuring vibration oscillators. Measuring vibration frequencies enables determination frequencies enables determination of force constants, bond strengths, of force constants, bond strengths, and properties of solids. and properties of solids.

Verify thatVerify that , where, where αα is a positive is a positive constant, is a solution of constant, is a solution of the the SchrSchrödinger ödinger equation equation for the harmonic oscillatorfor the harmonic oscillator

2

00 )( xeAx

)()(2

1)(

222

2

22

xExxmdx

xd

m

chapter 5 the wavelike properties of particles. de broglie wavesde broglie waves electron...

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