Wavelike Properties of Particles - With Solutions

165
Chapter 5 Chapter 5 The Wavelike Properties of The Wavelike Properties of Particles Particles

Transcript of Wavelike Properties of Particles - With Solutions

Page 1: Wavelike Properties of Particles - With Solutions

Chapter 5Chapter 5

The Wavelike Properties of The Wavelike Properties of ParticlesParticles

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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

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The de Broglie HypothesisThe de Broglie Hypothesis

As it was showed by As it was showed by ThomsonThomson the rays of the rays of a cathode tube can be deflected by electric and a cathode tube can be deflected by electric and magnetic fields and therefore must consist of magnetic fields and therefore must consist of electrically charged particles.electrically charged particles.

ThomsonThomson showed that all the particles showed that all the particles have the samehave the same charge-to-masscharge-to-mass ratio ratio q/mq/m. He . He also showed that particles with this also showed that particles with this charge-to-charge-to-massmass ratios can be obtained using any materials ratios can be obtained using any materials for the cathode, which means that these for the cathode, which means that these particles, now calledparticles, now called electronselectrons, are a , are a fundamental constituent of matter.fundamental constituent of matter.

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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.

In 1924, a French physics student, In 1924, a French physics student, Louis de BroglieLouis de Broglie, , suggested this idea in his doctoral dissertation.suggested this idea in his doctoral dissertation.

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.

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

E = pc = hf = hc/E = pc = hf = hc/λλUsing relativistic mechanics de Broglie Using relativistic mechanics de Broglie

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

The de Broglie HypothesisThe de Broglie Hypothesis

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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.

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The de Broglie WavelengthThe de Broglie Wavelength

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 thanIts wavelength is than

mK

h

p

h

2

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The de Broglie WavelengthThe de Broglie Wavelength

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 electronvolts.electronvolts.

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The de Broglie WavelengthThe de Broglie Wavelength

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

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The de Broglie WavelengthThe de Broglie Wavelength

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

Than the momentum of the molecule is:Than 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:nm

THe

26.1

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The de Broglie WavelengthThe de Broglie Wavelength

.

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 show, that for the acceleratedThis calculations show, 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. So, we can expect, that diffraction can be observed for . So, we can expect, that diffraction can be observed for this particlesthis particles

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(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

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(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

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)

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(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

1.00 keV 1000 eVnK (b)(b)

11132.87 10

m 9.07 10 m 907 fm1000

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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?

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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)

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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?

(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

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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.

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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.

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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.

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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..

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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.

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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

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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

/106.11022.92

106.61

2 EEeVJkg

sJ

Em

h

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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. .

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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. .

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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.

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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.

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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: ): )

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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

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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. .

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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

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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.)

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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

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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 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

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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?

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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

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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

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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 speeds?(d) At low particle speeds? approach at high particle speeds?(d) At low particle speeds?

(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

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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. Classical waves are the solution of the that waves. Classical waves are the solution of the classical wave equationclassical 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

)(2

cos2cos)cos( 000 txyT

txytkxyy

22

2 kandT

f

fvp

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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.

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

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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.

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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.

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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.

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

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

forfor n n = integer= integer

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

2nh

nmvr

orbitofncecircumferenp

nh

mv

nhr 2

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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.

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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.

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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.

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((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. .

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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

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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

Page 53: Wavelike Properties of Particles - With Solutions

BEATSBEATS

withwith

222121

2121

kkk

kkk

)sin(22

cos2)( txktxk

FxF

tx

kktx

kkFxF

22sin

22cos2)( 21212121

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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( txkF

txk

22cos2

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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

Page 56: Wavelike Properties of Particles - With Solutions

Gaussian-shaped wave packetsGaussian-shaped wave packets y(x)y(x) and the and the corresponding Gaussian distributions of wave numberscorresponding Gaussian distributions of wave numbers A(k)A(k).. (a) A narrow packet. (b) A wide packet. The standard (a) A narrow packet. (b) A wide packet. The standard deviations in each case are related bydeviations in each case are related by σσxxσσkk = ½ = ½..

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Wave packet for which Wave packet for which the group velocity is half the group velocity is half of phase velocity. Water of phase velocity. Water waves whose waves whose wavelengths are a few wavelengths are a few centimeters, but much centimeters, but much less than the water less than the water depth, have this depth, have this property. The arrow property. The arrow travels at the phase travels at the phase velocity, following a point velocity, following a point of constant phase for the of constant phase for the dominant wavelength. dominant wavelength. The cross at the center The cross at the center of the group travels at of the group travels at the group velocity. the group velocity.

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A three-dimensional wave packet representing a particle A three-dimensional wave packet representing a particle moving along the moving along the xx-axis. The dots indicate the position of -axis. The dots indicate the position of classical particle. Note that the particle spreads out in the classical particle. Note that the particle spreads out in the x x and and y y directions. This spreading is due to dispersion , resulting directions. This spreading is due to dispersion , resulting from the fact that the phase velocity of the individual wave from the fact that the phase velocity of the individual wave making up the packet depends on the wavelength of the making up the packet depends on the wavelength of the waves. waves.

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““Seeing an electron” with a gamma-ray microscope. Seeing an electron” with a gamma-ray microscope.

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Because of the size of the lens, the momentum of the Because of the size of the lens, the momentum of the scattered photons is uncertain byscattered photons is uncertain by ΔΔppxx ≈psin ≈psinθθ = hsin = hsinθθ// λλ. . Thus the recoil momentum of the electron is also Thus the recoil momentum of the electron is also uncertain uncertain by at least this amount. by at least this amount.

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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. .

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The Interpretation of the Wave FunctionThe Interpretation of the Wave Function

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

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

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

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The Interpretation of the Wave FunctionThe Interpretation of the Wave Function

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

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

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

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The Interpretation of the Wave FunctionThe Interpretation of the Wave Function

The The wave functionwave function for waves in a string is for waves in a string is the string displacementthe string displacement yy. The wave function for . The wave function for sound waves can be either the displacement of sound waves can be either the displacement of the air molecules, or the pressurethe air molecules, or the pressure PP. The wave . The wave function of the electromagnetic waves is the function of the electromagnetic waves is the electric fieldelectric 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 describes a single particle. The square of the wave function single particle. The square of the wave function for a particle describes the for a particle describes the probability densityprobability density, , which is the probability per unit volume, of which is the probability per unit volume, of finding the particle at a location. finding the particle at a location.

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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..

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The Interpretation of the Wave FunctionThe Interpretation of the Wave Function

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 approach is approach infinityinfinity..

12

dx

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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??

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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. .

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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

Page 70: Wavelike Properties of Particles - With Solutions

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

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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

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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.

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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))

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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

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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)(

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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) ….. …..

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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.

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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..

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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.

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Page 81: Wavelike Properties of Particles - With Solutions

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..

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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)..

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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

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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. .

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Complex NumbersComplex Numbers

A complex number has the form A complex number has the form a+iba+ib, with , with ii22=-1=-1 or or i=i=√-1√-1 – – imaginary unitimaginary unit..aa - - real part; real part; bb – – imaginary part; imaginary part; i i – – imaginary imaginary unitunit

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

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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

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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

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Fourier TransformFourier TransformIn quantum mechanics, our basic function is the pure In quantum mechanics, our basic function is the pure sinusoidal plane wave describing a free particle, given in sinusoidal plane wave describing a free particle, given in equation:equation:

We are not interest here in how things behave in time, so We are not interest here in how things behave in time, so we chose a convenient time of zero. Thus, our “building 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)()(

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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)..

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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)(

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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 function is actually infinite. of basic function is actually infinite.

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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..

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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)(

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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

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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.

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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

pUKEEtot

)(2

2

xUm

pE

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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

)( tkxie 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

Page 98: Wavelike Properties of Particles - With Solutions

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

Page 99: Wavelike Properties of Particles - With Solutions

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)(

Page 100: Wavelike Properties of Particles - With Solutions

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

Page 101: Wavelike Properties of Particles - With Solutions

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

Page 102: Wavelike Properties of Particles - With Solutions

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

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 particle , we will receive the standing wave condition for particle in the box:in the box:

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

Page 103: Wavelike Properties of Particles - With Solutions

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)(

Page 104: Wavelike Properties of Particles - With Solutions

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

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.

Page 105: Wavelike Properties of Particles - With Solutions

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

Page 106: Wavelike Properties of Particles - With Solutions

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

Page 107: Wavelike Properties of Particles - With Solutions

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

Page 108: Wavelike Properties of Particles - With Solutions

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

Page 109: Wavelike Properties of Particles - With Solutions

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. .

Page 110: Wavelike Properties of Particles - With Solutions

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

Page 111: Wavelike Properties of Particles - With Solutions

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

Page 112: Wavelike Properties of Particles - With Solutions

The Harmonic OscillatorThe Harmonic Oscillator

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

Page 113: Wavelike Properties of Particles - With Solutions

The Harmonic OscillatorThe Harmonic Oscillator

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

Page 114: Wavelike Properties of Particles - With Solutions
Page 115: Wavelike Properties of Particles - With Solutions

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.

Page 116: Wavelike Properties of Particles - With Solutions

The Harmonic OscillatorThe Harmonic Oscillator

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

Page 117: Wavelike Properties of Particles - With Solutions

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. .

Page 118: Wavelike Properties of Particles - With Solutions

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

Page 119: Wavelike Properties of Particles - With Solutions

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.

Page 120: Wavelike Properties of Particles - With Solutions

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

Page 121: Wavelike Properties of Particles - With Solutions

OperatorsOperators

As we have seen, for a particle in a state of definite energy As we have seen, for a particle in a state of definite energy the probability distribution is independent of time. The the probability distribution is independent of time. The expectation value of expectation value of xx is then given by is then given by

In general, the expectation value of any function f(x) is given In general, the expectation value of any function f(x) is given by by

dxxxxx )()(*

dxxxfxxf )()()()( *

Page 122: Wavelike Properties of Particles - With Solutions

OperatorsOperators

If we know the momentum If we know the momentum pp of the particle as function of the particle as function

of of xx, we can calculate the expectation value , we can calculate the expectation value ‹p›‹p›. However, it is . However, it is

impossible in principle to find impossible in principle to find pp as function of as function of xx since, since,

according to uncertainty principle, both according to uncertainty principle, both pp and and xx can not be can not be

determined at the same time. determined at the same time.

To find To find ‹p›‹p› we need to know the distribution function for we need to know the distribution function for

momentum. If we know momentum. If we know ΨΨ(x)(x), the distribution function can be , the distribution function can be

found by found by FourierFourier analysis. It can be shown that analysis. It can be shown that ‹p›‹p› can be can be

found fromfound from

dxxxi

p )(

Page 123: Wavelike Properties of Particles - With Solutions

OperatorsOperators

Similarly, Similarly, ‹p‹p22› › can be found from can be found from

Notice that in computing the expectation value the Notice that in computing the expectation value the

operator representing the physical quantity operates on operator representing the physical quantity operates on ΨΨ(x)(x), ,

not on not on ΨΨ**(x)(x). This is not important to the outcome when the . This is not important to the outcome when the

operator is simply some function of operator is simply some function of xx, but it is critical when the , but it is critical when the

operator includes a differentiation, as in the case of operator includes a differentiation, as in the case of

momentum operator. momentum operator.

dxxxixi

p )(2

Page 124: Wavelike Properties of Particles - With Solutions

Expectation Values for Expectation Values for pp and and pp22

Find Find ‹p›‹p› and and ‹p‹p22›› for the ground state wave for the ground state wave function of the infinity square well.function of the infinity square well.

Page 125: Wavelike Properties of Particles - With Solutions

In classical mechanics, the total energy written in terms of In classical mechanics, the total energy written in terms of position and momentum variables is called the Hamiltonian position and momentum variables is called the Hamiltonian functionfunction

Um

pH

2

2

If we replace the momentum by the momentum operator pIf we replace the momentum by the momentum operator popop

and note that U = U(x), we obtain the Hamiltonian operator and note that U = U(x), we obtain the Hamiltonian operator HHopop::

)(2

2

xUm

pH op

op

The time-independent SchrThe time-independent Schrödinger equation can then be ödinger equation can then be written:written:

EHop

Page 126: Wavelike Properties of Particles - With Solutions

EHop

The advantage of writing the SchrThe advantage of writing the Schrödinger equation in this ödinger equation in this formal way is that it allows for easy generalization to more formal way is that it allows for easy generalization to more complicated problems such as those with several particles complicated problems such as those with several particles moving in three dimensions.moving in three dimensions.

We simply write the total energy of the system in terms of We simply write the total energy of the system in terms of position and momentum and replace the momentum position and momentum and replace the momentum variables by the appropriate operators to obtain the variables by the appropriate operators to obtain the Hamiltonian operator for the system. Hamiltonian operator for the system.

Page 127: Wavelike Properties of Particles - With Solutions

SymbolSymbol Physical quantityPhysical quantity OperatorOperator

f(x)f(x) Any function of x (the position, x; the Any function of x (the position, x; the potential energy U(x), etc.potential energy U(x), etc.

f(x)f(x)

ppxx x component of momentumx component of momentum

ppyy y component of momentumy component of momentum

ppzz z component of momentumz component of momentum

EE Hamiltonian (time-independent)Hamiltonian (time-independent)

EE Hamiltonian (time-dependent)Hamiltonian (time-dependent)

EEkk Kinetic energyKinetic energy

LLzz Z component of angular Z component of angular momentummomentum

xi

i

yi

zi

ti

)(2

2

xUm

pop

2

22

2 xm

Page 128: Wavelike Properties of Particles - With Solutions

Minimum Energy of a Particle in a BoxMinimum Energy of a Particle in a Box

An important consequence of the An important consequence of the uncertainty uncertainty principleprinciple is that a particle confined to a finite space can not is that a particle confined to a finite space can not have zero kinetic energy. have zero kinetic energy.

Let’s consider a one-dimensional box of lengthLet’s consider a one-dimensional box of length LL. If . If we know that the particle is in the box,we know that the particle is in the box, ΔΔxx is not larger is not larger thanthan LL. This implies that. This implies that ΔΔpp is at leastis at least ħ/Lħ/L. Let us take the . Let us take the standard deviation as a measurestandard deviation as a measure ofof ΔΔpp::

222222 )2()()( ppppppppp avav

Page 129: Wavelike Properties of Particles - With Solutions

Minimum Energy of a Particle in a BoxMinimum Energy of a Particle in a Box

If the box is symmetric,If the box is symmetric, pp will be zero since the particle will be zero since the particle moves to the left as often as to the right. Thenmoves to the left as often as to the right. Then

and the average kinetic energy is:and the average kinetic energy is:

222222 )2()()( ppppppppp avav

222)(

L

pp

2

22

22 mLm

pE

Page 130: Wavelike Properties of Particles - With Solutions

Minimum Energy of a Particle in a BoxMinimum Energy of a Particle in a Box

The average kinetic energy of a particle in a box The average kinetic energy of a particle in a box is:is:

Thus, we see that the uncertainty principle Thus, we see that the uncertainty principle indicate that the minimum energy of a particle in indicate that the minimum energy of a particle in a box cannot be zero. This minimum energy is a box cannot be zero. This minimum energy is called called zero-point energy. zero-point energy.

2

22

22 mLm

pE

Page 131: Wavelike Properties of Particles - With Solutions

The Hydrogen AtomThe Hydrogen Atom

The energy of an electron of momentum The energy of an electron of momentum pp a distance a distance rr from a proton isfrom a proton is

If we take for the order of magnitude of the position If we take for the order of magnitude of the position uncertaintyuncertainty ΔΔx = rx = r, we have:, we have:

((ΔΔpp22) = p) = p22 ≥ ≥ ћћ22/r/r22

The energy is thenThe energy is then

r

ke

m

pE

22

2

r

ke

mrE

2

2

2

2

Page 132: Wavelike Properties of Particles - With Solutions

The Hydrogen AtomThe Hydrogen Atom

There is a radiusThere is a radius rrmm at whichat which EE is minimum. is minimum.

SettingSetting dE/dr = 0dE/dr = 0 yields yields rrmm andand EEmm::

r

ke

mrE

2

2

2

2

nmamke

rm 0529.002

2

eVmek

r

ke

mrEm 6.13

22 2

422

2

2

rrmm came out to came out to

be exactly the be exactly the radius of the first radius of the first Bohr orbitBohr orbit

The ground The ground state state energyenergy

Page 133: Wavelike Properties of Particles - With Solutions

The Hydrogen AtomThe Hydrogen Atom

The potential energy of the electron-proton The potential energy of the electron-proton system varies inversely with separation distancesystem varies inversely with separation distance

As in the case of gravitational potential As in the case of gravitational potential energy, the potential energy of the electron-proton energy, the potential energy of the electron-proton system is chosen to be system is chosen to be zerozero if the electron is an if the electron is an infinity distance from the proton. Then for all finite infinity distance from the proton. Then for all finite distances, the potential energy is negative.distances, the potential energy is negative.

r

qqkVqU pe

e

Page 134: Wavelike Properties of Particles - With Solutions

The Hydrogen AtomThe Hydrogen Atom

Like the energies of Like the energies of a particle in a box a particle in a box and of a harmonic and of a harmonic oscillator, the oscillator, the energy levels in the energy levels in the hydrogen atom are hydrogen atom are described by a described by a quantum number quantum number nn. . The allowed The allowed energies of the energies of the hydrogen atom are hydrogen atom are given bygiven by

EEnn = -13.6 eV/n = -13.6 eV/n22, ,

n = 1,2,3,……n = 1,2,3,……

Energy-level diagram for the hydrogen Energy-level diagram for the hydrogen atom. The energy of the ground state isatom. The energy of the ground state is -13.6 eV-13.6 eV. As. As nn approachesapproaches ∞∞ the energy the energy approachesapproaches 00..

Page 135: Wavelike Properties of Particles - With Solutions

Step PotentialStep Potential

Consider a particle of energyConsider a particle of energy EE moving in region in which the moving in region in which the potential energy is the step functionpotential energy is the step function

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

U(x) = VU(x) = V00,, x>0 x>0

What happened when What happened when

a particle moving from a particle moving from

left to right encountersleft to right encounters

the step?the step?

The classical answer isThe classical answer is

simple: to the left of the simple: to the left of the

step, the particle movesstep, the particle moves

with a speed with a speed v = v = √2E/m√2E/m

Page 136: Wavelike Properties of Particles - With Solutions

Step PotentialStep PotentialAtAt x =0x =0, , an impulsive force an impulsive force act on the particle. If the act on the particle. If the initial energyinitial energy EE is less thanis less than VV00, , the particle will be the particle will be

turned around and will then turned around and will then move to the left at its move to the left at its original speed; that is, the original speed; that is, the particle will be reflected by particle will be reflected by the step. Ifthe step. If EE is greater than is greater than VV00, the particle will continue , the particle will continue

to move to the right but with to move to the right but with reduced speed given byreduced speed given by

v = v = √2(E – U√2(E – U00)/m)/m

Page 137: Wavelike Properties of Particles - With Solutions

Step PotentialStep Potential

We can picture this classical problem as a We can picture this classical problem as a ball rolling along a level surface and coming to a ball rolling along a level surface and coming to a steep hill of heightsteep hill of height hh given bygiven by mgh=Vmgh=V00. .

If the initial kinetic energy of the ball is less If the initial kinetic energy of the ball is less thanthan mghmgh, the ball will roll part way up the hill and , the ball will roll part way up the hill and then back down and to the left along the lower then back down and to the left along the lower surface at it original speed. Ifsurface at it original speed. If EE is greater thanis greater than mghmgh, the ball will roll up the hill and proceed to the , the ball will roll up the hill and proceed to the right at a lesser speed. right at a lesser speed.

Page 138: Wavelike Properties of Particles - With Solutions

The quantum mechanical result is similar whenThe quantum mechanical result is similar when EE is is less thanless than VV00.. If If E<VE<V00 the wave function does not go to the wave function does not go to

zero at zero at x=0x=0 but rather decays exponentially. The wave but rather decays exponentially. The wave penetrates slightly into the classically forbidden regionpenetrates slightly into the classically forbidden region x>0x>0, but it is eventually completely reflected. , but it is eventually completely reflected.

Page 139: Wavelike Properties of Particles - With Solutions

Step PotentialStep PotentialThis problem is somewhat similar to that of This problem is somewhat similar to that of

total internal reflection in optics. total internal reflection in optics.

ForFor E>VE>V00, , the quantum mechanical result differs the quantum mechanical result differs

from the classical result. Atfrom the classical result. At x=0x=0, the wavelength , the wavelength changes from changes from

λλ11=h/p=h/p11 = h/√2mE = h/√2mE

toto

λλ22=h/p=h/p22 = h/√2m(E-V = h/√2m(E-V00).).

When the wavelength changes suddenly, part of When the wavelength changes suddenly, part of

the wave is reflected and part of the wave is the wave is reflected and part of the wave is transmitted.transmitted.

Page 140: Wavelike Properties of Particles - With Solutions

Reflection CoefficientReflection Coefficient

Since a motion of an electron (or other Since a motion of an electron (or other

particle) is governed by a wave equation, the particle) is governed by a wave equation, the

electron sometimes will be electron sometimes will be transmitted transmitted and and

sometimes will be sometimes will be reflectedreflected. .

The probabilities of reflection and The probabilities of reflection and

transmission can be calculated by solving the transmission can be calculated by solving the

SchrSchröödinger equationdinger equation in each region of space in each region of space

and comparing the amplitudes of transmitted and comparing the amplitudes of transmitted

waves and reflected waves with that of the waves and reflected waves with that of the

incident wave. incident wave.

Page 141: Wavelike Properties of Particles - With Solutions

Reflection CoefficientReflection Coefficient

This calculation and its result are similar to finding This calculation and its result are similar to finding

the fraction of light reflected from the air-glass interface. Ifthe fraction of light reflected from the air-glass interface. If

RR is the probability of reflection, called the is the probability of reflection, called the reflection reflection

coefficientcoefficient, this calculation gives:, this calculation gives:

wherewhere kk11 is the wave number for the incident wave andis the wave number for the incident wave and kk22

is the wave number for the transmitted wave. is the wave number for the transmitted wave.

221

221

)(

)(

kk

kkR

Page 142: Wavelike Properties of Particles - With Solutions

Transmission CoefficientTransmission Coefficient The result is the same as the result in optics for The result is the same as the result in optics for the reflection of light at normal incidence from the the reflection of light at normal incidence from the boundary between two media having different boundary between two media having different indexes indexes of refractionof refraction nn..

The probability of transmissionThe probability of transmission TT, called the , called the transmission coefficienttransmission coefficient, can be calculated from the , can be calculated from the reflection coefficient, since the probability of reflection coefficient, since the probability of transmission plus the probability of reflection must transmission plus the probability of reflection must equal 1:equal 1:

T + R = 1 T + R = 1 In the quantum mechanics, a localized particle is In the quantum mechanics, a localized particle is

represented by the wave packet, which has arepresented by the wave packet, which has a maximum maximum at the most probable position of the particle. at the most probable position of the particle.

Page 143: Wavelike Properties of Particles - With Solutions

Time development of a one dimensional wave packet Time development of a one dimensional wave packet representing a particle incident on a step potential forrepresenting a particle incident on a step potential for E>VE>V00. .

The position of a classical particle is indicated by the dot. Note The position of a classical particle is indicated by the dot. Note that part of the packet is transmitted and part is reflected.that part of the packet is transmitted and part is reflected.

Page 144: Wavelike Properties of Particles - With Solutions

Reflection coefficientReflection coefficient RR and transmission coefficientand transmission coefficient TT for a for a potential steppotential step VV00 high versus energyhigh versus energy EE (in units(in units VV00).).

Page 145: Wavelike Properties of Particles - With Solutions

A particle of energyA particle of energy EE00 traveling in a region in traveling in a region in

which the potential energy is zero is incident on which the potential energy is zero is incident on a potential barrier of heighta potential barrier of height VV00=0.2E=0.2E00. . Find the Find the

probability thatprobability that the particle will be reflected. the particle will be reflected.

Page 146: Wavelike Properties of Particles - With Solutions

Lets consider a rectangular potential barrier of heightLets consider a rectangular potential barrier of height VV00 and and

withwith aa given by:given by:U(x) = 0,U(x) = 0, x<0x<0U(x) = VU(x) = V00,, 0<x<a0<x<a

U(x) = 0,U(x) = 0, x>ax>a

Page 147: Wavelike Properties of Particles - With Solutions

Barrier PotentialBarrier PotentialWe consider a particle of We consider a particle of energyenergy EE , which is slightly , which is slightly less thanless than VV00, that is incident , that is incident

on the barrier from the left. on the barrier from the left. Classically, the particle Classically, the particle would always be reflected. would always be reflected. However, a wave incident However, a wave incident from the left does not from the left does not decrease immediately to decrease immediately to zero at the barrier, but itzero at the barrier, but itwill instead decay exponentially in the classically forbidden will instead decay exponentially in the classically forbidden regionregion 0<x<a0<x<a. Upon reaching the far wall. Upon reaching the far wall of the barrierof the barrier (x=a),(x=a), the wave function must join smoothlythe wave function must join smoothly to a sinusoidal to a sinusoidal wave function to the right of barrier.wave function to the right of barrier.

Page 148: Wavelike Properties of Particles - With Solutions

The potentials and the SchrThe potentials and the Schrödinger equations for the three ödinger equations for the three regions are as follows: regions are as follows:

Region I (x<0)Region I (x<0) V = 0,V = 0,

Region IIRegion II (0<x<a)(0<x<a) V = V V = V00,,

Region III (x>a) Region III (x>a) V = 0,V = 0,

0222

2

I

I Em

x

0)(2

022

2

II

II VEm

x

0222

2

III

III Em

x

Page 149: Wavelike Properties of Particles - With Solutions

Barrier PotentialBarrier PotentialIf we have a beam of particle incident from left, all with the If we have a beam of particle incident from left, all with the

same energysame energy E<VE<V00, the general solution of the wave equation are, , the general solution of the wave equation are,

following the example for a potential step,following the example for a potential step,

wherewhere kk11 =√2mE/ħ =√2mE/ħ and and αα = √2m(V = √2m(V00-E)/ħ-E)/ħ

This implies that there is some probability of the particle (which is This implies that there is some probability of the particle (which is represented by the wave function) being found on the far side of represented by the wave function) being found on the far side of the barrier even though, classically, it should never pass through the barrier even though, classically, it should never pass through the barrier. the barrier.

axGeFex

axDeCex

xBeAex

xikxik

xx

xikxik

11

11

)(

0)(

0)(

3

2

1

Page 150: Wavelike Properties of Particles - With Solutions

We assume that we have incident particles coming We assume that we have incident particles coming from the left moving along the from the left moving along the +x+x direction. In this case the direction. In this case the

termterm AeAeikik11xx in region I represents the incident particles. in region I represents the incident particles.

The term The term BeBe-ik-ik11xx represents the reflected particles moving represents the reflected particles moving in the in the –x–x direction. In region III there are no particles direction. In region III there are no particles initially moving along the initially moving along the -x-x direction. Thus direction. Thus G=0G=0, and the , and the

only term in region III is only term in region III is FeFeikik11xx. We summarize these wave . We summarize these wave functions:functions:

xikIII

xikI

xikI

Fedtransmitte

Bereflected

Aeincident

1

1

1

)(

)(

)(

Page 151: Wavelike Properties of Particles - With Solutions

Barrier PotentialBarrier Potential

For the case in which the quantityFor the case in which the quantity

ααa = √2maa = √2ma22(V(V00 – E)/ħ – E)/ħ22

is much greater thanis much greater than 1 1, the transmission coefficient , the transmission coefficient is proportionalis proportional toto ee-2-2ααaa, with, with

αα = √2m(V = √2m(V00 – E)/ħ – E)/ħ22

The probability of penetration of the barrier thus The probability of penetration of the barrier thus decreases exponentially with the barrier thicknessdecreases exponentially with the barrier thickness aa and with the square root of the relative barrier and with the square root of the relative barrier heightheight (V(V00-E)-E). This phenomenon is called barrier . This phenomenon is called barrier

penetration or tunneling. The relative probability of penetration or tunneling. The relative probability of its occurrence in any given situation is given by the its occurrence in any given situation is given by the transmission coefficient. transmission coefficient.

Page 152: Wavelike Properties of Particles - With Solutions

A wave packet representing a particle incident on two A wave packet representing a particle incident on two barriers of height just slightly greater than the energy of the barriers of height just slightly greater than the energy of the particle. At each encounter, part of the packet is transmitted particle. At each encounter, part of the packet is transmitted and part reflected, resulting in part of the packet being and part reflected, resulting in part of the packet being trapped between the barriers from same time. trapped between the barriers from same time.

Page 153: Wavelike Properties of Particles - With Solutions

A A 30-eV30-eV electron is incident on a square barrier of electron is incident on a square barrier of height height 40 eV40 eV. What is the probability that the electron will . What is the probability that the electron will tunnel through the barrier if its width is (a) tunnel through the barrier if its width is (a) 1.0 nm1.0 nm? ?

(b) (b) 0.1nm0.1nm??

Page 154: Wavelike Properties of Particles - With Solutions

The penetration of the barrier is not unique to quantum The penetration of the barrier is not unique to quantum mechanics. When light is totally reflected from the glass-air mechanics. When light is totally reflected from the glass-air interface, the light wave can penetrate the air barrier if a interface, the light wave can penetrate the air barrier if a second peace of glass is brought within a few wavelengths of second peace of glass is brought within a few wavelengths of the first, even when the angle of incidence in the first prism is the first, even when the angle of incidence in the first prism is greater than the critical angle. This effect can be demonstrated greater than the critical angle. This effect can be demonstrated with a laser beam and two with a laser beam and two 4545°° prisms. prisms.

Page 155: Wavelike Properties of Particles - With Solutions

αα- Decay- Decay

Potential well shown on the diagram for anPotential well shown on the diagram for an αα particle in a particle in a

radioactive nucleus approximately describes a strong attractive radioactive nucleus approximately describes a strong attractive

force whenforce when rr is less than the nuclear radiusis less than the nuclear radius RR.. Outside the Outside the

nucleus the strong nuclear force is negligible, and the potential nucleus the strong nuclear force is negligible, and the potential

is given by the is given by the Coulomb’s lawCoulomb’s law,, U(r) = +k(2e)(Ze)/rU(r) = +k(2e)(Ze)/r, , wherewhere ZeZe is is

the nuclear charge andthe nuclear charge and 2e2e is the charge of is the charge of αα particle.particle.

The theory of barrier The theory of barrier penetration was used by penetration was used by George GamovGeorge Gamov in 1928 to in 1928 to explain the enormous explain the enormous variation of the half-lives forvariation of the half-lives for αα decay of radioactive decay of radioactive nuclei.nuclei.

Page 156: Wavelike Properties of Particles - With Solutions

αα- Decay- Decay

An An αα-particle inside the nucleus oscillates back and forth, being -particle inside the nucleus oscillates back and forth, being

reflected at the barrier at reflected at the barrier at RR. Because of its wave properties, . Because of its wave properties,

when the when the αα-particle hits the barrier there is a small chance that it -particle hits the barrier there is a small chance that it

will penetrate and appear outside the well at will penetrate and appear outside the well at r = rr = r00. The wave . The wave

function is similar to that for a square barrier potential. function is similar to that for a square barrier potential.

Page 157: Wavelike Properties of Particles - With Solutions

The probability that an The probability that an αα-particle will tunnel through the -particle will tunnel through the barrier is given by barrier is given by

which is a very small number, i.e., the which is a very small number, i.e., the αα particle is usually particle is usually reflected. The number of times per second reflected. The number of times per second NN that the that the αα particle approaches the barrier is given byparticle approaches the barrier is given by

aEVm

eT)(22 0

R

vN

2

where where vv equals the particle’s speed inside the nucleus. equals the particle’s speed inside the nucleus.

The decay rate, or the probability per second that the nucleus The decay rate, or the probability per second that the nucleus will emit an will emit an αα particle, which is also the reciprocal of the mean particle, which is also the reciprocal of the mean life time , is given bylife time , is given by

aEVm

eR

vratedecay

)(22 0

2

1

Page 158: Wavelike Properties of Particles - With Solutions

The decay rate for emission ofThe decay rate for emission of αα particles from radioactive particles from radioactive nuclei ofnuclei of PoPo212212. . The solid curve is the prediction of equationThe solid curve is the prediction of equation

The points are the experimental results.The points are the experimental results.

aEVm

eR

vratedecay

)(22 0

2

1

Page 159: Wavelike Properties of Particles - With Solutions

Applications of TunnelingApplications of Tunneling

• NanotechnologyNanotechnology refers to the design and application of refers to the design and application of devices having dimensions ranging from devices having dimensions ranging from 11 to to 100 nm100 nm

• Nanotechnology uses the idea of trapping particles in Nanotechnology uses the idea of trapping particles in potential wellspotential wells

• One area of nanotechnology of interest to researchers is the One area of nanotechnology of interest to researchers is the quantum dotquantum dot– A quantum dot is a small region that is grown in a silicon A quantum dot is a small region that is grown in a silicon

crystal that acts as a potential wellcrystal that acts as a potential well

• Nuclear fusionNuclear fusion– Protons can tunnel through the barrier caused by their Protons can tunnel through the barrier caused by their

mutual electrostatic repulsionmutual electrostatic repulsion

Page 160: Wavelike Properties of Particles - With Solutions

Resonant Tunneling DeviceResonant Tunneling Device

• Electrons travel in the gallium arsenide Electrons travel in the gallium arsenide semiconductor semiconductor

• They strike the barrier of the quantum dot from the They strike the barrier of the quantum dot from the leftleft

• The electrons can tunnel through the barrier and The electrons can tunnel through the barrier and produce a current in the deviceproduce a current in the device

Page 161: Wavelike Properties of Particles - With Solutions

Scanning Tunneling MicroscopeScanning Tunneling Microscope

• An electrically conducting An electrically conducting probe with a very sharp probe with a very sharp edge is brought near the edge is brought near the surface to be studiedsurface to be studied

• The empty space The empty space between the tip and the between the tip and the surface represents the surface represents the “barrier”“barrier”

• The tip and the surface The tip and the surface are two walls of the are two walls of the “potential well”“potential well”

Page 162: Wavelike Properties of Particles - With Solutions

Scanning Tunneling MicroscopeScanning Tunneling Microscope

• The The STMSTM allows allows highly detailed highly detailed images of surfaces images of surfaces with resolutions with resolutions comparable to the comparable to the size of a single atomsize of a single atom

• At right is the surface At right is the surface of graphite “viewed” of graphite “viewed” with the with the STMSTM

Page 163: Wavelike Properties of Particles - With Solutions

Scanning Tunneling MicroscopeScanning Tunneling Microscope

• The The STMSTM is very sensitive to the distance from is very sensitive to the distance from the tip to the surfacethe tip to the surface– This is the thickness of the barrierThis is the thickness of the barrier

• STMSTM has one very serious limitation has one very serious limitation– Its operation is dependent on the electrical Its operation is dependent on the electrical

conductivity of the sample and the tipconductivity of the sample and the tip– Most materials are not electrically conductive at their Most materials are not electrically conductive at their

surfacessurfaces– The atomic force microscope (The atomic force microscope (AFMAFM) overcomes this ) overcomes this

limitation by tracking the sample surface maintaining limitation by tracking the sample surface maintaining a constant interatomic force between the atoms on a constant interatomic force between the atoms on the scanner tip and the sample’s surface atoms. the scanner tip and the sample’s surface atoms.

Page 164: Wavelike Properties of Particles - With Solutions

SUMMARYSUMMARY1. Time-independent Schr1. Time-independent Schrödinger equation:ödinger equation:

2.In the simple harmonic oscillator:2.In the simple harmonic oscillator:

the ground wave function is given:the ground wave function is given:

where where AA00 is the normalization constant andis the normalization constant and a=ma=mωω00/2ħ/2ħ. .

3. In a finite square well of height3. In a finite square well of height VV00, , there are only a finite there are only a finite number of allowed energies. number of allowed energies.

)()()()(

2 2

22

xExxUdx

xd

m

02

1

nEn

2

)(0axAex

Page 165: Wavelike Properties of Particles - With Solutions

SUMMARYSUMMARY

4.Reflection and barrier penetration:4.Reflection and barrier penetration:

When the potentials changes abruptly over a When the potentials changes abruptly over a small distance, a particle may be reflected even small distance, a particle may be reflected even thoughthough E>U(x)E>U(x).. A particle may penetrate a A particle may penetrate a region in whichregion in which E<U(x)E<U(x).. Reflection and Reflection and penetration of electron waves are similar for penetration of electron waves are similar for those for other kinds of waves. those for other kinds of waves.