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Quantum Mechanics

Fall 2018 Prof. Sergio B. Mendes 1

An essential theory to understand, from a microscopic perspective, properties of matter and light.

• Chemical• Electronic• Magnetic• Thermal• Optical• Etc.

Experimental Basis of Quantum Physics


CHAPTER 3

Topics


• 3.1 Discovery of the X Ray and the Electron• 3.2 Determination of Electron Charge• 3.3 Line Spectra• 3.4 Discrete Quantities• 3.5 Blackbody Radiation• 3.6 Photoelectric Effect• 3.7 X-Ray Production• 3.8 Compton Effect• 3.9 Pair Production and Annihilation

3.1 Discovery of the X Rays


Wilhelm Röntgen - 1895

Discovery of the Electron


J. J. Thomson - 1897

Cathode rays were charged particles

Thomson’s Experiment


𝐹𝐹𝑦𝑦 = 𝑞𝑞 𝐸𝐸 = 𝑚𝑚 𝑎𝑎𝑦𝑦 𝑡𝑡𝑎𝑎𝑎𝑎 𝜃𝜃 =𝑣𝑣𝑦𝑦𝑣𝑣𝑥𝑥

=𝑞𝑞 𝐸𝐸𝑚𝑚

𝑙𝑙𝑣𝑣𝑥𝑥 2

𝑣𝑣𝑦𝑦 = 𝑎𝑎𝑦𝑦 𝑡𝑡 =𝑞𝑞 𝐸𝐸𝑚𝑚

𝑙𝑙𝑣𝑣𝑥𝑥

𝑞𝑞𝑚𝑚

=𝑣𝑣𝑥𝑥 2 𝑡𝑡𝑎𝑎𝑎𝑎 𝜃𝜃

𝐸𝐸 𝑙𝑙

𝐵𝐵 = 0

𝑞𝑞 𝑬𝑬 + 𝑞𝑞 𝒗𝒗 × 𝑩𝑩 = 𝑚𝑚 𝒂𝒂

Magnetic Field turned ON and adjusted for:


𝑞𝑞 𝑬𝑬 + 𝑞𝑞 𝒗𝒗 × 𝑩𝑩 = 0

𝑣𝑣𝑥𝑥 =𝐸𝐸𝐵𝐵

𝑞𝑞𝑚𝑚

=𝑣𝑣𝑥𝑥 2 𝑡𝑡𝑎𝑎𝑎𝑎 𝜃𝜃

𝐸𝐸 𝑙𝑙=𝐸𝐸 𝑡𝑡𝑎𝑎𝑎𝑎 𝜃𝜃𝐵𝐵2 𝑙𝑙1.76 × 1011

𝐶𝐶𝑘𝑘𝑘𝑘

=

3.2 Millikan’s Experiment


Major Steps:


𝑚𝑚 𝑘𝑘 − 𝑏𝑏 𝑣𝑣 = 𝐹𝐹

𝑚𝑚 𝑘𝑘 − 𝑞𝑞 𝐸𝐸 − 𝑏𝑏 𝑣𝑣 = 𝐹𝐹

= 𝑚𝑚𝑑𝑑𝑣𝑣𝑑𝑑𝑡𝑡 = 0

= 𝑚𝑚𝑑𝑑𝑣𝑣𝑑𝑑𝑡𝑡 = 0

𝑚𝑚 𝑘𝑘 − 𝑏𝑏 𝑣𝑣𝑑𝑑 = 0

𝑚𝑚 𝑘𝑘 − 𝑞𝑞 𝐸𝐸 − 𝑏𝑏 𝑣𝑣𝑢𝑢 = 0

𝑞𝑞 𝐸𝐸 + 𝑏𝑏 𝑣𝑣𝑢𝑢 − 𝑏𝑏 𝑣𝑣𝑑𝑑 = 0

moving down, no electric field

moving down, with electric field 𝑦𝑦

(1)

(2)

(1) - (2):

Charges were multiples of unit “e”


𝑞𝑞 =𝑏𝑏 𝑣𝑣𝑑𝑑 − 𝑣𝑣𝑢𝑢

𝐸𝐸

𝑒𝑒 = 1.6 × 10−19 𝐶𝐶

𝑚𝑚 = 9.1 × 10−31 𝑘𝑘𝑘𝑘

= 𝑎𝑎 𝑒𝑒

3.3 Line Spectra


Color Separation by a Diffraction Grating


𝑑𝑑 𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃𝑚𝑚 = 𝑚𝑚 𝜆𝜆 grating equation

0 1 2𝑚𝑚 =

𝑚𝑚 = 1

Light Emission and Absorption by a Hot Body


Light Emission by Atoms and Molecules

Light Absorption by Atoms and Molecules

Emission Spectra of a Few Atoms


• Identification of Chemical Elements

• Composition of Materials

Balmer’s series for the Hydrogen atom, 1885


𝜆𝜆 = 364.56 𝑎𝑎𝑚𝑚𝑘𝑘2

𝑘𝑘2 − 4

𝑘𝑘 = 3, 4, 5, …

A simple rearrangement:


1𝜆𝜆

= 𝑅𝑅𝐻𝐻1

22−

1𝑘𝑘2

𝜆𝜆 = 364.56 𝑎𝑎𝑚𝑚𝑘𝑘2

𝑘𝑘2 − 4

𝑅𝑅𝐻𝐻 ≡4

364.56 𝑎𝑎𝑚𝑚

𝑘𝑘 = 3, 4, 5, …

=1.096776 × 107

𝑚𝑚

Rydberg equation, 1890


1𝜆𝜆

= 𝑅𝑅𝐻𝐻1𝑎𝑎2

−1𝑘𝑘2

𝑘𝑘 > 𝑎𝑎

3.4 Quantization


Discrete Quantities !!

• Electric charge

• Discrete spectral lines

• Atomic mass

• What else ??

3.5 Blackbody Radiation


Matter & Light (Electromagnetic Radiation)

in Thermal Equilibrium


• Constant temperature

• Absorption and emission are balanced

• Continuous spectral distribution

𝑻𝑻

Experimental Spectral Distribution


Empirical Laws:


Wien’s Law:

Stefan-Boltzmann Law:

𝜆𝜆𝑚𝑚𝑚𝑚𝑥𝑥 𝑇𝑇 ≅ 2.898 × 10−3 𝑚𝑚 𝐾𝐾

𝑅𝑅 𝑇𝑇 = �0

∞𝛪𝛪 𝜆𝜆,𝑇𝑇 𝑑𝑑𝜆𝜆 = 𝜖𝜖 𝜎𝜎 𝑇𝑇4

𝜎𝜎 = 5.6705 × 10−8 W/(m2 K4)

𝜖𝜖 = 0 , 1

PhET on Blackbody Radiation


https://phet.colorado.edu/sims/html/blackbody-spectrum/latest/blackbody-spectrum_en.html

Classical Calculation of Blackbody Radiation


𝑢𝑢 𝑓𝑓,𝑇𝑇 = 𝑁𝑁 𝑓𝑓 �𝐸𝐸𝑓𝑓(𝑇𝑇)

𝛪𝛪 𝑓𝑓,𝑇𝑇 = 𝑐𝑐 𝑢𝑢 𝑓𝑓,𝑇𝑇

∆𝐴𝐴

𝑐𝑐 ∆𝑡𝑡

𝑢𝑢

𝑐𝑐 ∆𝑡𝑡

𝑐𝑐 ∆𝑡𝑡14

Intensity = power / (per-unit-area × per-unit-frequency)

Appendix 1

Density of Standing Waves (Modes) of Electromagnetic Radiation


𝑁𝑁 𝑓𝑓 = 8 𝜋𝜋𝑓𝑓2

𝑐𝑐3

𝑁𝑁 𝑓𝑓 = ? ?

number of standing waves / (per-unit-volume × per-unit-frequency)

1D Standing Waves


𝑓𝑓 𝑥𝑥 = 𝐴𝐴 𝑠𝑠𝑠𝑠𝑎𝑎(𝑘𝑘 𝑥𝑥)

𝑘𝑘 𝐿𝐿 = 𝑚𝑚 𝜋𝜋 ∆𝑚𝑚 = ∆𝑘𝑘𝐿𝐿𝜋𝜋

𝑥𝑥

𝑓𝑓 𝑥𝑥

𝑚𝑚 = 1, 2, 3, …

𝐿𝐿

3D Standing Waves


𝑓𝑓 𝑥𝑥,𝑦𝑦, 𝑧𝑧 = 𝐴𝐴 𝑠𝑠𝑠𝑠𝑎𝑎(𝑘𝑘𝑥𝑥 𝑥𝑥) 𝑠𝑠𝑠𝑠𝑎𝑎 𝑘𝑘𝑦𝑦 𝑦𝑦 𝑠𝑠𝑠𝑠𝑎𝑎(𝑘𝑘𝑧𝑧 𝑧𝑧)

𝑘𝑘𝑥𝑥 𝐿𝐿𝑥𝑥 = 𝑚𝑚𝑥𝑥 𝜋𝜋 ∆𝑚𝑚𝑥𝑥 = ∆𝑘𝑘𝑥𝑥𝐿𝐿𝑥𝑥𝜋𝜋

𝑘𝑘𝑦𝑦 𝐿𝐿𝑦𝑦 = 𝑚𝑚𝑦𝑦 𝜋𝜋 ∆𝑚𝑚𝑦𝑦 = ∆𝑘𝑘𝑦𝑦𝐿𝐿𝑦𝑦𝜋𝜋

𝑘𝑘𝑧𝑧 𝐿𝐿𝑧𝑧 = 𝑚𝑚𝑧𝑧 𝜋𝜋 ∆𝑚𝑚𝑧𝑧 = ∆𝑘𝑘𝑧𝑧𝐿𝐿𝑧𝑧𝜋𝜋

∆𝑚𝑚𝑥𝑥 ∆𝑚𝑚𝑦𝑦 ∆𝑚𝑚𝑧𝑧 = ∆𝑘𝑘𝑥𝑥 ∆𝑘𝑘𝑦𝑦 ∆𝑘𝑘𝑧𝑧𝑉𝑉𝜋𝜋3

𝑀𝑀 =4 𝜋𝜋3𝑘𝑘3

𝑉𝑉𝜋𝜋3

𝑑𝑑𝑀𝑀𝑉𝑉

=1

2 𝜋𝜋2𝑘𝑘2 𝑑𝑑𝑘𝑘 = 4 𝜋𝜋 𝑓𝑓2

𝑐𝑐3 𝑑𝑑𝑓𝑓𝑘𝑘

2 𝜋𝜋=

1𝜆𝜆

=𝑓𝑓𝑐𝑐

18

𝑑𝑑𝑘𝑘2 𝜋𝜋

=𝑑𝑑𝑓𝑓𝑐𝑐


𝑁𝑁 𝑓𝑓 ≡𝑑𝑑𝑀𝑀𝑉𝑉 𝑑𝑑𝑓𝑓 = 2 × 4 𝜋𝜋

𝑓𝑓2

𝑐𝑐3= 8 𝜋𝜋

𝑓𝑓2

𝑐𝑐3

𝑑𝑑𝑀𝑀𝑉𝑉

= 4 𝜋𝜋𝑓𝑓2

𝑐𝑐3𝑑𝑑𝑓𝑓

2

Average Energy for Each Mode of the Electromagnetic Radiation


�𝐸𝐸 = ? ?

�𝐸𝐸 =∫0∞𝐸𝐸 𝑝𝑝 𝐸𝐸 𝑑𝑑𝐸𝐸

∫0∞𝑝𝑝 𝐸𝐸 𝑑𝑑𝐸𝐸

𝑝𝑝 𝐸𝐸 = 𝑒𝑒−𝐸𝐸𝑘𝑘 𝑇𝑇

= 𝑘𝑘 𝑇𝑇=∫0∞𝐸𝐸 𝑒𝑒−

𝐸𝐸𝑘𝑘 𝑇𝑇𝑑𝑑𝐸𝐸

∫0∞𝑒𝑒−

𝐸𝐸𝑘𝑘 𝑇𝑇 𝑑𝑑𝐸𝐸

�𝐸𝐸 = 𝑘𝑘 𝑇𝑇

Appendix 2

Bringing All the Pieces Together:


𝑢𝑢 𝑓𝑓,𝑇𝑇 = 𝑁𝑁 𝑓𝑓 �𝐸𝐸 𝑇𝑇


𝑐𝑐3

𝛪𝛪 𝑓𝑓,𝑇𝑇 =14𝑐𝑐 𝑢𝑢 𝑓𝑓,𝑇𝑇 = 2 𝜋𝜋 𝑓𝑓2

𝑐𝑐2𝑘𝑘 𝑇𝑇



�𝐸𝐸 𝑇𝑇 = 𝑘𝑘 𝑇𝑇


𝛪𝛪 𝑓𝑓,𝑇𝑇 = 2 𝜋𝜋 𝑓𝑓2


𝛪𝛪 𝜆𝜆,𝑇𝑇 𝑑𝑑𝜆𝜆 = 𝛪𝛪 𝑓𝑓,𝑇𝑇 𝑑𝑑𝑓𝑓

𝜆𝜆 𝑓𝑓 = 𝑐𝑐

𝛪𝛪 𝜆𝜆,𝑇𝑇 = 2 𝜋𝜋 𝑐𝑐𝜆𝜆4𝑘𝑘 𝑇𝑇

𝑓𝑓 𝑑𝑑𝜆𝜆 + 𝜆𝜆 𝑑𝑑𝑓𝑓 = 0𝑑𝑑𝑓𝑓𝑑𝑑𝜆𝜆

=𝑓𝑓𝜆𝜆

=𝑐𝑐𝜆𝜆2

Rayleigh-Jeans Formula

Theory and Experiment



Rayleigh-Jeans

Planck’s Assumption


�𝐸𝐸 = ? ?

�𝐸𝐸 =∑𝑛𝑛=0∞ 𝐸𝐸𝑛𝑛 𝑝𝑝 𝐸𝐸𝑛𝑛∑𝑛𝑛=0∞ 𝑝𝑝 𝐸𝐸𝑛𝑛

𝑝𝑝 𝐸𝐸 = 𝑒𝑒−𝐸𝐸𝑘𝑘 𝑇𝑇

�𝐸𝐸 =∫0∞𝐸𝐸 𝑝𝑝 𝐸𝐸 𝑑𝑑𝐸𝐸

∫0∞𝑝𝑝 𝐸𝐸 𝑑𝑑𝐸𝐸

𝐸𝐸𝑛𝑛 = 𝑎𝑎 ℎ 𝑓𝑓

=ℎ 𝑓𝑓

𝑒𝑒ℎ 𝑓𝑓𝑘𝑘 𝑇𝑇 − 1

Appendix 3

discrete exchange of energy

�𝐸𝐸 =ℎ 𝑓𝑓


Planck Theory of Blackbody Radiation


𝑢𝑢 𝑓𝑓,𝑇𝑇 = 𝑁𝑁 𝑓𝑓 �𝐸𝐸 𝑇𝑇


𝑐𝑐3

𝛪𝛪 𝑓𝑓,𝑇𝑇 =14𝑐𝑐 𝑢𝑢 𝑓𝑓,𝑇𝑇 =

2 𝜋𝜋 ℎ 𝑓𝑓3

𝑐𝑐2 𝑒𝑒ℎ 𝑓𝑓𝑘𝑘 𝑇𝑇 − 1


𝑐𝑐3ℎ 𝑓𝑓


�𝐸𝐸 𝑇𝑇 =ℎ 𝑓𝑓


Frequency and Wavelength Descriptions:


𝛪𝛪 𝑓𝑓,𝑇𝑇 =2 𝜋𝜋 ℎ 𝑓𝑓3


𝛪𝛪 𝜆𝜆,𝑇𝑇 =2 𝜋𝜋 ℎ 𝑐𝑐2

𝜆𝜆5 𝑒𝑒ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 − 1

𝛪𝛪 𝜆𝜆,𝑇𝑇 𝑑𝑑𝜆𝜆 = 𝛪𝛪 𝑓𝑓,𝑇𝑇 𝑑𝑑𝑓𝑓

𝑑𝑑𝑓𝑓𝑑𝑑𝜆𝜆 =

𝑓𝑓𝜆𝜆 =

𝑐𝑐𝜆𝜆2

𝜆𝜆 𝑓𝑓 = 𝑐𝑐

Comparison





Rayleigh-Jeans

Planck

ℎ = 6.6261 × 10−34 𝐽𝐽 𝑠𝑠

Classical Theory by

𝐸𝐸𝑛𝑛 = 𝑎𝑎 ℎ 𝑓𝑓discrete exchange of energy

Wien’s Law from Planck’s Theory


𝑑𝑑𝛪𝛪 𝜆𝜆,𝑇𝑇𝑑𝑑𝜆𝜆

= 0



ℎ 𝑐𝑐𝜆𝜆𝑚𝑚𝑚𝑚𝑥𝑥 𝑘𝑘 𝑇𝑇

≅ 4.966 𝜆𝜆𝑚𝑚𝑚𝑚𝑥𝑥 𝑇𝑇 ≅ℎ 𝑐𝑐

4.966 𝑘𝑘= 2.898 × 10−3 𝑚𝑚 𝐾𝐾

Appendix 4, Example 3.6

𝑥𝑥 𝑒𝑒 𝑥𝑥 = 5 𝑥𝑥 − 1 𝑥𝑥 ≡ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇

Stefan-Boltzmann Law from Planck’s Theory





∞𝛪𝛪 𝜆𝜆,𝑇𝑇 𝑑𝑑𝜆𝜆

𝜎𝜎 =2 𝜋𝜋5 𝑘𝑘4

15 ℎ3 𝑐𝑐2= 5.6705 × 10−8 W/(m2 K4)

= �0

∞ 2 𝜋𝜋 ℎ 𝑐𝑐2


𝑑𝑑𝜆𝜆 =2 𝜋𝜋5 𝑘𝑘4

15 ℎ3 𝑐𝑐2𝑇𝑇4

Example 3.7

3.6 Photoelectric Effect


PhET on Photoelectric Effect


https://phet.colorado.edu/en/simulation/legacy/photoelectric

Same frequency, different light intensities:


For light of a specific frequency, the retarding potential energy 𝑒𝑒 𝑉𝑉𝑜𝑜 required to stop the electric current is always the same,

regardless of the light intensity

𝑒𝑒 𝑉𝑉𝑜𝑜 =12𝑚𝑚 𝑣𝑣𝑚𝑚𝑚𝑚𝑥𝑥

2

Same light intensity,different frequencies:


For a given light intensity, the retarding potential energy 𝑒𝑒 𝑉𝑉𝑜𝑜 required to stop the current increases with the light frequency.

Experimental Facts and Puzzles:


• The maximum kinetic energy of the photoelectrons is independentof the light intensity. Why ?

• The maximum kinetic energy of the photoelectrons, for a given light intensity and emitting material, depends on the light frequency. Why ?

• To produce photoelectrons requires a minimum light frequency. Why ?

• The photoelectrons are emitted almost instantly following illumination of the photocathode, independent of the light intensity. Why ?

Einstein’s Article of 1905 on “Creation and Conversion of Light”


full article


12𝑚𝑚 𝑣𝑣2 = ℎ 𝑓𝑓 − 𝜙𝜙

ℎ 𝑓𝑓

𝜙𝜙

𝜙𝜙 is the work function of the material body


ℎ 𝑓𝑓 − 𝜙𝜙 =12𝑚𝑚 𝑣𝑣2 = 𝑒𝑒 𝑉𝑉𝑜𝑜

𝑉𝑉𝑜𝑜

Millikan on Photoelectric Effect:


Millikan utilized light of varying frequency on a sodium electrode and measured the maximum kinetic energies of the photoelectrons.

He found that no photoelectrons were emitted below a frequency of 4.39 X 1014

Hz (or longer than a wavelength of 683 nm).

The results were independent of the light intensity.

The slope of a straight line drawn through the data produced a value of Planck’s constant in excellent agreement with Einstein’s prediction.

Even though Millikan admitted his own data were sufficient proof of Einstein’s photo-electric effect equation, Millikan was not convinced of the photon concept for light and its role in quantum theory.

Robert Millikan (University of Chicago) published data on the

photoelectric effect, Phys. Rev. VII, 352-388 (1916).

Retarding potential energy versus frequency:


The retarding potential energy required to stop the current increases linearly with the light frequency. Regardless of

the material, the slope is always the same.


Work Function of Materials


The photoelectric effect can be described by 𝛾𝛾 + 𝑒𝑒𝑒 → 𝑒𝑒.


Can the reverse process 𝑒𝑒 → 𝑒𝑒𝑒 + 𝛾𝛾occur ?

And the answer is yes.

3.7 X-Ray Production


𝑒𝑒 𝑉𝑉𝑜𝑜 = 𝐸𝐸𝑖𝑖

𝐸𝐸𝑓𝑓 + ℎ 𝑓𝑓 = 𝐸𝐸𝑖𝑖

𝐸𝐸𝑓𝑓 + ℎ𝑐𝑐𝜆𝜆

= 𝐸𝐸𝑖𝑖0, 𝑒𝑒 𝑉𝑉𝑜𝑜 = 𝐸𝐸𝑓𝑓

Bremsstrahlung, from the German word for “braking radiation”

𝑒𝑒 → 𝑒𝑒𝑒 + 𝛾𝛾

Three Different Materials


The relative intensity of x-rays for an accelerating voltage of 35 kV.

𝐸𝐸𝑓𝑓 + ℎ𝑐𝑐𝜆𝜆

= 𝐸𝐸𝑖𝑖 𝐸𝐸𝑓𝑓 = 0, 𝑒𝑒 𝑉𝑉𝑜𝑜

𝐸𝐸𝑖𝑖 = 𝑒𝑒 𝑉𝑉𝑜𝑜

Notice that 𝜆𝜆𝑚𝑚𝑖𝑖𝑛𝑛 is the same for all three targets.

𝜆𝜆 =ℎ 𝑐𝑐

𝐸𝐸𝑖𝑖 − 𝐸𝐸𝑓𝑓𝜆𝜆𝑚𝑚𝑖𝑖𝑛𝑛 =

ℎ 𝑐𝑐𝑒𝑒 𝑉𝑉𝑜𝑜𝜆𝜆𝑚𝑚𝑖𝑖𝑛𝑛 = 0.0354 𝑎𝑎𝑚𝑚

3.8 Compton Effect


(Classical) Light-Matter Interaction and Scattered Light


• Electric field in the light wave drives the oscillation of electrons present in matter.

• Frequency of oscillation coincides with the frequency of the incident light.

• Re-emitted (scattered) light has the same frequency as the incident light.

• Thomson radiation

Compton Effect, 1923


𝛾𝛾 + 𝑒𝑒 → 𝛾𝛾𝑒 + 𝑒𝑒𝑒

• Conservation of electric charge

• Conservation of energy

• Conservation of linear momentum

• In addition to the collective interaction between light waves & electrons, Arthur Compton (University of Chicago) considered the photon-electron interaction.

• Hypothesized the phenomenon as a single-photon & single-electron collision.

Linear Momentum for Photons:


𝐸𝐸 = ℎ 𝑓𝑓 = 𝑝𝑝 𝑐𝑐

𝐸𝐸2 − 𝑝𝑝2 𝑐𝑐2 = 𝑚𝑚2 𝑐𝑐4

𝐸𝐸 = 𝑝𝑝 𝑐𝑐

𝑚𝑚 = 0

𝑝𝑝 = ℎ𝑓𝑓𝑐𝑐 =

ℎ𝜆𝜆

𝑝𝑝 =ℎ𝜆𝜆

𝑢𝑢 = 𝑐𝑐

Conservation of Energy:


ℎ 𝑓𝑓 + 𝐸𝐸𝑖𝑖 = ℎ 𝑓𝑓𝑒 + 𝐸𝐸𝑓𝑓

𝐸𝐸𝑖𝑖 = 𝑚𝑚 𝑐𝑐2

𝐸𝐸𝑓𝑓2 = 𝑚𝑚 𝑐𝑐2 2 + 𝑝𝑝𝑒𝑒2 𝑐𝑐2

ℎ 𝑓𝑓 = ℎ𝑐𝑐𝜆𝜆

ℎ 𝑓𝑓′ = ℎ𝑐𝑐𝜆𝜆𝑒

ℎ𝑐𝑐𝜆𝜆

+ 𝑚𝑚 𝑐𝑐2 = ℎ𝑐𝑐𝜆𝜆𝑒

+ 𝑚𝑚 𝑐𝑐2 2 + 𝑝𝑝𝑒𝑒2 𝑐𝑐2

Conservation of Linear Momentum:


ℎ𝜆𝜆

=ℎ𝜆𝜆𝑒𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 + 𝑝𝑝𝑒𝑒 𝑐𝑐𝑐𝑐𝑠𝑠 𝜙𝜙𝑥𝑥:

0 =ℎ𝜆𝜆𝑒𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃 − 𝑝𝑝𝑒𝑒 𝑠𝑠𝑠𝑠𝑎𝑎 𝜙𝜙𝑦𝑦:

From linear momentum equations:


ℎ𝜆𝜆−ℎ𝜆𝜆′𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑝𝑝𝑒𝑒 𝑐𝑐𝑐𝑐𝑠𝑠 𝜙𝜙

ℎ𝜆𝜆𝑒𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃 = 𝑝𝑝𝑒𝑒 𝑠𝑠𝑠𝑠𝑎𝑎 𝜙𝜙

ℎ𝜆𝜆

2

+ℎ𝜆𝜆𝑒

2

− 2ℎ𝜆𝜆

ℎ𝜆𝜆𝑒

𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑝𝑝𝑒𝑒2

(1)

(2)

(1)2 + (2)2 :

From energy equation:


ℎ𝑐𝑐𝜆𝜆

+ 𝑚𝑚 𝑐𝑐2 = ℎ𝑐𝑐𝜆𝜆𝑒

+ 𝑚𝑚 𝑐𝑐2 2 + 𝑝𝑝𝑒𝑒2𝑐𝑐2

ℎ𝑐𝑐𝜆𝜆− ℎ

𝑐𝑐𝜆𝜆𝑒

2+ 2 ℎ

𝑐𝑐𝜆𝜆− ℎ

𝑐𝑐𝜆𝜆𝑒

𝑚𝑚 𝑐𝑐2 = 𝑝𝑝𝑒𝑒2𝑐𝑐2

ℎ𝜆𝜆−ℎ𝜆𝜆𝑒

2

+ 2ℎ𝜆𝜆−ℎ𝜆𝜆𝑒

𝑚𝑚 𝑐𝑐 = 𝑝𝑝𝑒𝑒2

Combining the two equations for 𝑝𝑝𝑒𝑒2 :


ℎ𝜆𝜆

2

+ℎ𝜆𝜆𝑒

2

− 2ℎ𝜆𝜆

ℎ𝜆𝜆𝑒

𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑝𝑝𝑒𝑒2

ℎ𝜆𝜆−ℎ𝜆𝜆𝑒

2


𝑚𝑚 𝑐𝑐 = 𝑝𝑝𝑒𝑒2

−2ℎ𝜆𝜆

ℎ𝜆𝜆𝑒

𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = −2ℎ𝜆𝜆

ℎ𝜆𝜆𝑒


𝑚𝑚 𝑐𝑐

ℎ𝜆𝜆

ℎ𝜆𝜆𝑒

1 − 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 =ℎ𝜆𝜆−ℎ𝜆𝜆𝑒

𝑚𝑚 𝑐𝑐

Compton Equation:


ℎ𝑚𝑚 𝑐𝑐

1 − 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝜆𝜆𝑒 − 𝜆𝜆

ℎ𝜆𝜆

ℎ𝜆𝜆𝑒

1 − 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 =ℎ𝜆𝜆−ℎ𝜆𝜆𝑒

𝑚𝑚 𝑐𝑐

• Compton wavelength: ℎ𝑚𝑚𝑒𝑒 𝑐𝑐

= 0.00243 𝑎𝑎𝑚𝑚

• Compton equation

• Independent of wavelength

• Usually observed with x rays

Experimental Results by Compton:





3.9 Pair Production𝛾𝛾 → 𝑒𝑒− + 𝑒𝑒+

Conservation of electric charge

• Cannot satisfy simultaneously conservation of energy and linear momentum

Conservation of electric charge

Conservation of energy

Conservation of linear momentum

ℎ 𝑓𝑓 > 2 𝑚𝑚𝑒𝑒 𝑐𝑐21.022 𝑀𝑀𝑒𝑒𝑉𝑉

Positron-Electron Annihilation


𝑒𝑒− + 𝑒𝑒+ → 𝛾𝛾 + 𝛾𝛾𝑒 Conservation of electric charge

Conservation of energy

Conservation of linear momentum

2 𝑚𝑚𝑒𝑒 𝑐𝑐2 ≅ ℎ 𝑓𝑓1 + ℎ 𝑓𝑓2

Annihilation of a positronium atom (consisting of an electron and positron) producing two photons.

0 ≅ℎ 𝑓𝑓1𝑐𝑐

−ℎ 𝑓𝑓2𝑐𝑐

ℎ 𝑓𝑓 = 𝑚𝑚𝑒𝑒 𝑐𝑐2 ≅ 0.511 𝑀𝑀𝑒𝑒𝑉𝑉

Positron Emission Tomography (PET)

Fall 2018 67

A useful medical diagnostic tool to study the path and location of a positron-emitting radiopharmaceutical in the human body.

Appropriate radiopharmaceuticals are chosen to concentrate by physiological processes in

the region to be examined.

The positron travels only a few millimeters before annihilation, which produces two

photons that can be detected to give the positron position.

Summary


Blackbody Radiation


𝛪𝛪 𝑓𝑓,𝑇𝑇 =2 𝜋𝜋 ℎ 𝑓𝑓3




Wien’s Law

𝜆𝜆𝑚𝑚𝑚𝑚𝑥𝑥 𝑇𝑇 ≅ 2.898 × 10−3 𝑚𝑚 𝐾𝐾

Stefan-Boltzmann Law


∞𝛪𝛪 𝜆𝜆,𝑇𝑇 𝑑𝑑𝜆𝜆 = 𝜖𝜖 𝜎𝜎 𝑇𝑇4

𝐸𝐸𝑛𝑛 = 𝑎𝑎 ℎ 𝑓𝑓

𝜎𝜎 =2 𝜋𝜋5 𝑘𝑘4

15 ℎ3 𝑐𝑐2 = 5.6705 × 10−8 W/(m2 K4)𝜖𝜖 = 0 , 1

ℎ = 6.6261 × 10−34 𝐽𝐽 𝑠𝑠

𝛪𝛪 𝜆𝜆,𝑇𝑇

Photoelectric Effect



𝛾𝛾 + 𝑒𝑒 → 𝑒𝑒𝑒

X-Ray Production


𝑒𝑒 → 𝑒𝑒𝑒 + 𝛾𝛾

𝜆𝜆𝑚𝑚𝑖𝑖𝑛𝑛 =ℎ 𝑐𝑐𝑒𝑒 𝑉𝑉𝑜𝑜

𝑓𝑓𝑚𝑚𝑚𝑚𝑥𝑥 =𝑒𝑒 𝑉𝑉𝑜𝑜ℎ

Compton Effect


𝛾𝛾 + 𝑒𝑒 → 𝛾𝛾𝑒 + 𝑒𝑒𝑒



Positron-Electron Annihilation


𝑒𝑒− + 𝑒𝑒+ → 𝛾𝛾 + 𝛾𝛾𝑒

𝑚𝑚𝑒𝑒 𝑐𝑐2 ≅ ℎ 𝑓𝑓1 ≅ ℎ 𝑓𝑓2

𝑓𝑓1 ≅ 𝑓𝑓2

ℎ 𝑓𝑓 = 𝑚𝑚𝑒𝑒 𝑐𝑐2 ≅ 0.511 𝑀𝑀𝑒𝑒𝑉𝑉

Spectral Lines of Atoms


1𝜆𝜆

= 𝑅𝑅𝐻𝐻1𝑎𝑎2

−1𝑘𝑘2

𝑘𝑘 > 𝑎𝑎

𝑅𝑅𝐻𝐻 =1.096776 × 107

𝑚𝑚

Appendices


Appendix 1


𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 =∫02 𝜋𝜋 𝑑𝑑𝜑𝜑 ∫0

𝜋𝜋/2 𝑑𝑑𝜃𝜃 𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃

∫02 𝜋𝜋 𝑑𝑑𝜑𝜑 ∫0

𝜋𝜋/2 𝑑𝑑𝜃𝜃 𝑠𝑠𝑠𝑠𝑎𝑎 𝜃𝜃= �

0

𝜋𝜋/2𝑑𝑑𝜃𝜃

𝑠𝑠𝑠𝑠𝑎𝑎 2 𝜃𝜃4

=− 𝑐𝑐𝑐𝑐𝑠𝑠 2 𝜃𝜃

8 0

𝜋𝜋/2

=14

Appendix 2


∫0∞𝐸𝐸 𝑒𝑒−


∫0∞ 𝑒𝑒−


=𝐸𝐸 −𝑘𝑘 𝑇𝑇 𝑒𝑒−

𝐸𝐸𝑘𝑘 𝑇𝑇 𝐸𝐸=0

𝐸𝐸=∞− ∫0

∞ −𝑘𝑘 𝑇𝑇 𝑒𝑒−𝐸𝐸𝑘𝑘 𝑇𝑇 𝑑𝑑𝐸𝐸

∫0∞ 𝑒𝑒−


= 𝑘𝑘 𝑇𝑇

Appendix 3


�𝐸𝐸 =∑𝑛𝑛=0∞ 𝐸𝐸𝑛𝑛 𝑝𝑝 𝐸𝐸𝑛𝑛∑𝑛𝑛=0∞ 𝑝𝑝 𝐸𝐸𝑛𝑛

=∑𝑛𝑛=0∞ 𝑎𝑎 ℎ 𝑓𝑓 𝑒𝑒−

𝑛𝑛 ℎ 𝑓𝑓𝑘𝑘 𝑇𝑇

∑𝑛𝑛=0∞ 𝑒𝑒−𝑛𝑛 ℎ 𝑓𝑓𝑘𝑘 𝑇𝑇

= 𝑘𝑘 𝑇𝑇∑𝑛𝑛=0∞ 𝑎𝑎 𝛼𝛼 𝑒𝑒− 𝑛𝑛 𝛼𝛼

∑𝑛𝑛=0∞ 𝑒𝑒− 𝑛𝑛 𝛼𝛼 = 𝑘𝑘 𝑇𝑇 −𝛼𝛼𝑑𝑑 𝑙𝑙𝑎𝑎 ∑𝑛𝑛=0∞ 𝑒𝑒− 𝑛𝑛 𝛼𝛼

𝑑𝑑𝛼𝛼

= 𝑘𝑘 𝑇𝑇 −𝛼𝛼𝑑𝑑 𝑙𝑙𝑎𝑎 1

1 − 𝑒𝑒−𝛼𝛼𝑑𝑑𝛼𝛼

= 𝑘𝑘 𝑇𝑇 𝛼𝛼𝑑𝑑 𝑙𝑙𝑎𝑎 1 − 𝑒𝑒−𝛼𝛼

𝑑𝑑𝛼𝛼

= 𝑘𝑘 𝑇𝑇 𝛼𝛼𝑒𝑒−𝛼𝛼

1 − 𝑒𝑒−𝛼𝛼=

ℎ 𝑓𝑓


𝛼𝛼 ≡ℎ 𝑓𝑓𝑘𝑘𝑇𝑇

Appendix 4


−5

𝜆𝜆6 𝑒𝑒ℎ 𝑐𝑐𝜆𝜆𝑘𝑘𝑇𝑇 − 1

−ℎ 𝑐𝑐𝑘𝑘 𝑇𝑇 − 1

𝜆𝜆2 𝑒𝑒ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇


= 0

−5 +ℎ 𝑐𝑐𝑘𝑘 𝑇𝑇 𝑒𝑒

ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇

𝜆𝜆 𝑒𝑒ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 − 1

= 0

𝑑𝑑𝛪𝛪 𝜆𝜆,𝑇𝑇𝑑𝑑𝜆𝜆 = 0


𝜆𝜆5 𝑒𝑒ℎ 𝑐𝑐𝜆𝜆𝑘𝑘𝑇𝑇 − 1

ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 𝑒𝑒

ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 = 5 𝑒𝑒

ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 − 1

ℎ 𝑐𝑐𝜆𝜆 𝑘𝑘 𝑇𝑇 ≅ 4.966Numerical Solution Technique

quantum mechanics 300 fall... · quantum mechanics fall 2018 prof. sergio b. mendes 1 an essential...

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