CHAPTER 5 300 fall... · Fall 2018 Prof. Sergio B. Mendes 1 CHAPTER 5. Topics Fall 2018 Prof. Sergio B. Mendes 2 5.4 Wave Motion in Classical Physics 5.1 X-Ray Light Scattering 5.2

Wave Properties of Matter and Quantum Mechanics I

Fall 2018 Prof. Sergio B. Mendes 1

CHAPTER 5

Topics


5.4 Wave Motion in Classical Physics

5.1 X-Ray Light Scattering

5.2 De Broglie Matter Waves

5.3 Electron Scattering

5.5 Waves or Particles

5.6 Uncertainty Principle

5.7 Probability, Wave Functions, and Copenhagen Interpretation

5.8 Particle in a Box

5.4 Wave Motion in Classical Physics


4Prof. Sergio B. MendesFall 2018

Waves: propagation of energy, not particles

electric and magnetic fieldssurface height

pressure

rope height

𝜓(𝑥, 𝑡)


a pulse

a wave train

a continuous harmonic wave

Different Waveforms:


Two Snapshots of a Wave Pulse

𝑡 = 0

𝜓 𝑥, 𝑡 = 0 = 𝑓 𝑥

𝑡 ≥ 0

𝜓 𝑥, 𝑡 = 𝑓 𝑥 − 𝑣 𝑡

propagating with velocity v along the x-axis

𝑓 𝑥 − 𝑣 𝑡


Propagation towards Positive x-direction

𝜓 𝑥, 𝑡 = 𝑓 𝑥 − 𝑣 𝑡

𝜓 𝑥, 𝑡 = 𝑓 𝑥 + 𝑣 𝑡

Propagation towards Negative x-direction

𝑣 > 0

𝑣 > 0


𝜓(𝑥, 𝑡) = 𝑓 𝑥 − 𝑣 𝑡

Fingerprint of the Wave Phenomena: 𝑥 − 𝑣 𝑡

𝜓 𝑥, 𝑡 = 𝐴 𝑒−

𝑥 − 𝑣 𝑡𝜎2

2

𝜓 𝑥, 𝑡 = 0 = 𝐴 𝑒−𝑥2

𝜎2

𝜓 𝑥, 𝑡 = 0 = 𝐴 𝑒−𝑥𝜎 𝜓 𝑥, 𝑡 = 𝐴 𝑒−

𝑥 − 𝑣 𝑡𝜎

𝜓 𝑥, 𝑡 = 0 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 𝜓 𝑥, 𝑡 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 − 𝑣 𝑡


𝜓(𝑥, 𝑡) = 𝑓 𝑥 − 𝑣 𝑡

𝜕2𝜓

𝜕𝑥2=

1

𝑣2𝜕2𝜓

𝜕𝑡2

Appendix 1


Wave Equation: 𝜕2𝜓

𝜕𝑥2=

1

𝑣2𝜕2𝜓

𝜕𝑡2

• 2nd order partial differential equation

• Linear equation in 𝜓 : 𝜓, 𝜕𝛼𝜓, 𝜕𝛼2𝜓, …

• Homogeneous equation: no term involving independent variables

If 𝜓1 and 𝜓2 are solutions then a linear combination 𝛼 𝜓1 + 𝛽 𝜓2

is also a solution equation.


Harmonic Wave Solution:

𝜓 𝑥, 𝑡 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 − 𝑣 𝑡 + 𝜀

𝜓 𝑥, 𝑡 = 𝐵 𝑠𝑖𝑛 𝑘 𝑥 − 𝑣 𝑡 + 𝜀′

𝜓 𝑥, 𝑡 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 − 𝑣 𝑡 + 𝜀 + 𝐵 𝑠𝑖𝑛 𝑘 𝑥 − 𝑣 𝑡 + 𝜀′

or

or

𝜓 𝑥, 𝑡 = ℛℯ 𝐴 𝑒𝑖 𝑘 𝑥−𝑣 𝑡 + 𝜀

or


Wavelength in a Harmonic Wave:

𝑡 = 𝑓𝑖𝑥𝑒𝑑

𝑘 ≡2 𝜋

𝜆


𝑥


Period in a Harmonic Wave:

𝑥 = 𝑓𝑖𝑥𝑒𝑑

𝜏 ≡𝜆

𝑣


𝜏

𝜏

𝜏

𝑡

𝑘 𝑣 ≡2 𝜋

𝜏or


Wave Speed

𝑣 =𝜆

𝜏𝜏

𝜏

𝜏


A Few Definitions:

𝑘 ≡2 𝜋

𝜆

Wave Number


𝑓 ≡1

𝜏

Angular Frequency

= 2 𝜋 𝑓

𝜓 𝑥, 𝑡 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

𝜔 ≡2 𝜋

𝜏

Frequency

𝑣 =𝜆

𝜏

𝑣 =𝜔

𝑘


𝜓 𝑥, 𝑡 = 𝐴 𝑐𝑜𝑠 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

= 𝐴 𝑐𝑜𝑠 𝜑 𝑥, 𝑡

𝜑 𝑥, 𝑡 ≡ 𝑘 𝑥 − 𝜔 𝑡 + 𝜀Phase:

Amplitude: 𝐴


Constant Phase and Phase Velocity

𝜑 𝑥, 𝑡 ≡ 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

𝑑𝜑 𝑥, 𝑡 = 𝑘 𝑑𝑥 − 𝜔 𝑑𝑡

𝑑𝜑 𝑥, 𝑡 = 0

𝑑𝑥

𝑑𝑡=𝜔

𝑘=𝜆

𝜏

𝑘 𝑑𝑥 − 𝜔 𝑑𝑡 = 0

𝑣𝑝ℎ =


Adding Two Waves with Same Wavelength and Frequency:

(Almost Completely) In Phase:


Constructive Interference

𝜓1 = 𝑠𝑖𝑛 𝜃 𝜓2 = 𝑠𝑖𝑛 𝜃 +𝜋

20


(Almost Completely) Out of Phase:

Destructive Interference

𝜓1 = 𝑠𝑖𝑛 𝜃 𝜓2 = 𝑠𝑖𝑛 𝜃 +𝜋

20+ 𝜋


Interference

22

Wave Propagation Through Slits:

One slit:

Two slits:

𝐼

𝐼 ≡ 𝜓 2

𝜓 = 𝜓1 + 𝜓2

𝜓 =𝜓0 𝑒

𝑖 𝑘 𝑟 − 𝜔 𝑡

𝑟

𝜓1 =𝜓01 𝑒

𝑖 𝑘 𝑟1− 𝜔 𝑡

𝑟1

𝜓2 =𝜓02 𝑒

𝑖 𝑘 𝑟2−𝜔 𝑡

𝑟2


Interference of Waves:

Both slits open

Only one slit open

𝐼

𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 =𝑝𝑜𝑤𝑒𝑟

𝑎𝑟𝑒𝑎= 𝐼 Ԧ𝑟 ≡ 𝜓 Ԧ𝑟, 𝑡 2

24

𝐼 Ԧ𝑟 = 𝜓 Ԧ𝑟, 𝑡 2 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 𝑐𝑜𝑠 𝑘 𝑟1 − 𝑟2 + 𝜀1 − 𝜀2

ൗ−𝑎 2

𝑠

𝑌

𝑅

𝜃

Phase Difference in Young’s Interferometer

ൗ𝑎 2

Λ =𝜆 𝑅

𝑎

Λ

𝜓 Ԧ𝑟, 𝑡 = 𝜓1 Ԧ𝑟, 𝑡 + 𝜓2 Ԧ𝑟, 𝑡

𝑟1 − 𝑟2 ≅ 𝑎 𝑠𝑖𝑛 𝜃 = 𝑎𝑌

𝑅

𝑟1

𝑟2


Diffraction:

𝜓 = 𝜓1 + 𝜓2 +⋯ =

𝑖

𝜓𝑖

Interference:

𝜓 = ඵ

𝑎𝑝𝑒𝑟𝑡𝑢𝑟𝑒

𝜓𝑖 𝑑𝑎𝑖


𝜓 Ԧ𝑟, 𝑡 = ඵ


𝜓0 𝑒𝑖 𝑘 𝑟 − 𝜔 𝑡

𝑟𝑑𝑦 𝑑𝑧

27

Single Slit

𝑠

𝑌

𝑦

𝑅

𝑑

𝜓 𝑌, 𝑡 =𝜓0 𝑒

𝑖 𝑘 𝑅 −𝜔 𝑡

𝑅න

− ൗ𝑑 2

+ ൗ𝑑 2

𝑒− 𝑖𝑘 𝑌𝑅 𝑦𝑑𝑦

𝜃

𝑦

𝑧

𝑑

𝑟 = 𝑠2 + 𝑌 − 𝑦 2 = 𝑅2 + 𝑦2 − 2 𝑦 𝑌 ≅ 𝑅 −𝑦 𝑌

𝑅

𝑟

=𝜓0 𝑒

𝑖 𝑘 𝑅 −𝜔 𝑡

𝑅

𝑠𝑖𝑛𝑘 𝑌 𝑑2 𝑅

𝑘 𝑌 𝑑2 𝑅

28

Single Slit, cont.

𝐼 𝑌 = 𝜓 2 = 𝐼0

𝑠𝑖𝑛𝑘 𝑌 𝑑2 𝑅

𝑘 𝑌 𝑑2 𝑅

2

𝑘 𝑌𝑚 𝑑

2 𝑅= 𝑚 𝜋

𝑚 = ±1, ±2,±3

𝑌𝑚 = 𝑚𝜆 𝑅

𝑑

𝑌𝑚𝑅= 𝑠𝑖𝑛 𝜃𝑚 = 𝑚

𝜆

𝑑

𝑌(𝑚𝑚)

𝑌1

ൗ𝐼 𝑌 𝐼0

zeros at

geometrical shadow

𝑌−1

with

𝑠𝑖𝑛 𝜃1 =𝜆

𝑑


∆ 𝑠𝑖𝑛 𝜃 ≅ ∆𝜃 ≥𝜆

𝑑

After diffraction, the wave spreads over a range of angles:


PhET

https://phet.colorado.edu/en/simulation/wave-interference


𝑎

𝑌

𝑅

𝜃

Λ =𝜆 𝑅

𝑎

Λ

𝜓 Ԧ𝑟, 𝑡 2 ≡ 𝐼 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 𝑐𝑜𝑠 2 𝜋𝑎 𝑌

𝜆 𝑅+ 𝜀1 − 𝜀2

𝜓 Ԧ𝑟, 𝑡 = 𝜓1 Ԧ𝑟, 𝑡 + 𝜓2 Ԧ𝑟, 𝑡

Interference


PhET

https://phet.colorado.edu/en/simulation/wave-interference


Diffraction

𝜓 2 ≡ 𝐼 = 𝐼0

𝑠𝑖𝑛𝜋 𝑌 𝑑𝜆 𝑅

𝜋 𝑌 𝑑𝜆 𝑅

2

𝜓 Ԧ𝑟, 𝑡 = ඵ


𝜓0 𝑒𝑖 𝑘 𝑟 − 𝜔 𝑡

𝑟𝑑𝑦 𝑑𝑧

𝑑

𝑌

𝑅

∆𝜃 ≥𝜆

𝑑


Adding Two Waves of Different Frequencies

(therefore, Different Wavelengths)


Adding Two Waves:

𝑘 ≡1

2𝑘1 + 𝑘2

𝜔 ≡1

2𝜔1 + 𝜔2

𝜀 ≡1

2𝜀1 + 𝜀2

∆𝑘 ≡1

2𝑘1 − 𝑘2

∆𝜔 ≡1

2𝜔1 − 𝜔2

∆𝜀 ≡1

2𝜀1 − 𝜀2

𝜓 𝑥, 𝑡 = 2 𝜓0 𝑐𝑜𝑠 𝑘 𝑥 − 𝜔 𝑡 + 𝜀 𝑐𝑜𝑠 𝑥 ∆𝑘 − 𝑡 ∆𝜔 + ∆𝜀

𝜓0,1 = 𝜓0,2 = 𝜓0

Appendix 2

𝑘 𝑥 − 𝜔 𝑡 + 𝜀 = 𝜑

𝑥 ∆𝑘 − 𝑡 ∆𝜔 + ∆𝜀 = 𝜑𝑚

𝑣𝑝ℎ =𝜔

𝑘

𝑣𝑔𝑟 =𝑑𝜔

𝑑𝑘

phase velocity

group velocity36



Fourier Theory


Harmonic Function

𝜓 𝑡 = 𝐴 𝑒 𝑖 𝑘 𝑧 − 𝜔 𝑡

One harmonic function extends all the way from

−∞ to +∞

Can we use Harmonic Functions to build a Spatially Confined Function ?

∆ 𝑡


𝑓 𝑥 =1

2 𝜋න−∞

∞

𝐹 𝑘𝑥 𝑒− 𝑖 𝑘𝑥 𝑥 𝑑𝑘𝑥

𝑓(𝑡) =1

2 𝜋න−∞

∞

𝐹 𝜔 𝑒− 𝑖 𝜔 𝑡 𝑑𝜔

Fourier Analysis

𝐹 𝑘𝑥 = න−∞

∞

𝑓 𝑥′ 𝑒+ 𝑖 𝑘𝑥 𝑥′𝑑𝑥′

𝐹 𝜔 = න−∞

∞

𝑓 𝑡′ 𝑒+ 𝑖 𝜔 𝑡′𝑑𝑡′

𝑇

2

Example 1: Top Hat Function

𝑓 𝑡

𝑡

−𝑇

2≤ 𝑡 ≤

+𝑇

2when

otherwise

𝑓 𝑡 =

𝐴

0

40

𝜔𝑚 𝑇

2= 𝑚 𝜋

𝑚 = ±1,±2,±3, …

at 𝐹 𝜔 = 0

𝐹 𝜔 = 𝐴 𝑇 𝑠𝑖𝑛𝑐𝜔 𝑇

2

𝜔+1 − 𝜔−1 =4 𝜋

𝑇𝑇 ∆𝜔 ≈ 4 𝜋

𝐹 𝜔 = න−𝑇2

+𝑇2𝐴 𝑒+ 𝑖 𝜔 𝑡

′𝑑𝑡′

41

𝑓(𝑡) =1

2 𝜋න−∞

∞

𝐴 𝑇 𝑠𝑖𝑛𝑐𝜔 𝑇

2𝑒− 𝑖 𝜔 𝑡 𝑑𝜔

𝐹 𝜔 = 𝐴 𝑇 𝑠𝑖𝑛𝑐𝜔 𝑇

2

−𝑇

2≤ 𝑡 ≤

+𝑇

2when

otherwise𝑓 𝑡 =

𝐴

0

42

=1

𝜋න0

∞

𝐴 𝑇 𝑠𝑖𝑛𝑐𝜔 𝑇

2𝑐𝑜𝑠(𝜔 𝑡) 𝑑𝜔

even function


𝑓 𝑡

𝑡

𝐹 𝜔

𝜔

2 ∆𝜔2 ∆𝑡

Truncated Cosine Function:

∆𝑡 ∆𝜔 ≈ 𝜋𝑓 𝑡 =

−𝑇

2≤ 𝑡 ≤

+𝑇

2when

otherwise

𝐴 cos 𝜔𝑜 𝑡

0


Gaussian Function:

∆𝑥 ∆𝑘𝑥 ≈1

2

𝐹 𝑘𝑥

𝑓 𝑥

𝑓 𝑥 = 𝑒− ∆𝑘2 𝑥2 𝑐𝑜𝑠 𝑘𝑜 𝑥

𝑘𝑥


In Summary, Fourier Theory gives us:


∆𝑡 ∆𝜔 ≥1

2

𝑓 𝑡 = 𝜓 𝑧𝑜, 𝑡

𝐹 𝜔

𝜓 𝑧, 𝑡 =1

2 𝜋න−∞

∞

𝐹 𝜔 𝑒 𝑖 𝑘 𝑧 − 𝜔 𝑡 𝑑𝜔

𝑧

A wave of finite time duration can never be monochromatic !!

𝜓 𝑧, 𝑡 = 𝐴 𝑒 𝑖 𝑘 𝑧 − 𝜔 𝑡


∆𝑥 ∆𝑘𝑥 ≥1

2

𝜓 𝑥, 𝑧, 𝑡 =1

2 𝜋න−∞

∞

𝐹 𝑘𝑥 𝑒𝑖 𝑘𝑥 𝑥 + 𝑘𝑧 𝑧 − 𝜔 𝑡 𝑑𝑘𝑥

𝑘𝑥2 + 𝑘𝑧

2 = 𝑘2 =2 𝜋

𝜆

2

𝑧

𝑥

Lateral confinement of a wave leads to divergence !!

𝑘𝑥

𝑘𝑧

𝑘𝑥 =2 𝜋

𝜆𝑠𝑖𝑛 𝜃

𝜃

𝒌 =2 𝜋

𝜆ො𝒆𝑧

𝒌

𝜓 𝑧, 𝑡 = 𝐴 𝑒 𝑖 𝑘 𝑧 − 𝜔 𝑡

5.1 X-Ray Scattering


• Max von Laue suggested that if x rays were a form of electromagnetic radiation, interference effects should be observed.

• Crystals act as three-dimensional gratings, scattering the waves and producing observable interference effects.

X-Ray Scattering and Bragg’s Law


• William Lawrence Bragg interpreted the x-ray scattering as the re-emission of the incident x-ray beam from a unique set of planes of atoms within the crystal.

Bragg’s Law


• There are two conditions for constructive interference of the scattered x-rays by a crystal:

1) The angle of incidence must equal the angle of reflection of the outgoing wave.

2) The difference in path lengths must be an integral number of wavelengths:

𝑛 𝜆 = 2 𝑑 𝑠𝑖𝑛 𝜃 𝑛 = 1, 2, 3, …


𝑛 𝜆 = 2 𝑑 𝑠𝑖𝑛 𝜃

DNA


Rosalind Franklin produced the X-ray diffraction images of the DNA molecule that helped Watson and Crick unravel the DNA structure.

X-Ray Diffraction Nowadays


5.2 de Broglie Waves, 1920


For Electromagnetic Waves:


𝐸 = 𝑝 𝑐 = ℎ 𝑓

𝐸 = ℎ 𝑓

𝐸 = 𝑝 𝑐

𝑝 =ℎ 𝑓

𝑐=

ℎ

𝜆

• According to Special Relativity, and m = 0:

• According to Planck, Einstein, Compton, etc:

𝑝 =ℎ

𝜆


𝜆 =ℎ

𝑝

• Maybe matter also behaves like waves.

• With the wavelength given by:

de Broglie Hypothesis:

𝑝 =1

𝑐𝐸𝑡𝑜𝑡𝑎𝑙

2 − 𝑚 𝑐2 2

≅ 2𝑚 𝐾 𝐾 ≪ 𝑚 𝑐2when

Example 5.2


𝑚 = 57 𝑔

𝑣 = 25 𝑚/𝑠

𝜆 =ℎ

𝑝

=6.63 × 10−34 𝐽 𝑠

57 × 10−3 𝑘𝑔 25 𝑚/𝑠

= 4.7 × 10−34 𝑚

𝐾 = 50 𝑒𝑉

(56 mph) 𝜆 =ℎ

𝑝≅

ℎ

2 𝑚 𝐾

Tennis ball Electron

= 0.17 𝑛𝑚

=ℎ 𝑐

2 𝑚 𝑐2 𝐾

=1240 𝑒𝑉 𝑛𝑚

2 × 0.511 × 106 𝑒𝑉 50 𝑒𝑉


𝜆 =ℎ

𝑝

de Broglie Wavelength

𝑝 =1

𝑐𝐸𝑡𝑜𝑡𝑎𝑙

2 − 𝑚 𝑐2 2

≅ 2𝑚 𝐾 𝐾 ≪ 𝑚 𝑐2when

=𝐸𝑡𝑜𝑡𝑎𝑙𝑐

=ℎ 𝑓

𝑐𝑚 = 0when (e.g., photons)

(e.g., non-relativistic particles)

5.3 Electron Scattering, 1925


Clinton J. Davisson (1881–1958) isshown here in 1928 (right) lookingat the electronic diffraction tubeheld by Lester H. Germer (1896–1971). Davisson received hisundergraduate degree at theUniversity of Chicago and hisdoctorate at Princeton. Theyperformed their work at BellTelephone Laboratory located inNew York City. Davisson receivedthe Nobel Prize in Physics in 1937.

Electron Scattering


Transmission Electron Microscope diffraction photographs


120-keV electrons scattered on the

quasicrystal Al80Mn20.

Electron diffraction pattern on beryllium.

The dots in (a) indicate that the sample was a crystal, whereas the rings in (b) indicate that a randomly

oriented sample (or powder) was used.


Scanning Electron Microscope

Combining Bohr’s Quantization of the Angular Momentum 𝐿 = 𝑛 ℏ

& de Broglie Matter Waves 𝑝 =ℎ

𝜆


𝐿 = 𝑛 ℏ𝑝 𝑟 =ℎ

𝜆𝑟 =

2 𝜋 𝑟 = 𝑛 𝜆


5.5 Wave or Particle ?


Interference with Wave-Particle:


Young’s Interference with the Photon Wave-Particle:

𝜓 2

𝜓 = 𝜓1 + 𝜓2


Demonstration of electron interference using two slits similar in concept to Young’s double-slit experiment for light.

The result by Claus Jönsson(1961) clearly shows that electrons exhibit wave behavior.

Young’s Interference with the Electron Wave-Particle:


Experimental Facts:

wave-particle

• The interference pattern only appears when both slits are opened.

• The wave-particle (photon, electron, …) cannot be divided.

• One wave-particle passes through only one slit.

• An experiment to locate in which slit the wave-particle passed through destroys the interference pattern.


Copenhagen Interpretation


There is a wave-function 𝜓 associated with each possible path allowed for

the wave-particle.

The 𝜓 2 determines the probability of finding a particle at a given place and time.

The interference pattern comes from the interference of the individual wave-functions

associated with each path: 𝜓1 + 𝜓22.

The average number of wave-particles 𝑁 is proportional to 𝜓 2. It’s a statistical value,

therefore, non-deterministic.


Diffraction with Wave-Particles:

The wavelength 𝜆 of the wave-function 𝜓 is

determined by the de Broglie relation: 𝜆 =ℎ

𝑝.


∆ 𝑠𝑖𝑛 𝜃 ≅ ∆𝜃 ≥𝜆

𝐿

After diffraction, the wave-function spreads over a range of angles:

𝐿

𝜆∆𝜃 ≥ 1

𝐿ℎ

𝜆∆𝜃 ≥ ℎ𝐿 ∆𝑝 =

∆𝑥 ∆𝑝𝑥 ≥ ℎ

𝐿 𝑝 ∆𝜃 =

𝐿

𝜓, 𝜆

∆𝜃

𝑝 =ℎ

𝜆


There is a fundamental limitation in the simultaneous knowledge of

position and linear momentum

∆𝑥 ∆𝑝𝑥 ≥ ℎ

𝐿 = ∆𝑥

∆𝑝𝑥

𝑝


∆𝑝𝑥 = 2 𝑝 𝑠𝑖𝑛 𝜃

𝑝 =ℎ

𝜆

When we attempt to find the position of an electron:

∆𝑥 =𝜆

𝑠𝑖𝑛 𝜃′

∆𝑥 ∆𝑝𝑥 ≈ 2 ℎ

= 2ℎ

𝜆𝑠𝑖𝑛 𝜃


5.6 Heisenberg Uncertainty Principle

∆𝑡 ∆𝜔 ≥1

2

∆𝑥 ∆𝑘𝑥 ≥1

2∆𝑥 ∆𝑝𝑥 ≥

ℏ

2

∆𝑡 ∆𝐸 ≥ℏ

2


In Classical Physics:

• The position and linear momentum of a particle can be known simultaneously with absolute precision.

• The Classical Laws (e.g., Newton’s Law) unambiguously describe (deterministically) the evolution of a particular system.

• The combination of deterministic laws with the absolute knowledge of the initial conditions (e.g., position and linear momentum) allows one to completely and precisely predict the future.


In Quantum Physics:

• There is a fundamental limitation on the simultaneous knowledge (through measurement) of a particular experimental variable and its associated (conjugated) variable:

∆𝑥 ∆𝑝𝑥 ≥ℏ

2∆𝑡 ∆𝐸 ≥

ℏ

2


Consequences:

(Non-Relativistic) Minimum Kinetic Energy

∆𝑝𝑥 ≥ℏ

2 ∆𝑥

𝐾 =𝑝𝑥

2

2 𝑚≅

∆𝑝𝑥2

2 𝑚≥

ℏ2

8 𝑚 ∆𝑥2


5.7 Wave Functions and Probability


𝜖𝑜 𝑐 𝐸2 = 𝐼 = 𝑁 ℎ 𝑓

𝑁 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠

𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 × 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒

𝐸 2 ∝ 𝑁 ∝ 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦

𝑒𝑛𝑒𝑟𝑔𝑦

𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 × 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒= 𝐼 =

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠

𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 × 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒× ℎ 𝑓

For the Photon Wave-Particle:


𝜓 2 ∝ 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦

For Any Wave-Particle:

• The probability describes the likelihood of occurrence of an event, but it does not guarantee its occurrence.

• By using the Wave-Function, 𝜓(𝑥, 𝑡), we are adopting a non-deterministic description of nature.


5.8 Wave-Particle in a Box

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛(𝑘 𝑥)

𝑘 𝑙 = 𝑛 𝜋 𝑘 = 𝑛𝜋

𝑙

𝑝𝑛 =ℎ

𝜆

= 𝑛2ℎ2

8 𝑚 𝑙2

𝜓 𝑥 2 = 𝐴 2𝑠𝑖𝑛2(𝑘 𝑥)

= 𝑛ℏ 𝜋

𝑙

𝐾𝑛 =𝑝𝑛

2

2 𝑚

Linear Momentum

Kinetic Energy

= ℏ 𝑘


Appendix 1


𝜓(𝑥, 𝑡) = 𝑓 𝑥 − 𝑣 𝑡 ≡ 𝑓 𝛼

𝜕2𝜓

𝜕𝑥2=

1

𝑣2𝜕2𝜓

𝜕𝑡2

𝛼 ≡ 𝑥 − 𝑣 𝑡

𝜕𝜓

𝜕𝑥=𝜕𝛼

𝜕𝑥

𝜕𝜓

𝜕𝛼= 1

𝜕𝜓

𝜕𝛼=𝑑𝑓

𝑑𝛼

𝜕𝜓

𝜕𝑡=𝜕𝛼

𝜕𝑡

𝜕𝜓

𝜕𝛼= − 𝑣

𝜕𝜓

𝜕𝛼= − 𝑣

𝑑𝑓

𝑑𝛼

𝜕2𝜓

𝜕𝑥2=

𝜕

𝜕𝑥

𝜕𝜓

𝜕𝑥=𝜕𝛼

𝜕𝑥

𝜕

𝜕𝛼

𝑑𝑓

𝑑𝛼= 1

𝑑

𝑑𝛼

𝑑𝑓

𝑑𝛼=𝑑2𝑓

𝑑𝑥2

𝜕2𝜓

𝜕𝑡2=

𝜕

𝜕𝑡

𝜕𝜓

𝜕𝑡=𝜕𝛼

𝜕𝑡

𝜕

𝜕𝛼− 𝑣

𝑑𝑓

𝑑𝛼= − 𝑣

𝑑

𝑑𝛼− 𝑣

𝑑𝑓

𝑑𝛼= 𝑣2

𝑑2𝑓

𝑑𝑥2= 𝑣2

𝜕2𝜓

𝜕𝑥2


Appendix 2

𝜓1 𝑥, 𝑡 = 𝜓0,1 𝑐𝑜𝑠 𝑘1 𝑥 − 𝜔1 𝑡 + 𝜀1

𝜓2 𝑥, 𝑡 = 𝜓0,2 𝑐𝑜𝑠 𝑘2 𝑥 − 𝜔2 𝑡 + 𝜀2

𝜓 𝑥, 𝑡 = 𝜓1 𝑥, 𝑡 + 𝜓2 𝑥, 𝑡

88

Two waves with different frequencies (and wavelengths):

𝜓1 𝑥, 𝑡 = 𝑎 + 𝑏 𝑐𝑜𝑠 𝐴 + 𝐵

𝐴 ≡1

2𝑘1 + 𝑘2 𝑥 −

1

2𝜔1 + 𝜔2 𝑡 +

1

2𝜀1 + 𝜀2

𝐵 ≡1

2𝑘1 − 𝑘2 𝑥 −

1

2𝜔1 −𝜔2 𝑡 +

1

2𝜀1 − 𝜀2

𝜓2 𝑥, 𝑡 = 𝑎 − 𝑏 𝑐𝑜𝑠 𝐴 − 𝐵

𝑎 ≡1

2𝜓0,1 + 𝜓0,2

𝑏 ≡1

2𝜓0,1 − 𝜓0,2

89

𝜓 𝑥, 𝑡 = 𝜓1 𝑥, 𝑡 + 𝜓2 𝑥, 𝑡

= 𝑎 + 𝑏 𝑐𝑜𝑠 𝐴 + 𝐵 + 𝑎 − 𝑏 𝑐𝑜𝑠 𝐴 − 𝐵

= 𝑎 𝑐𝑜𝑠 𝐴 + 𝐵 + 𝑐𝑜𝑠 𝐴 − 𝐵 + 𝑏 𝑐𝑜𝑠 𝐴 + 𝐵 − 𝑐𝑜𝑠 𝐴 − 𝐵

= 2 𝑎 𝑐𝑜𝑠 𝐴 𝑐𝑜𝑠 𝐵 − 2 𝑏 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 𝐵

90

Same Amplitude

𝑎 ≡1

2𝜓0,1 + 𝜓0,2 = 𝜓0

𝑏 ≡1

2𝜓0,1 − 𝜓0,2 = 0

𝜓0,1 = 𝜓0,2 = 𝜓0

𝜓 𝑥, 𝑡 = 2 𝑎 𝑐𝑜𝑠 𝐴 𝑐𝑜𝑠 𝐵

= 2 𝜓0 𝑐𝑜𝑠1

2𝑘1 + 𝑘2 𝑥 −

1

2𝜔1 + 𝜔2 𝑡 +

1

2𝜀1 + 𝜀2

𝑐𝑜𝑠1

2𝑘1 − 𝑘2 𝑥 −

1

2𝜔1 − 𝜔2 𝑡 +

1

2𝜀1 − 𝜀2

91

𝑘 ≡1

2𝑘1 + 𝑘2 𝜔 ≡

1

2𝜔1 + 𝜔2 𝜀 ≡

1

2𝜀1 + 𝜀2

∆𝑘 ≡1

2𝑘1 − 𝑘2 ∆𝜔 ≡

1

2𝜔1 − 𝜔2 ∆𝜀 ≡

1

2𝜀1 − 𝜀2


92

𝜓0,1 = 𝜓0,2 = 𝜓0


Appendix 3


Group Velocity of Wave-Particles


𝑣𝑔 =𝑑𝜔

𝑑𝑘

𝐸 = 𝐾 +𝑚 𝑐2

𝑝 =ℎ

𝜆= ℏ 𝑘

𝐸 = ℎ 𝑓 = ℏ 𝜔

=𝑑 ℏ 𝜔

𝑑 ℏ 𝑘=𝑑𝐸

𝑑𝑝

𝐸𝑛𝑟 ≅𝑝2

2 𝑚+𝑚 𝑐2

=𝑝

𝑚

Group Velocity: non-relativistic case

≅𝑑𝐸𝑛𝑟𝑑𝑝


𝑣𝑔 =𝑑𝜔

𝑑𝑘=𝑑 ℏ 𝜔

𝑑 ℏ 𝑘=𝑑𝐸

𝑑𝑝

𝐸2 = 𝑝2 𝑐2 +𝑚2 𝑐4

= 𝛽 𝑐=𝑝 𝑐2

𝐸

2 𝐸 𝑑𝐸 = 2 𝑝 𝑐2 𝑑𝑝

Group Velocity: general case

= 𝛾2 𝑚2 𝑐4

1 =𝑝2 𝑐2

𝐸2+

1

𝛾21 −

1

𝛾2=𝑝2 𝑐2

𝐸2𝛽2 =

𝑝2 𝑐2

𝐸2

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CHAPTER 5 300 fall... · Fall 2018 Prof. Sergio B. Mendes 1 CHAPTER 5. Topics Fall 2018 Prof. Sergio B. Mendes 2 5.4 Wave Motion in Classical Physics 5.1 X-Ray Light Scattering 5.2