PHYS 405 - Fundamentals of Quantum Theory Icsrri.iit.edu/~segre/phys405/16F/lecture_01.pdfC. Segre...

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PHYS 405 - Fundamentals of Quantum Theory I

Term: Fall 2016Meetings: Monday & Wednesday 11:25-12:40Location: 212 Stuart Building

Instructor: Carlo SegreOffice: 106A Life SciencesPhone: 312.567.3498email: segre@iit.edu

Book: Introduction to Quantum Mechanics, 2nd ed.,D. Griffiths (Pearson, 2005)

Web Site: http://phys.iit.edu/∼segre/phys405/16F

C. Segre (IIT) PHYS 405 - Fall 2016 August 22, 2016 1 / 24

Course Objectives

1 Understand the interpretation of the Schrodinger equation and thewave function.

2 Understand the solution of the time-independent Schrodingerequation for one-dimensional potentials.

3 Understand quantum formalism including operators and the Diracnotation.

4 Understand the solution of three-dimensional potentials.

5 Understand how systems of identical particles are solved.

6 Be able to solve quantum mechanics problems in one and threedimensions and with identical particles.

Course Objectives

Course Grading

15% – Homework assignments

Weekly or bi-weeklyDue at beginning of classMay be turned in via Blackboard

50% – Two mid-term exams

35% – Final examination (TBA)

Grading scaleA – 88% to 100%B – 75% to 88%C – 62% to 75%D – 50% to 62%E – 0% to 50%

Course Grading

15% – Homework assignmentsWeekly or bi-weekly

Due at beginning of classMay be turned in via Blackboard

Course Grading

15% – Homework assignmentsWeekly or bi-weeklyDue at beginning of class

May be turned in via Blackboard

Course Grading

15% – Homework assignmentsWeekly or bi-weeklyDue at beginning of classMay be turned in via Blackboard

Course Grading

Topics to be Covered (Chapter titles)

1 The wave function

2 Time-independent Schrodinger equation

3 Quantum formalism

4 Three dimensional quantum mechanics

5 Identical particles

6 Other topics as appropriate

1 The wave function

3 Quantum formalism

1 The wave function

3 Quantum formalism

1 The wave function

3 Quantum formalism

1 The wave function

3 Quantum formalism

1 The wave function

3 Quantum formalism

Tips for success

1 Do the reading assignments before lecture, you willunderstand them better.

2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook.

TAKE NOTES!

3 Ask questions in class, it’s likely that others have thesame ones.

4 Go through the derivations yourself, kill some trees!

5 Do the homework the “right” way, only use the solutionsmanual as a last resort.

Struggling is good and helps youlearn!

6 Come to office hours with questions, I’ll be less lonelyand it will help you too!

Tips for success

2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook.

TAKE NOTES!

Tips for success

2 Attend class or really view the lectures completely, thereare things discussed which are not on the slides or thebook. TAKE NOTES!

Tips for success

5 Do the homework the “right” way, only use the solutionsmanual as a last resort. Struggling is good and helps youlearn!

Tips for success

5 Do the homework the “right” way, only use the solutionsmanual as a last resort. Struggling is good and helps youlearn!

Course Schedule

Focus on “mechanics” but will bring in some originalarticles if there is time available.

Exam #1 – Monday, October 03, 2016Exam #2 – Wednesday, November 09, 2016

Up-to-date schedule athttp://phys.iit.edu/∼segre/phys405/16F/schedule.html

We have 25 class sessions,

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Course Schedule

2 mid-term exams,

250 pages to cover,

and we’re online.

Let’s start!

Today’s Outline - August 22, 2016

• Black-body radiation

• Photoelectric effect

• Compton scattering

• Davisson-Germer experiment

• The 1-D Schrodinger equation

Reading Assignment: Chapter 1.1–1.6

Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 31, 2016

Black Body Radiation

The radiation spectrum of ablack-body depends on thetemperature of the object.

For example, T=5000 K.

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

The radiation spectrum of ablack-body depends on thetemperature of the object.

For example, T=5000 K.

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

The maximum wavelengthλm is seen to scale inverselywith temperature such that

λmT = 2.898× 10−3m · K3

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

λmT = 2.898× 10−3m · K3

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

λmT = 2.898× 10−3m · K3

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

This proves to be a universalcurve.

However, the classical the-oretical model (Rayleigh–Jeans), is unable to describethe low wavelength cutoffobserved.∫ ∞0

u(λ)dλ ∝∫ ∞0

λ−4dλ→∞

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

However, the classical the-oretical model (Rayleigh–Jeans), is unable to describethe low wavelength cutoffobserved.

∫ ∞0

u(λ)dλ ∝∫ ∞0

λ−4dλ→∞

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

However, the classical the-oretical model (Rayleigh–Jeans), is unable to describethe low wavelength cutoffobserved.∫ ∞0

u(λ)dλ ∝∫ ∞0

λ−4dλ→∞

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

Rayleigh-Jeans(5000 K)

Planck’s Solution

By assuming that the modesof oscillation in the black-body cavity were quantized.

The resulting function forthe energy distribution is

which cuts off properly asλ→ 0.

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Planck’s Solution

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Planck’s Solution

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Planck’s Solution

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Planck’s Solution

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Planck’s Solution

Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

Photoelectric Effect

In the photoelectric effect, the emission of electrons depends on the colorof the incident light rather than its intensity.

Einstein (1905) explained this by reasoning that light must be quantizedaccording to its frequency, thereby acting as a particle.

max = hν − φ

Compton Scattering Experiment

In 1923, Arthur Compton mea-sured the scattering of x-raysfrom a carbon foil.

He observed x-rays at lower en-ergies than the incident energyand that the energy dependedon the observation angle.

This could be explained bytreating the x-rays as particlescolliding with the electrons inthe carbon atoms of the foil.

Compton Scattering Phenomenon

A photon-electron collision

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

Treat the electron relativistically and conserve energy and momentum

mc2 +hc

λ′+ γmc2 (energy)

λ′cosφ+ γmv cos θ (x-axis)

λ′sinφ+ γmv sin θ (y-axis)

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λ

p′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

mc2 +hc

Compton Scattering Derivation

squaring the momentumequations

λ− h

λ′cosφ

= γ2m2v2 cos2 θ(− h

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1, expand the squares,and sin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

γ2m2v2(sin2 θ + cos2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

= γ2m2v2 cos2 θ

(− h

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together,

substitute sin2 θ + cos2 θ = 1, expand the squares,and sin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1, expand the squares,

and sin2 φ+ cos2 φ = 1, then rearrange and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

now add them together, substitute sin2 θ + cos2 θ = 1, expand the squares,and sin2 φ+ cos2 φ = 1, then rearrange

and substitute v = βc

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

λ− h

λ′cosφ

λ′sinφ

= γ2m2v2 sin2 θ

λ− h

λ′cosφ

(− h

λ′sinφ

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

λ′2sin2 φ+

λ′2cos2 φ

m2c2β2

1− β2=

λ2− 2h2

λλ′cosφ+

λ′2

Now take the energy equation and square it,

then solve it for β2 which issubstituted into the equation from the momentum.

(mc2 +

λ− hc

= γ2m2c4 =m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ′2− 2h2

λλ′cosφ =

m2c2β2

1− β2

(mc2 +

λ− hc

−m2c2

Now take the energy equation and square it, then solve it for β2

which issubstituted into the equation from the momentum.

(mc2 +

λ− hc

= γ2m2c4 =m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ′2− 2h2

λλ′cosφ =

m2c2β2

1− β2

(mc2 +

λ− hc

−m2c2

Now take the energy equation and square it, then solve it for β2 which issubstituted into the equation from the momentum.(

mc2 +hc

λ− hc

= γ2m2c4 =m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ′2− 2h2

λλ′cosφ =

m2c2β2

1− β2

(mc2 +

λ− hc

−m2c2

Now take the energy equation and square it, then solve it for β2 which issubstituted into the equation from the momentum.(

mc2 +hc

λ− hc

= γ2m2c4 =m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

λ′2− 2h2

λλ′cosφ =

m2c2β2

1− β2

(mc2 +

λ− hc

−m2c2

After expansion, cancellation, and rearrangement, we obtain

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λλ′−��m2c2

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

��λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

(λ′ − λλλ′

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

After expansion,

cancellation, and rearrangement, we obtain

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λλ′−m2c2

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

��λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

After expansion, cancellation,

and rearrangement, we obtain

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

λ′2− 2h2

λλ′

��λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

��h2

λ2+��h2

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

��λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

After expansion, cancellation, and rearrangement, we obtain

��h2

λ2+��h2

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

λλ′(1− cosφ) = 2m

λ− hc

= 2mhc

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

��h2

λ2+��h2

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

λ− hc

)= 2mhc

=2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

��h2

λ2+��h2

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

λ− hc

)= 2mhc

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)

��h2

λ2+��h2

λ′2− 2h2

λλ′cosφ =

λ− h

−m2c2

= ��m2c2 +h2

λ′2− 2mch

λ− 2mch

λ′+

λ− hc

��h2

λ2+��h2

λ′2− 2h2

λλ′

��λλ′(1− cosφ) = 2m

λ− hc

)= 2mhc

2mhc∆λ

��λλ′

∆λ =h

mc(1− cosφ)

Compton Scattering Equation

∆λ =h

mc(1− cosφ)

This explains the change in en-ergy of the broader peak withincreasing angle

It also provides the basis for un-derstanding why the Comptonpeak is so broad . . .

. . . since the electrons are notreally “stationary”, there will bea spread in energy and momen-tum of the outgoing photon.

∆λ =h

mc(1− cosφ)

∆λ =h

mc(1− cosφ)

∆λ =h

mc(1− cosφ)

Davisson-Germer Experiment

In 1928, Davisson & Germershowed that DeBroglie’s hy-pothesis of the wave nature ofparticles was correct.

By measuring the electronsscattered at various energiesfrom a metal foil, the observa-tion of Bragg’s Law for elec-trons was made.

This could only be explained byinterference between electrons.

1D Schrodinger equation

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

∂x2= Kinetic Energy

V Ψ = Potential Energy

where the wave function,Ψ(x , t) is a function of bothtime and space

this equation can be viewed asan expression of conservation ofenergy

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

i~∂Ψ

∂t= − ~2

∂2Ψ

∂x2+ V Ψ

i~∂Ψ

∂t= Total Energy

− ~2

∂2Ψ

Deriving the Schrodinger equation

“When Schrodinger first wrote it down, he gave a kind of derivation basedon some heuristic arguments and some brilliant intuitive guesses. Some ofthe arguments he used were even false, but that does not matter; the onlyimportant thing is that the ultimate equation gives a correct description ofnature.” - Richard Feynman

Inspired by wave optics,Schrodinger started with thewave equation for electromagneticradiation with solution

E (x , t) = E0e i(kx−ωt)

taking the derivatives and substi-tuting results in the dispersion re-lation for photons

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

Inspired by wave optics,Schrodinger started with thewave equation for electromagneticradiation

with solution

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

Inspired by wave optics,Schrodinger started with thewave equation for electromagneticradiation

with solution

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

taking the derivatives and substi-tuting

results in the dispersion re-lation for photons

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

taking the derivatives and substi-tuting

results in the dispersion re-lation for photons

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

−→ E = pc

0 =∂2E

∂x2− 1

(−k2 +

)E0e i(kx−ωt)

c−→ E = pc

Deriving the Schrodinger equation (cont.)

This works for a relativistic massless particle like the photon, but for anon-relativistic particle with mass, we have to start with a differentdispersion relation.

The dispersion relation for a non-relativistic particle must be consis-tent with the classical energy andDeBroglie’s relation

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

Therefore, we need a wave equation which gives this dispersion relationwhen applied to a traveling matter plane wave, Ψ(x , t) = ψ0e i(kx−ωt)

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

when including the possibility of a potential, V (x)

The dispersion relation for a non-relativistic particle must be consis-tent with the classical energy

andDeBroglie’s relation

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

The dispersion relation for a non-relativistic particle must be consis-tent with the classical energy

andDeBroglie’s relation

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

2m= ~ω

p = ~k

−→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t)

+ V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t)

+ V (x)Ψ(x , t)

2m= ~ω

p = ~k −→ ~2k2

2m= ~ω

∂tΨ(x , t) = − ~2

∂x2Ψ(x , t) + V (x)Ψ(x , t)

Meaning of the wave function

The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)

a complex quantity but itsphase is meaningless

spatial integral gives probabilityof the particle being found inthe interval from a to b

Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!

a|Ψ(x , t)|2 dx

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j .

Thetotal number of people, N, is

The probability of an individual chosen at randomfrom the crowd having the age j is

The sum of all the probabilities must be 1

The average value of the age (not the most probable)is given by

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Probability review

N =∞∑j=0

P(j) =N(j)

1 =∞∑j=0

〈j〉 =

∑jN(j)

=∞∑j=0

Expectation values

In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as

and given thename, expectation value

One particular quantity, the variance, de-scribes the “width” of the distributionand is given by

Where σ is called the standard deviationof the distribution

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2

Expectation values

In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2

Expectation values

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2

Expectation values

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

√〈j2〉 − 〈j〉2

Expectation values

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

√〈j2〉 − 〈j〉2

Expectation values

〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2

Computing the variance

σ2 =⟨(∆j)2

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

expanding the square

dividing into threesums

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2

=⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

σ2 =⟨(∆j)2

⟩=∑

(∆j)2 P(j)

(j − 〈j〉)2 P(j)

j2 − 2j 〈j〉+ 〈j〉2)

j2P(j) +∑

2j 〈j〉P(j) +∑〈j〉2 P(j)

=⟨j2⟩− 2 〈j〉〈j〉+ 〈j〉2 =

⟨j2⟩− 〈j〉2

∆j = (j − 〈j〉)

Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2

PHYS 405 - Fundamentals of Quantum Theory Icsrri.iit.edu/~segre/phys405/16F/lecture_01.pdfC. Segre...

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