PHYS 405 - Fundamentals of Quantum Theory Icsrri.iit.edu/~segre/phys405/13F/lecture_01.pdf6 Be able to solve quantum mechanics problems in one and three dimensions and with identical

PHYS 405 - Fundamentals of Quantum Theory I

Term: Fall 2013Meetings: Monday & Wednesday 11:25-12:40Location: 220 Stuart Building

Instructor: Carlo SegreOffice: 166A Life SciencesPhone: 312.567.3498email: [email protected]

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

Web Site: http://phys.iit.edu/∼segre/phys405/13F

C. Segre (IIT) PHYS 405 - Fall 2013 August 19, 2013 1 / 23

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 Objectives








Course Objectives








Course Objectives








Course Objectives








Course Grading

10% – Homework assignments

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

5% – Class participation (work in progress)

50% – Two mid-term exams

35% – Final examination (08:00 Dec. 03)

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


Course Grading

10% – Homework assignmentsWeekly or bi-weekly

Due at beginning of classMay be turned in via Blackboard






Course Grading

10% – Homework assignmentsWeekly or bi-weeklyDue at beginning of class

May be turned in via Blackboard






Course Grading

10% – Homework assignmentsWeekly or bi-weeklyDue at beginning of classMay be turned in via Blackboard






Course Grading







Course Grading







Course Grading







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





Course Schedule

Up-to-date schedule athttp://phys.iit.edu/∼segre/phys405/13F/schedule.html

27 class sessions

2 mid-term exams

∼250 pages to cover

∼20 pages/week

Focus on “mechanics” but will bring in some philosophy as well.


Course Schedule


27 class sessions

2 mid-term exams


∼20 pages/week



Course Schedule


27 class sessions

2 mid-term exams


∼20 pages/week



Course Schedule


27 class sessions

2 mid-term exams


∼20 pages/week



Course Schedule


27 class sessions

2 mid-term exams


∼20 pages/week



Course Schedule


27 class sessions

2 mid-term exams


∼20 pages/week



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.

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.

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


Tips for success








Tips for success








Tips for success








Tips for success








Tips for success








Today’s Outline - August 19, 2013

• Black-body radiation

• Photoelectric effect

• Compton scattering

• Davisson-Germer experiment

• The 1-D Schrodinger equation

Reading Assignment: Chapter 1.1–1.6


















































Black Body Radiation

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

For example, T=5000 K.

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

Wavelength (µm)

5000 K



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

For example, T=5000 K.

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

Wavelength (µm)

5000 K



The maximum wavelengthλm is seen to scale inverselywith temperature such that

λmT = 2.898× 10−3m · K3

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

Wavelength (µm)

5000 K




λmT = 2.898× 10−3m · K3

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

Wavelength (µm)

5000 K

4000 K

3000 K




λmT = 2.898× 10−3m · K3

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

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

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

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

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

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

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Inte

nsity (

arb

.)

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

λ5


Planck’s Solution




Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

λ5


Planck’s Solution




Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

λ5


Planck’s Solution




Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

λ5


Planck’s Solution




Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

λ5


Planck’s Solution




Em = mhν, m = 0, 1, 2, 3, · · ·

u(λ) ∝ λ−5

ehc/λkT − 1

limλ→0

u(λ) =e−hc/λkT

λ5


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.

1

2mv2max = hν − φ





1






1



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

ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣

Treat the electron relativistically and conserve energy and momentum

mc2 +hc

λ=

hc

λ′+ γmc2 (energy)

h

λ=

h

λ′cosφ+ γmv cos θ (x-axis)

0 =h

λ′sinφ+ γmv sin θ (y-axis)




ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λ

p′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h





ϕ

θ

λ

v

λ’

p = ~k = 2π~/λp′ = ~k′ = 2π~/λ′

|k| 6=∣∣k′∣∣


mc2 +hc

λ=

hc


h

λ=

h


0 =h



Compton Scattering Derivation

squaring the momentumequations

(h

λ− h

λ′cosφ

)2

= γ2m2v2 cos2 θ(− h

λ′sinφ

)2

= 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

)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2

= γ2m2v2 cos2 θ

(− h

λ′sinφ

)2

= γ2m2v2 sin2 θ



)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2


λ′sinφ

)2

= γ2m2v2 sin2 θ



)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2


λ′sinφ

)2

= 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


)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2


λ′sinφ

)2

= 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


)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2


λ′sinφ

)2

= 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


)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2




(h

λ− h

λ′cosφ

)2


λ′sinφ

)2

= γ2m2v2 sin2 θ



)=

(h

λ− h

λ′cosφ

)2

+

(− h

λ′sinφ

)2

γ2m2v2 =h2

λ2− 2h2

λλ′cosφ+

h2

λ′2sin2 φ+

h2

λ′2cos2 φ

m2c2β2

1− β2=

m2v2

1− β2=

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2



Now take the energy equation and square it,

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

(mc2 +

hc

λ− hc

λ′

)2

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

)2h2

λ2− 2h2

λλ′cosφ+

h2

λ′2=

m2c2β2

1− β2



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

which issubstituted into the equation from the momenta.

(mc2 +

hc

λ− hc

λ′

)2

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

)2

h2

λ2− 2h2

λλ′cosφ+

h2

λ′2=

m2c2β2

1− β2



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

mc2 +hc

λ− hc

λ′

)2

= γ2m2c4=m2c4

1− β2

β2 = 1− m2c4(mc2 + hc

λ −hcλ′

)2h2

λ2− 2h2

λλ′cosφ+

h2

λ′2=

m2c2β2

1− β2



After substitution, expansion and cancellation, we obtain

h2

λ2+

h2

λ′2− 2h2

λλ′cosφ = 2m

(hc

λ− hc

λ′

)+

h2

λ2+

h2

λ′2− 2h2

λλ′

2h2

λλ′(1− cosφ) = 2m

(hc

λ− hc

λ′

)= 2mhc

(λ′ − λλλ′

)=

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)




h2

λ2+

h2

λ′2− 2h2

λλ′cosφ = 2m

(hc

λ− hc

λ′

)+

h2

λ2+

h2

λ′2− 2h2

λλ′

2h2


(hc

λ− hc

λ′

)= 2mhc


)=

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)




h2

λ2+

h2

λ′2− 2h2

λλ′cosφ = 2m

(hc

λ− hc

λ′

)+

h2

λ2+

h2

λ′2− 2h2

λλ′

2h2


(hc

λ− hc

λ′

)= 2mhc


)=

2mhc∆λ

λλ′

∆λ =h

mc(1− cosφ)


Compton Scattering Equation

∆λ =h

mc(1− cosφ)


Davisson-Germer Experiment

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

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

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ

∂x2= Kinetic Energy

VΨ = Potential Energy

where the wave function,Ψ(x , t) is a function of bothtime and spacethis equation can be viewed asan expression of conservationof energy



i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ



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

this equation can be viewed asan expression of conservationof energy



i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ






i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ






i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ






i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ





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!

∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b


Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)

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

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)

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

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






Probability review

N =∞∑j=0

N(j)

P(j) =N(j)

N

1 =∞∑j=0

P(j)

〈j〉 =

∑jN(j)

N

=∞∑j=0

jP(j)






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

⟩

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

⟩σ =

√〈j2〉 − 〈j〉2


Computing the variance

σ2 =⟨(∆j)2

⟩

=∑

(∆j)2 P(j)

=∑

(j − 〈j〉)2 P(j)

=∑(

j2 − 2j 〈j〉+ 〈j〉2)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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)P(j)

=∑

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


Continuous variables

We can extend all of these quantities to a system of continuous variables

with the introduction of the probability density, ρ(x) = |Ψ|2

P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2



We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2

P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

N

ρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j)

1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j)

〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2

σ2 ≡⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2




P(j) =N(j)

Nρ(x)

1 =∞∑j=0

P(j) 1 =

∫ +∞

−∞ρ(x)dx

〈f (j)〉 =∞∑j=0

f (j)P(j) 〈f (x)〉 =

∫ +∞

−∞f (x)ρ(x)dx

σ2 ≡⟨(∆j)2

⟩=⟨j2⟩− 〈j〉2 σ2 ≡

⟨(∆x)2

⟩=⟨x2⟩− 〈x〉2


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PHYS 405 - Fundamentals of Quantum Theory Icsrri.iit.edu/~segre/phys405/13F/lecture_01.pdf6 Be able to solve quantum mechanics problems in one and three dimensions and with identical