phys.iit.eduphys.iit.edu/~segre/phys406/19s/lecture_01.pdf · phys 406 - fundamentals of quantum...

PHYS 406 - Fundamentals of Quantum Theory II

Term: Spring 2019Meetings: Tuesday & Thursday 10:00-11:15Location: 212 Stuart Building

Instructor: Carlo SegreOffice: 136A Pritzker SciencePhone: 312.567.3498email: [email protected]

Book: Introduction to Quantum Mechanics, 3rd ed.,D. Griffiths & D. Schroeter (Cambridge Univ Press, 2018)

Web Site: http://phys.iit.edu/∼segre/phys406/19S

C. Segre (IIT) PHYS 406 - Spring 2019 January 15, 2019 1 / 23

Course Objectives

1. Understand the connection between symmetry andconservation laws.

2. Understand time-independent perturbation theory.

3. Understand the variational method.

4. Understand the WKB approximation and scatteringtheory.

5. Understand dynamical effects in quantum mechanics.

6. Be able to solve quantum mechanics problems using theapproximation method appropriate to the situation.

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

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. Symmetry & conservation laws

2. Time-independent perturbation theory

3. Variational method

4. WKB approximation

5. Scattering theory

6. Quantum dynamics

7. Quantum paradoxes

8. Other topics as appropriate

6. Quantum dynamics

Course Schedule

Up-to-date schedule athttp://phys.iit.edu/∼segre/phys406/19S/schedule.html

28 class sessions

2 mid-term exams

∼190 pages to cover

∼15 pages/week

Focus on approximate methods for solving real problemsin quantum mechanics and actual quantum mechanicsresearch.

Course Schedule

28 class sessions

2 mid-term exams

∼15 pages/week

Course Schedule

28 class sessions

2 mid-term exams

∼15 pages/week

Course Schedule

28 class sessions

2 mid-term exams

∼15 pages/week

Course Schedule

28 class sessions

2 mid-term exams

∼15 pages/week

Course Schedule

28 class sessions

2 mid-term exams

∼15 pages/week

Today’s Outline - January 15, 2019

• Tips for success

• The big picture

• Transformations

• Translation operator

• Parity operator

Reading Assignment: Chapter 6.1-6.5

Homework Assignment #01:Chapter 6: 1,3,4,8,9,10due Tuesday, January 22, 2019

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

• The big picture

• Transformations

• Parity operator

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!

Why approximate methods?

In the first semester of this course, we learned the “me-chanics” of quantum physics

The problems we could solve, however, were very limited

Approximate methods permit us to approach a widerrange of phenomena

This semester we will use our toolbox to understand“real” quantum physics phenomena and connect themto experiment

Quantum physics is the foundation of the discipline andis part of the day-to-day work of a professional physicist

A bit more about me. . .

1976 – B.S. in Physics & Chemistry, University of Illinois atUrbana-Champaign

1981 – Ph.D. in Physics, University of California, San Diego

1983 – joined Illinois Tech Faculty

2006 – elected Fellow International Center for DiffractionData

2011 – appointed Duchossois Professor of Physics

2013 – elected Fellow American Association for theAdvancement of Science

2014 – Co-founder and current CTO of Influit Energy startup

The company . . . Influit Energy, LLC

Translation operator

The translation operator, T (a), can be expressed in terms of themomentum operator by starting with the Taylor series expansion forψ(x − a) about x

T (a)ψ(x) = ψ′(x) = ψ(x − a)

=∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

momentum is thus the generator of translations and the translationoperator is clearly unitary

T (a)−1 = T (−a) = T†(a) e+iap/~ = e+iap/~ = e+iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x)

=∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x)

=∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x)

−→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)−1

= T (−a) = T†(a)

e+iap/~

= e+iap/~ = e+iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

T (a)−1 = T (−a)

= T†(a)

e+iap/~ = e+iap/~

= e+iap/~

T (a)ψ(x) = ψ′(x) = ψ(x − a) =∞∑n=0

n![(x − a)− x ]n

dnψ(x − a)

d(x − a)n

∣∣∣∣x−a=x

=∞∑n=0

n!(−a)n

dnψ(x)

∣∣∣∣x

=∞∑n=0

n!(−a)n

dxnψ(x)

=∞∑n=0

(−ia~

)nψ(x) =

∞∑n=0

(−ia~

)nψ(x)

= e−iap/~ψ(x) −→ T (a) = e−iap/~

Operator transformations

Clearly we can translate functions, but the concept of translation and theT operator are much more general.

It is also possible to translate an operator

the translated operator Q′

is de-fined in terms of the translatedwave function

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

The translated operator Q′

gives the same expectation value in theuntranslated state as the untranslated operator Q gives in the translatedstate

This effectively equates the shifting of the wavefuction, an activetransformation, with the shifting of the coordinate system, a passivetransformation⟨

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

This effectively equates the shifting of the wavefuction, an activetransformation, with the shifting of the coordinate system, a passivetransformation

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉

= 〈ψ| Q ′ |ψ〉 → Q′ ≡ T

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉

→ Q′ ≡ T

⟨ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| Q ′ |ψ〉

ψ′∣∣ Q ∣∣ψ′⟩ = 〈ψ| T †QT |ψ〉 = 〈ψ| Q ′ |ψ〉 → Q

′ ≡ T†QT

Example 6.1

Find the operator x ′ obtained by applying a translation through a distancea to the operator x . That is, what is the action of x ′ on an arbitraryfunction f (x)?

Begin with the definition of the trans-lated operator applied to a test func-

tion, but T†(a) = T (−a) so

now apply T (−a) to x and f (x − a)according to its definition

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

The transformed operator corresponds to shifting the coordinate system by−a so positions in this transformed coordinate system are greater by a,just as occurs when directly translating the function.

in the same way, we find that p′ = p and once we know how x and ptranslate, we can translate any other operator

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

but T†(a) = T (−a) so

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p)

= T†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T

= Q(x ′, p′) = Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′)

= Q(x + a, p)

Example 6.1

x ′f (x) = T†(a)x T (a)f (x)

= T (−a)x T (a)f (x)

= T (−a)xf (x − a)

x ′f (x) = (x + a)f (x)

Q′(x , p) = T

†Q(x , p)T = Q(x ′, p′) = Q(x + a, p)

Continuous translational symmetry

We discussed the Bloch theorem previously in which there is a discretetranslational symmetry. In a system with continuous translationalsymmetry, any choice of a is possible

for an infintesmal translation δ

for continuous translational sym-metry, the Hamiltonian must com-mute with this operator

Thus the Hamiltonian commuteswith the momentum operator

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

and according to Ehrenfest’s Theorem, this leads to conservation ofmomentum

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

thus, symmetries imply conservation laws

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~

≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

[H, T (δ)

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

[H, T (δ)

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

T (δ) = e−iδp/~ ≈ 1− iδ

H, T (δ)]

[H, 1− i

dt〈p〉 =

⟨[H, p

⟨∂p

⟩= 0

thus, symmetries imply conservation lawsC. Segre (IIT) PHYS 406 - Spring 2019 January 15, 2019 14 / 23

Conservation laws

If an operator Q commutes with the Hamiltonian, then by the Ehrenfestrelation its expectation value 〈Q〉 is independent of time if ∂Q/∂t = 0.

Let’s see where this definition leads us.

The probability of getting qn whenmeasuring Q in state |Ψ(t)〉 at atime t is where Q|fn〉 = qn|fn〉

given the time dependence of thewave function where |ψm〉 are theeigenfunctions of the Hamiltonian

since [H,Q] ≡ 0 we can find simul-taneous eigenfunctions of the two:so we choose |ψn〉 = |fn〉

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

e−iEmt/~cm〈fn|ψm〉∣∣∣2

=∣∣∣∑

e−iEmt/~cm〈ψn|ψm〉∣∣∣2

= |cn|2

the probability of obtaining a particular value qn is independent of time

the two definitions of conservation of Q are equivalent

Conservation laws

If an operator Q commutes with the Hamiltonian, then by the Ehrenfestrelation its expectation value 〈Q〉 is independent of time if ∂Q/∂t = 0.Let’s see where this definition leads us.

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

The probability of getting qn whenmeasuring Q in state |Ψ(t)〉 at atime t is

where Q|fn〉 = qn|fn〉

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

The probability of getting qn whenmeasuring Q in state |Ψ(t)〉 at atime t is

where Q|fn〉 = qn|fn〉

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

given the time dependence of thewave function

where |ψm〉 are theeigenfunctions of the Hamiltonian

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

given the time dependence of thewave function

where |ψm〉 are theeigenfunctions of the Hamiltonian

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Conservation laws

P(qn) = |〈fn|Ψ(t)〉|2

|Ψ(t)〉 =∑m

e−iEmt/~cm|ψm〉

P(qn) =∣∣∣∑

=∣∣∣∑

= |cn|2

Parity in 1D

In one dimension, the parity opera-tor, Π inverts space

this operator is its own inverse, isHermitian, and thus unitary

operators transform under spatialinversion as

specifically the position and mo-mentum operators are odd underparity thus any operator Q musttransform under parity as

a system has inversion symmetry ifthe Hamiltonian is unchanged by aparity transformation

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

specifically the position and mo-mentum operators are odd underparity

thus any operator Q musttransform under parity as

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −x

p′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

Π†HΠ

→ ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

ΠH = ΠΠ†HΠ

→ ΠH = HΠ → [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

ΠH = ΠΠ†HΠ → ΠH = HΠ

→ [H, Π] = 0

Parity in 1D

Πψ(x) = ψ′(x) = ψ(−x)

Π−1 = Π = Π†

= Π†QΠ

x ′ = Π†xΠ = −xp′ = Π†pΠ = −p

Q′(x , p) = Q(−x ,−p)

= Π†HΠ = H

ΠH = ΠΠ†HΠ → ΠH = HΠ → [H, Π] = 0

Parity in 1D

For a Hamiltonian which describes a particle of mass m in aone-dimensional potential V (x), inversion symmetry means thatV (x) ≡ V (−x), an even function of position

Because Π and H commute, we can find common eigenfunctions ψn(x)such that

Πψn(x) = ψn(−x)

= ±ψn(x)

since the eigenvalues of parity can only be ±1

Thus the eigenfunctions of such a Hamiltonian are either even or oddunder parity and by Eherenfest’s Theorem, we have

dt〈Π〉 =

~〈[H, Π]〉 = 0

which means that parity is conserved in time, that is an even functionunder parity will remain even for all time and an odd function under paritywill remain odd for all time

Parity in 1D

= ±ψn(x)

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 1D

= ±ψn(x)

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 1D

Πψn(x) = ψn(−x) = ±ψn(x)

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 1D

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 1D

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 1D

dt〈Π〉 =

~〈[H, Π]〉 = 0

Parity in 3D

In three dimensions, the parity op-erator inverts the system throughthe origin

the r and p operators and any arbi-trary operator transform as

Πψ(r) = ψ′(r) = ψ(−r)r′ = Π†rΠ = −rp′ = Π†pΠ = −p

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

The Hamiltonian in three dimensions will have parity when V (−r) = V (r)which is true for all central potentials

The eigenstates of the hydrogen atom are in fact, also eigenstates of parity

Πψnlm(r , θ, φ) = (−1)lψnlm(r , θ, φ), ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)

Parity in 3D

Πψ(r) = ψ′(r) = ψ(−r)

r′ = Π†rΠ = −rp′ = Π†pΠ = −p

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Πψ(r) = ψ′(r) = ψ(−r)

r′ = Π†rΠ = −rp′ = Π†pΠ = −p

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity in 3D

Q ′(r, p) = Π†Q(r, p)Π = Q(−r,−p)

Parity selection rules

Selection rules, which will be very important when we talk abouttime-dependent phenomena, indicate when a matrix element which couplestwo states, 〈a|Q|b〉, is zero based on symmetry.

A particularly important operator is the electric dipole operator, pe = qrwhose selection rules determine which atomic transitions are allowed andwhich are forbidden.

It is evident that pe is odd under parity because r is odd.

Π†peΠ = −peConsider the matrix elements of the electric dipole operator between twoatomic states ψnlm, and ψn′l ′m′

〈n′l ′m′|pe |n l m〉 = −〈n′l ′m′|Π†peΠ|n l m〉

= −〈n′l ′m′|(−1)l′pe(−1)l |n l m〉

= (−1)l+l ′+1〈n′l ′m′|pe |n l m〉 = 0 if l + l ′ = 2n

= −〈n′l ′m′|(−1)l′pe(−1)l |n l m〉

Π†peΠ = −pe

Consider the matrix elements of the electric dipole operator between twoatomic states ψnlm, and ψn′l ′m′

= −〈n′l ′m′|(−1)l′pe(−1)l |n l m〉

= (−1)l+l ′+1〈n′l ′m′|pe |n l m〉

= 0 if l + l ′ = 2n

= −〈n′l ′m′|(−1)l′pe(−1)l |n l m〉

Rotation about the z axis

The operator Rz(ϕ) rotates the function an angle ϕ about the z axis

Rz(φ)ψ(r , θ, φ) = ψ′(r , θ, φ) = ψ(r , θ, φ− ϕ)

just as with the translation operator we can determine the generator ofrotations starting with a Taylor series (hiding the r and θ variables)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

= e−iϕLz/~ψ(φ) −→ Rz(ϕ) = e−iϕLz/~

Rz(φ)ψ(r , θ, φ) = ψ′(r , θ, φ)

= ψ(r , θ, φ− ϕ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ)

=∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

= e−iϕLz/~ψ(φ)

−→ Rz(ϕ) = e−iϕLz/~

Rz(ϕ)ψ(φ) = ψ(φ− ϕ) =∞∑n=0

n![(φ− ϕ)− φ]n

dnψ(φ− ϕ)

d(φ− ϕ)n

∣∣∣∣φ−ϕ=φ

=∞∑n=0

n!(−ϕ)n

dnψ(φ)

∣∣∣∣φ

=∞∑n=0

n!(−ϕ)n

dφnψ(φ)

=∞∑n=0

(−iϕ~

)nψ(φ) =

∞∑n=0

(−iϕ~

)nψ(φ)

Transformations under Rz

Assuming an infinitesmal rotation δ we have that Rz(δ) ≈ 1− iδ~ Lz

andthe x operator thus transforms as

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

[Lz , x ] = [xpy − ypx , x ] = [xpy , x ]− [ypx , x ] = 0− y(−i~) = i~y

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

Assuming an infinitesmal rotation δ we have that Rz(δ) ≈ 1− iδ~ Lz and

the x operator thus transforms as

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz

1 +iδ

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

≈ x +iδ

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ]

≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ]

≈ x − δy

[Lz , x ] = [xpy − ypx , x ]

= [xpy , x ]− [ypx , x ] = 0− y(−i~) = i~y

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ]

≈ x − δy

[Lz , x ] = [xpy − ypx , x ] = [xpy , x ]− [ypx , x ]

= 0− y(−i~) = i~y

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ]

≈ x − δy

[Lz , x ] = [xpy − ypx , x ] = [xpy , x ]− [ypx , x ] = 0− y(−i~)

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ]

≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz

1 +iδ

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

≈ y +iδ

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ]

≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz

1 +iδ

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

≈ z +iδ

~[Lz , z ] ≈ z

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ]

x ′ = R†z x Rz ≈

(1− iδ

)≈ x +

~[Lz , x ] ≈ x − δy

y ′ = R†z y Rz ≈

(1− iδ

)≈ y +

~[Lz , y ] ≈ y + δx

z ′ = R†z z Rz ≈

(1− iδ

)≈ z +

~[Lz , z ] ≈ z

These three results can be combined into a matrix equation

1 −δ 0δ 1 00 0 1

for infinitesmal rotations, cosϕ→ 1 and sinϕ→ δ so the matrix is morecorrectly written as x ′

cosϕ − sinϕ 0sinϕ cosϕ 0

These three results can be combined into a matrix equation x ′

1 −δ 0δ 1 00 0 1

for infinitesmal rotations, cosϕ→ 1 and sinϕ→ δ so the matrix is morecorrectly written as

1 −δ 0δ 1 00 0 1

Rotations in 3D

The result for Rz can be generalized as arotation about an arbitrary direction n

Rn(ϕ) = e−iϕn·L/~

any 3D operator, V that transforms in the same way as the positionoperator under rotations, that is with [Li , V j ] = i~εijk V k , is called a vectoroperator and the transformation is V

A scalar operator, f is one that is unchanged by rotations, that is, itcommutes with L, [Li , f ] ≡ 0

operators can thus be classified as scalar or vector operators based on theircommutation relations with L and as true or pseudo quantities based onhow they transform under parity

Rotations in 3D

The result for Rz can be generalized as arotation about an arbitrary direction n Rn(ϕ) = e−iϕn·L/~

Rotations in 3D

any 3D operator, V that transforms in the same way as the positionoperator under rotations, that is with [Li , V j ] = i~εijk V k , is called a vectoroperator and the transformation is

Rotations in 3D

phys.iit.eduphys.iit.edu/~segre/phys406/19s/lecture_01.pdf · phys 406 - fundamentals of quantum...

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