Introduction to Quantum Mechanics I Lecture 13 ...

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Introduction to Quantum Mechanics I

Lecture 13: Eigenvalues and eigenfunctions

Eigenvalues and eigenfunctions

The schedule…

Part I Introduction: The Schrödinger equation and fundamental quantum systems

Part II The formalism

Part III Quantum mechanics of atoms and solids

Exam I Part I

Exam II Part II + the hydrogen atom

Final exam All material covered in the course

Last time… 𝐴 𝐵

𝐴 + 𝐵

vectors?

vector spaces obey a simple set of rules

the polynomials of degree 2

the even functions

all possible sound waves

the complex numbers

arithmetic progressions

the solutions of the Schrödinger equation

Examples:

a Hilbert space is a vector space with a norm, and it is ‘complete’(large enough).

The solutions of the Schrödinger equation (the ‘wave functions’) span a vector space

... much larger than Hilbert’s Grand Hotel

ℕ, ℤ, and ℚ are ‘equally large’, but ℝ is larger (much larger!)

(e. g. ℝ, ℝ3, 𝑃∞, 𝑓 )

Last time…

transcendental numbers are not lonely

OperatorsToday:

What are operators?

Observables?

Hermitian operators?

Determinate states?

What is a degenerate spectrum?

a map 𝑇 between two vector spaces 𝑉, 𝑊𝑇: 𝑉 𝑊

such that

a) 𝑇(𝑣1 + 𝑣2) = 𝑇(𝑣1) + 𝑇(𝑣2)b) 𝑇(𝛼𝑣2) = 𝛼𝑇(𝑣2)

a linear transformation:

such that

𝐴𝐵

𝐴 + 𝐵𝑇( 𝐴) + T(𝐵) = 𝑇( 𝐴 + 𝐵)

𝑇(𝐵)

𝑇( 𝐴)

𝑽𝑾

such that

𝛼 𝐴𝑇( 𝐴)

𝑽𝑾

𝛼𝑇 𝐴 = 𝑇 𝛼 𝐴𝑇

such that

𝑽𝑽

𝐸Ψ 𝐻

𝐻Ψ = 𝐸Ψ

such that

𝜓𝑛

𝑽𝑽

𝑛 + 1 𝜓𝑛+1

𝑎+𝜓𝑛 = 𝑛 + 1 𝜓𝑛+1

such that

𝐻Ψ = 𝐸Ψ

𝑎+𝜓𝑛 = 𝑛 + 1 𝜓𝑛+1

𝑥 𝑝

[ 𝑥, 𝑝]

other operators:

In Quantum Mechanics

Observables are represented by linear Hermitian operators

In Quantum Mechanics:

What is an observable?

Who is observing?

What do you need to satisfy to be an observer?

What does ‘Hermitian’ imply?

𝐴 is Hermitian 𝐴 is real 𝐴 = 𝐴∗

𝐴 = Ψ∗ 𝐴 Ψ d𝑥 𝐴∗

= Ψ∗ 𝐴 Ψ d𝑥

= 𝐴Ψ∗Ψ d𝑥

Ψ| 𝐴Ψ 𝐴Ψ|Ψ

In a finite dimensional vector space:

operators can be represented as a matrix – with respect to a certain basis:

𝐴𝑖𝑗 = 𝑒𝑖| 𝐴|𝑒𝑗

(so the form of the matrix depends on the choice of basis)

Determinate states return the same value 𝑞 after each measurement 𝑄

(e.g. ) 𝐻Ψ = 𝐸Ψ

“Eigenfunction of the Hamiltonian”

“(corresponding) Eigenvalue”

If two eigenfunctions have the same eigenvalue,

we say that “the spectrum is degenerate”

For determinate states 𝜎 = 0

𝐻Ψ = 𝐸Ψ

𝐸Ψ

𝐻 does not change the ‘direction’ of its eigenvectors

(it does not change the state)

Question

Find the eigenvectors of the following operators:

The operator 𝑂 that mirrors vectors in the 𝑥-𝑦 plane in the 𝑥-axis

Question

The operator 𝑂 that mirrors vectors in the 𝑥-𝑦 plane in the 𝑥-axis

𝜆1 = 1

𝜆2 = −1

Question

The operator 𝑂 that mirrors vectors in the 𝑥-𝑦 plane in the 𝑦-axis

Question

The operator 𝑂 that mirrors vectors in the 𝑥-𝑦 plane in the 𝑦-axis

𝜆1 = −1

𝜆2 = 1

Question

The operator 𝑂 that projects vectors in ℝ3 onto the 𝑥-𝑦 plane

Question

𝜆2 = 𝜆3 = 1

𝜆1 = 0

Question

𝜆2 = 𝜆3 = 1

𝜆1 = 0

(all the vectors in the 𝑥-𝑦 plane)

Question

The operator 𝑂 that mirrors vectors in ℝ3 into the origin

Question

𝜆1 = 𝜆2 = 𝜆3 = −1

Question

𝜆1 = 𝜆2 = 𝜆3 = −1(all the vectors in ℝ3)

An operator does not change the ‘direction’ of its eigenvector

Conclusion:

In quantum mechanics: An operator does not change the state of its eigenvectors (‘eigenstates’)

Conclusion:

In quantum mechanics: An operator does not change the state of its eigenvectors (‘eigenstates’,

‘eigenfunctions’, ‘eigenkets’ …)

Conclusion:

𝐻Ψ = 𝐸Ψ

Conclusion:

𝐻Ψ = 𝐸Ψ

𝑎+𝜓𝑛 = 𝑛 + 1 𝜓𝑛+1

Conclusion:

𝐻Ψ = 𝐸Ψ

𝑎+𝜓𝑛 = 𝑛 + 1 𝜓𝑛+1

not an eigenstate of 𝑎+

Conclusion:

How to find eigenvectors:

Conclusion:

(in finite dimensional vector space) – solve the characteristic equation

det 𝐴 − 𝜆𝐼 = 0𝐴𝑣 = 𝜆𝑣

Conclusion:

(in high dimensional Hilbert space) – e.g. by solving a differential equation

𝐻Ψ = 𝐸Ψ

Conclusion:

𝐻Ψ = 𝐸Ψ

if the spectrum is non-degenerate then the eigenfunctions are orthogonal

Conclusion:

𝐻Ψ = 𝐸Ψ

if the spectrum is non-degenerate then the eigenfunctions are orthogonal

if the spectrum is discrete, then the Ψ’s are normalizable

if the spectrum is continuous, then the Ψ’s are not normalizable

Reading: Sections 3.3

Summarize section 3.3

Homework due Thursday 9 March :

Introduction to Quantum Mechanics I Lecture 13 ...

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