Slide 1 Chapter 2 Quantum Theory. Slide 2 Outline Interpretation and Properties of Operators and...

Chapter 2

Quantum Theory

Outline

• Interpretation and Properties of

• Operators and Eigenvalue Equations

• Normalization of the Wavefunction

• Operators in Quantum Mechanics

• Math. Preliminary: Even and Odd Integrals

• Eigenfunctions and Eigenvalues

• The 1D Schrödinger Equation: Time Depend. and Indep. Forms

• Math. Preliminary: Probability, Averages and Variance

• Expectation Values (Application to HO wavefunction)

• Hermitian Operators

Continued on Second Page

Outline (Cont’d.)

• Commutation of Operators

• Differentiability and Completeness of the Wavefunctions

• Dirac “Bra-Ket” Notation

• Orthogonality of Wavefunctions

First Postulate: Interpretation of

One Dimension

Postulate 1: (x,t) is a solution to the one dimensional Schrödinger Equation and is a well-behaved, square integrable function.

x x+dx

The quantity, |(x,t)|2dx = *(x,t)(x,t)dx, representsthe probability of finding the particle betweenx and x+dx.

x

y

z

dxdy

dz

Three Dimensions

Postulate 1: (x,y,z,t) is a solution to the three dimensional Schrödinger Equation and is a well-behaved, square integrable function.

The quantity, |(x,y,z,t)|2dxdydz = *(x,y,z,t)(x,y,z,t)dxdydz, represents the probability of finding the particle between x and x+dx, y and y+dy, z and z+dz.

( , , , ) ( , )x y z t r t

Shorthand Notation

Two Particles

2

1 2 1 1

1 2 1 1 1 2 2 2

( , )r r dr dr

where dr dr dx dy dz dx dy dz

Required Properties of

Finite X

Single Valued

x

(x)

Continuous

x

(x)

And derivatives mustbe continuous

Required Properties of

Vanish at endpoints(or infinity)

0 as x ± y ± z ±

Must be “Square Integrable”

2( )r dxdydz

2

( )r d

or

Shorthand notation

Reason: Can “normalize” wavefunction

2( ) 1r d

Which of the following functions would be acceptablewavefunctions?

2xe x OK

xe x No - Diverges as x -

1s i n x No - Multivalued i.e. x = 1, sin-1(1) = /2, /2 + 2, ...

xe x No - Discontinuous first derivativeat x = 0.

Outline











Operators and Eigenvalue Equations

One Dimensional Schrödinger Equation2 2

2( )

2

dV x E

m dx

2 2

2( )

2

dV x E

m dx

H E 2 2

2( )

2

dH V x

m dx

This is an “Eigenvalue Equation”

A f a f

Operator

Operator

Eigenvalue

Eigenvalue

Eigenfunction

Eigenfunction

Linear Operators

A quantum mechanical operator must be linear

ˆ ˆ ˆA c d c A d A

Operator Linear ?

x2•

log

sin

d

dx

2

2

d

dx

Yes

No

No

No

Yes

Yes

Operator Multiplication

ˆ ˆˆ ˆ( )A B A B

First operate with B, and then operate on the result with A.^ ^

2ˆ ˆ ˆ( )A A A Note:

Example

ˆ ˆdA and B x

dx

ˆ ˆ ?A B

ˆ ˆ dAB x

dx

2ˆ( )A

Operator Commutation

ˆ ˆˆ ˆA B B A ?

Not necessarily!! If the result obtained applying two operatorsin opposite orders are the same, the operatorsare said to commute with each other.

Whether or not two operators commute has physical implications,as shall be discussed later, where we will also give examples.

Eigenvalue Equations

A f a f

A f Eigenfunction? Eigenvalue

3 x2 Yes 3

x sin(x) No

sin(x)d

dxNo

sin(x)2

2

d

dxYes -2 (All values of

allowed)2

224

dx

dx

2xe Only for = ±1

2 (i.e. ±2)

Outline











Operators in Quantum Mechanics

Postulate 2: Every observable quantity has a corresponding linear, Hermitian operator.

The operator for position, or any function of position,is simply multiplication by the position (or function)

ˆ. .i e x x 2 2. .i e x x^ etc.

The operator for a function of the momentum, e.g. px, isobtained by the replacement:

ˆ. .i e pi x

I will define Hermitian operators and their importance inthe appropriate context later in the chapter.

“Derivation” of the momentum operator

( )i k x tC e 2hp

Wavefunction for a free particle (from Chap. 1)

where 2k

p

( )i kx tikC ex

i k p

i

pi x

ˆ. .i e pi x

Some Important Operators (1 Dim.) in QM

Quantity Symbol Operator

Position x x

Potential Energy V(x) V(x)

Momentum px (or p)i x

2 2

22m x

Kinetic Energy

2

2xp

m

Total Energy2

( )2xp V xm

2 2

2( )

2H V x

m x

Some Important Operators (3 Dim.) in QM

Quantity Symbol Operator

Potential Energy V(x,y,z) V(x,y,z)

Kinetic Energy2 2 2 2

2 2 22m x y z

2

2

p

m

Total Energy2

( , , )2

pV x y z

m

22 ( , , )

2H V x y z

m

Position r x i y j z k

r x i y j z k

Momentum x y zp p i p j p k i j k

i x y z

i

i j kx y z

22

2m

2 2 22

2 2 2x y z

Outline











The Schrödinger Equation (One Dim.)

Postulate 3: The wavefunction, (x,t), is obtained by solving the time dependent Schrödinger Equation:

( , )( , )

x ti H x t

t

2 2

2( , ) ( , )

2V x t x t

m x

If the potential energy is independent of time, [i.e. if V = V(x)],then one can derive a simpler time independent form of the Schrödinger Equation, as will be shown.

In most systems, e.g. particle in box, rigid rotator, harmonicoscillator, atoms, molecules, etc., unless one is consideringspectroscopy (i.e. the application of a time dependent electricfield), the potential energy is, indeed, independent of time.

( , )( , )

x ti H x t

t

2 2

2( , ) ( , )

2V x t x t

m x

If V is independent of time, then so is the Hamiltonian, H.

( , )( , )

x ti H x t

t

Assume that (x,t) = (x)f(t)

On Board

The Time-Independent Schrödinger Equation (One Dimension)

I will show you the derivation FYI. However, you are responsibleonly for the result.

1 ( ) 1( )

( ) ( )

df ti H xf t d t x

= E (the energy, a constant)

1 ( ) 1( )

( ) ( )

df ti H xf t d t x

= E (the energy, a constant)

On Board

( ) ( )H x E x

Time IndependentSchrödinger Equation

( , ) ( ) ( ) ( )i E t

x t x f t x e

Note that *(x,t)(x,t) = *(x)(x)

Outline











Math Preliminary: Probability, Averages & Variance

Probability

Discrete Distribution: P(xJ) = Probability that x = xJ

If the distribution is normalized: P(xJ) = 1

Continuous Distribution: P(x)dx = Probability that particle has positionbetween x and x+dx

x x+dx

P(x)If the distribution is normalized:

( ) 1P x dx

Positional Averages

Discrete Distribution: ( )J Jx x x P x If normalized

( )

( )J J

J

x P x

P x

If not normalized

2 ( )

( )J J

J

x P x

P x

If not normalizedIf normalized

2 2 2 ( )J Jx x x P x

Continuous Distribution: ( )x x xP x d x

If normalized

( )

( )

xP x dx

P x dx

If not normalized

If normalized

2 2 2 ( )x x x P x d x

If not normalized

2 ( )

( )

x P x dx

P x dx

Continuous Distribution: ( )x x xP x d x

If normalized If normalized

2 2 2 ( )x x x P x d x

Note: <x2> <x>2

Example: If x1 = 2, P(x1)=0.5 and x2 = 10, P(x2) = 0.5

Calculate <x> and <x2>

2 0 . 5 1 0 0 . 5 6x

2 22 2 0 .5 1 0 0 .5 5 2x

Note that <x>2 = 36

It is always true that <x2> <x>2

Variance

One requires a measure of the “spread” or “breadth” of a distribution.This is the variance, x

2, defined by:

22x x x 2 ( )x x P x d x

2 2 22 ( )x x x x x P x d x

2 2 2( ) 2 ( ) ( )x x P x d x x xP x d x x P x d x

2 2 22x x x x x

2 2 2x x x

Variance

2 2 2x x x x

Standard Deviation

Example

P(x) = Ax 0x10P(x) = 0 x<0 , x>10

Calculate: A , <x> , <x2> , x

1 0

0( )P x d x A xd x

5 0A 1

1

50A

1 0

0( )x xP x d x xA xd x

103

03

xA

3101

50 3 6 . 6 7

Note: 22 6 .6 7 4 4 .4x

1 02 2 2

0( )x x P x d x x A xd x

104

04

xA

4101

50 4 5 0 . 0

2 2x x x 5 0 . 0 4 4 . 4 2 . 3 7

102

02

xA

Outline











Normalization of the Wavefunction

For a quantum mechanical wavefunction: P(x)=*(x)(x)

For a one-dimensional wavefunction to be normalized requires that:

* ( ) ( ) 1x x d x

For a three-dimensional wavefunction to be normalized requires that:

* ( , , ) ( , , ) 1x y z x y z d x d y d z

In general, without specifying dimensionality, one may write:

* 1d

Example: A Harmonic Oscillator Wave Function

Let’s preview what we’ll learn in Chapter 5 about theHarmonic Oscillator model to describe molecular vibrationsin diatomic molecules.

= reduced massk = force constant

The Hamiltonian:2 2

22

1

2 2

dH kx

dx

A Wavefunction:2 / 2( ) xx A e x

2k

Outline











Math Preliminary: Even and Odd Integrals

Integration Limits: - Integration Limits: 0

2

0

1

2xe dx

2

0

1

2xxe dx

22

0

1

4xx e dx

2320

1

2xx e dx

2420

3

8xx e dx

2 2

02x xe d x e d x

2 22 2

02x xx e d x x e d x

2 24 4

02x xx e d x x e d x

0

0

21 * d x d x

Find the value of A that normalizes the Harmonic Oscillator

oscillator wavefunction:2 / 2( ) xx A e x

2

0

1

2xe dx

2 2/ 2xA e d x

22 xA e dx

22

02 xA e dx 2 1

22

A

1/ 2

2 1A

1/ 22A

1/4

A

Outline











Eigenfunctions and Eigenvalues

Postulate 4: If a is an eigenfunction of the operator Â with eigenvalue a, then if we measure the property A for a system whose wavefunction is a, we always get a as the result.

Example

2 22

2

1

2 2

dH kx

dx

2 / 2( ) xx A e x

2k

The operator for the total energy of a system is the Hamiltonian.Show that the HO wavefunction given earlier is an eigenfunctionof the HO Hamiltonian. What is the eigenvalue (i.e. the energy)

Preliminary: Wavefunction Derivatives2 / 2( ) xx A e

2 / 2xd dA e

dx dx 2

2/ 2 ( / 2)x d x

Aedx

2 / 2 ( 2 / 2 )xAe x

2 / 2xdA xe

dx

22

/ 22

xd d d dA xe

dx dx dx dx

2

2/ 2

/ 2x

xde dxA x e

dx dx

2 2/ 2 / 2( 2 / 2 )x xA x e x e

2 22

2 / 2 / 22

x xdA x e e

dx

2 22

2

1

2 2

dH kx

dx

2 2 2

22 / 2 / 2 2 / 21

2 2x x xA x e e kx Ae

2 2 22 2 2

2 / 2 / 2 2 / 21

2 2 2x x xx Ae Ae kx Ae

2 2 22 21

2 2 2x kx

2 2 2

2 1

2 2 2x k

To end up with a constant times ,this term must be zero.

2 22

2 / 2 / 22

x xdA x e e

dx

2 22

2

1

2 2

dH kx

dx

2 / 2( ) xx A e x

2k

2 2 22 1

2 2 2H x k

2 2

22

2

2

( / )k

2

k

2 22

2

1

2 2 2

kH x k

22 1 1

2 2 2x k k

2

2

1

2H

22

h

h

1

2h

E = ½ħ = ½h

Because the wavefunction is aneigenfunction of the Hamiltonian,the total energy of the systemis known exactly.

2 22

2 / 2 / 22

x xdA x e e

dx

2 22

2

1

2 2

dH kx

dx

2 / 2( ) xx A e x

2k

2 22

2

1

2 2

dH kx

dx

2 / 2( ) xx A e x

Is this wavefunction an eigenfunction of the potential energy operator?

21ˆ ( )2

V V x kx No!! Therefore the potential energy cannotbe determined exactly.

One can only determine the “average” value of a quantity if thewavefunction is not an eigenfunction of the associated operator.

The method is given by the next postulate.

Is this wavefunction an eigenfunction of the kinetic energy operator?2 2 2

2

ˆ

2 2

p dKE

dx

No!! Therefore the kinetic energy cannotbe determined exactly.

Eigenfunctions of the Momentum Operator

pi x

Recall that the one dimensional momentum operator is:

2 / 2( ) xx A e

Is our HO wavefunction an eigenfunction of the momentum operator?

No. Therefore the momentum of an oscillator in this eigenstate cannot be measured exactly.

( )i k x tC e 2

k

The wavefunction for a free particle is:

Is the free particle wavefunction an eigenfunction of the momentumoperator?

Yes, with an eigenvalue of h \ , which is just the de Broglie expression for the momentum.

Thus, the momentum is known exactly. However, the position iscompletely unknown, in agreement with Heisenberg’sUncertainty Principle.

Outline











Expectation Values

Postulate 5: The average (or expectation) value of an observable with the operator Â is given by

ˆ*

*

A da

d

ˆ* A d

If is normalized

Expectation values of eigenfunctions

It is straightforward to show that If a is eigenfunction of Â with eigenvalue, a, then:

<a> = a

<a2> = a2

a = 0 (i.e. there is no uncertainty in a)

Expectation value of the position

ˆ*

*

x dxx

dx

*

*

x dx

dx

*

*

x dx

dx

( )

( )

xP x dxx

P x dx

This is just the classical expression for calculating theaverage position.

The differences arise when one computes expectation valuesfor quantities whose operators involve derivatives, suchas momentum.

Calculate the following quantities:

<p>

<p2>

p2

<x>

<x2>

x2

xp (to demo. Unc. Prin.)

<KE>

<PE>

Consider the HO wavefunction we have been using in

earlier examples:2 / 2( ) xx A e x

2k

Preliminary: Wavefunction Derivatives2 / 2( ) xx A e

2 / 2xd dA e

dx dx 2

2/ 2 ( / 2)x d x

Aedx

2 / 2 ( 2 / 2 )xA e x

2 / 2xdA xe

dx

22

/ 22

xd d d dA xe

dx dx dx dx

2

2/ 2

/ 2x

xde dxA x e

dx dx

2 2/ 2 / 2( 2 / 2 )x xA x e x e

2 22

2 / 2 / 22

x xdA x e e

dx

<x>

*x x d x

2 2/ 2 / 2x xA e x A e d x

22 xA xe dx

0

<x2>

2 2*x x d x

2 2/ 2 2 / 2x xA e x A e d x

22 2 xA x e d x

1/ 21

24

2 / 2( ) xx A e 1/4

A

1

222 2

02 xA x e d x

22

0

1

4xx e dx

2 1

2x

Also:

2 / 2( ) xx A e 2 / 2xd

A xedx

<p>

ˆd

pi dx

*d

p dxi dx

2 2/ 2 / 2( )x xA e A xe dxi

22

xAxe dx

i

0p

<p2>

22 2

2* ( )

dp dx

dx

2 / 2( ) xx A e 2 22

2 / 2 / 22

x xdA x e e

dx

22 2

2

dp

dx

^

2 2 2/ 2 2 2 / 2 / 2( ) ( )x x xA e A x e e d x

2 22 2 2 x xA x e e d x

2 22 2 2 x xA x e d x e d x

2 2 1 12 24 2

A

2 2 11

2A

2 2 1

2p

2

2

22

0

1

4xx e dx

2

0

1

2xe dx

22

2p

Also:

Uncertainty Principle

22

2p

2 1

2x

22 2 1

2 2x p

2

4

2 2

2x p x p

<KE>

22

2 2

ppKE

21

2 2

2

4

1

4

1

4 2

h

1

4h

<PE>

2 21 1

2 2PE kx k x 1 1

2 2k

4

k

4 /

k

1

4

k

2 1

4

1

4 1

4h

22

2p

2 1

2x

k

Calculate the following quantities:

<x>

<x2>

<p>

<p2>

p2x

2

xp

<KE>

<PE>

Consider the HO wavefunction we have been using in

earlier examples:2 / 2( ) xx A e x

2k

= 0 = 0

= 1/(2)

= 1/(2)

= ħ2/2

= ħ2/2

= ħ/2 (this is a demonstration of the Heisenberg uncertainty principle)

= ¼ħ = ¼h

= ¼ħ = ¼h

Outline











Hermitian Operators

GeneralDefinition: An operator Â is Hermitian if it satisfies the relation:

ˆ ˆ( ) * *A d A d “Simplified”Definition (=): An operator Â is Hermitian if it satisfies the relation:

ˆ ˆ( ) * *A d A d

It can be proven that if an operator Â satisfies the “simplified” definition,it also satisfies the more general definition.(“Quantum Chemistry”, I. N. Levine, 5th. Ed.)

So what? Why is it important that a quantum mechanical operator be Hermitian?

The eigenvalues of Hermitian operators must be real.

Proof: ˆ ˆ( ) * *A d A d A a and

a* = a

i.e. a is real

In a similar manner, it can be proven that the expectation values<a> of an Hermitian operator must be real.

( ) * *a d a d

* * *a d a d

* * *a d a d

Is the operator x (multiplication by x) Hermitian?

ˆ ˆ( ) * *A d A d

Yes.

Is the operator ix Hermitian? No.

ˆd

pi dx

Is the momentum operator Hermitian?

Math Preliminary: Integration by Parts

( )d uv dv duu v

dx dx dx

( )dv d uv duu vdx dx dx

( )d v d u v duu d x d x v d xd x d x d x

dv duu dx u v v d xd x d x

You are NOT responsible for the proof outlined below, butonly for the result.

Yes: I’ll outline the proof

ˆ ˆ( ) * *A d A d dv duu dx u v v d xd x d x

ˆd

pi dx

Is the momentum operator Hermitian?

**

d ddx dx

i dx i dx

?The question is whether:

**

d ddx dx

dx dx

?

or:

The latter equality can be proven by using Integration by Partswith: u = and v = *, together with the fact that both and * arezero at x = . Next Slide

Thus, the momentum operator IS Hermitian

?*??*

d ddx dx

i dx i dx

*

* ??d d

dx dxdx dx

or:

dv duu dx u v v d xd x d x

Let u = and v = *: ** *

0 *

d ddx dx

dx dxddx

dx

Because and *vanish at x = ±∞

Therefore:*

*d d

dx dxdx dx

?

?

**

d ddx dx

i dx i dx

By similar methods, one can show that:

d

dxis NOT Hermitian (see last slide)

IS Hermitiandidx

2

2

d

dxIS Hermitian (proven by applying integration by parts twice successively)

The Hamiltonian:2 2

2( )

2

dH V x

m dx

IS Hermitian

Outline (Cont’d.)





Orthogonality of Eigenfunctions

Assume that we have two different eigenfunctions of the sameHamiltonian: i i i j j jH E a n d H E

If the two eigenvalues, Ei = Ej, the eigenfunctions (aka wavefunctions)are degenerate. Otherwise, they are non-degenerate eigenfunctions

We prove below that non-degenerate eigenfunctions are orthogonal to each other.

* *j i i jH d H d Because the Hamiltonianis Hermitian

*( ) 0i j j iE E d

Proof:

* *j i i i j jE d E d * *i j jE d

* * *i j i j i jE d E d *

j j iE d

*( ) 0i j j iE E d

Thus, if Ei Ej (i.e. the eigenfunctions are not degenerate,

then: * 0j i d We say that the two eigenfunctions are orthogonal

If the eigenfunctions are also normalized, then we can say thatthey are orthonormal.

*j i i jd

ij is the Kronecker Delta, defined by:

1

0

ij

ij

if i j

if i j

Linear Combinations of Degenerate Eigenfunctions

Assume that we have two different eigenfunctions of the sameHamiltonian: i i i j j jH E a n d H E

If Ej = Ei, the eigenfunctions are degenerate. In this case, any linearcombination of i and j is also an eigenfunction of the Hamiltonian

i ja b

( ) ( )i j i jH a b a H b H

iH E

Thus, any linear combination of degenerate eigenfunctions is alsoan eigenfunction of the Hamiltonian.

If we wish, we can use this fact to construct degenerate eigenfunctionsthat are orthogonal to each other.

Proof:

( )i i j jaE bE

( ) ( )i j i i jH a b E a b o r If Ej = Ei ,

Outline (Cont’d.)





Commutation of Operators

ˆ ˆˆ ˆA B B A ?

Not necessarily!! If the result obtained applying two operatorsin opposite orders are the same, the operatorsare said to commute with each other.

Whether or not two operators commute has physical implications,as shall be discussed below.

One defines the “commutator” of two operators as:

ˆ ˆ ˆˆ ˆ ˆ,A B AB BA

If ˆ ˆ, 0A B for all , the operators commute.

A B ˆ ˆ,A B

x x2 0 Operators commute

0 Operators commute3d

dx

-iħ Operators DO NOT commuteˆd

pi dx

x x

And so??

Why does it matter whether or not two operators commute?

Significance of Commuting Operators

ˆ ˆn n n n n nA a and B b

Let’s say that two different operators, A and B, have thesame set of eigenfunctions, n:

^ ^

This means that the observables corresponding to both operators can be exactly determined simultaneously.

Conversely, it can be proven that if two operators do notcommute, then the operators cannot have simultaneouseigenfunctions.

This means that it is not possible to determine bothquantities exactly; i.e. the product of the uncertaintiesis greater than zero.

Then it can be proven**

that the two operators commute; i.e. ˆ ˆ, 0A B

**e.g. Quantum Chemistry (5th. Ed.), by I. N. Levine, Sect. 5.1

We just showed that the momentum and position operators do not

commute: ˆ ˆ, 0xp x i

This means that the momentum and position of a particle cannotboth be determined exactly; the product of their uncertainties isgreater than 0.

0xp x

If the position is known exactly ( x=0 ), then the momentumis completely undetermined ( px ), and vice versa.

This is the basis for the uncertainty principle, which we demonstratedabove for the wavefunction for a Harmonic Oscillator, wherewe showed that px = ħ/2.

Outline (Cont’d.)





Differentiability and Completenessof the Wavefunction

Differentiability of

It is proven in in various texts** that the first derivative of the wavefunction, d/dx, must be continuous.

** e.g. Introduction to Quantum Mechanics in Chemistry, M. A. Ratner and G. C. Schatz, Sect. 2.7

x

This wavefunction would not be acceptablebecause of the sudden change in thederivative.

The one exception to the continuous derivative requirement isif V(x).

We will see that this property is useful when setting “BoundaryConditions” for a particle in a box with a finite potential barrier.

Completeness of the Wavefunction

The set of eigenfunctions of the Hamiltonian, n , form a “complete set”.

n nH E

This means that any “well behaved” function defined over thesame interval (i.e. - to for a Harmonic Oscillator, 0 to a for a particle in a box, ...) can be written as a linear combinationof the eigenfunctions; i.e.

1

( ) n nn

f x c

We will make use of this property in later chapters when wediscuss approximate solutions of the Schrödinger equation formulti-electron atoms and molecules.

Outline (Cont’d.)





Dirac “Bra-Ket” Notation

A standard “shorthand” notation, developed by Dirac, and termed“bra-ket” notation, is commonly used in textbooks andresearch articles.

In this notation: ˆ ˆ* A d A

is the “bra”: It represents the complex conjugate partof the integrand

is the “ket”: It represents the non-conjugate partof the integrand

In Bra-Ket notation, we have the following:

“Scalar Product” of two functions: 1 2 1 2* d

ˆ ˆ* *A d A d HermitianOperators:

Orthogonality: * 0i j i jd

Normalization: * 1i i i id

ˆ ˆA A

ˆ ˆ*

*

A d Aa

d

ExpectationValue:

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