Today’s Outline - September 24, 2012csrri.iit.edu/~segre/phys405/12F/lecture_10.pdf · 2012. 9. 29. · Today’s Outline - September 24, 2012 Problem 2.27 Problem 2.33 Hilbert

Today’s Outline - September 24, 2012

• Problem 2.27

• Problem 2.33

• Hilbert space

Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators providedReading Assignment: Chapter 3.3-3.5

C. Segre (IIT) PHYS 405 - Fall 2012 September 24, 2012 1 / 11


• Problem 2.27

• Problem 2.33

• Hilbert space




• Problem 2.27

• Problem 2.33

• Hilbert space




• Problem 2.27

• Problem 2.33

• Hilbert space




• Problem 2.27

• Problem 2.33

• Hilbert space

Exam #1: Monday, October 1, 2012 Covers Chapters 1 & 2, closedbook, calculators provided

Reading Assignment: Chapter 3.3-3.5



• Problem 2.27

• Problem 2.33

• Hilbert space



Problem 2.27

Consider the double delta-function potential

V (x) = −α[δ(x + a) + δ(x − a)],

where α and a are positive constants.

How many bound states does this potential posess? Find the allowedenergies for α = ~2/4ma and for α = ~2/4ma, and sketch the wavefunctions.


Problem 2.27

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

z

e-z

cz-1

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

z

e-z

1-(0.3)z

1-(1.0)z


Problem 2.27

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

z

e-z

cz-1

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

z

e-z

1-(0.3)z

1-(1.0)z


Problem 2.33

Determine the transmission coefficient for a rectangular barrier (same asEq. 2.145, only with V (x) = +V0 > 0 in the region −a < x < a). Treatseparately the three cases E < V0, E = V0, and E > V0 (note that thewave function in the region of the barrier is different in the three cases).


Problem 2.33

0

1

0 2V0 4V0

T

E

Finite barrier

0

1

0 2V0 4V0

T

E

Finite well


Problem 2.33

0

1

0 2V0 4V0

T

E

Finite barrier

0

1

0 2V0 4V0

T

E

Finite well


Problem 2.33

0

1

0 2V0 4V0

T

E

Finite barrier

0

1

0 2V0 4V0

T

E

Finite well


Similarity to linear algebra

Wave functions are math-ematically identical tovectors in linear algebra.

Wave functions have in-ner products.

Operators act on wavefunctions as linear trans-formations.

|α〉

→ a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN

However, wave functions are not defined on an N-dimensional space buton infinite dimensional spaces of the continuous variable x .






|α〉 → a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN







|α〉 → a =

a1a2...aN

〈α|β〉

= a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN







|α〉 → a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN







|α〉 → a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉

→ Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN







|α〉 → a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN







|α〉 → a =

a1a2...aN

〈α|β〉 = a∗1b1 + a∗2b2 + · · ·+ a∗NbN

T |α〉 → Ta =

t11 · · · t1Nt21 · · · t2N...

...tN1 · · · tNN

a1a2...aN



Vector properties: addition

|α〉+ |β〉 = |γ〉

addition is commutative

addition is associative

the null vector exists

every vector has an inverse

|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉



|α〉+ |β〉 = |γ〉





|α〉+ |β〉 = |β〉+ |α〉

|α〉+ (|β〉+ |γ〉) = (|α〉+ |β〉) + |γ〉

|α〉+ |0〉 = |α〉

|α〉+ | − α〉 = |0〉


Vector properties: scalar multiplication

a|α〉 = |γ〉

scalar multiplication is distributive

scalar multiplication is associative

a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉

(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉

0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉

1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉

| − α〉 = (−1)|α〉 = −|α〉



a|α〉 = |γ〉



a(|α〉+ |β〉) = a|α〉+ a|β〉(a + b)|α〉 = a|α〉+ b|α〉

a(b|α〉) = (ab)|α〉0|α〉 = |0〉1|α〉 = |α〉| − α〉 = (−1)|α〉 = −|α〉


Hilbert space

Therefore, we need to apply certain restrictions when we use the languageof linear algebra in quantum mechanics.

A physically meaningful wave functionmust be normalized.

Our vector space is therefore the set ofall square integrable functions f (x) on aspecified interval, such that

In quantum mechanics we call this aHilbert space and all wave functions ex-ist in this space.

The inner product of two functions is

∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Hilbert space






∫|Ψ|2 dx = 1

∫ b

a|f (x)|2 dx <∞

〈f |g〉 ≡∫ b

af (x)∗g(x) dx


Properties of Hilbert space

The inner product is guaranteed to exist in a Hilbert space if all functionsare square integrable.

Show that the sum of any two functions in Hilbert space is also a functionin the space

h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f

Thus the square integral of h(x) is given by∫|h|2 dx =

∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx

The first two terms are finite since both f (x) and g(x) are defined asbeing part of the Hilbert space. The last two terms are finite because ofthe Schwartz inequality:∣∣∣∣∫ b

af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g)

= |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f

Thus the square integral of h(x) is given by

∫|h|2 dx =

∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx

The first two terms are finite since both f (x) and g(x) are defined asbeing part of the Hilbert space.

The last two terms are finite because ofthe Schwartz inequality:∣∣∣∣∫ b

af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx

The first two terms are finite since both f (x) and g(x) are defined asbeing part of the Hilbert space. The last two terms are finite because ofthe Schwartz inequality:

∣∣∣∣∫ b

af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx





h(x) = f (x) + g(x)

|h|2 = (f + g)∗(f + g) = |f |2 + |g |2 + f ∗g + g∗f


∫|f |2 dx +

∫|g |2 dx +

∫f ∗g dx +

∫g∗f dx


af (x)∗g(x) dx

∣∣∣∣ ≤√∫ b

a|f (x)|2 dx

∫ b

a|g(x)|2 dx


Inner products

〈g |f 〉 =

∫ b

ag∗f dx

=

(∫ b

af ∗g dx

)∗= 〈f |g〉∗

The inner product of f (x) with itself is real, non-negative, and zero onlywhen f (x) = 0.

〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx

f (x) is normalized if the inner product with itself is 1 and two functionsare orthogonal when their inner product is 0. Two functions areorthonormal when

〈fm|fn〉 = δmn

A set of functions is complete if any other function in the Hilbert spacecan be written as

f (x) =∞∑n=1

cnfn(x) , cn = 〈fn|f 〉


Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗

The inner product of f (x) with itself is

real, non-negative, and zero onlywhen f (x) = 0.

〈f |f 〉 =

∫ b

af ∗f dx

=

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗

The inner product of f (x) with itself is

real, non-negative, and zero onlywhen f (x) = 0.

〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx

f (x) is normalized if the inner product with itself is 1 and two functionsare orthogonal when their inner product is 0.

Two functions areorthonormal when

〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1

cnfn(x)

, cn = 〈fn|f 〉


Inner products

〈g |f 〉 =

∫ b

ag∗f dx =

(∫ b

af ∗g dx

)∗= 〈f |g〉∗


〈f |f 〉 =

∫ b

af ∗f dx =

∫ b

a|f |2 dx


〈fm|fn〉 = δmn


f (x) =∞∑n=1



Documents

Today’s Outline - September 24, 2012csrri.iit.edu/~segre/phys405/12F/lecture_10.pdf · 2012. 9. 29. · Today’s Outline - September 24, 2012 Problem 2.27 Problem 2.33 Hilbert