Today’s outline - September 23, 2015phys.iit.edu/~segre/phys405/15F/lecture_09.pdf · Today’s outline - September 23, 2015 Problem 2.20 Problem 2.38 Problem 2.49 Determinate states

Today’s outline - September 23, 2015

• Problem 2.20

• Problem 2.38

• Problem 2.49

• Determinate states

• Discrete and continuous spectra

Exam #1: Monday, October 05, 2015Room 111 Stuart BuildingCovers Chapters 1 & 2, closed book, calculators provided

Reading Assignment: Chapter 3.3-3.5

Homework Assignment #04:Chapter 2: 27, 29, 33, 34, 40, 52due Monday, September 28, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 September 23, 2015 1 / 19


• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49








• Problem 2.20

• Problem 2.38

• Problem 2.49







Problem 2.20 – Proof of Plancherel’s theorem

(a) Dirchlet’s theorem says that “any” function f (x) on the interval[−a,+a] can be expanded as a Fourier series. Show the equivalence:

f (x) =∞∑n=0

[an sin(nπx/a) + bn cos(nπx/a)] =∞∑

n=−∞cne

inπx/a

(b) Show that

cn =1

2a

∫ +a

−af (x)e−inπx/a

(c) Using k = (nπ/a) and F (k) =√

2/πacn show that:

f (x) =1√2π

∞∑n=−∞

F (k)e ikx∆k F (k) =1√2π

∫ +a

−af (x)e−ikxdx

where ∆k is the increment in k from one n to the next

(d) Take the limit a→∞ to obtain Plancherel’s theorem


Problem 2.20 – solution part (a)

Start with the initial form for f (x) and expand the sine and cosine interms of the complex exponentials

f (x) =∞∑n=0

[an sin(nπx/a) + bn cos(nπx/a)]

=

b0 +∞∑n=1

an2i

(e inπx/a − e−inπx/a

)+∞∑n=1

bn2

(e inπx/a + e−inπx/a

)

= b0

+∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1

12(−ian + bn)e inπx/a +

−∞∑n=−1

12(ia−n + b−n)e inπx/a

= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


=

b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0

+∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


=

b0 +

∞∑n=1

an2i


)

+∞∑n=1

bn2


)= b0

+∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)

= b0

+∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0

+∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a

+∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0

+∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0 +∞∑n=1

cneinπx/a

+−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0 +∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a

=∞∑

n=−∞cne

inπx/a




f (x) =∞∑n=0


= b0 +∞∑n=1

an2i


)+∞∑n=1

bn2


)= b0 +

∞∑n=1

(an2i

+bn2

)e inπx/a +

∞∑n=1

(−an

2i+

bn2

)e−inπx/a

= b0 +∞∑n=1


−∞∑n=−1


= c0 +∞∑n=1

cneinπx/a +

−∞∑n=−1

cneinπx/a =

∞∑n=−∞

cneinπx/a


Problem 2.20 – solution part (b)

Start with the integral

and expand the function f (x) according to part (a)

∫ a

−af (x)e−imπx/a dx

=∞∑

n=−∞cn

∫ a

−ae inπx/ae−imπx/a dx

for n 6= m, the integral becomes∫ a

−ae i(n−m)πx/a dx

=e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0

for n = m, the integral becomes∫ a

−ae i(m−m)πx/a dx

=

∫ a

−adx = 2a

∫ a

−af (x)e−imπx/a dx = 2acm

−→ cm =1

2a

∫ a




Start with the integral and expand the function f (x) according to part (a)∫ a

−af (x)e−imπx/a dx =

∞∑n=−∞

cn

∫ a




=e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




=e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a



−ae i(n−m)πx/a dx =

e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=e i(n−m)π − e−i(n−m)π

i(n −m)π/a

= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a

= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



=

∫ a

−adx = 2a∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0


−ae i(m−m)πx/a dx =

∫ a

−adx

= 2a∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



∫ a

−adx = 2a

∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



∫ a

−adx = 2a∫ a


−→ cm =1

2a

∫ a






∞∑n=−∞

cn

∫ a




e i(n−m)πx/a

i(n −m)π/a

∣∣∣∣∣a

−a

=(−1)n−m − (−1)n−m

i(n −m)π/a= 0



∫ a

−adx = 2a∫ a

−af (x)e−imπx/a dx = 2acm −→ cm =

1

2a

∫ a



Problem 2.20 – solution parts (c) & (d)

Start with the definition off (x)

substitute k = (nπ/a) andF (k) =

√2/πacn

if ∆k = π/a

using the result of part (b)

cn =1

2a

∫ a

−af (x)e−inπx/a dx

f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx

As a→∞, k becomes a continuous variable

f (x)→ 1√2π

∫ ∞−∞

F (k)e ikx dk F (k)→ 1√2π

∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞

F (k)e ikx dk

F (k)→ 1√2π

∫ ∞−∞

f (x)e−ikx dx





√2/πacn

if ∆k = π/a


cn =1

2a

∫ a


f (x) =∞∑

n=−∞cne

inπx/a

=∞∑

n=−∞

√π

2

1

aF (k)e ikx

=1√2π

∞∑n=−∞

F (k)e ikx∆k

F (k) =

√2

πa

1

2a

∫ a

−af (x)e−ikx dx

=1√2π

∫ a

−af (x)e−ikx dx


f (x)→ 1√2π

∫ ∞−∞


∫ ∞−∞

f (x)e−ikx dx


Problem 2.38

A particle of mass m is in the ground state of the infinite square well.Suddenly the well expands to twice its original size – the right wall movingfrom a to 2a – leaving the wave function momentarily undisturbed. Theenergy of the particle is now measured.

(a) What is the most probable result? What is the probability of gettingthat result?

(b) What is the next most probable result, and what is its probability?

(c) What is the expectation value of the energy?

0 a

x

−→

0 a 2a

x


Problem 2.38





0 a

x

−→

0 a 2a

x


Problem 2.38





0 a

x

−→

0 a 2a

x


Problem 2.38





0 a

x

−→

0 a 2a

x


Problem 2-38 – solution part (a)

The starting wave func-tion is the ground state ofthe well of width a

the new eigenfuctions ofthe well of width 2a are

When the energy is mea-sured, the results can onlybe one of the new station-ary states such that

Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉

=

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx






Ψ(x , 0) =

√2

asin(πax),Et=0 =

π2~2

2ma2

ψn(x) =

√2

2asin(nπ

2ax),En =

n2π2~2

2m(2a)2

Ψ(x , 0) =∞∑n=1

cnψn

cn = 〈ψn(x)|Ψ(x , 0)〉 =

√2

a

∫ a

0sin(πax)

sin(nπ

2ax)dx

=

√2

2a

∫ a

0

{cos(nπ

2ax − π

ax)− cos

(nπ2a

x +π

ax)}

dx



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx

first solve for the special caseof n = 2

now solve for the generalcase n 6= 2

c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2

=1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }

=

{0, n even

± 4√2

π(n2−4) , n odd



cn =

√2

2a

∫ a

0

{cos[(n

2− 1) πx

a

]− cos

[(n2

+ 1) πx

a

]}dx



c2 =

√2

a

∫ a

0sin2

(πxa

)dx

=

√2

a

a

2=

1√2

cn =1√2a

{sin[(

n2 − 1

)πxa

](n2 − 1

)πa

−sin[(

n2 + 1

)πxa

](n2 + 1

)πa

∣∣∣∣∣a

0

=1√2π

{sin[(

n2 − 1

)π](

n2 − 1

) −sin[(

n2 + 1

)π](

n2 + 1

) }=

{0, n even

± 4√2

π(n2−4) , n odd


Problem 2.38 – solution part (a), (b), (c)

The probability of measuringEn is thus

The most probable energymeasured is for n = 2

the next most probable is forn = 1

Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx

=2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx =

2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx =

2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2

= Et=0 unchanged!






Pn =

12 , n = 2

32π2(n2−4)2 , n odd

0, otherwise

E2 =π2~2

2ma2, P2 =

1

2

E1 =π2~2

8ma2, P1 =

32

9π2≈ 0.36

〈H〉 =

∫Ψ∗HΨ dx =

2

a

∫ a

0sin(πax)(− ~2

2m

d2

dx2

)sin(πax)dx

=π2~2

2ma2= Et=0 unchanged!


Problem 2.49

(a) Show that

Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt

)]satisfies the time-dependent Schrodinger equation for the harmonicoscillator potential. Where a is any real constant with the dimensions oflength.

(b) Find |Ψ(x , t)|2, and describe the motion of the wave packet.

(c) Compute 〈x〉 and 〈p〉, and check that Ehrenfest’s theorem is satisfied

d〈p〉dt

= 〈−∂V∂x〉



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt

)]The Schrodinger equation for aharmonic oscillator is

− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+

i~m− 2ax(−iω)e−iωt

]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1

2~ω + maxω2e−iωt

]Ψ

∂Ψ

∂x=[(−mω

2~

)(2x − 2ae−iωt)

]Ψ = −mω

~(x − ae−iωt)Ψ

∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~

)2(x − ae−iωt)2

]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt

)]The Schrodinger equation for aharmonic oscillator is − ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ

= −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Ψ(x .t) =(mωπ~

)1/4exp

[−mω

2~

(x2 +

a2

2(1 + e−2iωt) +

i~tm− 2axe−iωt


2m

∂2Ψ

∂x2+

1

2mω2x2Ψ = i~

∂Ψ

∂t

∂Ψ

∂t=(−mω

2~

)[a22

(−2iωe−2iωt

)+


]Ψ

i~∂Ψ

∂t=

[−1

2ma2ω2e−2iωt +

1


]Ψ

∂Ψ

∂x=[(−mω

2~


]Ψ = −mω


∂2Ψ

∂x2= −mω

~Ψ− mω

~(x − ae−iωt)

∂Ψ

∂x

=

[−mω

~+(mω

~


]Ψ



Putting all the pieces together in the Schrodinger equation we have:

− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1

2mω2(x2 − 2axe−iωt + a2e−i2ωt) +

1

2mω2x2

]Ψ

=

[1

2~ω + maxω2e−iωt − 1

2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t




− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1


1

2mω2x2

]Ψ

=

[1


2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t




− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1


1

2mω2x2

]Ψ

=

[1


2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t




− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1


1

2mω2x2

]Ψ

=

[1


2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t




− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1


1

2mω2x2

]Ψ

=

[1


2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t




− ~2

2m

∂2Ψ

∂x2+

1

2mω2x2Ψ

= − ~2

2m

[−mω

~+(mω

~


]Ψ +

1

2mω2x2Ψ

=

[1

2~ω − 1


1

2mω2x2

]Ψ

=

[1


2ma2ω2e−i2ωt

]Ψ

= i~∂Ψ

∂t



|Ψ|2 =

√mω

π~e−mω

2~

[x2+ a2

2(1+e−2iωt)+ i~t

m−2axe−iωt

]·

e−mω

2~

[x2+ a2

2(1+e+2iωt)− i~t

m−2axe+iωt

]

=

√mω

π~e−mω

2~

[2x2+a2+ a2

2(e+2iωt+e−2iωt)−2ax(e+iωt+e−iωt)

]

=

√mω

π~e−

mω2~ [2x2+a2+a2 cos(2ωt)−4ax cos(ωt)]

=

√mω

π~e−

mω2~ [2x2+2a2 cos2(ωt)−4ax cos(ωt)]

=

√mω

π~e−

mω~ [x+a cos(ωt)]2



|Ψ|2 =

√mω

π~e−mω

2~

[x2+ a2

2(1+e−2iωt)+ i~t

m−2axe−iωt

]·

e−mω

2~

[x2+ a2

2(1+e+2iωt)− i~t

m−2axe+iωt

]

=

√mω

π~e−mω

2~

[2x2+a2+ a2


]

=

√mω

π~e−


=

√mω

π~e−


=

√mω

π~e−




|Ψ|2 =

√mω

π~e−mω

2~

[x2+ a2

2(1+e−2iωt)+ i~t

m−2axe−iωt

]·

e−mω

2~

[x2+ a2

2(1+e+2iωt)− i~t

m−2axe+iωt

]

=

√mω

π~e−mω

2~

[2x2+a2+ a2


]

=

√mω

π~e−


=

√mω

π~e−


=

√mω

π~e−




|Ψ|2 =

√mω

π~e−mω

2~

[x2+ a2

2(1+e−2iωt)+ i~t

m−2axe−iωt

]·

e−mω

2~

[x2+ a2

2(1+e+2iωt)− i~t

m−2axe+iωt

]

=

√mω

π~e−mω

2~

[2x2+a2+ a2


]

=

√mω

π~e−


=

√mω

π~e−


=

√mω

π~e−




|Ψ|2 =

√mω

π~e−mω

2~

[x2+ a2

2(1+e−2iωt)+ i~t

m−2axe−iωt

]·

e−mω

2~

[x2+ a2

2(1+e+2iωt)− i~t

m−2axe+iωt

]

=

√mω

π~e−mω

2~

[2x2+a2+ a2


]

=

√mω

π~e−


=

√mω

π~e−


=

√mω

π~e−




|Ψ|2 =

√mω

π~e−


this is a Gaussian wave packetwhose position varies as a functionof time as x = a cos(ωt)

0-5 0 +5

x



|Ψ|2 =

√mω

π~e−



0-5 0 +5

x



|Ψ|2 =

√mω

π~e−



0-5 0 +5

x



|Ψ|2 =

√mω

π~e−



0-5 0 +5

x


Problem 2.49 – solution part (c)

Start with the expectation value of x

this can be made into a Gaussian centered on the origin by substitutingy = x + a cos(ωt)

〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞

xe−mω~ [x+a cos(ωt)]2dx

=

√mω

π~

∫ ∞−∞

[y − a cos(ωt)] e−mωy2

~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that

〈p〉 = md〈x〉dt

〈p〉 = −maω sin(ωt)

d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx

=

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy

= a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)





〈x〉 =

∫x |Ψ|2 dx =

√mω

π~

∫ ∞−∞


=

√mω

π~

∫ ∞−∞


~ dy

= 0−√

mω

π~a cos(ωt)

∫ ∞−∞

e−mωy2

~ dy = a cos(ωt)

recalling that



d〈p〉dt

= −maω2 cos(ωt)



〈x〉 = a cos(ωt) 〈p〉 = −maω sin(ωt)d〈p〉dt

= −maω2 cos(ωt)

The potential for the harmonic os-cillator is

its spatial derivative is

thus, we can write the expectationvalue

. . . confirming Ehrenfest’s Theorem

V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt




= −maω2 cos(ωt)





V =1

2mω2x2

dV

dx= mω2x

⟨−dV

dx

⟩= −mω2〈x〉

= −mω2a cos(ωt)

=d〈p〉dt


Problem 3.5

The hermitian conjugate (or adjoint) of an op-erator Q is the operator Q† such that

〈f |Qg〉 = 〈Q†f |g〉Q = Q†

Find the hermitian conjugates of x , i , d/dx , a+, and QR.

〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−

〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉 (QR)† = R†Q†


Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉

Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx

=

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx

= 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx

=

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx

= 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx

=��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx = f ∗g

∣∣∣∞−∞−∫ (

df ∗

dx

)g dx

= −〈 ddx

f |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx

= −〈 ddx

f |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(−ip + mωx)†

=1√

2~mω(i

p + mωx

) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(i

p + mωx

)

= a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip

+ mωx

)

= a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx)

= a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx) = a−



Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx) = a−

〈f |(QR)g〉

= 〈Q†f |Rg〉 = 〈R†Q†f |g〉 (QR)† = R†Q†


Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx) = a−

〈f |(QR)g〉 = 〈Q†f |Rg〉

= 〈R†Q†f |g〉 (QR)† = R†Q†


Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx) = a−

〈f |(QR)g〉 = 〈Q†f |Rg〉 = 〈R†Q†f |g〉

(QR)† = R†Q†


Problem 3.5


〈f |Qg〉 = 〈Q†f |g〉Q = Q†


〈f |xg〉 =

∫f ∗(xg) dx =

∫(xf )∗g dx = 〈xf |g〉

〈f |ig〉 =

∫f ∗(ig) dx =

∫(−if )∗g dx = 〈−if |g〉

〈f | ddx

g〉 =

∫f ∗

dg

dxdx =

��

f ∗g∣∣∣∞−∞−∫ (

df ∗

dx

)g dx = −〈 d

dxf |g〉

a†+ =1√

2~mω(− ip + mωx)† =

1√2~mω

(ip + mωx) = a−



Determinate states

Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system.

If there is adeterminate state for the system, we will always get the same answer, q.

In this case, the standard deviation of Q must be exactly zero.

Since Q is hermitian, then Q − q must be as well.

σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0

The only function whose square in-tegral vanishes is zero

(Q − q)Ψ ≡ 0

QΨ = qΨ

This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q and q being its eigenvalue.

Thus, determinate states are eigenfunctions of Q.


Determinate states

Suppose that we have many systems all prepared in the same way and wemeasure the expectation value of Q on each system. If there is adeterminate state for the system, we will always get the same answer, q.



σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉

= 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉

= 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉

= 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉

≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ

This is the eigenvalue equation for the Q operator with Ψ being aneigenfunction of Q

and q being its eigenvalue.



Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




Determinate states




σ2 = 〈(Q − 〈Q〉)2〉 = 〈Ψ|(Q − q)2Ψ〉 = 〈(Q − q)Ψ|(Q − q)Ψ〉 ≡ 0


(Q − q)Ψ ≡ 0

QΨ = qΨ




All about eigenfunctions...

An eigenfunction multiplied by a constant gives the same eigenvalue:

Q(aΨ) = q(aΨ)

Zero cannot be an eigenfunction

The collection of all eigenvalues of an operator is its spectrum

If two linearly independent eigenfunctions have the same eigenvalue, theyare said to be degenerate.

Determinate states of the total energy are thus, eigenfunctions of theHamiltonian:

Hψ = Eψ

If the spectrum of an operator is discrete (eigenvalues are “quantized”),then the eigenfunctions lie in Hilbert space and are physical states

If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states. But linear combinations of them(wavepackets) might.




Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ






Q(aΨ) = q(aΨ)





Hψ = Eψ


If the spectrum is continuous, then the eigenfunctions are not normalizableand cannot represent physical states.

But linear combinations of them(wavepackets) might.




Q(aΨ) = q(aΨ)





Hψ = Eψ




Documents

Today’s outline - September 23, 2015phys.iit.edu/~segre/phys405/15F/lecture_09.pdf · Today’s outline - September 23, 2015 Problem 2.20 Problem 2.38 Problem 2.49 Determinate states