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Quantum Theory I: Final Exam for Phys 321 (Fall 2006), Exams of Physics

The final exam for phys 321: quantum theory i, which was held on december 13, 2006. The exam covers various topics related to quantum mechanics, including physical constants, angular momentum, spherical harmonics, and spin. It consists of five questions and includes instructions, physical constants, and useful formulae.

Typology: Exams

Pre 2010

Uploaded on 08/16/2009

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Phys 321
Fall 2006
Quantum Theory I : Final Exam
13 December 2006
Name: Total: /100
Instructions
There are 5 questions on 11 pages.
Show your reasoning and calculations and always motivate your answers.
Physical constants and useful formulae
Charge of an electron e=1.60 ×1019 C
Planck’s constant h= 6.63 ×1034 Js ~= 1.05 ×1034 Js
Mass of electron me= 9.11 ×1031 kg = 511 ×103eV/c2
Mass of proton mp= 1.673 ×1027 kg = 938.3×106eV/c2
Mass of neutron mn= 1.675 ×1027 kg = 939.6×106eV/c2
Spherical coordinates ˆn= sin θcos φˆx+ sin θsin φˆy+ cos θˆz
Spin 1/2 state |+ˆni= cos (θ/2) |+ˆzi+e sin (θ/2) |−ˆzi
Spin 1/2 state |−ˆni= sin (θ/2) |+ˆzi e cos (θ/2) |−ˆzi
Zsin (ax) sin (bx) dx=sin ((ab)x)
2(ab)sin ((a+b)x)
2(a+b)if a6=b
Zsin (ax) cos (ax) dx=1
2asin2(ax)
Zsin2(ax) dx=x
2sin (2ax)
4a
Zxsin2(ax) dx=x2
4xsin (2ax)
4acos (2ax)
8a2
Zx2sin2(ax) dx=x3
6x2
4asin (2ax)x
4a2cos (2ax) + 1
8a3sin (2ax)
Formulae continued . . .
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Phys 321

Fall 2006

Quantum Theory I : Final Exam

13 December 2006

Name: Total: /

Instructions

  • There are 5 questions on 11 pages.
  • Show your reasoning and calculations and always motivate your answers.

Physical constants and useful formulae

Charge of an electron e = − 1. 60 × 10

− 19 C

Planck’s constant h = 6. 63 × 10

− 34

Js ℏ = 1. 05 × 10

− 34

Js

Mass of electron m e

= 9. 11 × 10

− 31

kg = 511 × 10

3

eV/c

2

Mass of proton m p

= 1. 673 × 10

− 27 kg = 938. 3 × 10

6 eV/c

2

Mass of neutron m n

= 1. 675 × 10

− 27 kg = 939. 6 × 10

6 eV/c

2

Spherical coordinates nˆ = sin θ cos φxˆ + sin θ sin φyˆ + cos θ ˆz

Spin 1/2 state |+nˆ〉 = cos (θ/2) |+ˆz〉 + e

iφ sin (θ/2) |−zˆ〉

Spin 1/2 state |−nˆ〉 = sin (θ/2) |+zˆ〉 − e

iφ cos (θ/2) |−zˆ〉

sin (ax) sin (bx) dx =

sin ((a − b)x)

2(a − b)

sin ((a + b)x)

2(a + b)

if a 6 = b

sin (ax) cos (ax) dx =

2 a

sin

2 (ax)

sin

2

(ax) dx =

x

sin (2ax)

4 a

x sin

2

(ax) dx =

x

2

x sin (2ax)

4 a

cos (2ax)

8 a

2

x

2

sin

2

(ax) dx =

x

3

x

2

4 a

sin (2ax) −

x

4 a

2

cos (2ax) +

8 a

3

sin (2ax)

Formulae continued...

Angular Momentum

L

x

= iℏ

sin φ

∂θ

cos θ cos φ

sin θ

∂φ

Ly = iℏ

− cos φ

∂θ

cos θ sin φ

sin θ

∂φ

L

z

= −iℏ

∂φ

L

|l, m〉 = ℏ

l(l + 1) − m(m + 1) |l, m + 1〉

L

|l, m〉 = ℏ

l(l + 1) − m(m − 1) |l, m − 1 〉

Spherical Harmonics Y 0 , 0

(θ, φ) =

4 π

Y

1 , 0

(θ, φ) =

4 π

cos θ

Y

1 ,± 1

(θ, φ) = ∓

8 π

e

±iφ sin θ

sin (ax) sin (bx) dx =

sin ((a − b)x)

2(a − b)

sin ((a + b)x)

2(a + b)

if a 6 = b

sin (ax) cos (ax) dx =

2 a

sin

2 (ax)

sin

2 (ax) dx =

x

sin (2ax)

4 a

x sin

2

(ax) dx =

x

2

x sin (2ax)

4 a

cos (2ax)

8 a

2

x

2

sin

2

(ax) dx =

x

3

x

2

4 a

sin (2ax) −

x

4 a

2

cos (2ax) +

8 a

3

sin (2ax)

−∞

e

−αx

2 +βx dx =

π

α

e

β

2 / 4 α

−∞

xe

−αx

2 +βx dx =

β

π

2 α

3 / 2

e

β

2 / 4 α

−∞

x

2 e

−αx

2 +βx dx =

2

  • 2α)

π

4 α

5 / 2

e

β

2 / 4 α

−∞

x

3 e

−αx

2 +βx dx =

β(β

2

  • 6α)

π

8 α

7 / 2

e

β

2 / 4 α

c) Although |Ψ i

〉 is unknown, the fact that the SG ˆn measurement gave S n

= +ℏ/2 does

restrict the possible initial states. Determine one state in which the particle could not

have been prior to the measurement and express this in terms of the {|+zˆ〉 , |−zˆ〉} basis.

/XX

Question 2

A spin-1/2 particle of charge q and g-factor g is placed in a magnetic field B = B ˆz for time t.

The resulting Hamiltonian is

H =

ℏω

σˆ z

where ω = −gqB/ 2 m and in the basis {|+zˆ〉 , |−zˆ〉} ,

a) Find the matrix representation, in the basis {|+zˆ〉 , |−zˆ〉} , of the evolution operator,

U (t).

Question 2 continued...

Question 3

A particle of mass m is in an infinite square well potential

V (x) =

0 0 6 x 6 L

∞ otherwise,

for which the energy eigenstates are

ψ n

(x) =

L

sin

nπx

L

0 6 x 6 L

0 otherwise.

corresponding to energy eigenvalue E n

2 n

2 π

2

2 mL

2

for n = 1, 2 ,.... The particle is in the state

ψ(x) =

L

sin

2 πx

L

0 6 x 6

L

0 otherwise.

Note that the wavefunction is only non-zero in the region 0 6 x 6

L

The energy of the particle is measured. Determine the probability that the outcome of this

measurement is E 2

/XX

Question 4

Consider an ensemble of charged harmonic oscillators, each of mass m and frequency ω.

Suppose that the oscillators are initially each in the state

|Ψ(t = 0)〉 =

where |n〉 is the n

th energy eigenstate.

a) Prove that at a later time t, the state of each oscillator is

|Ψ(t)〉 = e

−iωt/ 2

| 0 〉 + e

−i 2 ωt

Question 4 continued...

Question 5

An ensemble of rigid rotators, such as the CO molecule, are each prepared so that each

ensemble member is in the state

|ψ〉 =

where the notation denotes standard angular momentum states |l, m〉. The ensemble is sub-

jected to measurements of Lx.

a) Determine the expectation value, 〈L x

Question 5 continued...

b) Determine the uncertainty, ∆Lx.

c) Suppose that one more molecule in the state |Ψ〉 is subjected to a measurement of L x

What can you predict regarding the outcome of the measurement?

/XX