Download 1.

Classical Mechanics
1.
The Lagrangian of a one-dimensional system is given by
1
L = e2αt ẋ2 − ω02 x2 .
2
Here x = x(t) is the coordinate, ẋ is the velocity, t is time and α is a constant parameter.
(a) [10 pts.] Derive the differential equation of motion for the coordinate x(t) from the above Lagrangian.
(b) [5 pts.] Discuss the physical meaning of the obtained equation of motion.
(c) [10 pts.] Take x(t) = Ae−λt cos[Ωt + ϕ] as a trial solution and determine λ and Ω as a function of
system’s parameters α and ω0 .
Thermodynamics 2.
One mole of a mono-atomic ideal gas is
contained in a sealed, compressible cylinder.
The gas expands along a path that makes a
straight-line in a pressure vs. volume (P vs.
V) plot as shown, from initial values of
volume and pressure (V0, P0) to the final
values (4V0, ½P0).
(a) [4 pts.] Find an expression for ∆T (including the sign), the change in temperature of the gas
through the expansion (from the initial to final state) in terms of V0, P0, and the gas constant R.
(b) [5 pts.] Find an expression for ∆U, the change in the internal energy of the gas
(c) [6 pts.] Use the First Law of Thermodynamics to find an expression for ∆Q, the heat flow
INTO the gas.
(d) [10 pts.] Find an expression for ∆S, the change in the entropy of the gas
Solution:
(a) One mole of an ideal gas has the equation of state: 𝑃𝑉 = 𝑅𝑇, so that
(1𝑃0 )(4𝑉0 )
𝑃0 𝑉0
2𝑃0 𝑉0
𝑃0 𝑉0
, 𝑇𝑓 = 2
=
→ ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 =
𝑇𝑖 =
𝑅
𝑅
𝑅
𝑅
(b) The internal energy of one mole mono-atomic gas is given by 𝑈 = 32𝑅𝑇, so that we have
𝑃0 𝑉0
2𝑃0 𝑉0
3
3
� = 2𝑃0 𝑉0 , 𝑈𝑖 = 2𝑅 �
� = 3𝑃0 𝑉0 →
𝑅
𝑅
3
𝑈𝑖 = 2𝑅 �
3
∆𝑈 = 𝑈𝑓 − 𝑈𝑖 = 2𝑃0 𝑉0
(c) The (integral form) of the 1st Law of Thermodynamics states: ∆𝑈 = ∆𝑊 + ∆𝑄, where ∆𝑊 is the work done ON
the gas, and ∆𝑄 is the heat flowing INTO the gas
Work done ON the gas is given by the integral of 𝑑𝑊 = −𝑃𝑑𝑉
The easy way to do this part is to say that the work done ON the gas is the negative of the area under the expansion
line, which is a trapezoid:
1
1 3
1 1
9
Area = �𝑃𝑓 + 𝑃𝑖 ��𝑉𝑓 − 𝑉𝑖 � = � 𝑃0 + 𝑃0 � (4𝑉0 − 𝑉0 ) = � 𝑃0 � (3𝑉0 ) = 4𝑃0 𝑉0
2
2 2
2 2
9
→
So we have
∆𝑊 = −4𝑃0 𝑉0
3
9
∆𝑄 = ∆𝑈 − ∆𝑊 = 2𝑃0 𝑉0 − �−4𝑃0 𝑉0 � =
(d) Here we must integrate
𝑑𝑆 =
15
𝑃𝑉
4 0 0
𝑑𝑄
𝑇
We will recap the derivation for the differential 𝑑𝑄 for one mole of ideal, mono-atomic gas
3
3
𝑑𝑄 = 𝑑𝑈 − 𝑑𝑊, 𝑑𝑈 = 𝑑 �2𝑅𝑇� = 2𝑅𝑑𝑇, 𝑇 =
3
→ 𝑑𝑈 = 2𝑅
We also have
𝑑𝑊 = −𝑃𝑑𝑉 →
𝑓
𝑑𝑆 =
∆𝑆 = � 𝑑𝑆 =
𝑖
𝑑𝑄
,
𝑇
1
(𝑃𝑑𝑉 + 𝑉𝑑𝑃) =
𝑅
3
𝑃𝑉
1
→ 𝑑𝑇 = (𝑃𝑑𝑉 + 𝑉𝑑𝑃)
𝑅
𝑅
3
2
(𝑃𝑑𝑉 + 𝑉𝑑𝑃)
5
3
𝑑𝑄 = 2 (𝑃𝑑𝑉 + 𝑉𝑑𝑃) + 𝑃𝑑𝑉 = 2𝑃𝑑𝑉 + 2𝑉𝑑𝑃
1
𝑅
=
→
𝑇 𝑃𝑉
5
𝑑𝑆 = 2𝑅
𝑃𝑑𝑉 3 𝑉𝑑𝑃 𝑅 𝑑𝑉
𝑑𝑃
+ 2𝑅
= �5
+3 �
𝑃𝑉
𝑃𝑉
2
𝑉
𝑃
4𝑉0
𝑃0 /2
𝑅
𝑑𝑉
𝑑𝑃
𝑅
𝑅
1
4𝑉
𝑃 /2
�5 �
+ 3�
� = �5[𝑙𝑛𝑉]𝑉0 0 + 3[𝑙𝑛𝑃]𝑃00 � = �5𝑙𝑛4 + 3𝑙𝑛 �
2
𝑉
𝑃
2
2
2
𝑉0
𝑃0
𝑅
𝑅
7
= (5𝑙𝑛22 + 3𝑙𝑛2−1 ) = (10𝑙𝑛2 − 3𝑙𝑛2) = �2𝑙𝑛2� 𝑅
2
2
(a) Alternate solution,
we can integrate
𝑑𝑇 =
1
(𝑃𝑑𝑉 + 𝑉𝑑𝑃)
𝑅
But here the integration path follows the line
𝑃0
7𝑃0
−
𝑉 →
𝑃=
6
6𝑉0
𝑑𝑃 = −
So we have
and
𝑑𝑇 =
∆𝑇 =
𝑃0
𝑑𝑉
6𝑉0
1 7𝑃0
𝑃0
1 𝑃0
𝑃0 7
𝑉
𝑉
𝑃0 7
𝑉
�
−
𝑉� 𝑑𝑉 − 𝑉
𝑑𝑉 =
� −
−
� 𝑑𝑉 =
� −
� 𝑑𝑉
𝑅 6
6𝑉0
𝑅 6 6𝑉0 6𝑉0
𝑅 6 3𝑉0
𝑅 6𝑉0
4𝑉0
𝑉
𝑃0 7
𝑉2
𝑃0 4𝑉0 7
� � −
� 𝑑𝑉 = � 𝑉 −
�
𝑅 𝑉0 6 3𝑉0
𝑅 6
6𝑉0 𝑉
=
0
(4𝑉0 )2
(𝑉0 )2
𝑃0 7
7
�� (4𝑉0 ) −
� − � (𝑉0 ) −
��
𝑅 6
6𝑉0
6𝑉0
6
𝑃0 28 − 16 − 7 + 1
6 𝑃0 𝑉0 𝑃0 𝑉0
= �
� 𝑉0 =
=
𝑅
𝑅
6
6 𝑅
(b) Alternate Solution:
we can integrate
3
3
𝑑𝑈 = 2𝑅𝑑𝑇 = 2𝑅
𝑃0 7
𝑉
𝑃0 7 𝑉
� −
� 𝑑𝑉 =
� − � 𝑑𝑉
𝑅 6 3𝑉0
2 2 𝑉0
4𝑉0
𝑃0 4𝑉0 7 𝑉
𝑃0 7
𝑉2
∆𝑈 = � � − � 𝑑𝑉 = � 𝑉 −
�
2 𝑉0 2 𝑉0
2 2
2𝑉0 𝑉
0
=
(4𝑉0 )2
(𝑉0 )2
𝑃0 7
7
�� (4𝑉0 ) −
� − � (𝑉0 ) −
��
2 2
2𝑉0
2𝑉0
2
𝑃0 28 − 16 − 7 + 1
6
3
= �
� 𝑉0 = 𝑃0 𝑉0 = 2𝑃0 𝑉0
2
2
4
(c) Alternate Solution:
∆𝑊 = −𝑃0 �
4𝑉0
𝑉0
𝑑𝑊 = −𝑃𝑑𝑉 = − �
7𝑃0
𝑃0
7
𝑉
−
𝑉� 𝑑𝑉 = −𝑃0 � −
� 𝑑𝑉
6
6𝑉0
6 6𝑉0
4𝑉0
7
𝑉
7
𝑉2
� −
� 𝑑𝑉 = −𝑃0 � 𝑉 −
�
12𝑉0 𝑉
6 6𝑉0
6
0
= −𝑃0 ��
(4𝑉0 )2
(𝑉0 )2
14
14
(4𝑉0 ) −
� − � (𝑉0 ) −
��
12𝑉0
12𝑉0
12
12
56 − 16 − 14 + 1
27
9
= −𝑃0 �
� 𝑉0 = − 𝑃0 𝑉0 = −4𝑃0 𝑉0
12
12
Alternately, from parts (b) and (c) we have for 𝑑𝑈 and 𝑑𝑊:
3
7
𝑉
7
𝑉
35 2𝑉
𝑑𝑄 = 𝑑𝑈 − 𝑑𝑊 = 𝑃0 � −
� 𝑑𝑉 − �−𝑃0 � −
� 𝑑𝑉� = 𝑃0 � −
� 𝑑𝑉
2
6 3𝑉0
6 6𝑉0
12 3𝑉0
∆𝑄 = 𝑃0 �
4𝑉0
𝑉0
4𝑉0
35 2𝑉
35
𝑉2
−
� 𝑑𝑉 = 𝑃0 � 𝑉 −
�
3𝑉0 𝑉
12 3𝑉0
12
�
0
= 𝑃0 ��
(4𝑉0 )2
(𝑉0 )2
35
35
(4𝑉0 ) −
� − � (𝑉0 ) −
��
3𝑉0
3𝑉0
12
12
140 − 64 − 35 + 4
45
15
= 𝑃0 �
� 𝑉0 =
𝑃 𝑉 = 4 𝑃0 𝑉0
12
12 0 0
(d) Alternate solution
we can integrate through the path as we had done before. From part (d)
35 2𝑉
35 2𝑉
𝑃0 � −
𝑃0 � −
� 𝑑𝑉
� 𝑑𝑉
𝑑𝑄
𝑑𝑄
12 3𝑉0
12 3𝑉0
𝑑𝑆 =
=𝑅
=𝑅
=𝑅
7
1
𝑃𝑉
𝑇
𝑃𝑉
𝑃0 � −
𝑉� 𝑉
6 6𝑉0
1
𝑉
𝑉
�35 − 8 � 𝑑𝑉
𝑅 �35 − 8 𝑉0 � 𝑑𝑉
12
𝑉0
=𝑅
=
1
𝑉
2 �7 − 𝑉 � 𝑉
�7 − � 𝑉
𝑉0
6
𝑉0
35 2𝑉
𝑑𝑄 = 𝑃0 � −
� 𝑑𝑉 →
12 3𝑉0
And so
𝑉
𝑅 4𝑉0 �35 − 8 𝑉0 � 𝑑𝑉
∆𝑆 = �
𝑉
2 𝑉0
�7 − � 𝑉
𝑉0
Making a change in variable:
𝑥=
𝑉
𝑑𝑉
, 𝑑𝑥 =
→
𝑉0
𝑉0
∆𝑆 =
For the integral: we rewrite the rational function integrand as
𝑅 4 (35 − 8𝑥)𝑑𝑥
�
(7 − 𝑥)𝑥
2 1
𝐴
𝐵 𝐴𝑥 + 7𝐵 − 𝐵𝑥 7𝐵 − (𝐵 − 𝐴)𝑥
35 − 8𝑥
=
+ =
=
(7 − 𝑥)𝑥
(7 − 𝑥)𝑥
(7 − 𝑥)𝑥
7−𝑥 𝑥
�
4 (35
1
→ 7𝐵 = 35, 𝐵 − 𝐴 = 8 → 𝐵 = 5, 𝐴 = 𝐵 − 8 = −3
4
4
− 8𝑥)𝑑𝑥
𝑑𝑥
𝑑𝑥
−3
= −3 �
+5�
= 3[𝑙𝑛 (𝑥 − 7)]14 + 5[𝑙𝑛𝑥]14 = 3𝑙𝑛 � � + 5𝑙𝑛4 = −3𝑙𝑛2 + 10𝑙𝑛2
(7 − 𝑥)𝑥
−6
1 7−𝑥
1 𝑥
= 7𝑙𝑛2
And we get the same answer of ∆𝑆 = �72𝑙𝑛2�𝑅
General and Modern Physics
3.
Figure 1
Consider a two inertial frames of
references: S, at rest with respect to the
lab, and S’ moving along the +x
direction relative to S at constant speed
v. The two origins coincide at t=t’=0.
An electromagnetic wave is
propagating in the +x direction as
illustrated in Figure 1.
Given an event that occurs in frame S at the time and location (𝑡, 𝑥, 𝑦, 𝑧) as seen in S, write down
the time and spatial coordinates 𝑡′, 𝑥′, 𝑦′ and 𝑧′ of this event, a seen in frame S’, in terms of 𝑡, 𝑥, 𝑦
and 𝑧, first,
(a) [5 pts.] according to the classical mechanics—these are known as the Galilean
Transformations, and then,
(b) [5 pts.] according to the Special Theory of Relativity—these are known as the Lorentz
transformations: remember that in the limit 𝑣/𝑐 ≪ 1, the Lorentz transformations should reduce
to Galilean Transformations.
(c) [15 pts.] Assuming Lorentz transformation: Find an expression for the ratio 𝜆′ /𝜆 of the
wavelength measured in the S' frame to that in the S (lab) frame. Write your answer in terms of
the relative speed 𝑣 between the frames and speed of light 𝑐. Follow the recipe given below.
Consider the following two events in figure 2:
Figure 2
Event 1: Wave front A hits a telescope on
Earth at time and location (𝑡1 , 𝑥1 ), as
measured in frame S (lab frame).
Event 2: Wave front B (one wavelength/period following wave front A) hits the
telescope at (𝑡2 , 𝑥2 ) = (𝑡1 + ∆𝑡, 𝑥2 + ∆𝑥),
where ∆𝑡 and ∆𝑥 are the time and space
intervals between the two events as seen in
frame S .
Now use the Lorentz Transformations to find the corresponding time and space intervals
between these two events in frame S’.
Solution:
(a) Galilean Transformation:
𝑡 ′ = 𝑡,
𝑥 ′ = 𝑥 − 𝑣𝑡,
(b) Lorentz Transformation
𝑡′ =
1
2
�1−𝑣2
𝑐
�𝑡 −
𝑣
𝑐2
𝑦 ′ = 𝑦,
𝑥′ =
𝑥�,
1
𝑧′ = 𝑧
2
�1−𝑣2
𝑐
𝑦 ′ = 𝑦,
(𝑥 − 𝑣𝑡),
𝑧′ = 𝑧
(c) The relevant transformation for the time and space intervals have the same form as those for 𝑡 ′ and 𝑥 ′ :
∆𝑡 ′ =
1
2
�1−𝑣2
𝑐
�∆𝑡 −
𝑣
𝑐2
∆𝑥�,
∆𝑥 ′ =
1
2
�1−𝑣2
𝑐
(∆𝑥 − 𝑣∆𝑡)
Note, however, that since both event occurs at the front of the
telescope, which is the same location in frame S’, we have
∆𝑥 ′ = 0 →
∆𝑥 = 𝑣∆𝑡
And applying this result to ∆𝑡 ′ , which is by definition the period
of the wave as seen in the S’ frame, 𝑇′, then
𝑇 ′ = ∆𝑡 ′ =
1
2
�1−𝑣2
𝑐
�1 −
1
2
�1−𝑣2
𝑐
𝑣2
𝑐2
�∆𝑡 −
𝑣
𝑐2
� ∆𝑡 = �1 −
∆𝑥� =
𝑣2
𝑐2
∆𝑡
1
2
�1−𝑣2
𝑐
�∆𝑡 −
𝑣
𝑐2
… (1)
𝑣∆𝑡� =
One might be tempted to identify ∆𝑡as the period of the wave in
frame S. This would not be correct, because in time ∆𝑡, the
wave front B traveled (at speed 𝑐) further than one wavelength, by an amount equal to the amount by which the
telescope had moved in the same time interval. So from the diagram, we can write:
𝑐∆𝑡 = 𝜆 + 𝑣∆𝑡
→
𝑣
𝜆 = (𝑐 − 𝑣)∆𝑡 = �1 − � 𝑐∆𝑡 →
𝑐
∆𝑡 =
1
𝑣
𝑐
�1− �
𝜆�
𝑐
Now in frame S’, we have the relation: 𝑐𝑇 ′ = 𝜆′ and so Equation (1) above can be re-written:
𝑇 ′ = �1 −
→
𝑣2
∆𝑡 →
𝑐2
2
𝜆′� = �1 − 𝑣
𝑐
𝑐2
1
𝜆
𝑣 �𝑐
�1 − �
𝑐
2
𝑣
𝑣
𝑣
𝑣
𝑣
�1 − 𝑣2
��1 − � �1 + �
��1 − � �1 + �
��1 + �
𝑐
𝑐
𝑐
𝑐
𝑐
𝑐
𝜆′ =
𝜆=
𝜆=
𝜆
𝑣 𝜆=
𝑣
𝑣
𝑣
𝑣
�1 − �
�1 − �
��1
��1
−
−
�
�1
−
�
�
𝑐
𝑐
𝑐
𝑐
𝑐
→
𝑣
�1 + �
𝜆′
𝑐
=�
𝑣
𝜆
�1 − �
𝑐
Electromagnetism
4.
We consider a zero resistance coaxial cable of inner radius a and outer radius b connected to a constant voltage
source V . Both the inner and outer conductors are hollow cylindrical shells of negligible thickness. The variable
r is used to represent distances from the cable axis z. The coaxial cable is very long so you can neglect edge
effects.
r
ϕ
z
b
r
a
z
V
R
~
(a) [6 pts.] Calculate and specify the magnitude and direction of the electric field E(r)
in the three regions
r < a; a < r < b and r > b.
(b) [6 pts.] A resistance R is now connected between the two conductors at the other end of the line with
respect to the voltage source as on the Figure. The electric field remains unchanged but a magnetic field
appears as a result of the current flow along both conductors. Calculate and specify the magnitude and
~
direction of the magnetic field B(r)
in the three regions r < a; a < r < b and r > b.
~
(c) [5 pts.] Calculate and specify the magnitude and direction of the Poynting vector S(r)
in the three
regions r < a; a < r < b and r > b.
(d) [3 pts.] How is the direction of the Poynting vector changed when the polarity of the voltage source is
reversed?
(e) [5 pts.] Using your earlier results, calculate the flux of the Poynting vector through a surface perpendicular to the z axis. What is the physical meaning of this quantity?
Solution
We consider a zero resistance coaxial cable of inner radius a and outer radius b connected to a constant voltage
source V . Both the inner and outer conductors are hollow cylindrical shells of negligible thickness. The variable
r is used to represent distances from the cable axis z. The coaxial cable is very long so you can neglect edge
effects.
r
ϕ
z
b
r
a
z
V
R
~
(a) [6 pts.] Calculate and specify the magnitude and direction of the electric field E(r)
in the three regions
r < a; a < r < b and r > b.
Because of the symmetry of the problem, as long as the edge effect are neglected, the electric field must
be oriented along r̂. The potential difference must be associated with a linear charge density +λ on the
central conductor and −λ on the outer conductor. We can choose a cylindrical Gauss surface of radius
r and length L. The flux of the electric field through this Gauss surface is the flux of the electric field
through the curved side of the cylinder.
• We see that E(r) = 0 for both r < a and r > b, as the enclosed charge is zero.
~
• For a < r < b, from Gauss’s law, we obtain E(r)
=
~ = V r̂.
λ = 2π0 V so that in the end, E
ln(b/a)
λ
r̂.
2π0 r
Using this and
Rb
a
~ · r̂dr = V , we see that
E
r ln(b/a)
(b) [6 pts.] A resistance R is now connected between the two conductors at the other end of the line with
respect to the voltage source as on the Figure. The electric field remains unchanged but a magnetic field
appears as a result of the current flow along both conductors. Calculate and specify the magnitude and
~
direction of the magnetic field B(r)
in the three regions r < a; a < r < b and r > b.
A current I = V /R flows along ẑ in the central conductor and along −ẑ in the outer conductor. Because
of the symmetry of the problem, the magnetic field must be oriented along ϕ̂ We choose circular Amper
loops of axis z and radius r.
• We see that B(r) = 0 for both r < a and r > b as the enclosed current is zero in both cases.
~
• For a < r < b, from Amper’s law and applying the right hand rule, B(r)
=
µ0 V
ϕ̂
2πr R
~
(c) [5 pts.] Calculate and specify the magnitude and direction of the Poynting vector S(r)
in the three
regions r < a; a < r < b and r > b.
• From our results above, S(r) = 0 for both r < a and r > b.
~=
• For a < r < b, we can use the fact that S
1 ~
E
µ0
~ Since r̂ × ϕ̂ = ẑ, we get S(r)
~
× B.
=
V2
ẑ
2πr2 R ln(b/a)
(d) [3 pts.] How is the direction of the Poynting vector changed when the polarity of the voltage source is
reversed?
When the polarity of the voltage source is reversed (that is when the sign of V is changed), both the
~ and B
~ are reversed and consequently, the Poynting vector is unaffected.
direction of E
(e) [5 pts.] Using your earlier results, calculate the flux of the Poynting vector through a surface perpendicular to the z axis. What is the physical meaning of this quantity?
The flux P of the Poynting vector through a surface perpendicular to z is P =
Rb
a
2
2πrdr 2πr2 RVln(b/a) =
This is the power dissipated in the resistor. It is the rate at which energy is carried along the line.
V2
.
R
5.
Quantum Mechanics
We consider a spin 1/2 particle in a static magnetic field B0 in the ẑ direction. The hamiltonian is then
H0 = −γSz B0 . You will use the basis {|+iz ; |−iz } with Sz |±iz = ± h̄2 |±iz ( See foot note [1]) .
(a) [5 pts.] Write the eigenvalues E0+ and E0− as well as the eigenstates |+i0 and |−i0 of H0 .
(b) [6 pts.] At time t = 0, the system is prepared in the state |ψi = α|+iz + β|−iz with α and β, two
complex numbers. Write the state of the system |ψ(t)i at an arbitrary time t > 0.
(c) [7 pts.] An additional small static magnetic field B1 B0 is applied in the x̂ direction. The hamiltonian
becomes H = H0 + H1 with H1 = −γSx B1 . Write de general first order perturbation relation providing
and estimate of the change in energy resulting from the perturbation and apply it to the present problem
regarding H1 as a perturbation. Explain your result qualitatively.
(d) [7 pts.] Without assuming B1 B0 , find the exact energy levels E− and E+ and the Hamiltonian H
eigenstates |−iH and |+iH . You do not need to normalize the eigenstates.
[1] If you are not familiar with Dirac notation, just consider |+iz =
this basis, Sx ≡
h̄
2
01
10
, Sy ≡
h̄
2
0 −i
i 0
and Sz ≡
h̄
2
1 0
0 −1
.
1
0
and |−iz =
0
1
. Also it is reminded that in
Problem n – Quantum Mechanics with solution
We consider a spin 1/2 particle in a static magnetic field B0 in the ẑ direction. The hamiltonian is then
H0 = −γSz B0 . You will use the basis {|+iz ; |−iz } with Sz |±iz = ± h̄2 |±iz ( See foot note [1]) .
(a) [5 pts.] Write the eigenvalues E0+ and E0− as well as the eigenstates |+i0 and |−i0 of H0 .
From the given information, it is clear that H0 |+iz = −γB0 Sz |+iz = − γB20 h̄ |+iz so |−i0 = |+iz and
E0− = − γB20 h̄ . In the same way, |+i0 = |−iz and E0+ = + γB20 h̄
(b) [6 pts.] At time t = 0, the system is prepared in the state |ψi = α|+iz + β|−iz with α and β, two
complex numbers. Write the state of the system |ψ(t)i at an arbitrary time t > 0.
+
The Schrödingier equation is linear and ih̄ dtd |±i0 = H0 |±i0 = E0± |±i0 so |ψ(t)i = αe−iE0 t/h̄ |+iz +
−
βe−iE0 t/h̄ |−iz
(c) [7 pts.] An additional small static magnetic field B1 B0 is applied in the x̂ direction. The hamiltonian
becomes H = H0 + H1 with H1 = −γSx B1 . Write de general first order perturbation relation providing
and estimate of the change in energy resulting from the perturbation and apply it to the present problem
regarding H1 as a perturbation. Explain your result qualitatively.
To the first order perturbation theory, the eigenvalues of the unperturbed Hamiltonian will shift by
δE± =0h±|H1 |±i0 . Since H1 ∝ Sx , it is clear that we get δE± = 0. So to the level of the first order
perturbation theory, the eigenvalues of the Hamiltonian are not changed by the perturbation. This is
q
~ = B02 + B12 =
not surprising as the exact eigenvalue of the Hamiltonian H will be proportional to |B|
B0
q
2
~ ≈ B0 1 +
1 + (B1 /B0 ) and for B1 B0 , we have |B|
1
2
B1
B0
2 in which there is no linear term in
B1
.
B0
(d) [7 pts.] Without assuming B1 B0 , find the exact energy levels E− and E+ and the Hamiltonian H
eigenstates |−iH and |+iH . You do not need to normalize the eigenstates.

 B0
H = −γ (B0 Sz + B1 Sx ) = −γ h̄2 



B1 
 B0 B1 
 The eigenvalues B+ and B− of B = 
 are
B1 −B0
B1 −B0
q
solutions of −(B0 − B± )(B0 + B± ) − B12 = 0 or B± = ± B02 + B12 . So the eigen values of H are
E± =
±γ h̄2
q
B02 + B12 .
q
Eigenvectors of H are eigenvectors of B. With B± = ± B02 + B12 we need to solve

 B0





B1   1 
1
   = B±   which gives B0 + βB1 = B± so β =
B1 −B0
β
β
B± −B0
B1
in such a way that:
√
|−iH ∝ |+iz −
|+iH ∝ |+iz +
B02 +B12 +B0
|−iz
√ 2 B1 2
B0 +B1 −B0
|−iz
B1
[1] If you are not familiar with Dirac notation, just consider |+iz =
this basis, Sx ≡
h̄
2
01
10
, Sy ≡
h̄
2
0 −i
i 0
and Sz ≡
h̄
2
1 0
0 −1
.
1
0
and |−iz =
0
1
. Also it is reminded that in
1
Astrophysics:
6.
(a) [8 points] Assume that the Sun is a uniform-density sphere of mass M and radius R.
Calculate the total gravitational binding energy of the Sun in terms M , R, and Newton’s
constant G. (Hint: consider the total energy associated with assembling the Sun by successive spherical shells brought in from infinity.)
(b) [7 points] The Sun has an approximate observed luminosity of L = 4 × 1026 watts.
Calculate to within an order of magnitude the age in years that the Sun would have if
gravitational contraction were the source of this luminosity, and if the luminosity were constant over the period of contraction. You may take M = 2 × 1030 kg, R = 7 × 108 m,
G = 7 × 10−11 m3 kg−1 s−2 , and 1 year = 3 × 107 s.
(c) [10 points] The process of nuclear fusion of hydrogen into helium converts 0.7% of the
original rest mass of the hydrogen into energy. Calculate to within an order of magnitude
the lifetime in years of the Sun assuming it starts as pure hydrogen and ends as pure helium,
with a constant nuclear burning rate yielding the current observed luminosity.
2
Astrophysics (solution):
(a) The mass density of the Sun in this approximation is
ρ=
3M
.
4πR3
The mass enclosed within a radius r (for r < R) is
r
M (r) = M
R
3
.
The differential mass of a spherical shell of radius r is
dM = ρ dV =
3M r2 dr
.
R3
The differential binding energy associated with a shell relative to the mass interior to it is:
dU =
GM (r) dM
3GM 2 r4 dr
=
r
R6
Integrating this from the center of the Sun out to its radius R, we get the total binding
energy as:
Z
Z R
3GM 2 r4 dr
3 GM 2
=
.
U = dU =
R6
5 R
0
(b) The age of the Sun under these assumptions (which is called the “Kelvin-Helmholtz”
time) would be
TKH =
U
3GM 2
3(7 × 10−11 )(2 × 1030 )2
=
'
s = 6 × 1014 s ' 2 × 107 years.
L
5RL
5(7 × 108 )(4 × 1026 )
(c) The lifetime of the Sun under these assumption would be
TNuc =
0.007M c2
(7 × 10−3 )(2 × 1030 )(3 × 108 )2
'
' 3 × 1018 s ' 1011 years.
26
L
4 × 10
1
Classical Mechanics
7.
M
g
R m m
µ
v
(wheels not to scale)
A commercial jetliner of total mass M is landing with speed v relative to the ground. It
has N main landing-gear wheels, which you may model as uniform-density disks of mass m
and radius R. The coefficient of friction between the rubber of the wheels and the runway
is µ, which you may take to be the same for both the sliding and the static case. Before
landing, the wheels are not rotating. Make the approximation that landing is instantaneous:
i.e., that the plane goes from having no weight on its wheels to all weight on its wheels in
an instant.
(a) [10 points] Find a symbolic expression for the time t that it takes the wheels to spin
up to rolling without slipping. You can assume that the weight of the plane is distributed
equally across all wheels after landing, and that the speed of the plane does not change
during spin-up.
(b) [5 points] The Boeing 747-400 has a maximum landing weight of 2.96 × 105 kg. It has
16 wheels in its main landing gear, each with a diameter of 62 cm and a mass of 184 kg. It
lands at a speed of 95 m s−1 , with a coefficient of friction between the wheels and the runway
of µ = 0.50. Use these values to find the spin-up time of the wheels in seconds, to two
significant digits.
(c) [5 points] What fraction f of the 747’s translational kinetic energy, to two significant
digits, is transferred to the rotational kinetic energy of the wheels once they have reached
the state of rolling without slipping?
(d) [5 points] Assuming the wheel brakes on the 747 can take maximal advantage of friction
between the rubber and the runway, what is the shortest length of runway d (to two significant
digits) that would be needed to bring the plane to a stop in the absence of any other arresting
forces?
2
Classical Mechanics (solution):
(a) The moment of inertia of each
wheel about its axis is I = 21 mR2 , which can either
R 2
be recalled or derived from I = r dm. The normal force on each wheel after landing is
M g/N , so the frictional force on each wheel is µM g/N and the torque on each wheel is
τ = µRM g/N . The angular acceleration can thus be found from
τ =I
dω
µRM g
1
dω
N mR
N mR
⇒
= mR2
⇒ dt =
dω ⇒ t =
ωf .
dt
N
2
dt
2µM g
2µM g
The final angular speed of the wheels upon rolling without slipping is ωf = v/R. So, the
time for the wheels to spin up is
N mv
t=
.
2µM g
(It is of course not realistic to assume an instantaneous landing, because the lift from the
wings decreases gradually during landing).
(b) Putting in numerical values:
t=
(16)(184)(95)
s = 9.6 × 10−2 s.
(2)(0.50)(2.96 × 105 )(9.8)
(c) The kinetic energy of the wheels after spin-up is
Kwh
1
1 2
1
=N
Iω = N
mR2
2
2
2
v
R
2
1
= (16)mv 2 = 4mv 2 ,
4
while the original energy of the plane is Ktot = 12 M v 2 . So, the fraction of the initial kinetic
energy that ends up in the wheels is
f=
m
Kwh
(184)
4mv 2
=8
=8
= 5.0 × 10−3
= 1
5
2
Ktot
M
(2.96 × 10 )
Mv
2
(i.e., 0.50 %.)
(d) The maximum frictional arresting force during the braking phase (static friction) will be
F = µM g. The minimum runway distance d is found from setting
1 v2
1 (95)2
1
=
m = 9.2 × 102 m.
F d = µM gd = Ktot = M v 2 ⇒ d =
2
2 µg
2 (0.50)(9.8)
Problem 8: Quantum Mechanics
A particle of mass m is confined in a one - dimensional harmonic potential with potential energy
1
2
given by V ( x) = mω 2 x 2 . A wave function of the ground state in this potential has a Gaussian
mω
mω 
and the normalization constant A = 
 .
2
 π 
1/4
form=
ϕ ( x) A exp(−α x 2 ) , where α =
(a) [7 pts.] Find x 2 , the expectation value of a squared displacement of the particle.
1
2
(b) [8 pts.] At t = 0 , the confining potential instantaneously changes to V1 ( x) = mω12 x 2 . What
is the probability for finding the particle now in the ground state of the new Hamiltonian
2 d 2
H1 =
−
+ V1 ( x) ?
2m dx 2
(c) [10 pts.] The particle is initially confined by the potential V ( x) and is in the ground state
ϕ ( x,=
t 0)= A exp(−α x 2 ) . At t = 0 , the confining potential is completely removed ( V2 ( x) = 0 ).
Find the wave function ϕ ( x, t ) as a function of time for t > 0.
To answer part (c), proceed in two steps.
(i) Compute the wave function of the particle at t = 0 in the wave vector representation, a(k ) . For
this step please complete the integration as needed.
(ii) Write down an expression connecting ϕ ( x, t ) and a(k ) that shows the time evolution of the
wave function. You do not need to compute the integral in this step.
Sample solution
You may need one or more of the following Gaussian Integrals (similar forms given in the exam).
∞
1 π
∫ exp(−α x )dx =
2 α
2
0
∞
1
∫ x exp(−α x )dx =
2α
2
0
∞
∫x
2
0
∞
∫x
3
0
1 π
exp(−α x 2 )dx =
4 α3
1
exp(−α x 2 )dx =
2α 2
a. Mean of the squared displacement.
+∞
+∞
2
2
2
2
∫ ϕ * ( x) x ϕ ( x)dx= A ∫ x exp(−2α x )dx=
x 2=
−∞
−∞
−1 2 ∂
A
∂α
2
+∞
∫ exp(−2α x
2
)dx
−∞
π
−1 2 ∂
A2
1 
π α −3/ 2
=
=
A
∂α 2α 4 2
2
2 mω
b. Probability
Instantaneous change of the potential does not change the wave function. In the basis of eigenstates
2 d 2
1
−
+ V1 ( x) with V1 ( x) = mω12 x 2 , the wave function can be written as a
ψ n ( x) of H1 =
2
2m dx
2
=
x2
∞
superposition ϕ ( x) = ∑ Cnψ n ( x) . Probability to find a particle in the ground state ψ 0 ( x) is C0 .
2
n =0
+∞
+∞
−∞
−∞
*
2
2
∫ ψ 0 ( x)ϕ ( x)dx= AB ∫ exp(−β x ) exp(−α x )dx= AB
C0=
π
α +β
,
mω1
 mω1 
and B = 
 . After substitution we find
2
 π 
2ω1/2ω11/2
.
=
ω + ω1
1/4
where β =
C0
2
c. Time evolution
Eigenstates of the particle in free space ( V2 ( x) = 0 ) are plane waves. The wave function of the particle
at t = 0 in k-representation is computed by taking Fourier transform
+∞
+∞
1
A
2
=
a(k )
ϕ
(
x
)
exp(
−
ikx
=
)
dx
))dx
∫
∫ exp(−(α x + ikx=
2π −∞
2π −∞
2
2
+∞
 
 ik   ik   
A
2


=
exp
−
α
x
+
ikx
+
−
=

 
   dx
∫  
2π −∞
2
α
2
α



  
 
2
 
 k 2  +∞
A
ik  
exp  −
exp
α
x
−
+

∫

  dx =
2π
2 α  
 
 4α  −∞
a(k )
=
 k2 
exp  −
.
2α
 4α 
A
The time evolution of ϕ ( x, t ) can be found by the reverse Fourier transform that includes timedependent factor for the Fourier components
=
ϕ ( x, t )
1
2π
+∞
∫ a(k ) exp(ikx − iω (k )t )dk ,
−∞
where the dispersion ω (k ) is for non-relativistic particle of mass m in free space, ω (k ) =
k 2
.
2m
Electromagnetism
9.
(a) [10 pts.] A thin loop of radius R carries current I in the +φ (i.e.
counter-clock-wise) direction (see Figure 1). The loop lies in the xy-plane,
centered on the origin. Find an expression for the magnitude of the
magnetic field B(z) along the z axis (as a function of position z).
Figure 1
(b) [10 pts.] A thin, non-conducting, disk of radius R carries charge Q distributed uniformly over
its surface. The disk lies in the xy-plane and rotates in the +φ direction at a constant angular speed
���⃗ of the disk in terms of ω , R and Q.
ω. Find an expression for the magnetic moment 𝒎
(c) [5 pts.] The rotating disk from part (b) above is now placed in a uniform external magnetic
field B0 which makes angle θ with the axis of rotation (z axis). Express the potential energy U of
the system as a function of θ and other relevant parameters.
Problem 10: Statistical Mechanics
Two atoms each carrying a spin ½ forms a molecule. The axis of the molecule lies along the zaxis. Due to strong anisotropy each spin points also along the z-axis either in the positive or
negative direction. One of the possible configurations of the spins is shown in the figure below.
The spins interact with each other. The interaction energy between two spins is E = JSi S j , where
J > 0 is the interaction constant and Si = ±1 corresponds to the two possible orientations of the
spin. The system is in thermal equilibrium at temperature T .
+1
-1
Z-axis
(a) [5 pts.] Write the expression for the partition function of the system.
(b) [5 pts.] Write the expression for the average internal energy of a molecule.
(c) [5 pts.] Write the expression for the average entropy of a molecule.
(d) [10 pts.] Consider now a situation when magnetic field B is applied in positive z direction.
Each spin carries magnetic moment m ; the energy of the spin in magnetic field is EB = −mB if
the spin and field are aligned and EB = mB if they are anti-aligned. The system is still in thermal
equilibrium at temperature T .
Now write the expression for the partition function of the system. Using either the partition
function or otherwise compute the average magnetic moment of a molecule in the limit
mB << kT << J , where k is the Boltzmann constant.
Solutions
(a) Compute partition function of the system.
There are total 4 configurations of the spins (++,- - , + - , - +),
 J 
 J 
Partition function is Z = 2exp  −  + 2exp  + 
 kT 
 kT 
(b) Compute internal energy of the system.
1
 J 
 J 
The internal energy of the system=
is U
 2 J exp  −  − 2 J exp   
Z
 kT 
 kT  
 ∂F 
(c) The entropy of the system can be found as S = − 
 , where free energy F = − kT ln Z .
 ∂T 
.
J
+
 − kTJ

J
kT
+
 − kTJ

J
e
e
−


kT
S k ln  2e + 2e  +  J
=
J 
−
+
T


 e kT + e kT 


(d) Compute partition function for the system in the magnetic field.
 J 2mB 
 J 2mB 
 J 
Z B = exp  −
+
+ exp  −
−
+ 2exp  + 


kT 
kT 
 kT
 kT
 kT 
Compute average magnetic moment the molecule.
 ∂G 
The average magnetic moment can be found as M = − 
 , where the Gibbs free energy is
 ∂B T
G = − kT ln Z B .
−
J
e kT
After taking derivative =
we have M 2m
ZB
mB
−
 mB

kT
kT
−
e
e

.


m 2 B − 2kTJ
e . The average magnetic moment is
kT
exponentially suppressed compare to the case of non-interacting spins.
Then in the limit of mB << kT << J we have M = 4