MITES 2010: Physics III
Survey of Modern Physics
Final Exam Solutions
Exercises
1. Problem 1. Consider a particle with mass m that moves in one-dimension. Its position
at time t is x(t). As a function of its position, Its potential energy is U (x) = C, where C
is a constant.
(a) Derive the equation of motion for the particle using Newtons second law.
(b) Derive the equation of motion for the particle, but this time, using the Lagrangian.
(c) Solve the equation of motion (i.e., what are x(t) and dx
dt ?). Define all the free
parameters in your solution (i.e. what are their physical meanings?)
Solution: (a)From Newton’s Second Law we find
mẍ = F (x)
d
= − U (x)
dx
d
= − C
dx
= 0
=⇒
ẍ = 0
(b) From the definition of the Lagrangian we find that the Lagrangian of this system has the form
L = KE − PE
1
=
mẋ2 − C
2
From this Lagrangian we may derive equation of motion using the Euler-Lagrange (E-L) equation.
∂L
∂x
∂
(−C)
∂x
0
d ∂ 1
2
mẋ
dt ∂ ẋ 2
d
(mẋ)
dt
mẍ
=
=
d ∂L
dt ∂ ẋ
=
=
=
and so
0=
∂L
d ∂L
−
= 0 − mẍ
∂x
dt ∂ ẋ
1
=⇒
ẍ = 0
(c) Our equation of motion ẍ = 0 is a simple second order differential equation so we may solve it by
separation of variables.
ẍ = 0
dẋ
= 0
dt
dẋ = 0 dt
Z
ẋ(t)
dẋ = 0
ẋ0
ẋ(t) − ẋ0 (t) = 0
=⇒
ẋ(t) = ẋ0
This equation for velocity provides us with another differential equation for the positon x(t).
ẋ = ẋ0
dx
= ẋ0
dt
dẋ = ẋ0 dt
Z x(t)
Z t
dx =
ẋ0 dt0
x0
0
x(t) − x0 = ẋ0 t
=⇒
x(t) = x0 + ẋ0 t
In our solutions, the free parameter ẋ0 represents the initial velocity of particle
and x0 represents the initial position.
2. Problem 2. Hanging mass The potential energy for a mass hanging from a spring is
V (y) = ky 2 /2 + mgy, where y = 0 corresponds to the position of the spring when nothing
is hanging from it.
(a) What is the net force acting on the particle?
(b) Write down the Lagrangian for this system.
(c) Use the Euler-Lagrange equation to derive the equation of motion for this system.
DO NOT use Newtons second law to derive it. What kind of motion does this correspond to?
(d) Solve the equation of motion, thereby obtaining y(t). What do all the free parameters in your solution physically mean?
Solution: (a)
F (y) = −V 0 (y)
d
= − (ky 2 /2 + mgy)
dy
=
−ky − mg
(b)
L = KE − PE
1
1
=
mẋ2 − ky 2 − mgy
2
2
2
(c)
∂L
∂y
d ∂L
dt ∂ ẏ
0=
∂
1
(− ky 2 − mgy)
∂y 2
= −ky − mg
d ∂ 1
2
=
mẏ
dt ∂ ẏ 2
= mẍ
=
∂L
d ∂L
−
= −ky − mg − mÿ
∂y
dt ∂ ẏ
=⇒
ÿ +
k
y+g =0
m
If we let δ = y + mg/k then we have δ̈ + ω 2 δ = 0 where ω 2 = k/m. This is the simple harmonic
oscillator equation which models all free undamped oscilatory motion.
d) The solution to the SHO EOM, δ̈ + ω 2 δ = 0, is
δ(t) = A sin ωt + B cos ωt
And from our definition δ = y + mg/k, we find that y(t) is
mg
y(t) = A sin ωt + B cos ωt −
k
r
ω=
k
m
. In order to find what A and B represent, we solve for the initial conditions of our mass. Setting
t = 0 in our position equation, we find
y0 = y(t = 0) = A sin 0 + B cos 0 − mg/k = B − mg/k
=⇒ B = y0 +
mg
k
B represents the initial displacement from the equilibrium position yeq = −mg/k. Similarly, differentiating y(t) and then plugging in t = 0 gives us
v0 = ẏ(t = 0) = Aω cos 0 − Bω sin 0 = Aω
=⇒ A =
v0
ω
A represents the inital velocity of the bead divided by the angular frequency. All of this information
gives us a general solution x(t) of the form
v0
mg y(t) =
sin ωt + y0 +
cos ωt
ω
k
3. Problem 3. Zipper problem (Model for unzipping of DNA). A zipper has N links; each
link has a state in which it is closed with energy 0 and a state in which it is open with
energy . We require, however, that the zipper can only unzip from the left end, and
that the link number s can only open if all links to the left (1, 2, . . . , s − 1) are already
open. (a.) Show that the partition function can be summed in the form
h
i
1 − exp − (NkB+1)
T
h
i
Z=
1 − exp − kB T
3
(b.) In the limit >> kB T , find the average number of open links. This model is a very
simplified model of the unwinding of two-stranded DNA molecules.
Solution: (a)To obtain a feel for the problem we will begin by considering the case of N = 2. If
we have two links connected successively, then there are three possible configurations: one with both
links closed; one with the left link open and the right link closed; one with both links open. We do
not have a state with the right link open and the left link closed because we require that the links
open from the left. Considering the fact that each open link has an energy associated with it we
know that the energy of the first configuration is 0, the energy of the second configuration is , and
the energy of the last configuration is 2. This collection of configurations and associated energies
generates a partition function of the form
Z = e−β·0 + e−β + e−2β
We can easily generalize this example to our case of arbitrary N . We realize the partition function
must then take on the form
Z = e−β·0 + e−β + e−2β + · · · + e−(N −1)β + e−N β
= 1 + e−β + e−2β + · · · + e−(N −1)β + e−N β
which is a finite geometric series. We can sum the series using the formula
1 + r + r2 + · · · + rN −1 + rN =
1 − rN +1
1 − rN
with r = e−β . Therefore, we find that
Z=
1 − e−β(N +1)
1 − e−β
Z=
N
X
e−βi .
i=0
(b) To calculate the average number of open links, we will calculate the average energy of our various
configurations and divide this result by , the energy of one open link. This procedure works because
the only nonzero energy in our various configurations comes from the number of open links.
hni =
4
hEi
Computing the average energy, we find
∂ log Z
∂β
1 ∂Z
−
Z ∂β
1 ∂ 1 − e−β(N +1)
−
Z ∂β 1 − e−β
"
#
1 (N + 1)e−(N +1)β (1 − e−β ) − e−β (1 − e−(N +1)β )
−
Z
(1 − e−β )2
"
#
1 − e−β
(N + 1)e−(N +1)β (1 − e−β ) − e−β (1 − e−(N +1)β )
−
(1 − e−β )2
1 − e−β(N +1)
hEi = −
=
=
=
=
= −
(N + 1)e−(N +1)β
e−β
+
1 − e−β
1 − e−(N +1)β
Now, if we take β >> 1, then we can take the following approximations
1 − e−β
−β
e
≈
1
>> e−(N +1)β
So that our average energy becomes
hEi ≈ e−β
and the average number of open links is hni = e−β .
(c.) Give a physical meaning of what >> kB T means. In particular, give a physical justification
for what the approximation >> kB T means and why you can ignore some of the terms in your
partition function in (b) under this approximation. Recall that kB T is the typical thermal energy
in a system. [Hint: How accessible are the states with energy much larger than the typical thermal
energy?]
Solution: is defined as the energy necessary to open one link in our zipper, so >> kB T means
that it requires a lot of energy to open one link. Consequently, it will take a lot more energy to open
multiple links. Therefore, if we consider this problem as a describing a state with a limited amount
of energy, then the approximation >> kB T restricts the number of states the system can access.
Since it requires such a large amount of energy to open mutliple states, the most probable states will
be the ones with the lowest energy. This allows us to shorten our partition function so that it only
contains the configurations with the lowest energy
Z = 1 + e−β + e−2β + · · · + e−(N −1)β + e−N β
≈ 1 + e−β
4. Problem 4: Applying the rules of Quantum Mechanics.
Consider a particle whose energy can only take on the following values:
E1 = 3h̄ω, E2 = 6h̄ω, E3 = 9h̄ω, E4 = 0
5
These are the only allowed energies that the particle can have. They have corresponding
eigenstates:
|E1 i, |E2 i, |E3 i, |E4 i. Ĥ is the Hamiltonian operator - operator representing measurement
of the particle’s energy.
Consider a particle in the
r following indeterminate state.
iθ
−1
e
2 iϕ
|ψi = √ |E1 i + √ |E2 i +
e |E4 i
3
6
6
(i.) If you measure the energy of 10,000 identical particles, all in the above ’indeterminate’ state just before your measurement, how many of them do you expect to yield
energy value of E2 ?
iθ 2
e 1
Solution: The probability of obtaining E2 is √ = . So, out of 10,000 measurements, we expect
6
6
10, 000
≈ 1667 particles
6
to yield energy E2 .
(ii.) On average, what value of energy would your measurement yield?
Solution:
Eavg = |c1 |2 E1 + |c2 |2 E2 + |c3 |2 E3 + |c4 |2 E4
1
1
2
=
3h̄ω + 6h̄ω + 0 · 9h̄ω + · 0
6
6
3
3
=
h̄ω
2
(iii.) For which value of θ is the probability of a particle having energy E2 zero?
Solution: There is no value of θ for which the probability of obtaining E2 is zero . First of all, this
is because the coefficient of the state |E2 i is proportional to eiθ and eiθ never equals zero. Second of
all, the magnitude of√eiθ is always 1 regardless of θ, so the probability of being in E2 is independent
of θ Prob(E2 )=|eiθ / 6|.
(iv.) Suppose you measure the energy of a particle. You find that it has energy E1 . Does
this mean that the particle was in state |E1 i just before your measurement? Suppose
you measure the energy of the same particle for the second time. What is the probability that you get E4 as the energy this second time?
Solution: No, the particle was not necessarily in the state |E1 i before the measurement . A postulate of quantum mechanics states that a quantum state is in a superposition of many states before
a measurement and only collapses to one state after the measurement. Also, since the state is now
in |E1 i then there is zero probability that it is in any other state .
6
(v.) Suppose you measure the energy of the particle and find that it has energy E2 .
Write down the ket-vector representing the particle’s state immediately after this measurement.
Solution: Since we measured energy and obtained the value E2 we know that the particle’s state
has now collapsed to the state corresponding to E2 . That is, the particle is in state |ψi = |E2 i .
5. Problem 5. Sequential measurements
An operator  representing observable A, has two normalized eigenstates ψ1 and ψ2 ,
with eigenvalues a1 and a2 respectively. Operator  representing observable B, has two
normalized eigenstates φ1 and φ2 , with eigenvalues b1 and b2 . The eigenstates are related
by
3φ1 + 4φ2
4φ1 − 3φ2
ψ1 =
ψ2 =
(1)
5
5
(a.) How do we know that observables  and B̂ do not commute (i.e.  and B̂ are incompatible
observables)?
Solution: We know that the observables  and B̂ do not commute1 because they do not have the
same eigenstates (i.e. φ1 6= ψ1 and φ2 6= ψ2 ). Compatible observables (observables with commuting
operators) must have the same eigenstate because the order of their operation does not matter. For
example, let |B1 i be an eigenstate of B̂ with eigenvalue b1 . Then we can perform the following
operation for commuting operators B̂ and Ĉ
0 = [Ĉ, B̂]|B1 i
= Ĉ B̂|B1 i − B̂ Ĉ|B1 i
B̂ Ĉ|B1 i = Ĉ B̂|B1 i
B̂(Ĉ|B1 i) = b1 (Ĉ|B1 i)
So we see that Ĉ|B1 i must be an eigenvector of B̂ with eigenvalue b1 . This means that we can write
Ĉ|B1 i as a multiple of |B1 i such as
Ĉ|B1 i = k|B1 i
But, this is just the definition of an eigenvector |B1 i of Ĉ with eigenvalue k. So we see that commuting operators share the same eigenvectors. Conversely, noncommuting operators will have different
eigenvectors.
(b.) Observable A is measured, and the value a1 is obtained. What is the state of the
system (immediately) after this measurement?
Solution: By one of the postulates of quantum mechanics, the state has collapsed into the state
corresponding to eigenvalue a1 . So, the state is in ψ1 .
(c.) If B is now measured, what are the possible results, and what are their probabilities?
Solution: The state is currently in ψ1 .
3
4
ψ 1 = φ1 + φ2
5
5
1
Commuting observables satisfy ÂB̂ = B̂ Â or [Â, B̂] = 0
7
From this state, there are two possible results upon measurement of B. These results and their
associated probabilities are
Results
Probability
2
3
9
=
5
25
2
16
4
=
5
25
b1
b2
which fits our normalization requirement because 9/25 + 16/25 = 25/25 = 1
(d.) Right after the measurement of B, A is measured again. What is the probability of
getting a1 ? (Note that the answer would be quite different if I had told you the outcome
of the B measurement.)
Solution: There are two possible ways to get to a1 from our initial a1 measurement. We can obtain
b1 from the B measurement and then obtain a1 from the A measurement. Or, we can obtain b2 from
the B measurement and then obtain a1 from the A measurement. In order to calculate the total
probability of obtaining a1 , we must calculate the probability of these independent paths and then
add them. But first, we must write φ1 and φ2 in terms of ψ1 and ψ2 in order to obtain the probabilities
for the second measurement of A after our measurement of B. Solving Eq (1)by elimination, we find
for φ1
ψ1 =
5ψ1 − 4φ2
3
3φ1 + 4φ2
5
= φ1
Substituting this result into our other equation, we find
ψ2 =
5ψ2 =
=
=
25φ2 =
φ2 =
4φ1 − 3φ2
5
4
(5ψ1 − 4φ2 ) − 3φ2
3
20
16
ψ1 − φ2 − 3φ2
3
3
20
25
ψ1 − φ1
3
3
20
ψ1 − 5ψ2
3
4ψ1 − 3ψ2
5
Now, solving for φ1
4
3
5ψ1 = 3φ1 + 4( ψ1 − ψ2 )
5
5
16
12
= 3φ1 + ψ1 − ψ2
5
5
9
12
ψ1 = 3φ1 − ψ2
5
5
3ψ1 + 4ψ2
φ1 =
5
8
So, in summary, we have
φ1 =
3ψ1 + 4ψ2
5
φ2 =
4ψ1 − 3ψ2
5
From this result, we may now construct our probabilities. There are two paths to get to a1 from a
measurement of B. These paths and their associated probabilities are
Path
Probability
9
9
81
×
=
25 25
625
16 16
256
×
=
25 25
625
1) From b1 to a1
2) From b2 to a1
For each path, we multiplied the probability of obtaining our first measured value of B by the
probability of obtaining our second measured value of A. The total probability of obtaining a1 is
81
256
337
the sum of these probabilities and is therefore
+
=
We can check that this procedure
625 625
625
for calculating probability is consistent with normalization by calculating the probability other two
possible paths. Instead of obtaining a1 , we could have obtained a2 in our second measurement and
the paths and probabilities associated with this value are
Path
Probability
9
16
144
×
=
25 25
625
16
9
144
×
=
25 25
625
3) From b1 to a2
4) From b2 to a2
We have listed all possible outcomes of our measurements. The sum of the probabilities for theses
outcomes must be one.
Prob(Path 1) + Prob(Path 2) + Prob(Path 3) + Prob(Path 4) =
81
256 144 144
625
+
+
+
=
=1
625 625 625 625
625
Our outcomes are normalized, so we know that our procedure is correct.
6. Problem 6. Eigenstates and Eigenvalues of a two level system.
The Hamiltonian Ĥ of a particle with two possible energy states is represented by the
2× 2 matrix
1 0
Ĥ = h̄ω
0 2
(a.) What are the possible values of energy for this particle (i.e. eigenvalues of Ĥ)?
Solution: We employ the standard procedure for the calculation of eigenvalues.
ˆ
0 = det(Ĥ − E I)
h̄ω − E
0
= det
0
2h̄ω − E
= (h̄ω − E)(2h̄ω − E)
=⇒
E1 = h̄ω, E2 = 2h̄ω
(b.) For each eigenvalue of Ĥ you found in (a), find the corresponding eigenstate of Ĥ.
9
Solution: E1 = h̄ω
ˆ 1i
0 = (Ĥ − E1 I)|E
h̄ω − h̄ω
0
=
|E1 i
0
2h̄ω − h̄ω
0 0
=
|E1 i
0 h̄ω
The last line suggests that |E1 i cannot have a second component. So we have
|E1 i =
α1
0
=⇒
|E1 i =
1
0
Where we set α1 = 1 in order to normalize our eigenstate.
For the other eigenvalue, E2 = 2h̄ω, we have
ˆ 2i
0 = (Ĥ − E2 I)|E
h̄ω − 2h̄ω
0
|E2 i
=
0
2h̄ω − 2h̄ω
−h̄ω 0
|E2 i
=
0
0
The last line suggests that |E2 i cannot have a first component. So we have
|E1 i =
0
α2
=⇒
|E2 i =
0
1
Where we set α2 = 1 in order to normalize our eigenstate.
(c.) The two eigenstates you found in (b,) represent two orthogonal vectors. What is the physical
significance of this fact? Explain your answer in terms of the measurement of energy of a particle
that is in an arbitrary state before your measurement.
Solution: The fact that the two eigenstates are orthogonal (i.e. hE1 |E2 i = hE2 |E1 i = 0) signifies
the fact that when we are in one eigenstate of an observable then we can only be in that eigenstate.
There is no overlap between the states so that if we measure the energy and obtain eigenvalue E1
then when we perform another measurement we will obtain E1 again because the probability of the
particle being in state E2 is zero.
Probability of measuring E2 after obtaining E1 = |hE2 |E1 i|2 = 0
10