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Lecture Notes of Quals Reviewfor Physics PhD Candidates
Zhuoran HeColumbia University
February, 2019
Preface
The qualifying exams (Quals) in the department of physics of Columbia University are arequired qualification before PhD students begin their research. As a former PhD studentof Columbia University, I wrote these lecture notes when I was invited by the department ofphysics to give tutorial lectures to the Quals takers in 2014. The exams mainly focus on in-depth topics of undergrad physics, including mechanics, electrodynamics, relativity, quantumphysics, thermodynamics and statistical physics. These lecture notes were recreated when Igraduated from Columbia University with some new thoughts incorporated, in memory ofthe wonderful times during my PhD studies. The lecture notes are a good summary andrevision of undergrad physics and can be useful for future Quals takers and for PhDs andpostdocs as a quick mind refresher of the fundamentals.
The lecture notes contain 6 equal-length lectures, each can be covered in about 2 hours.The notes are typically a combination of formalism derivation and sample problems pickedfrom past qualifying exams with solutions and explanations. The 6 lecture notes in thisbook correspond to the 6 qualifying exam sections that take place on 3 separate days withminor reorganizations. Among all 6 lectures, Lectures 4 and 5 on quantum mechanics havethe richest and most diversified content. I can only cover some important selected topicsand encourage the reader to also go over Griffith’s Introduction to Quantum Mechanics fora more comprehensive revision. Other lectures tend to be more systematic and completebut no less interesting. There are also order-of-magnitude estimation, dimension analysis,and astrophysics problems that I did not have time to sufficiently talk about, but the tech-niques used in astrophysics, e.g., planetary motion, radiation, relativity, quantum physics,thermodynamics, fluid dynamics, etc., are covered in these lecture notes.
I acknowledge Columbia University for giving me the chance to teach the Quals andshare my notes to future students. I wish all students to perform well in the Quals!
Zhuoran HeFebruary, 2019
i
Contents
1 Classical mechanics and Lagrangian mechanics 11.1 Derivation of the Lagrange equation . . . . . . . . . . . . . . . . . . . . . . 11.2 Conservation laws of Lagrangian mechanics . . . . . . . . . . . . . . . . . . 41.3 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Electrostatics and radiation 142.1 Poisson equation in electrostatics . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Radiation using multipole expansion . . . . . . . . . . . . . . . . . . . . . . 21
3 Special relativity and covariant electrodynamics 273.1 Special relativity theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Covariant form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.3 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4 Quantum mechanics - theory 394.1 Simple harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.3 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.4 Electron in an electromagnetic field . . . . . . . . . . . . . . . . . . . . . . . 434.5 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.6 Perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Quantum mechanics - applications 525.1 Galilean transformation of QM . . . . . . . . . . . . . . . . . . . . . . . . . 525.2 Dirac equation (relativistic QM) . . . . . . . . . . . . . . . . . . . . . . . . . 545.3 Spin-orbit interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.4 Solid-state physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.5 Josephson junction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
6 Thermodynamics and other topics 636.1 Statistical physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636.2 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.3 Fluid dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
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Lecture 1 Classical Mechanics andLagrangian Mechanics
1 Derivation of the Lagrange equation1.1 Virtual work and the d’Alembert principleFrom a purely classical mechanical point of view, Lagrangian mechanics is introduced asa way to study particles moving under constraints. It is more convenient than Newtonianmechanics as it cancels the constraint forces and directly gives the motion of the particlesgiven only the applied forces acting on them. Suppose we have N constrained interactingparticles at positions given by
ri = ri(q, t), i = 1, 2, . . . , N, (1)
where q = (q1, q2, . . . , qs) is a set of independent real parameters called the generalizedcoordinates. The above N mapping equations map the set of generalized coordinates q intothe N Cartesian coordinates of the particles. The particles will be subject to forces of twotypes: applied forces FA
i , which are active and drive the particles to move, and constraintforces Ni, which are passive and keep the particles onto the constraint surface. The explicittime dependence t in the mapping equations allows for the possibility of a time-dependentconstraint surface, which changes the mapping rules ri(q, t) as time goes by. The Newtonianequations of the particles are given by
FAi + Ni = mi
¨ri, i = 1, 2, . . . , N. (2)
In case of a frictionless constraint, the constraint forces Ni will be normal to the surface butcan still do work if the surface moves. In order to cancel the constraint forces, we considera virtual displacement
δri =s∑
α=1
∂ri∂qα
δqα, i = 1, 2, . . . , N, (3)
of the N particles with the constraint surface fixed at a certain time. This is like taking asnapshot of the constraint surface and imagining that the particles move on it. And then itis clear to say that frictionless constraint forces do no virtual work, i.e.,
N · δri = 0, i = 1, 2, . . . , N. (4)
1
We therefore arrive at the d’Alembert principleN∑i=1
FAi · δri =
N∑i=1
mi¨ri · δri. (5)
The constraint forces Ni are now all canceled, with only the active forces FAi and particle
accelerations left in the equation.
1.2 Generalized forces and the Lagrange equationThe left-hand side of Eq. (5) can be written as
LHS =s∑
α=1
δqα
N∑i=1
FAi · ∂ri
∂qα=
s∑α=1
Qαδqα, (6)
where we have introduced the generalized forces defined as
Qα =N∑i=1
FAi · ∂ri
∂qα, α = 1, 2, . . . , s. (7)
The generalized forces do not act on a single particle, but on the generalized coordinates qα,which can describe collective motion. The right-hand side of Eq. (5) becomes
RHS =s∑
α=1
δqα
N∑i=1
mi¨ri ·
∂ri∂qα
=s∑
α=1
δqα
N∑i=1
mi
[d
dt
(˙ri ·
∂ri∂qα
)− ˙ri ·
d
dt
(∂ri∂qα
)]. (8)
From the mapping equations ri = ri(q, t), we have
˙ri =s∑
α=1
∂ri∂qα
qα +∂ri∂t
= ˙ri(q, q, t). (9)
Therefore we can directly read from Eq. (9) that
∂ ˙ri∂qα
=∂ri∂qα
, (10)
and by acting ∂/∂qα onto Eq. (9) obtain
∂ ˙ri∂qα
=s∑
α=1
qβ∂
∂qβ
(∂ri∂qα
)+
∂
∂t
(∂ri∂qα
)=
d
dt
(∂ri∂qα
). (11)
Using Eqs. (10) and (11), we rewrite the right-hand side of Eq. (5) into
RHS =s∑
α=1
δqα
N∑i=1
mi
[d
dt
(˙ri ·
∂ ˙ri∂qα
)− ˙ri ·
∂ ˙ri∂qα
]. (12)
2
Since the total kinetic energy of the mechanical system is given by
T =N∑i=1
1
2mi
˙r 2i = T (q, q, t), (13)
and the particle masses mi remain constant over time, we have
RHS =s∑
α=1
δqα
[d
dt
(∂T
∂qα
)− ∂T
∂qα
]=
s∑α=1
Qαδqα = LHS. (14)
Finally, from our assumption that the generalized coordinates qα are independent, we obtain
d
dt
(∂T
∂qα
)− ∂T
∂qα= Qα =
N∑i=1
FAi · ∂ri
∂qα, α = 1, 2, . . . , s. (15)
These s equations are called the Lagrange equations. They govern the motion of the particleswithout involving the constraint forces.
One may wonder why we need to assume that the virtual variations δqα of the generalizedcoordinates are independent. If they are not, why can’t we just cancel some of the generalizedcoordinates qα and keep only the rest? However, cancellation of generalized coordinatesusing the constraints works for holonomic constraints only, i.e., when the dependence of thevariations δqα is integrable. In case of a non-holonomic constraint, the δqα’s are inter-relatedbut not integrable, so that there is still no one-to-one correspondence between the generalizedcoordinates qα. The quantum-mechanical analogue of this situation is the Berry phase. Wewill not go into the details here.
1.3 Lagrange equation of a conservative systemIn case that the applied forces FA
i are conservative, we may introduce a potential functionV (r1, . . . , rN , t), such that
FAi = −∇iV (r1, . . . , rN , t). (16)
In this case, the generalized forces can be found to be
Qα =N∑i=1
FAi · ∂ri
∂qα= −
N∑i=1
∂ri∂qα
· ∇iV (r1(q, t), . . . , rN(q, t), t) = −∂V (q, t)
∂qα, (17)
where we have plugged in the mapping equations and used the chain rule. Since the potentialV (q, t) does not depend on the generalized velocities q, we have ∂V/∂qα = 0. Therefore, theLagrange equations become
d
dt
(∂L
∂qα
)=
∂L
∂qα, α = 1, 2, . . . , s, (18)
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where L = T − V is introduces as the Lagrangian of the system. Eq. (18) is more frequentlyreferred to as the Lagrange equation than Eq. (15).
In some systems, we may introduce a potential function V = V (q, q, t) that also dependson the generalized velocities q. If the Lagrangian L = T − V constructed yields the correctclassical motion, we still accept the validity of the above Lagrangian equation, in a moregeneralized sense. For example, the Lagrangian of a non-relativistic particle moving in anelectromagnetic field can be constructed as
L =1
2m ˙r 2 − e
[ϕ(r, t)− ˙r · A(r, t)
]. (19)
The vector potential A(r, t) is coupled with the particle’s velocity ˙r so as to give a velocity-dependent potential. Pluggin this Lagrangian into the Lagrange equation, we obtain
d
dt(∇ ˙rL)−∇rL =
d
dt
(m ˙r + eA
)+ e∇r ϕ+ e∇r( ˙r · A)
=d
dt(m ˙r) + e
(∇ϕ+
∂A
∂t
)− e ˙r × (∇× A) =
d
dt(m ˙r)− e(E + ˙r × B) = 0, (20)
which gives precisely the Lorentz force law.
2 Conservation laws of Lagrangian mechanics2.1 Cyclic coordinatesIn case that ∂L/∂qα = 0, qα is called a cyclic coordinate of the system. In such case, fromthe Lagrange equation (18), we have
d
dt
(∂L
∂qα
)= 0 ⇒ ∂L
∂qα= const. (21)
Therefore, whenever the system is invariant under translation along qα, the correspondingconjugate momentum pα = ∂L/∂qα is conserved.
2.2 Energy conservationIf ∂L/∂t = 0, the system exhibits time translation invariance. We can find another conservedquantity in this case. Multiplying qα onto the Lagrange equation (18), we obtain
s∑α=1
qαd
dt
(∂L
∂qα
)=
s∑α=1
qα∂L
∂qα+
∂L
∂t, (22)
where the added term on the right-hand side ∂L/∂t = 0 is zero by assumption. The left-hand
4
side of Eq. (22) can be written ass∑
α=1
qαd
dt
(∂L
∂qα
)=
s∑α=1
[d
dt
(qα
∂L
∂qα
)− qα
∂L
∂qα
]. (23)
Equating this to the right-hand side of Eq. (22), we have
d
dt
(s∑
α=1
qα∂L
∂qα− L
)= 0 ⇒ H =
s∑α=1
qα∂L
∂qα− L = const, (24)
where H is introduced as the Hamiltonian of the system. In case of a time-independentconstraint surface, since
˙ri =s∑
α=1
∂ri∂qα
qα +∂ri∂t
=s∑
α=1
∂ri∂qα
qα (25)
is a linear homogeneous function of the generalized velocities qα, the total kinetic energyT will be quadratic homogeneous while V is independent of qα. In this case, from Euler’stheorem of homogeneous functions, we have
H =s∑
α=1
qα∂L
∂qα− L = 2T − (T − V ) = T + V, (26)
which is the total energy of the system. Therefore time translation invariance implies energyconservation. The situations discussed in § 2.1 and § 2.2 are two special cases of the moregeneral Noether’s theorem, which states that any differentiable symmetry of a system has acorresponding conservation law.
3 Sample problems3.1 Problem 4 in Sec. 1, 2013A mass m slides down a circularly curved surface on an object with mass M as shown in thediagram below. Mass M is free to slide on a frictionless surface. What are the final speedsof the two masses after m separates from M?
5
Answer: To find the Lagrangian, we have the kinetic and potential energies
T =1
2Mx2
M +1
2m(x2M + 2RθxM sin θ +R2θ2
), V = −mgR sin θ. (27)
Therefore, the Lagrangian is given by
L = T − V =1
2(M +m)x2
M +mRθxM sin θ +1
2mR2θ2 +mgR sin θ. (28)
Since ∂L/∂xM = 0, we have
∂L
∂xM
= (M +m)xM +mRθ sin θ = const, (29)
which is the conservation of horizontal momentum. And since ∂L/∂t = 0, we have
H = xM∂L
∂xM
+ θ∂L
∂θ− L = T + V (30)
=1
2(M +m)x2
M +mRθxM sin θ +1
2mR2θ2 −mgR sin θ = const, (31)
which is the conservation of energy. Plugging in the initial and final values, the two conservedquantities give us two equations
MxM +m(xM +Rθ) = 0, (32)1
2Mx2
M +1
2m(xM +Rθ)2 −mgR = 0, (33)
which leads to the speeds
vM = |xM | = m√2gR√
M(M +m), vm = |xM +Rθ| = M
√2gR√
M(M +m). (34)
Lagrangian mechanics is an overskill for this simple problem, because we knew the energyand horizontal momentum were conserved. We don’t need the Lagrangian to tell us that.But the next problem will be different.
3.2 Problem 5 in Sec. 1, 2013A point particle of mass m slides without friction within a hoop of radius R and mass M .The hoop is free to roll without slipping along a horizontal surface. What is the frequencyof small oscillations of the point mass, when it is close to the bottom of the hoop?
Answer: Since the hoop rolls on the horizontal surface instead of sliding frictionlessly,the horizontal momentum is not conserved. But the system definitely exhibits translationalinvariance along the horizontal direction. Now you see sometimes it’s not so easy to find a
6
crucial conserved quantity without using Lagrangian mechanics. To find the Lagrangian, wehave the kinetic energy
T =1
2Mx2
M +1
2IMω2
M +1
2m(x2M + 2RθxM cos θ +R2θ2
), (35)
and the potential energy V = −mgR cos θ. Since the hoop M rolls without slipping, we havexM = ωMR. The moment of inertia of a hoop is IM = MR2. Therefore, the Lagrangian
L = T − V =1
2(2M +m)x2
M +mRθxM cos θ +1
2mR2θ2 +mgR cos θ. (36)
Since ∂L/∂xM = 0, we have
∂L
∂xM
= (2M +m)xM +mRθ cos θ = const, (37)
which is the conserved quantity we couldn’t easily find using Newtonian mechanics. Andsince ∂L/∂t = 0, we have again the conservation of total energy
H = xM∂L
∂xM
+ θ∂L
∂θ− L = T + V (38)
=1
2(2M +m)x2
M +mRθxM cos θ +1
2mR2θ2 −mgR cos θ = const. (39)
Since the system does an oscillation with no slippage in the horizontal direction, the conservedquantity ∂L/∂xM = 0. This gives a constraint between xM and θ so we can cancel one ofthem, say xM . Then the Hamiltonian becomes
H =1
2
(m− m2 cos2 θ
2M +m
)R2θ2 −mgR cos θ. (40)
Using the approximation θ ≪ 1 and keeping only quadratic terms of θ and θ, we obtain
H ≈ 1
2
(2Mm
2M +m
)R2θ2 +
1
2mgRθ2 −mgR. (41)
7
The last term is a constant and can be dropped without changing the physics. What we nowhave is the Hamiltonian of a simple harmonic oscillator. The angular frequency of oscillationcan be found to be
ω =
√(1 +
m
2M
) g
R. (42)
3.3 Problem 2 in Sec. 1, 2011A flyball governor consists of two masses m connected to arms of length l and a mass Mas shown below. The assembly is constrained to rotate around a shaft on which the massM can slide up and down without friction. Neglect the mass of the arms, air friction andassume that the diameter of mass M is small. Suppose that the shaft is constrained to rotateat angular velocity ω0. Now answer the following questions
(a) Calculate the equilibrium height of the mass M .
(b) Calculate the frequency of small oscillations around this value.
Answer: This problem can be solved using Newtonian mechanics in the rotating frameand introducing the centrifugal (inertial) forces. And once the system oscillates, there willalso be Coriolis forces. But these forces will be perpendicular to the velocities of the massesand will not affect their motion under the constraints. But still, using Lagrangian mechanicsin the lab frame, we needn’t worry about any of these inertial forces. All we need is writedown the kinetic and potential energies to obtain the Lagrangian.
(a) For Part (a) we have the kinetic energy
T =1
2m[(ω0l sin θ)
2 + (lθ)2]× 2 +
1
2M
[d
dt(2l cos θ)
]2(43)
= (m+ 2M sin2 θ)l2θ2 +mω20l
2 sin2 θ, (44)
8
and the potential energy
V = −2mgl cos θ − 2Mgl cos θ = −2(M +m)gl cos θ. (45)
Therefore, the Lagrangian is given by
L = T − V = (m+ 2M sin2 θ)l2θ2 +mω20l
2 sin2 θ + 2(M +m)gl cos θ. (46)
Since ∂L/∂t = 0, we have the conserved quantity
H = θ∂L
∂θ− L = (m+ 2M sin2 θ)l2θ2 −mω2
0l2 sin2 θ − 2(M +m)gl cos θ. (47)
Notice that the system in this problem is subject to a time-dependent constraint (because thesystem is forced to rotate at a fixed angular frequency ω0), so the Hamiltonian is not equal toT+V any more. The term −mω2
0l2 sin2 θ in H can be understood as the centrifugal potential
energy in the rotating frame, and Coriolis force is not involved. To find the equilibrium angleθ0, we set θ = 0 and minimize the total energy
H = Veff(θ) = −mω20l
2 sin2 θ − 2(M +m)gl] cos θ, (48)
by taking the derivative
dVeff(θ)
dθ= −2mω2
0l2 sin θ cos θ + 2(M +m)gl sin θ = 0 (49)
⇒ sin θ = 0, or cos θ =(M +m)g
mω20l
. (50)
If ω0 <√
(M +m)g/ml ≡ ωc, we have (M +m)g/mω20l > 1, so the only solution is θ = 0.
The two arms will stay vertically down if the system rotates too slowly. Once ω0 > ωc, onemay take the second-order derivative to find that the equilibrium position θ = 0 becomesunstable, so the system will stay at the other equilibrium angle
θ0 = cos−1
(ω2c
ω20
), ω0 > ωc =
√(M +m)g
ml. (51)
(b) Now we need to consider the motion of θ around θ0. We need to approximate Veff(θ)as a quadratic function by taking its second-order derivative at θ0 to obtain
d2Veff(θ)
dθ2
∣∣∣∣θ0
= −2mω20l
2(2 cos2 θ0 − 1) + 2(M +m)gl cos θ0. (52)
Then using cos θ0 = (M +m)g/mω20l, we have
d2Veff(θ)
dθ2
∣∣∣∣θ0
= 2
[mω2
0l2 − (M +m)2g2
mω20
]= 2ml2
(ω20 −
ω4c
ω20
). (53)
9
Assuming ω0 > ωc, we have V ′′eff(θ0) ≡ d2Veff(θ)/dθ
2|θ0 > 0. Therefore, the system oscillatesaround θ0. The Hamiltonian under small angle approximation |θ − θ0| ≪ 1 is given by
H ≈ (m+ 2M sin2 θ0)l2θ2 + Veff(θ0) +
1
2V ′′eff(θ0)(θ − θ0)
2 (54)
=
[m+ 2M
(1− ω4
c
ω40
)]l2θ2 +ml2
(ω20 −
ω4c
ω20
)(θ − θ0)
2 + const. (55)
The angular frequency of small angle oscillation is therefore
ω = ω0
√√√√√(1− ω4
c
ω40
)1 + 2M
m
(1− ω4
c
ω40
) , ω0 > ωc =
√(M +m)g
ml. (56)
This problem is nearly the hardest one involving small oscillations with only one degree offreedom. The Lagrangian method can also be used for oscillations with multiple degrees offreedom. The next problem gives an example.
3.4 Problem 5 in Sec. 1, 2011A pendulum consisting of a massless rod of length L with a mass M at its end hangs froma fixed point. A second pendulum of the same construction hangs from the end of the firstpendulum. The pendulums are constrained to move in the same plane. Find the frequenciesof small oscillation and clearly describe the corresponding motions assuming that neitherpendulum moves far from the vertical configuration.
Answer: Let the angle between the upper pendulum and the vertical direction be α,and the angle between the lower pendulum and the vertical direction be β. Given that bothα and β are small, the total kinetic energy of the system can be found to be
T =1
2ML2
[α2 + (α + β)2
]. (57)
And the total potential energy is
V = −MgL cosα−MgL(cosα + cos β) (58)
≈ MgL
(α2 +
1
2β2
)+ const. (59)
Therefore, dropping the constant, the Lagrangian is given by
L = T − V = ML2
(α2 + αβ +
1
2β2
)−MgL
(α2 +
1
2β2
). (60)
We recognize that the system is a harmonic oscillator with two degrees of freedom. The
10
Lagrangian and the two Lagrange equations can be written in matrix form
L =1
2ML2
(α β
) [2 11 1
](α
β
)− 1
2MgL
(α β
) [2 00 1
](αβ
). (61)
The two Lagrange equations yield the matrix equation[2 11 1
](α
β
)= − g
L
[2 00 1
](αβ
). (62)
This is a generalized eigenvalue problem. The eigen-equation we get is given by
det
[ω2
(2 11 1
)− g
L
(2 00 1
)]= det
[2(ω2 − ω2
0) ω2
ω2 ω2 − ω20
](63)
= ω4 − 4ω2ω20 + 2ω4
0 = 0, (64)
where ω0 =√
g/L. The two eigen-frequencies are therefore
ω =
√2±
√2ω0 =
√(2±
√2)
g
L. (65)
The two corresponding eigenmodes satisfy β = ∓√2α. The mode β = −
√2α has a higher
frequency√
2 +√2ω0 and the two pendulums oscillate in opposite directions. The mode
β =√2α has a lower frequency
√2−
√2ω0 and the two pendulums oscillate in the same
direction. The general small oscillation can be a superposition of the two eigenmodes. Ina typical experiment, the high-frequency mode decays away faster due to air resistance andthe oscillation decays towards the low-frequency mode.
The Lagrangian method is very useful and important in solving classical mechanicalproblems. Its advantage mainly reveals when the system is subject to complicated butfrictionless constraints and the constraint forces aren’t relevant to the problem. Followingthe Lagrangian method, the procedure is standard and the result is reliable. But one stillneeds to be aware of the limitations of the method. It is not applicable, for example, todissipative systems (with friction) or open systems (with changing mass). These problemsneed to be solved using other methods.
3.5 Problem 5 in Sec. 1, 2012Find the net acceleration of a raindrop falling through a cloud. Assume that the raindrop isa tiny sphere, and the cloud consists of similar small droplets that are uniformly suspendedat rest; when the falling drop hits a droplet within the cloud, the two merge into a combined,still spherical, drop, which continues falling.
11
Answer: There are three parameters that fully describe the status of the raindrop: itsradius r, its mass m and its falling speed v. Notice that this is a changing mass problem.Neither Newton’s second law nor Lagrangian mechanics is applicable here. To establishequations for r, m and v, we start with the most obvious ones:
m =4
3πr2ρw,
dm
dt= πr2vρ, (66)
where ρw is the density of water and ρ is the density of the cloud. For each tiny bit of massdm added to the raindrop, we have from the conservation of vertical momentum
(m+ dm)(v + dv)−mv − dm · 0 = mgdt, (67)
because the velocity of dm before hitting the raindrop is zero. Therefore we have
d(mv)
dt= mg. (68)
You may feel that this equation is intuitively true. But it should not be taken for granted,because it depends on that the velocity of dm before hitting the raindrop be zero. Usingm = 4
3πr3ρw, we may cancel the mass m and only keep two dynamical variables r and v.
We have from the other two equations
r =ρ
4ρwv, v = g − 3v
rr. (69)
To solve the two coupled ODEs for r and v, we may plug the first ODE into the second oneto cancel v and obtain
r =1
2
dr2
dr=
ρg
4ρw− 3
rr2. (70)
Now we have a first-order linear ODE, the general form of which is given by
dy
dx= P (x)y +Q(x). (71)
Using the similarity relationship of differential operators
d
dx− P (x) = exp
[∫ x
x0
P (x′)dx′]
d
dxexp
[−∫ x
x0
P (x′)dx′], (72)
we obtain the general solution formula
y(x) =
∫ x
x0
Q(x′)dx′ exp
[∫ x
x′P (x′′)dx′′
]+ C exp
[∫ x
x0
P (x′)dx′], (73)
12
where C = y(x0) is an integration constant to be determined by the initial condition. Usingthis general solution formula, we obtain from Eq. (70) that
r2 =ρg
14ρwr +
C
r6. (74)
Given that initially the raindrop has r = 0 and v = 0 so that r = ρ4ρw
v = 0, we get C = 0.Therefore, we have
dr
dt=
√ρg
14ρwr ⇒ r =
ρg
56ρwt2, (75)
given that initially r = 0 at t = 0. Therefore,
v =4ρwρ
r =1
7gt, (76)
which means the acceleration of the raindrop is 17g. Interestingly the result is independent
of the cloud density ρ/ρw relative to the density of condensed raindrops. Compared withthe standard Lagrangian procedure to follow for studying closed conservative systems, thedynamics of open systems can be tricky but interesting to solve.
13
Lecture 2 Electrostatics and Radiation
1 Poisson equation in electrostaticsThe most fundamental equation in electrostatics is the Poisson equation. Given the totalcharge distribution ρ(r), the electrostatic potential ϕ(r) satisfies
∇2ϕ(r) = −ρ(r)
ϵ0. (1)
The solution can be obtained using Fourier transform:
ϕ(r) =1
4πϵ0
∫d3r′
ρ(r ′)
|r − r ′|, (2)
in case that we know the charge distribution ρ(r) in the whole 3D space. Unfortunately,in most electrostatic problems, we only know ρ(r) in a finite region and some boundaryconditions of ϕ. For general shapes of boundary, solution can be extremely hard. Analyticallysolvable cases are mostly 3 types: the box, the sphere, and the cylinder.
1.1 The box1. (Problem 4 in Sec. 2, 2011) Consider a box with side lengths a, b, and c along the x, y,and z axes. Suppose there is no electric charge in the box and that ϕ = 0 on the surface ofthe box except at z = 0 where ϕ = V1 sin
πxasin πy
band at z = c where ϕ = V2 sin
2πxa
sin 2πyb
.Find ϕ everywhere in the box.
Answer: Electrostatic problems in the box are comparatively easier than the sphereand the cylinder as they involve separation of variables along different axes and no special
14
functions (only trigonometry and exponentials). In this problem, since there is no chargeinside the box, the inhomogeneous Poisson equation (1) becomes the homogeneous Laplaceequation ∇2ϕ(r) = 0. In Cartesian coordinates, we have
∂2ϕ
∂x2+
∂2ϕ
∂y2+
∂2ϕ
∂z2= 0. (3)
And the boundary conditions are given by
ϕ(x = 0, y, z) = ϕ(x = a, y, z) = 0, (4)
ϕ(x, y = 0, z) = ϕ(x, y = b, z) = 0, (5)
ϕ(x, y, z = 0) = V1 sinπx
asin
πy
b, (6)
ϕ(x, y, z = c) = V2 sin2πx
asin
2πy
b. (7)
The boundary conditions on the x and y faces are homogeneous while on the z face they areinhomogeneous. We may use the separation of variables
ϕ(x, y, z) = X(x)Y (y)Z(z). (8)
Then, from the Laplace equation and the boundary conditions, we have
X(x) = sinmπx
a, Y (y) = sin
nπy
b, m, n = 1, 2, 3, . . . , (9)
and along the z direction we have
d2Z
dz2= π2
(m2
a2+
n2
b2
)Z, (10)
which has the general solution
Z(z) = amn exp(πzab
√n2a2 +m2b2
)+ bmn exp
(−πz
ab
√n2a2 +m2b2
). (11)
Therefore, the general solution of ϕ(x, y, z) is
ϕ(x, y, z) =∞∑
m=1
∞∑n=1
(amne
πzab
√n2a2+m2b2 + bmne
−πzab
√n2a2+m2b2
)sin
mπx
asin
nπy
b. (12)
Using Eqs. (6) & (7), we obtain for the a11 and b11 modes thata11 + b11 = V1,
a11eπ√
a2+b2
abc + b11e
−π√
a2+b2
abc = 0,
(13)
(14)
15
and for the a22 and b22 modes thata22 + b22 = 0,
a22e2π
√a2+b2
abc + b22e
− 2π√
a2+b2
abc = V22.
(15)
(16)
From the orthogonality of different eigenfunctions sin mπxa
sin nπyb
, all other coefficients amn
and bmn are zero. We may now find the 4 nonzero coefficients a11, b11, a22, b22 to determinethe specific solution:
ϕ =V1 sinh
[π√a2+b2
ab(c− z)
]sinh
[π√a2+b2
abc] sin
πx
asin
πy
b+
V2 sinh[2π
√a2+b2
abz]
sinh[2π
√a2+b2
abc] sin
2πx
asin
2πy
b. (17)
2. (Problem 5 in Sec. 2, 2012) Consider an infinite pipe with a square cross section, as isdrawn in the figure, with three sides grounded and one side at potential V0. Calculate thepotential everywhere inside the pipe.
Answer: In this problem, ϕ(x, y, z) = ϕ(x, y) is z-independent. The problem is thereforeessentially 2D. Using the separation of variables ϕ(x, y) = X(x)Y (y), we obtain the generalsolution based on ϕ(x, y = 0 or a) = 0:
ϕ(x, y) =∞∑n=1
(ane
nπxa + bne
−nπxa
)sin
nπy
a. (18)
To satisfy ϕ(x = 0, y) = 0, we require an + bn = 0 for all n. This leads to
ϕ(x, y) =∞∑n=1
cn sinhnπx
asin
nπy
a. (19)
To determine the coefficients cn, we need the fourth edge x = a. We have
ϕ(x = a, y) = V0 =∞∑n=1
cn sinh(nπ) sinnπy
a. (20)
16
Using the orthogonality of sin nπya
, we obtain∫ a
0
V0 sinnπy
ady = cn sinh(nπ)
∫ a
0
sin2 nπy
ady. (21)
Therefore, the coefficients are found to be
cn =2V0[1− (−1)n]
nπ sinh(nπ). (22)
Since cn is only nonzero for odd n, the final result can be simplified to
ϕ =4V0
π
∞∑n=0
sinh(2n+1a
πx)sin(2n+1a
πy)
(2n+ 1) sinh[(2n+ 1)π]. (23)
At x = a, the two sinh functions in the numerator and denominator cancel and you recognizethe familiar trigonometric expansion of a square wave. So everything works. Later in thequestion, you are asked to find ϕ(x, y) when the boundary value ϕ(x, y = a) = V0 is alsononzero. This seems to have complicated the problem a lot, since we no longer have a pairof homogeneous boundary conditions in either x or y direction. But we can actually solveeach nonzero edge separately and add up the two solutions by linearity.
1.2 The sphere1. (Problem 2 in Sec. 2, 2012) A dielectric sphere of radius R is hollowed out in the region0 ≤ r ≤ s and a thin, grounded, conducting shell is inserted at r = s. The sphere is placedin a uniform, external E-field E = E0z along the z axis. The dielectric constant of thehollowed sphere is ϵr. Calculate the potential in the region r ≥ R.
Answer: Unless they are spherically symmetric, electrostatic problems of the spheretypically involve the Legendre polynomials. The potential ϕ(r, θ) in this problem has cylin-drical symmetry around the z axis. Inside the grounded conducting shell, we have ϕ(r, θ) = 0for r ≤ s. Outside the conducting shell, we have the Laplace equation ∇2ϕ = 0 for bothregions of s < r < R and r > R. At r = R, due to the continuity condition of the normal
17
component of the electric displacement D, we have
ϵr∂ϕ
∂r
∣∣∣∣R−0
=∂ϕ
∂r
∣∣∣∣R+0
. (24)
This will cause discontinuity of ∂ϕ/∂r at r = R, while the continuity of ϕ at r = R impliesthe continuity of the tangential component of the electric field E. Now we need to solve theLaplace equation using the separation of variables in spherical coordinates. We have
1
r2∂
∂r
(r2
∂ϕ
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂ϕ
∂θ
)= 0. (25)
Plugging in ϕ(r, θ) = R(r)Θ(θ), we find that Θ(θ) = Pl(cos θ) is a Legendre polynomial ofcos θ and the corresponding radial function R(r) satisfies
r2d2R
dr2+ 2r
dR
dr− l(l + 1)R = 0. (26)
This is an Euler differential equation. By doing a change of variable r = eu, the varyingcoefficients become constant. We have
d2R
du2+
dR
du− l(l + 1)R = 0. (27)
There are two linearly independent solutions that form the general radial function
R = aelu + be−(l+1)u = arl +b
rl+1. (28)
The general solution of ϕ(r, θ) is therefore given by
ϕ(r, θ) =
∞∑l=0
(alr
l +blrl+1
)Pl(cos θ), r ≥ R,
∞∑l=0
(clr
l +dlrl+1
)Pl(cos θ), s ≤ r ≤ R.
(29)
At r = s, we require that ϕ(r, θ) = 0. Therefore, we have
clsl +
dlsl+1
= 0, l = 0, 1, 2, . . . (30)
At r = R, we require that ϕ is continuous. So we have
alRl +
blRl+1
= clRl +
dlRl+1
, l = 0, 1, 2, . . . (31)
18
And the continuity condition in Eq. (24) gives
lalRl−1 − (l + 1)bl
Rl+2= ϵr
[lclR
l−1 − (l + 1)dlRl+2
]. (32)
Finally, by looking at the asymptotic behavior of
ϕ(r, θ) ≈ −E0z = −E0r cos θ = −E0P1(cos θ), r → ∞, (33)
we have al = −E0δl1. Only the l = 1 component is nonzero. We now have four equationsfor a1, b1, c1, d1. We find that
a1 = −E0, b1 =(ϵr − 1)R3 + (2ϵr + 1)s3
(2 + ϵr)R3 + 2(ϵr − 1)s3E0R
3, (34)
c1 = − 3E0R3
(2 + ϵr)R3 + 2(ϵr − 1)s3, d1 =
3E0R3s3
(2 + ϵr)R3 + 2(ϵr − 1)s3. (35)
The potential ϕ outside the dielectric sphere is then given by
ϕ = −E0r cos θ
[1− (ϵr − 1)R3 + (2ϵr + 1)s3
(2 + ϵr)R3 + 2(ϵr − 1)s3· R
3
r3
], r ≥ R. (36)
2. (Problem 2 in Sec. 2, 2013) A linear uniform charge distribution of λ coulomb/meterextends along the z axis from z = −a to z = a. Find a series expansion for V (r, θ) in termsof Legendre polynomials valid to all orders for r > a.
Answer: The solution is a straightforward integral of the Coulomb potential
ϕ(r, θ) =1
4πϵ0
∫ a
−a
λdz√r2 + z2 − 2rz cos θ
. (37)
In the region r > a, we may Taylor expand the integrand using the generating function ofthe Legendre polynomials
1√1− 2xt+ t2
=∞∑n=0
Pn(x)tn, |x| < 1, |t| < 1. (38)
19
Physically this mathematical formula gives us the multipole expansion
ϕ =λ
4πϵ0r
∞∑n=0
Pn(cos θ)
∫ a
−a
(zr
)ndz =
λ
2πϵ0
∞∑n=0
P2n(cos θ)
2n+ 1
(ar
)2n+1
. (39)
It is recommended to review these related topics to prepare for the Quals: the first feworders of Legendre polynomials, how they appear in the Poisson equation, their recursiverelations, series expansions, Rodriguez formula, orthogonality relations, etc. This problemrequires the generating function of Legendre polynomials.
1.3 The cylinder1. (Problem 3 in Sec. 2, 2010) Consider an infinitely long, grounded conducting cylinder,of radius a, which is introduced into a uniform electric field E0. The axis of the cylinder isperpendicular to E0.
(a) Find an expression for the external potential after insertion of the cylinder;
(b) Find an expression for the surface charge induced on the cylinder.
Answer: (a) The problem involves solving the Laplace equation ∇2ϕ = 0 in cylindricalcoordinates, which is given by
1
r
∂
∂r
(r∂ϕ
∂r
)+
1
r2∂2ϕ
∂θ2= 0. (40)
In the region r > a, using the separation of variables ϕ(r, θ) = R(r)Θ(θ), we get Θ(θ) =sinnθ or cosnθ and the radial function R(r) satisfies
rd
dr
(rdR
dr
)= n2R ⇒ R = arn +
b
rn, n ≥ 1. (41)
In case that n = 0, we have R = a+ b ln r. The general solution is therefore given by
ϕ = a0 + b0 ln r +∞∑n=1
[(anr
n +a−n
rn
)cosnθ +
(bnr
n +b−n
rn
)sinnθ
]. (42)
20
Defining θ = 0 as the direction of E0, we have the asymptotic behavior
ϕ(r, θ) ≈ −E0r cos θ, r → ∞. (43)
Therefore an = −E0δn1, so the only nonzero terms will be a1 and a−1. At r = a, we haveϕ = 0. This gives us a−1 = −a2a1 = a2E0, so we have
ϕ(r, θ) = −E0 cos θ
(r − a2
r
). (44)
(b) The surface charge induced can be found by Gauss’s law:
σ(r = a, θ) = −ϵ0∂ϕ
∂r
∣∣∣∣a+0
= 2ϵ0E0 cos θ. (45)
Positive charge gets repelled to the θ = 0 side while negative charge gets pulled to theθ = π side. Not every cylinder involves Bessel functions. This problem is special becausethe potential ϕ has no z dependence. Bessel functions will typically not appear on Quals.
2 Radiation using multipole expansion2.1 Fields in the radiation zoneIn Gaussian units, Maxwell’s equations in vacuum are given in Gaussian units by
∇ · E = 4πρ, ∇ · B = 0, (46)
∇× E = −1
c
∂B
∂t, ∇× B =
4π
cj +
1
c
∂E
∂t. (47)
After introducing the electromagnetic potentials A and ϕ, the fields are expressed asf
E = −∇ϕ− 1
c
∂A
∂t, B = ∇× A. (48)
Using the Lorentz gauge condition
∇ · A+1
c
∂ϕ
∂t= 0, (49)
the potentials A and ϕ satisfy the d’Alembert wave equations
1
c2∂2ϕ
∂t2−∇2ϕ = 4πρ,
1
c2∂2A
∂t2−∇2A =
4π
cj. (50)
21
Radiation theory will focus on the vector potential A. The retarded solution of the d’Alembertequation is given by
A(x, t) =1
c
∫d3x′
|x− x ′|j
(x ′, t′ = t− |x− x ′|
c
). (51)
In the radiation zone, based on |x| ≫ |x ′| we may do two approximations
t′ = t− |x− x ′|c
≈ t− |x|c
+
(∇|x|
c
)· x ′ = t− r
c+ r · x
′
c, (52)
where r = |x| and r = x/|x|, and the other approximation being 1|x−x ′| ≈
1|x| , to obtain the
vector potential in the radiation zone given by
A(x, t) ≈ 1
rc
∫j
(x ′, t′ = t− r
c+
r · x ′
c
)d3x′. (53)
Q Then the E and B fields and the energy flow are given by
B = −1
cr × ∂A
∂t, E =
1
cr ×
(r × ∂A
∂t
), S =
c
4πE × B =
r
4πc
∣∣∣∣∣r × ∂A
∂t
∣∣∣∣∣2
. (54)
We can now calculate the fields and the power distribution from these formulas.
2.2 Electric dipole radiationIf the typical length scale of the radiating source is much smaller than the wave length ofthe radiation, we can do a Taylor expansion of r · x ′/c and obtain from Eq. (53) that
A(x, t) =1
rc
∫d3x′
∞∑n=0
1
n!
∂nj(x ′, t′)
∂t′n
∣∣∣∣∣t′=t− r
c
(r · x ′
c
)n
(55)
=1
rc
∞∑n=0
1
n!
∂n
∂t′n
∣∣∣∣t′=t− r
c
∫j(x ′, t′)
(r · x ′
c
)n
d3x′. (56)
This simplifies the retardation to t′ = t − rc
and all we need is find the nth order momentof the current distribution j(x ′, t′). In most radiation problems in the uals, we only need tolook at the leading order term in the above expansion, which is the n = 0 term
A0(x, t) =1
rc
∫j(x ′, t′) d3x′. (57)
This term is called the electric dipole radiation, because the integral of current density overthe whole 3D space is equal to the change rate of the electric dipole moment, which can be
22
obtained from charge conservation. We have
d
dt′
∫x ′ρ(x ′, t′)d3x′ =
∫x ′∂ρ(x
′, t′)
∂t′d3x′ = −
∫x ′∇′ · j(x ′, t′)d3x′ (58)
= −∫
∇′ · [j(x ′, t′)x ′]d3x′ +
∫j(x ′, t′) · (∇′x ′)d3x′ =
∫j(x ′, t′)d3x′, (59)
where we have used the formulas in vector calculus
∇ · (AB) = ∂i(AiBj)ej = ∂i(Ai)Bj ej + Ai∂i(Bj ej) = (∇ · A)B + A · ∇B, (60)
∇x = ei∂ixj ej = eiδij ej = eiei =↔1 , (61)
and that the total divergence term integrated over the whole 3D space is zero because∫∇′ · [j(x ′, t′)x ′]d3x′ = lim
V→∞
∮∂V
dS · j(x ′, t′)x ′ = 0, (62)
following from our locality assumption of the radiating source. We define the electric dipolemoment of the source as
p(t′) =
∫x ′ρ(x ′, t′)d3x′, (63)
so that the leading-order term of the vector potential becomes
A0(x, t) =1
rc
∫j(x ′, t′)d3x′ =
1
rc
dp(t′)
dt′
∣∣∣∣t′=t− r
c
. (64)
Since t′ = t − rc
only differs from t by a constant, we have dt = dt′. Therefore, the vectorpotential under the electric dipole approximation can be written as
A(x, t) ≈ A0(x, t) =1
rc˙p |t′=t− r
c. (65)
From this, the E and B fields are found to be
E(x, t) =1
cr ×
(r × ∂A
∂t
)=
1
rc2r × (r × ¨p)|t′=t− r
c, (66)
B(x, t) = −1
cr × ∂A
∂t= − 1
rc2(r × ¨p)|t′=t− r
c. (67)
Finally, the angular distribution of radiation power is given by
dP
dΩ=
c
4πr2(E × B) =
1
4πc3|r × ¨p|2
∣∣∣t′=t− r
c
. (68)
The next order of expansion n = 1 gives us the magnetic dipole and electric quadruple terms,and the radiation power can have cross terms between different orders.
23
2.3 Sample problems1. (Problem 1 in Sec. 2, 2012) Two point particles, each having the same electric charge +e,move in the xy plane around the circumference of a circle with radius R. Both charges travelat the same angular velocity ω but maintain a fixed angular separation α throughout themotion. Assume that the motion of the particles is non-relativistic (ωR ≪ c). The particlesradiate energy at distance r far from the circular orbit.
Find the time-averaged power radiated per solid angle in the (θ, ϕ) direction in the radiationzone (r ≫ R). Hint: use multipole expansion.
Answer: The condition ωR ≪ c enables us to do a multipole expansion in the radiationzone r ≫ R. So long as the angular separation α = π, there’s going to be change in theelectric dipole moment of the system as the particles move. Therefore an electric dipoleapproximation will be enough to solve the problem. We have
|p| = 2eR cosα
2, (69)
and this dipole moment p rotates about the origin at angular frequency ω. Therefore,
| ¨p| = 2eRω2 cosα
2, (70)
and ¨p also rotates about the origin at angular frequency ω. The component of ¨p that isparallel to r can be found to be
(¨p)∥ = ¨p · r = | ¨p| cosϕ′ sin θ, (71)
where ϕ′ is the angle between ¨p and the projection of r in the xy plane. This componentgenerates no radiation power in the r direction. Therefore,
⟨(¨p)⊥⟩ = ⟨| ¨p× r|2⟩ = | ¨p|2 − ⟨(¨p)2∥⟩ = | ¨p|2(1− 1
2sin2 θ
), (72)
using ⟨cos2 ϕ′⟩ = 12. Then from Eqs. (68), (70) & (72), we obtain the power distribution
dP
dΩ=
e2R2
πc3ω4 cos2
α
2
(1− 1
2sin2 θ
). (73)
24
2. (Problem 1 in Sec. 2, 2013) Consider two very tiny oscillating electric dipoles, which arelocated on the z axis, as is shown in the figure below.
The two dipoles is located at z = ∓L/2 and have time-varying electric dipole moments
D1,2 = D0z cos(ωt∓ α
2
)= Re
[D0ze
−i(ωt∓α2)]. (74)
The two dipoles therefore oscillate in the same z direction with the same amplitude D0 andthe same angular frequency ω but are out of phase by phase angle α. Consider a field atpoint (r, θ, ϕ) in the radiation zone r → ∞ and keep only the leading 1/r terms.
(a) Find the electric field E at (r, θ, ϕ) in the radiation zone.
(b) Find the time-averaged power detected per unit solid angle dP/dΩ as a function ofthe direction (θ, ϕ) in the radiation zone.
Answer: (a) From the electric dipole radiation formula in Eq. (65), the vector potentialof this problem in the radiation zone is given by
A(x, t) =1
rc˙D1|
t′1=t−|x−x ′
1|c
+1
rc˙D2|
t′2=t−|x−x ′
2|c
, (75)
where x ′1 = −L
2z and x ′
2 =L2z are the positions of the two dipoles. There are two retarded
times t′1 and t′2 because the two dipoles are at different positions. We need to consider Lbecause we are not told whether ωL ≪ c or not. But we do know r = |x| ≫ L since we arein the radiation zone. Therefore, the retarded times can be approximated as
t′1 ≈ t− r
c+
r · x ′1
c= t− r
c− L
2ccos θ. (76)
t′2 ≈ t− r
c+
r · x ′2
c= t− r
c+
L
2ccos θ. (77)
25
Putting together Eqs. (74)–(77), we have the vector potential
A(x, t) =1
rc
d
dtRe[D0ze
−i(ωt′1−α2) +D0ze
−i(ωt′2+α2)]
(78)
=2D0z
rc
d
dtRe
[e−iω(t− r
c) cos
(ωL
2ccos θ +
α
2
)](79)
= −2ωD0z
rcsin[ω(t− r
c
)]cos
(ωL
2ccos θ +
α
2
). (80)
The E and B fields are then obtained from Eqs. (66) & (67) to be
E(x, t) =1
cr ×
(r × ∂A
∂t
)= −2ω2D0θ
rc2cos[ω(t− r
c
)]cos
(ωL
2ccos θ +
α
2
)sin θ, (81)
B(x, t) = −1
cr × ∂A
∂t= −2ω2D0ϕ
rc2cos[ω(t− r
c
)]cos
(ωL
2ccos θ +
α
2
)sin θ, (82)
where we have used r × z = −ϕ sin θ and r × (r × z) = θ sin θ.
(b) The instantaneous radiation power is given by the Poynting vector
S =c
4πE × B =
ω4D20
πr2c3r cos2
[ω(t− r
c
)]cos2
(ωL
2ccos θ +
α
2
)sin2 θ. (83)
Therefore, the time-averaged radiation power is given by⟨dP
dΩ
⟩= r2⟨S · r⟩ = ω4D2
0
2πc3cos2
(ωL
2ccos θ +
α
2
)sin2 θ, (84)
where ⟨cos2[ω(t− r
c)]⟩ = 1
2when averaged over many periods. The above power distribution
exhibits an interference pattern of two electric dipole radiators.
26
Lecture 3 Special Relativity andCovariant Electrodynamics
1 Special relativity theory1.1 Lorentz transformationConsider a 1D space, in which there are two inertial reference frames (x, t) and (x′, t′).Suppose at t = t′ = 0, the origins of the two frames x = 0 and x′ = 0 coincide. Due to thetranslational symmetry of spacetime, the spatiotemporal points of the two reference framesshould be related by a linear transformation
x′ = Ax+Bt, t′ = Cx+Dt. (1)
Suppose the origin x′ = 0 of frame (x′, t′) moves in the (x, t) frame at velocity v. ThereforeB = −vA, so we have
x′ = A(x− vt). (2)
Then if the length and time scales of the two coordinate systems are set equal and thepositive directions are the same, the origin x = 0 of frame (x, t) should be moving in the(x′, t′) frame at velocity −v in the low speed limit v ≪ c. We can make an intuitive guessthat this symmetry of speed measurement v ↔ −v between two relatively moving framesstill holds for high speed and therefore obtain D = A.
Here comes the crucial assumption: the constancy of speed of light. When the wavefrontof a beam of light travels in frame (x, t) according to x = ct, its motion in the (x′, t′) framewill be given by x′ = ct′, where the speed of light c is the same. This crucial assumptiongives us C = −vA/c2, so we have
t′ = A(t− v
c2x). (3)
No speed measurements can determine the last parameter A, because it is a universal scalingfactor that equally affects x′ and t′. We can only determine A by some symmetry argument.We notice that when frame (x′, t′) moves relative to frame (x, t) at velocity v, so doesframe (−x, t) (a frame that remains stationary relative to frame (x, t) but has an x axisin the opposite direction) relative to frame (−x′, t′). The transformation from (−x′, t′) to(−x, t) should therefore be the same as the transformation from (x, t) to (x′, t′), i.e., the
27
transformation should be given by
−x = A(−x′ − vt′), t = A
(t′ +
vx′
c2
). (4)
These equations are consistent with Eqs. (2) & (3) if the scaling factor A satisfies
A2
[1 −v
− vc2
1
] [1 vvc2
1
]= A2
[1− v2
c20
0 1− v2
c2
]= I2×2, (5)
or equivalently,
A2
(1− v2
c2
)= 1. (6)
The equation has no real solution if |v| ≥ c, in which case the whole theory breaks down.To get out of trouble, we have to boldly assume |v| < c for all moving objects. Then we canprove using the relativistic velocity-addition formula that all physically relevant referenceframes, in which there are moving objects, also satisfy |v| < c. Remember relativity statesthat a reference frame with no moving objects in it is meaningless and metaphysical.
In the low speed limit v ≪ c, we should return to the Galilean transformation. HenceA(v ≪ c) ≈ 1, which lets us keep the positive square root by continuity and obtain
x′ = γ(x− vt), t = γ(t− vx
c2
), (7)
where γ = 1/√
1− v2/c2 is the Lorentz factor. We have changed the symbol A to the morecommon symbol γ. Eq. (7) is called the Lorentz transformation. To generalize the resultsto 3D, we intuitively assume y = y′ and z = z′ and find that the constancy of speed of lightis preserved, since we have
x2 + y2 + z2 = c2t2 ⇔ x′2 + y′2 + z′2 = c2t′2. (8)
We have outlined a brief derivation, or more precisely, a construction of Einstein’s theoryof special relativity. Building physical theories is unlike proving mathematical theorems, inthat we always have to make intuitive and reasonable assumptions, and extrapolations fromwhat daily life experiences and experiments have confirmed at the time. Einstein’s theory isof no exception. What makes him exceptional is his brilliance of guessing everything rightin a beautifully consistent way.
1.2 Formulas in 1D relativistic kinematics1.2.1 Length contraction
To measure the length of a moving ruler, we need to read its two endpoint coordinates x1, x2
at the same time t. According to the Lorentz transformation in Eq. (7), in the co-movingframe of the ruler that moves at velocity v relative to our lab frame, the rest length L0 of
28
the ruler is longer than the measured moving length L. We have
L0 = x′2 − x′
1 = γ(x2 − x1) = γL. (9)
Therefore, the length of a moving ruler L is shorter than its rest length L0 by a factor of1/γ. This effect is called length contraction.
1.2.2 Relativity of simultaneity
Two simultaneous events (x1, t) and (x2, t) can happen at different times
t′2 − t′1 = −γv
c2(x2 − x1), (10)
in a different reference frame (x′, t′) moving at velocity v relative to frame (x, t). But thisdoes not violate causality because
x′2 − x′
1 = γ(x2 − x1), (11)
so we always have ∣∣∣∣x′2 − x′
1
t′2 − t′1
∣∣∣∣ = c2
v> c (12)
for any physically accessible reference frames (v < c). This means no causal link has enoughtime to get from (x′
1, t′1) to (x′
2, t′2), or from (x′
2, t′2) to (x′
1, t′1), whichever is earlier.
1.2.3 Time dilation
The time interval between two consecutive events (x, t1) and (x, t2) at the same position isviewed larger in a different reference frame. We have
∆t′ = t′2 − t′1 = γ(t2 − t1) = γ∆t. (13)
The interval ∆ is called the proper time. It is often measured in the co-moving frame by aclock that travels with the moving object.
1.2.4 Velocity-addition formula
One may verify that two Lorentz transformations along x still give a Lorentz transformation.The resultant velocity is given by
v =v1 + v21 + v1v2
c2
. (14)
The derivation is a straightforward matrix multiplication. We have from multiplying two
29
Lorentz transformation matrices of Eq. (7) that
L(v1)L(v2) =1√
1− v21c2
(1 −v1
−v1c2
1
)1√
1− v22c2
(1 −v2
−v2c2
1
)(15)
=1√
(1− v21c2)(1− v22
c2)
(1 + v1v2
c2−(v1 + v2)
−v1+v2c2
1 + v1v2c2
)= L(v), (16)
with the resultant velocity v given in Eq. (14). The result velocity v of two subluminalvelocities |v1| < c and |v2| < c is therefore always less than c. This proves that if all movingobjects are subluminal in all reference frames, then all reference frames that have movingobjects in them are also subluminal to us. This rescues our theory for now. But we willdefinitely need a new theory if a superluminal object is found someday.
2 Covariant formWe have now established a new set of rules for spacetime. Galilean transformation is replacedby Lorentz transformation and the speed of light c in vacuum now becomes reference frameindependent. Consequently, old physical laws like Newton’s laws of motion that are covariantunder Galilean transformation have to be modified into new laws that are Lorentz covariant.The correctness of the new laws will then have to be tested experimentally. Other physicallaws like Maxwell’s equations that are already Lorentz covariant will be kept. To make thissifting of physical laws easier, we introduce the covariant form of 4-vectors.
2.1 4-vectorsThe spacetime vector x = (ct, x) is a 4-vector. It transforms under a Lorentz boost of velocityv in the x direction as
ct′
x′
y′
z′
=
γ −βγ 0 0
−βγ γ 0 00 0 1 00 0 0 1
ctxyz
, (17)
where β = v/c, γ = 1/√
1− β2. All 4-vectors transform exactly the same way. We mayconstruct may other 4-vectors from the spacetime vector x = (ct, x). We have, for example,the 4-momentum of a particle
p = m0dx
dτ= γm0
d(ct, x)
dt= (γm0c, γm0v), (18)
where the rest mass m0 and the proper time dτ are Lorentz scalars, i.e., quantities thatremain unchanged under Lorentz transformations. The proper time dτ is measured in theco-moving frame of the particle, and that dt = γdτ is due to time dilation.
30
In Newtonian mechanics, the total m0v of two colliding particles is conserved. But since(m0c,m0v) without γ is not a 4-vector, the conservation of total m0v and total m0 is in generalnot true in a different reference frame. We therefore modify the law to the conservation ofthe total 4-momentum p = (γm0c, γm0v) and this new law is justified experimentally. Thenew law will mean that the mass of a moving particle is γm0, greater than its rest mass m0
by a factor of γ that depends on the particle’s velocity v.The 4-current (density) given by
j = ρ0dx
dτ= γρ0
d(ct, x)
dt= (ρc, j) (19)
is a 4-vector, where ρ0 is the proper charge density in the co-moving frame of the charge.Charge is scalar, but volume is subject to Lorentz contraction. Therefore ρ = γρ0. The4-vectors j and p are constructed in very similar ways.
Finally, in the next section, from Maxwell’s equations, we will find that the 4-potentialA = (ϕ, A) in Gaussian units is a 4-vector. The derivation uses the pseudo inner product(length is not positive definite)
a · b = a0b0 − a · b (20)
of two 4-vectors a = (a0, a) and b = (b0, b). This inner product is useful because Lorentztransformation leaves the speed of light invariant, i.e.,
c2t2 − x2 − y2 − z2 = 0 ⇔ c2t′2 − x′2 − y′2 − z′2 = 0. (21)
This condition can be written as x · x = x′ · x′ using the inner product, which, to a lineartransformation, will mean that it leaves the inner product
a · b =1
2[(a + b) · (a + b)− a · a− b · b] (22)
of any two 4-vectors a and b invariant, so that a ·b is Lorentz scalar. From this technique youcan derive, for example, that the 4-wavenumber k = (ω
c, k) of a beam of light is a 4-vector,
because the phase k · x = ωt− k · x should be a Lorentz scalar.
2.2 Covariant electrodynamicsIn Gaussian units, Maxwell’s equations in vacuum are given by
∇ · E = 4πρ, ∇ · B = 0, (23)
∇× E = −1
c
∂B
∂t, ∇× B =
4π
cj +
1
c
∂E
∂t. (24)
We have solved Maxwell’s equations in the radiation zone in Lecture 2. Let’s briefly review
31
the procedure here and this time we want to rewrite the equations in covariant form. Afterintroducing the electromagnetic potentials A and ϕ, the fields are expressed as
E = −∇ϕ− 1
c
∂A
∂t, B = ∇× A. (25)
Using the Lorentz gauge condition
∇ · A+1
c
∂ϕ
∂t= 0, (26)
the potentials A and ϕ satisfy d’Alembert wave equations as given by
1
c2∂2ϕ
∂t2−∇2ϕ = 4πρ,
1
c2∂2A
∂t2−∇2A =
4π
cj. (27)
We notice that if we define A = (ϕ, A), from the 4-current j = (ρc, j) we have
A =4π
cj, =
1
c2∂2
∂t2−∇2, (28)
where is called the d’Alembertian operator and can be proven to be a Lorentz scalar. Theproof uses the 4-gradient
∂ =
(1
c
∂
∂t,−∇
), (29)
which can be shown to be a 4-vector because from Eq. (17) we have using the chain rule∂ct∂x∂y∂z
=
γ −βγ 0 0
−βγ γ 0 00 0 1 00 0 0 1
∂ct′∂x′
∂y′∂z′
. (30)
The symmetric Lorentz matrix is unchanged after transposition. Then the sign of β changestwice via inversion and the negative sign of the spatial gradient in Eq. (29). So the 4-gradient∂ is a 4-vector and the d’Alembertian = ∂ · ∂ is a Lorentz scalar. This means if we havefound A in one reference frame by solving the d’Alembert equation
A =4π
cj (31)
under the Lorentz gauge condition ∂ · A = 0, we may transform A and j like 4-vectors wewe go to a different frame and both the equation and the gauge condition will be satisfied,which uniquely determine A under certain boundary conditions. We do not need to solvethe equation again. This proves that A is a 4-vector.
32
2.3 Transformation of E and B fieldsThe transformation of E and B fields can be derived using the transformation of the 4-potential A = (ϕ, A) and the 4-gradient ∂ = (∂ct,−∇). They satisfy
∂ct′∂x′
∂y′∂z′
=
γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1
∂ct∂x∂y∂z
,
ϕ′
A′x
A′y
A′z
=
γ −βγ 0 0
−βγ γ 0 00 0 1 00 0 0 1
ϕAx
Ay
Az
. (32)
From Eq. (25), we obtain the E field in the new frame given by
E ′x = −
(∂x′ ∂ct′
)( ϕ′
A′x
)= −
(∂x ∂ct
) [ γ βγβγ γ
] [γ −βγ
−βγ γ
](ϕAx
)= Ex. (33)
E ′y = −∂y′ϕ
′ − ∂ct′A′y = −∂y(γϕ− βγAx)− (γ∂ct + βγ∂x)Ay (34)
= γ [(−∂yϕ− ∂ctAy)− β(∂xAy − ∂yAx)] = γ(Ey − βBz), (35)
E ′z = −∂z′ϕ
′ − ∂ct′A′z = −∂z(γϕ− βγAx)− (γ∂ct + βγ∂x)Az (36)
= γ [(−∂zϕ− ∂ctAz) + β(∂zAx − ∂xAz)] = γ(Ez + βBy), (37)
and the B field in the new frame is given by
B′x = ∂y′A
′z − ∂z′A
′y = ∂yAz − ∂zAy = Bx, (38)
B′y = ∂z′A
′x − ∂x′A′
z = ∂z(−βγϕ+ γAx)− (βγ∂ct + γ∂x)Az (39)
= γ [(∂zAx − ∂xAz) + β(−∂zϕ− ∂ctAz)] = γ(By + βEz), (40)
B′z = ∂x′A′
y − ∂y′A′x = (βγ∂ct + γ∂x)Ay − ∂y(−βγϕ+ γAx) (41)
= γ [(∂xAy − ∂yAx)− β(−∂ctAy − ∂yϕ)] = γ(Bz − βEy). (42)
The equations tell us that even in case of a non-relativistic limit, the E and B fields arestill reference frame dependent. This makes the electromagnetic field quite bizarre from thepoint of view of classical physics and also inspired Einstein to establish his great theory ofrelativity. The calculations of these formulas can be a bit lengthy to go through. But youare allowed to put them on your own formula sheet to prepare for the Quals.
3 Sample problems3.1 Problem 1 in Sec. 4, 2013Object A has a rest mass m and moves at constant velocity v = 0.8c. Object A inelasticallycollides with another object B, which is at rest and has the same mass m.
33
(a) Find their common velocity u after the collision.
(b) Suppose the heat released in this collision is radiated in photons distributed isotropi-cally in the rest frame of A+B. How much energy is radiated in the lab frame?
(c) Would the answer change if the emission were anisotropic?
Answer: This problem may be a bit confusing. In relativity, inelastic collisions, unlikewhat the name suggests, also satisfy energy conservation, and thus the total 4-momentum isconserved. In classical physics, we say that in an inelastic collision, part of the kinetic energyis dissipated into heat. In relativity, that heat corresponds to an increase in the rest mass(energy). An inelastic collision therefore converts part of the moving mass into rest mass,but the total mass (energy) is still conserved before the heat radiates away into photons,which can take away both energy and momentum.
(a) From the conservation of mass and momentum, we have(m√
1− v2/c2+m
)u =
mv√1− v2/c2
. (43)
Plugging in v = 0.8c, we find u = 0.5c. Results of the collision are plotted below:
Once the relativistic description of an inelastic collision is understood, the calculation of thecommon velocity u is not difficult.
(b) We may then find the rest mass M of the composite object A+B. It satisfies
M√1− 0.52
=8
3m ⇒ M =
4√3m = 2.309m > 2m. (44)
In the rest frame of A + B, the emitted photons have zero total momentum and a totalmass of ∆M = M − 2m = 0.309m, as the composite object A+B cools down and its massreturns to 2m. We may transform the photon energy back to the lab frame to obtain thetotal energy emitted
∆E =∆Mc2√1− 0.52
=
(8
3− 4√
3
)mc2 = 0.357mc2. (45)
Recall that the energy formula is also applicable to multiple mass points (photons).
34
(c) If the photons are not emitted isotropically in the rest frame of A+B, there will bea recoil of A + B in the opposite direction of the total momentum of the emitted photonsbecause of momentum conservation. This will change the energy radiated. For a quantitativediscussion, we consider two extreme situations. The most convenient calculates are done inthe lab frame.
In case that all photons are emitted in the same direction of u = 0.5c, we have(8
3mc− p
)2
=
(4
3mc− p
)2
+ (2mc)2 ⇒ ∆E = pc = 0.5mc2 > 0.357mc2, (46)
where we require that the rest mass of A+B returns to 2m when it is fully cooled down.
In case that all photons are emitted in the opposite direction of u = 0.5c, we have(8
3mc− p
)2
=
(4
3mc+ p
)2
+ (2mc)2 ⇒ ∆E = pc =1
6mc2 < 0.357mc2. (47)
In general the emitted energy ∆E will be in between 16mc2 and 1
2mc2. The recoil affects
the final speed of the A + B composite. The three cases satisfy 513c < 0.5c < 3
5c. Thus the
available energies of photon emission satisfy 12mc2 > 0.357mc2 > 1
6mc2.
3.2 Problem 2 in Sec. 4, 2013Consider the reaction π+ + n → K+ + Λ0. The rest masses of the particles are mπc
2 =140 MeV, mnc
2 = 940 MeV, mKc2 = 494 MeV, mΛc
2 = 1115 MeV. What is the thresholdkinetic energy of the π+ to create a K+ moving at an angle of 90 in the initial direction ofthe π+ in the lab frame, in which the n is at rest?
35
Answer: This type of 2D scattering problems are the commonest relativistic collisionproblems in the Quals. The standard method is to use the conservation of 4-momentum
(γπmπc, pπ) + (mnc, 0) = (γKmKc, pK) + (γΛmΛc, pΛ). (48)
Since we know the angle between pK and pπ is 90, and we don’t know the direction of pΛ,we may use the inner product to cancel pΛ to obtain
m2Λc
2 = (γΛmΛc, pΛ) · (γΛmΛc, pΛ) = ((γπmπ +mn − γKmK)c, pπ − pK)2 (49)
= (γπmπ +mn − γKmK)2c2 − (p2π + p2K) (50)
= (γ2πm
2πc
2 − p2π) + (γ2Km
2Kc
2 − p2K)− 2γπγKmπmKc2
+ 2mn(γπmπ − γKmK)c2 +m2
nc2 (51)
= m2πc
2 +m2Kc
2 +m2nc
2 − 2γπγKmπmKc2 + 2mn(γπmπ − γKmK)c
2. (52)
Therefore, we have
γπmπ =m2
Λ −m2π −m2
n −m2K + 2γKmKmn
2(mn − γKmK). (53)
The right-hand side monotonically increases with γK . The minimum is found by pluggingin γK = 1, which means we generate a K+ at rest. Generating a moving K+ will definitelycost more energy. Therefore, we get the threshold kinetic energy
Eπ = (γπ − 1)mπ ≥ m2Λ −m2
π
2(mn −mK)− 1
2(mn −mK)−mπ = 1009MeV. (54)
3.3 Problem 3 in Sec. 2, 2013Suppose we have a point charge q in uniform motion with relativistic velocity v in the xdirection. What is the electric field configuration sourced by this charge? To be concrete,suppose the charge position as a function of time t is xq(t) = vtx, and we are interested inthe electric field at some arbitrary location and time.
Answer: This problem can be solved using either the retarded potential (no radiationzone approximation) or the relativistic transformation of the E and B fields. The secondmethod will be simpler if you remember the formulas. In the co-moving frame, the E-fieldis just the Coulomb field of a charge q (Lorentz scalar) at the origin x ′
q(t′) = 0 and there is
no B-field. The fields are thus given by
E ′(x ′, t′) =qx ′
|x ′|3, B ′(x ′, t′) = 0. (55)
36
We may then get the components of the E and B fields in the lab frame one by one, usingthe formulas we derived for the relativistic transformation of E and B fields. We get
Ex(x, t) = E ′x(x
′, t′) =qx′
(x′2 + y′2 + z′2)3/2=
qγ(x− vt)
[γ2(x− vt)2 + y2 + z2]3/2, (56)
Ey(x, t) = γE ′y(x
′, t′) =qγy′
(x′2 + y′2 + z′2)3/2=
qγy
[γ2(x− vt)2 + y2 + z2]3/2, (57)
Ez(x, t) = γE ′z(x
′, t′) =qγz′
(x′2 + y′2 + z′2)3/2=
qγz
[γ2(x− vt)2 + y2 + z2]3/2, (58)
because B ′(x ′, t′) = 0. However, B(x, t) is nonzero, which can be understood as beinggenerated by the moving charge in the lab frame. Electricity and magnetism are closelyrelated in relativity.
3.4 Problem 3 in Sec. 4, 2009A monochromatic beam of light is incident onto a flat mirror. In the lab frame, the mirroris traveling at the speed v in the +x direction. The plane of the mirror is perpendicular tothe x axis. The incident light beam has frequency f and is propagating at angle θ from thex axis. Find the frequency fR and the angle θR of the reflected light beam as measured inthe lab frame.
Answer: The problem can be solved using the Lorentz transformation of either the 4-wavevector k = (ω/c, k) if we think of light as a wave or the 4-momentum p = (E/c, p) if wethink of light as photons. Since both k and p are 4-vectors, they transform the same wayand give the same answer. The 4-wavevectors of the incident and reflected beams are
k = (k,−k cos θ, k sin θ, 0)T , k =2πf
c, (59)
kR = (kR, kR cos θR, kR sin θR, 0)T , kR =
2πfRc
. (60)
37
In the co-moving frame of the mirror, the 4-wavevector of incident beam becomes
k′ =
γ −βγ 0 0
−βγ γ 0 00 0 1 00 0 0 1
k−k cos θk sin θ
0
=
γk(1 + β cos θ)−γk(β + cos θ)
k sin θ0
, (61)
and the 4-wavevector of reflected beam becomes
k′R =
γ −βγ 0 0
−βγ γ 0 00 0 1 00 0 0 1
kRkR cos θRkR sin θR
0
=
γkR(1− β cos θR)γkR(cos θR − β)
kR sin θR0
, (62)
In the coming frame, k′ and k′R are symmetric about the normal direction (x axis) and equalin frequency. So we have
γk(1 + β cos θ) = γkR(1− β cos θR),
−γ(β + cos θ) + γkR(cos θR − β) = 0,
k sin θ = kR sin θR.
(63)
(64)
(65)
Since k′ · k′ = k′R · k′R = 0, only two of the above three equations are independent, whichis consistent with the two unknowns kR and θR to be solved for. Nevertheless, we can stilluse all three equations to double check. The first two equations are simultaneous linearequations of kR and kR cos θR, which give us kR = γ2k(1 + β2 + 2β cos θ),
kR cos θR = γ2k[(1 + β2) cos θ + 2β
].
(66)
(67)
One may get from these equations k2R sin2 θR = k2 sin2 θ, which is consistent with Eq. (65).
We may therefore write our answer as
2πfRc
= γ22πf
c(1 + β2 + 2β cos θ) ⇒ fR = γ2f(1 + β2 + 2β cos θ), (68)
sin θR =k
kRsin θ =
sin θ
γ2(1 + β2 + 2β cos θ)=
(1− β2) sin θ
1 + β2 + 2β cos θ. (69)
The law of reflection θR = θ is recovered if β = 0. Relativity generalizes physical laws tohigh speed scenarios, enriches our understanding of mass and energy and reveals the essenceof electricity, magnetism, light and spacetime.
38
Lecture 4 Quantum Mechanics - Theory
There are various topics that can appear in Secs. 3 (only QM) and 4 (with relativity) ofthe Quals. A thorough revision of the topics in Griffith’s Introduction to Quantum Mechanicscan be very helpful and necessary. This lecture only covers some important selected topics.
1 Simple harmonic oscillatorThe time-independent Shcrödinger equation of a 1D simple harmonic oscillator is given by
Hψ(x) =
(− ~2
2m
d2
dx2+
1
2mω2x2
)ψ(x) = Eψ(x). (1)
We define the lowering and raising operators as
a =
√mω
2~x+
ip√2m~ω
=
√mω
2~x+
√~
2mω
d
dx, (2)
a† =
√mω
2~x− ip√
2m~ω=
√mω
2~x−
√~
2mω
d
dx, (3)
so that in terms of them, the Hamiltonian is written as
H = ~ω(a†a+
1
2
), (4)
and we can prove the commutation relations
[a, a†] = 1, [H, a†] = ~ωa†, [H, a] = −~ωa. (5)
From these commutation relations, together with the fact that the Hamiltonian H is boundedbelow, we deduce that there is a ground state ψ0(x) such that
aψ0(x) =
√mω
2~xψ0(x) +
√~
2mω
dψ0(x)
dx= 0 ⇒ ψ0(x) = Ce−
mω2~ x2
, (6)
where the arbitrary coefficient C can be determined by the normalization condition∫ +∞
−∞|ψ0(x)|2 = 1 ⇒ C =
(mωπ~
)1/4, (7)
39
up to an arbitrary phase. Once the ground state ψ0(x) is found, we can use the raisingoperator a† to find the excited state ψn(x) by acting a† onto the ground state n times, i.e.,ψn(x) ∝ (a†)nψ0(x). Since we have
∥a|n⟩∥2 = ⟨n|a†a|n⟩ = n⟨n|n⟩ = n, (8)
∥a†|n⟩∥2 = ⟨n|aa†|n⟩ = ⟨n|a†a+ 1|n⟩ = n+ 1, (9)
where |n⟩ is the Dirac ket of the nth normalized eigenstate ψn(x), we obtain
a|n⟩ =√n |n− 1⟩, a†|n⟩ =
√n+ 1 |n+ 1⟩. (10)
Therefore, the wave function of the nth eigenstate can be found to be
ψn(x) =(a†)n√n!ψ0(x) =
1√2nn!
(√mω
~x−
√~mω
d
dx
)n (mωπ~
)1/4e−
mω2~ x2
. (11)
2 Hydrogen atom2.1 General central force fieldThe Schrödinger equation in a general central potential V (r) in 3D, where V (r) is only afunction of the distance r from the origin, is given in spherical coordinates by
− ~2
2me
[1
r2∂
∂r
(r2∂
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂
∂θ
)+
1
r2 sin2 θ
∂2
∂ϕ2
]ψ + V (r)ψ = Eψ. (12)
For separable solutions ψ(r, θ, ϕ) = R(r)Y (θ, ϕ), we have
2me
~2[E − V (r)]r2 +
1
R
∂
∂r
(r2∂R
∂r
)= − 1
Y sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)− 1
Y sin2 θ
∂2Y
∂ϕ2. (13)
The left-hand side is a function of r only, and the right-hand side is a function of θ and ϕ.Both of them have to equal a constant. Let’s call that constant l(l + 1), so we have
1
sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
1
sin2 θ
∂2Y
∂ϕ2+ l(l + 1)Y = 0. (14)
The solutions to this equation under proper boundary conditions are the spherical harmonicsYlm(θ, ϕ), where l = 0, 1, 2, . . . and m = 0,±1,±2, . . . ,±l. The angular part of the wavefunction ψ is always Ylm(θ, ϕ). Only the radial equation
1
r2d
dr
(r2dR
dr
)=
1
r
d2
dr2(rR) =
(l(l + 1)
r2+
2me
~2[V (r)− E]
)R (15)
40
may depend on the specific form of V (r). Let u(r) = rR(r) be the effective radial wavefunction, so we have an effective 1D Schrödinger equation
− ~2
2me
d2u
dr2+
[V (r) +
~2
2me
l(l + 1)
r2
]u = Eu. (16)
Apart from the central potential V (r), the electron is also subject to a centrifugal potentialwhich propels it away from the origin. Bound states can form if V (r) is attractive and itsattractive part at r → 0 is less singular than 1/r2. The Coulomb potential satisfies theseconditions, so the electron can form bound states in the hydrogen atom.
2.2 Coulomb potentialIn the hydrogen atom, an electron orbits around a proton due to the Coulomb attractionbetween them. Assuming the more massive proton is approximately at rest when the electronmoves around it, the electron is subject to a central Coulomb potential given by
V (r) = − e2
4πϵ0r∝ −1
r. (17)
Plugging this into Eq. (16) and by introducing dimensionless variables
ρ =
√−2meE
~r, ρ0 =
e2
4πϵ0~
√2me
−E, (18)
the effective radial Schrödinger equation (16) becomes
d2u
dρ2=
[1− ρ0
ρ+l(l + 1)
ρ2
]u. (19)
A full solution of this equation can be found in § 4.2.1 of Griffith’s book, which is a bitlengthy to go through and will not be duplicated here. We will only give an estimate of theenergy and length scales of the hydrogen atom using dimension analysis. Given an integerquantum number l, there should be a set of discrete values that the dimensionless variableρ0 can take in order to make u(ρ→ ∞) converge. Since ρ0 is dimensionless, the ground stateenergy can be estimated by setting ρ0 ∼ 1, so that
E ≈ −2me
(e2
4πϵ0~
)2
= −2mec2α2, (20)
which deviates from the exact ground state energy −12mec
2α2 = −13.6 eV of the hydrogenatom by a positive dimensionless factor 4. In expressing the results, we have used the finestructure constant α = e2
4πϵ0~c =1
137. The length scale, or atomic size, of the hydrogen atom
41
can be obtained by setting ρ ∼ 1, which leads to
r ≈ ~√−2meE
≈ ~2mecα
=1
2aB, (21)
where aB = ~mecα
= 0.529Å is the Bohr radius. Dimension analysis is an important techniquetested in the Quals. Though inaccurate, it gives the correct energy and length scales of thehydrogen atom and is applicable to other problems that are not analytically solvable.
3 Angular momentum
In quantum mechanics, the three components Lx, Ly, Lz of an angular momentum ˆL do not
commute. The commutation relations are given by
[Ly, Lz] = i~Lx, [Lz, Lx] = i~Ly, [Lx, Ly] = i~Lz. (22)
Defining L2 = L2x + L2
y + L2z, we obtain
[L2, Lx] = [L2, Ly] = [L2, Lz] = 0. (23)
We conventionally use the eigenvalues L2 = l(l + 1)~2 and Lz = m~ to completely describea basis state |l,m⟩ with a definite angular momentum. We may further define raising andlowering operators L± = Lx ± iLy in the z direction. They satisfy
[L2, L±] = 0, [Lz, L±] = ±~L±. (24)
Since we have for the raising operator L+ that
∥L+|l,m⟩∥2 = ⟨l,m|L−L+|l,m⟩ = ⟨l,m|L2x + L2
y + i[Lx, Ly]|l,m⟩ (25)= ⟨l,m|L2 − L2
z − ~Lz|l,m⟩ = (l −m)(l +m+ 1)~2, (26)
and for the lowering operator L− we have
∥L−|l,m⟩∥2 = ⟨l,m|L+L−|l,m⟩ = ⟨l,m|L2x + L2
y − i[Lx, Ly]|l,m⟩ (27)= ⟨l,m|L2 − L2
z + ~Lz|l,m⟩ = (l +m)(l −m+ 1)~2, (28)
the normalization constants of L±|l,m⟩ are determined and we obtain
L±|l,m⟩ =√
(l ∓m)(l ±m+ 1)~ |l,m± 1⟩. (29)
These normalization constants are useful for calculating the Clebsch-Gordan coefficients byhand. The formalism applies to both orbital and spin angular momenta.
42
4 Electron in an electromagnetic field4.1 Spinless theoryRecall that in Lecture 1, when a non-relativistic electron (with charge −e and mass me)moves in an electromagnetic field, the classical Lagrangian is given by
L =1
2mev
2 + e[ϕ(r, t)− v · A(r, t)
]. (30)
The conjugate momentum of r is
p = ∇vL = mev − eA(r, t). (31)
Therefore the classical Hamiltonian is given by
H = p · v − L =1
2mev
2 − eϕ(r, t) =1
2me
(p+ eA)2 − eϕ. (32)
The quantized Hamiltonian operator is therefore given by
H =1
2me
(−i~∇+ eA)2 − eϕ. (33)
The time-dependent Schrödinger equation is then
i~∂
∂tψ(r, t) =
[1
2me
(−i~∇+ eA)2 − eϕ
]ψ(r, t). (34)
4.2 Probability fluxFrom the above Schrödinger equation, we have
i~ψ∗∂ψ
∂t= ψ∗
[1
2me
(−i~∇+ eA)2 − eϕ
]ψ, (35)
−i~ψ∂ψ∗
∂t= ψ
[1
2me
(i~∇+ eA)2 − eϕ
]ψ∗. (36)
Taking the difference of the two equations, we therefore obtain
i~∂
∂t(ψ∗ψ) =
1
2me
[ψ∗ (−i~∇+ eA)2 ψ − ψ (i~∇+ eA)2 ψ∗
](37)
= − ~2
2me
(ψ∗∇2ψ − ψ∇2ψ∗)− i~me
∇ · (eA ψ∗ψ) (38)
= − i~me
∇ ·[~ Im(ψ∗∇ψ) + eA ψ∗ψ
]. (39)
43
Since we know ψ∗ψ is the probability density, the probability flux is given by
Jp =~me
Im(ψ∗∇ψ) + eA
me
ψ∗ψ. (40)
The extra term due to A in Jp can be understood from the classical conjugate momentump = mev − eA, from which we obtain v = (p + eA)/me. The probability flux Jp and theelectric current j = −eJp are therefore still determined by v rather than p.
4.3 Spin-1/2 theoryIn quantum mechanics, electrons are spin-1/2 particles. This leads to an extra term in H
due to the Zeeman coupling of the spin magnetic moment with B = ∇× A. We have
H =1
2me
(−i~∇+ eA)2 − eϕ+gse
2me
~2ˆσ · B, (41)
where gs ≈ 2 is the electron spin g-factor, and ˆσ = exσx + eyσy + ezσz is the Pauli vectorwith its three components being the Pauli matrices
σx =
(0 11 0
), σy =
(0 −ii 0
), σz =
(1 00 −1
). (42)
In Pauli representation, the time-evolution of a spin wave function is given by
i~∂
∂t
(ψ↑ψ↓
)=
[1
2me
(−i~∇+ eA)2 − eϕ+gse~4me
(Bz Bx − iBy
Bx + iBy −Bz
)](ψ↑ψ↓
). (43)
The total probability density n(r) = |ψ↑(r)|2 + |ψ↓(r)|2 in real space is independent of thespin representation (try σx-representation, for example, the total density n(r) is the same).One may verify that the probability flux Jp corresponding to n(r) still satisfies Eq. (40)because the magnetic field B(r) only flips the spins locally. So the spatial probability fluxstill agrees with the spinless theory.
5 Identical particles(Problem 4 in Sec. 3, 2012) Consider two identical electrons with spin-1/2 and mass m thatare confined to a 1D box with length L. The potential energy is 0 inside the box and infinityoutside. The interaction between two particles can be expressed by the potential energyV (x1, x2), where x1 and x2 are the coordinates of the two particles.
(a) We first consider noninteracting particles, i.e., V (x1, x2) = 0. What are the groundstate energy and wave function, considering spatial and spin state symmetry?
44
(b) What are the energy and degeneracy of the first excited state? Write down explicitwave function considering spatial and spin state symmetry.
(c) Now consider a simple form of electron-electron interaction V (x1, x2) = V0 δ(x1 − x2).Assuming V0 is small, find the correction for the ground state energy in (a) due tothis e-e interaction.
(d) Describe how degeneracy in the excited state discussed in (b) would be lifted.
Answer: We use a sample problem to help review our knowledge of identical fermions.Assuming you are familiar with single-electron eigenstates of an infinite square well, wedirectly begin with placing two electrons into two spin states to satisfy the Pauli exclusionprinciple. Then we require some perturbation theory for parts (c) and (d).
(a) The ground state is reached by putting both electrons in the ground state n = 1 ofthe infinite square well. Since electrons are fermions, the spins of the two electrons will haveto be opposite. This gives us a singlet spinor times a symmetric spatial wave function as thetwo-body ground-state spin wave function
Ψ0(x1, x2) = ψ1(x1)ψ1(x2)×↑1↓2 − ↓1↑2√
2, (44)
where ψn(x) =√
2Lsin nπx
L, n = 1, 2, 3, . . . are the single-electron eigenstates of an infinite
square well. The ground state energy is given by
E0 =1
2me
(π~L
)2
× 2 =π2~2
meL2. (45)
(b) In the first excited state, one of the electrons gets excited to n = 2, while the otherstays in n = 1. When the two-body spinor is a singlet, the spatial part of the two-bodywave function should be symmetric. When the spinor is a triplet, the spatial part should beantisymmetric. Therefore, we have a four-fold degeneracy given by
Ψsinglet(x1, x2) =ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)√
2× ↑1↓2 − ↓1↑2√
2, (46)
Ψtriplet(x1, x2) =ψ1(x1)ψ2(x2)− ψ2(x1)ψ1(x2)√
2×
↑1↑2
↑1↓2 + ↓1↑2√2
↓1↓2
. (47)
All four states have the same energy
E1 =1
2me
(π~L
)2
+1
2me
(2π~L
)2
=5π2~2
2meL2. (48)
45
(c) Since the interaction strength V0 is small and the singlet ground state Ψ0 is non-degenerate, we can use the first-order non-degenerate perturbation theory to obtain theenergy correction of E0, which is given by
∆E0 = ⟨Ψ0|V |Ψ0⟩ = V0
∫∫dx1dx2 |ψ1(x1)|2|ψ1(x2)|2 δ(x1 − x2) (49)
= V0
∫dx |ψ1(x)|4 =
4V0L2
∫ L
0
sin4 πx
Ldx =
3V02L
. (50)
(d) In the triplet states, the two electrons never meet, because the spatial wave functionΨsinglet(x1, x2) is antisymmetric so that Ψtriplet(x, x) = 0. Therefore, the matrix elements ofthe perturbation V (x1, x2) = V0 δ(x1 − x2) between any triplets or a triplet and a singletare zero. The perturbation matrix is thus block-diagonal and only the energy of the singletstate gets shifted up by
∆E1s = V0
∫∫dx1dx2
∣∣∣∣ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)√2
∣∣∣∣2 δ(x1 − x2) (51)
= 2V0
∫ L
0
dx |ψ1(x)ψ2(x)|2 =8V0L2
∫ L
0
sin2 πx
Lsin2 2πx
Ldx =
2V0L. (52)
The four-fold degeneracy of the first excited states are therefore split into 3 + 1 due tothe electron-electron repulsion V0 δ(x1 − x2). The singlet state becomes higher in energythan the triplet states. When the repulsion strength V0 is large enough to make the singletground state energy E0 higher than the triplet excited states, the triplet states become thenew ground states, which leads to a ferromagnetic phase transition when there are manyidentical copies of the system. This problem tells us that magnetism can arise from spin-independent electron-electron repulsion once the repulsion is strong enough.
6 Perturbation theoryThe problem in the previous section uses perturbation theory to deal with electron-electroninteractions. Here we give a revision of the general formalism of both time-independentand time-dependent perturbation theories and include some sample problems to apply thesetheories to. Perturbation theory is a powerful and important technique.
6.1 Time-independent perturbation theoryFor simplicity we consider the non-degenerate case up to first and second order. Considerthe Hamiltonian
H = H0 + λH1, (53)
where H0 is the unperturbed system, H1 is the perturbation applied, and λ is a tuningparameter that controls the strength of perturbation. Suppose the energy of the eigenstate
46
|ψ(0)n ⟩ of H0 is E(0)
n . When λ = 1, the perturbation is fully turned on, |ψ(0)n ⟩ becomes |ψn⟩
and E(0)n becomes En. We expand |ψn⟩ and En to all orders of λ as follows:
|ψn⟩ = |ψ(0)n ⟩+ λ|ψ(1)
n ⟩+ λ2|ψ(2)n ⟩+ · · · , (54)
En = E(0)n + λE(1)
n + λ2E(2)n + · · · . (55)
We plug the two series into the eigen-equation H|ψn⟩ = En|ψn⟩ and use the normalizationcondition ⟨ψn|ψn⟩ = 1 and the phase condition Im ⟨ψ(0)
n |ψn⟩ = 0 for all λ ∈ [0, 1] to uniquelyfix the eigenstate |ψn⟩. We have
(H0 + λH1)(|ψ(0)n ⟩+ λ|ψ(1)
n ⟩+ · · · ) = (E(0)n + λE(1)
n + · · · )(|ψ(0)n ⟩+ λ|ψ(1)
n ⟩+ · · · ), (56)
(⟨ψ(0)n |+ λ⟨ψ(1)
n |+ λ2⟨ψ(2)n |+ · · · )(|ψ(0)
n ⟩+ λ|ψ(1)n ⟩+ λ2|ψ(2)
n ⟩+ · · · ) = 1, (57)
⟨ψ(0)n |(|ψ(0)
n ⟩+ λ|ψ(1)n ⟩+ · · · ) = (⟨ψ(0)
n |+ λ⟨ψ(1)n |+ · · · )|ψ(0)
n ⟩. (58)
Comparing the terms on both sides of Eq. (56) in each order of λ, we obtain
H0 |ψ(0)n ⟩ = E(0)
n |ψ(0)n ⟩, (59)
H0 |ψ(1)n ⟩+ H1|ψ(0)
n ⟩ = E(0)n |ψ(1)
n ⟩+ E(1)n |ψ(0)
n ⟩, (60)
H0 |ψ(2)n ⟩+ H1|ψ(1)
n ⟩ = E(0)n |ψ(2)
n ⟩+ E(1)n |ψ(1)
n ⟩+ E(2)n |ψ(0)
n ⟩. (61)
Applying the same technique to Eqs. (57) & (58), we have
⟨ψ(0)n |ψ(0)
n ⟩ = 1, ⟨ψ(0)n |ψ(1)
n ⟩ = 0, ⟨ψ(0)n |ψ(2)
n ⟩ = −1
2⟨ψ(1)
n |ψ(1)n ⟩. (62)
One can keep going to higher orders. Projecting Eqs. (60) & (61) onto ⟨ψ(0)n | and acting H0
to the left to cancel the E(0)n term, we obtain
⟨ψ(0)n |H0|ψ(1)
n ⟩+ ⟨ψ(0)n |H1|ψ(0)
n ⟩ = E(0)n ⟨ψ(0)
n |ψ(1)n ⟩+ E(1)
n ⟨ψ(0)n |ψ(0)
n ⟩ (63)
⇒ E(1)n = ⟨ψ(0)
n |H1|ψ(0)n ⟩, (64)
⟨ψ(0)n |H0|ψ(2)
n ⟩+ ⟨ψ(0)n |H1|ψ(1)
n ⟩ = E(0)n ⟨ψ(0)
n |ψ(2)n ⟩+ E(1)
n ⟨ψ(0)n |ψ(1)
n ⟩+ E(2)n (65)
⇒ E(2)n = ⟨ψ(0)
n |H1|ψ(1)n ⟩. (66)
The first-order energy correction E(1)n in Eq. (64) is easy to calculate. The perturbation H1
averaged over the unperturbed state |ψ(0)n ⟩ gives the energy correction E(1)
n to first order.The second-order energy E(2)
n requires the first-order correction ψ(1)n of the wave function,
which we now compute. Eq. (60) can be rearranged to become
(E(0)n − H0)|ψ(1)
n ⟩ = (H1 − E(1)n )|ψ(0)
n ⟩. (67)
47
From Eq. (62), we know |ψ(1)n ⟩ is orthogonal to |ψ(0)
n ⟩, which in the non-degenerate casemeans that |ψ(1)
n ⟩ is in the invertible eigenspace of E(0)n − H0. We may then expand both
sides in the eigenstate basis of H0 and obtain
|ψ(1)n ⟩ =
∑′
m
|ψ(0)m ⟩⟨ψ(0)
m |H1|ψ(0)n ⟩
E(0)n − E(0)
m
, (68)
where the prime ′ means to exclude the m = n term. Plugging the expansion of |ψ(1)n ⟩ into
Eq. (66), we have the second-order energy correction
E(2)n =
∑′
m
⟨ψ(0)n |H1|ψ(0)
m ⟩⟨ψ(0)m |H1|ψ(0)
n ⟩E(0)
n − E(0)m
=∑′
m
|⟨ψ(0)m |H1|ψ(0)
n ⟩|2
E(0)n − E(0)
m
. (69)
Therefore E(2)n is always negative if ψ(0)
n is the ground state of H0. Aside from eigenstateexpansion, Eq. (67) can also sometimes be solved as a differential equation using the Green’sfunction method as in scattering theory. Generally speaking, the expansion is only recom-mended if it truncates, otherwise the infinite series can be hard to sum.
6.2 Sample problem(Problem 3 in Sec. 3, 2011) Consider a 1D harmonic oscillator. The Hamiltonian is
H0 = − ~2
2m
∂2
∂x2+
1
2mω2
0x2. (70)
We now consider the effect of a perturbation Hamiltonian
H ′ = λxf(t), (71)
where λ is a constant, and f(t) is a time-dependent dimensionless function.
(a) Assume that the perturbation Hamiltonian is time-independent, i.e., f(t) = 1 at alltime t. Find the energy shifts of the ground state to the first nonzero order in f .
Answer: Since the ground state ψ0(x) = ⟨x|ψ0⟩ is an even function of x, while H ′ isodd in x, the mean value ⟨ψ0|H ′|ψ0⟩ = 0. So there is no first-order correction to the groundstate energy due to H ′. We therefore have to go to the second order. From Eq. (69), thesecond-order correction of ground-state energy is given by
∆E(2)0 =
∞∑n=1
|⟨ψ(0)n |H1|ψ(0)
0 ⟩|2
E(0)0 − E(0)
n
= −λ2∞∑n=1
|⟨n|x|0⟩|2
n~ω0
, (72)
where |n⟩ = |ψ(0)n ⟩ is the nth eigenstate of the unperturbed simple harmonic oscillator H0.
The matrix element can be evaluated using the raising and lowering operators. We have
48
from Eqs. (2) & (3) in this lecture that
x =
√~
2mω0
(a+ a†). (73)
Therefore, the matrix elements can be evaluated using
⟨n|a+ a†|0⟩ = ⟨n|a†|0⟩ = ⟨n|1⟩ = δn1, (74)
not to forget checking the normalization constants from Eq. (10). Therefore,
∆E(2)0 = − λ2
2mω0
∞∑n=1
δn1nω0
= − λ2
2mω20
. (75)
We will do a degenerate perturbation problem in solid-state physics in Lecture 5. The part (b)of this problem involves the time-dependent perturbation theory. Instead of going throughthe general formalisms of Fermi’s golden rule, Rabi oscillations and adiabatic theorem, etc.,we only briefly review this part by working out the sample problem.
6.3 Time-dependent perturbation theory(Problem 3 in Sec. 3, 2012) (Continued)
(b) Now assume the perturbation is time-dependent, such that the power spectral densityof the time-stationary signal f(t) is given by
ρ(ω) ≡ limT→+∞
1
2πT
∣∣∣∣∫ T
0
f(t)eiωtdt
∣∣∣∣2 = ρ(ω) + ρ(−ω), (76)
where the line shape is a Gaussian function
ρ(ω) =
√1
πω20
e−(ω−ω0)2/ω20 , (77)
with the constant ω0 given in the harmonic oscillator. You can verify that∫ ∞
−∞ρ(ω)dω = lim
T→+∞
1
T
∫ T
0
|f(t)|2dt, (78)
which gives the total time-averaged power. Assume that the oscillator is in its ground stateat t = 0. What are the excited states of this oscillator that the ground state can make atransition into? Find the transition rate into these states using Fermi’s golden rule.
Answer: In this problem the total Hamiltonian H(t) = H0 + λxf(t) is time-dependent.We are first asked to derive the selection rule and then determine the transition rates. The
49
time-independent part H0 is a 1D simple-harmonic oscillator. At any time t, we may expandthe state |ψ(t)⟩ in terms of the eigenstates of H0 as
|ψ(t)⟩ =∑k
ck(t)|k⟩e−iEkt/~. (79)
Therefore, from the Schrödinger equation of the state |ψ(t)⟩, we obtain
i~∑k
ck(t)|k⟩e−iEkt/~ = λf(t)∑k
ck(t) x |k⟩e−iEkt/~. (80)
Projecting both sides onto a harmonic eigenstate ⟨j|, we have
dcj(t)
dt= −iλ
~f(t)
∑k
ck(t)⟨j|x|k⟩ei(Ej−Ek)t/~. (81)
Initially the system is in the ground state |0⟩, so we have ck(t = 0) = δk0. In case λ is verysmall, the coefficients will evolve very slowly. We may therefore solve this problem iteratively.We start from a zeroth order approximation c(0)k (t) = δk0, which means the coefficients neverevolve. We plug this into the right-hand side and get a first-order corrected set of coefficientsc(1)j (t) for all j = 0 (the initial state) as given by
c(1)j (t) = −iλ
~
∫ t
0
dt′f(t′)∑k
δk0⟨j|x|k⟩ei(Ej−Ek)t′/~ (82)
= −iλ~⟨j|x|0⟩
∫ t
0
dt′f(t′)eijω0t′ . (83)
From the results in Part (a), we know ⟨j|x|0⟩ is nonzero for j = 1 only, which tells us that toa first-order approximation, the system can only transition to the first excited state n = 1.This is the so-called selection rule. Transitions to other states are called forbidden transitions(although higher-order terms may allow the system to get to those states eventually). Theprobability of transitioning to n = 1 is therefore given by
P(1)0→1(t) = |c(1)1 (t)|2 = λ2
~2|⟨1|x|0⟩|2
∣∣∣∣∫ t
0
dt′f(t′)eiω0t′∣∣∣∣2 . (84)
In case that λ is small, the only way of getting a large transition probability P(1)0→1(t) is to
wait for a long time. Since we have the power spectral density ρ(ω) defined in Eq. (76), theasymptotic behavior of P (1)
0→1(t) as t→ +∞ is given by
P(1)0→1(t) ≈
2πλ2
~2|⟨1|x|0⟩|2ρ(ω0)t =
πλ2
m~ω0
[ρ(ω0) + ρ(−ω0)]t, t→ +∞. (85)
This is Fermi’s golden rule. Perturbation theory breaks down if t is so large that P (1)0→1(t)
is no longer small. Nevertheless, it will give us the right transition rate on an intermediate
50
timescale, which is much longer than 1/ω0 but much shorter than the time required forP
(1)0→1(t) to get large (which means comparable to 1). Such a timescale definitely exists if λ
is small. On the intermediate timescale, we have
P(1)0→1(t) =
dP(1)0→1(t)
dt≈ πλ2
m~ω0
[ρ(ω0) + ρ(−ω0)] =
√πλ2
m~ω20
(1 +
1
e4
), (86)
which is the transition rate from the ground state n = 0 to the first excited state n = 1.The 1/e4 = 1.83% term can be neglected as an approximation.
Note: The original problem was formulated in a less rigorous way. It said
f(t) =
∫ ∞
−∞ρ(ω)(eiωt + e−iωt)dω, (87)
where the line shape function is the same as Eq. (77). If the problem is formulated this way,then f(t) would be a Gaussian pulse (or wavelet) in the time domain
f(t) =
√1
πω20
· 2Re∫ ∞
−∞exp
[−(ω − ω0)
2
ω20
+ iωt
]dω = 2e−
14ω20t
2
cosω0t, (88)
so we would not get Fermi’s golden rule but some steady-state probability P(1)0→1(t → +∞)
instead. This would make the problem very different.
51
Lecture 5 Quantum Mechanics - Applications
In this lecture, we talk about some more advanced topics of quantum mechanics thatare not covered in Griffith’s book. There will be more sample problems than derivations ofgeneral theories in this lecture. We try to be broad to cover various topics from high energyphysics, particle physics, condensed matter physics, etc.
1 Galilean transformation of QM(Problem 1 in Sec. 3, 2010) Two observers in different inertial frames will need different wavefunctions to describe the same physical system. To make things simple, consider how it worksin the non-relativistic case. The first observer uses coordinates (x, t) and a wave functionψ(x, t), while the second observer uses (x ′, t) and ψ(x ′, t). From Galilean transformation,x ′ = x − v t, where v is a constant velocity. The wave functions for the two observers aresaid to be related as follows:
ψ(x ′, t) = ψ(x, t) exp
(− i
~
[mv · x− mv2
2t
]). (1)
Despite the innocuous look (it’s just a phase!), this transformation has interesting effects.
(a) Let us first verify that it makes sense. Suppose ψ(x, t) is the wave function of a freeparticle of momentum p = (px, py, pz). Show that ψ(x ′, t) indeed describes a free particlewith the proper momentum.
Answer: Since we are already given the transformation from ψ(x, t) to ψ(x ′, t), whichcan be derived from the Galilean covariance of the Schrödinger equation, solving the problemnow becomes some straightforward math work. The wave function of a free particle in thex frame with momentum p is given by
ψ(x, t) = exp
(i
~
[p · x− p2
2mt
]), (2)
up to an arbitrary normalization constant with an arbitrary phase. In the new frame ofx ′ = x− v t, from Eq. (1), the wave function becomes
ψ(x ′, t) = exp
(i
~
[(p−mv) · (x ′ + v t)− p2
2mt+
mv2
2t
])(3)
52
= exp
(i
~
[(p−mv) · x ′ − (p−mv)2
2mt
]), (4)
which is the wave function of a free particle with momentum p ′ = p−mv.
(b) Now let’s put this to work. Suppose we have a hydrogen atom, which at t < 0 wasat rest with the electron in the ground state |1s⟩ described by the wave function
ψ(x) = ψ1s(x) =1√πa3B
exp
(− r
aB
), r = |x|, (5)
where aB is the Bohr radius. Suppose at t = 0 the proton suddenly starts to move (e.g.due to a collision with a neutron) in the z direction with the velocity v. Let the change inthe velocity be so abrupt that the electron wave function remains the same. What is theprobability at t > 0 to find the moving hydrogen atom with the electron to be still in theground state ψ1s(x)?
Answer: After t = 0, the proton moves at a constant velocity v relative to the labframe. Therefore, in its co-moving inertial frame, we still have a time-independent Coulombpotential, in which the probabilities of finding the electrons in different |nlm⟩ states areconstant. To go into that frame, we transform the wave function ψ(x, t = 0) to
ψ(x ′, t = 0) =1√πa3B
exp
(− r
aB
)exp
(− i
~mv · x
). (6)
Since the velocity v is in the z direction, we have at t = 0 that
v · x = v · x ′ = vr′ cos θ′, (7)
so that the wave function is written as
ψ(x ′, t = 0) =1√πa3B
exp
(− r′
aB
)exp
(− i
~mvr′ cos θ′
). (8)
Suppose now we measure the |nlm⟩ state of ψ(x ′, t = 0). The probability amplitude offinding the electron in the ground state ψ1s(x
′) is given by∫d3x′ψ∗
1s(x′)ψ(x ′, t = 0) =
∫d3x′
πa3Bexp
(−2r′
aB
)exp
(− i
~mvr′ cos θ′
)(9)
= 2π
∫ ∞
0
r′2dr′
πa3Bexp
(−2r′
aB
)∫ π
0
sin θ′dθ′ exp
(− i
~mvr′ cos θ′
)(10)
=4~
mva3B
∫ ∞
0
r′dr′ exp
(−2r′
aB
)sin
mvr′
~=
(4~2
4~2 +m2v2a2B
)2
. (11)
53
The above integral is most conveniently calculated by taking the imaginary part of a gammaintegral using Euler’s formula eiθ = cos θ + i sin θ. The ground state probability is thereforethe modulus squared of the probability amplitude given by
P =
(4~2
4~2 +m2v2a2B
)4
. (12)
The probability P → 1 if mvaB ≪ ~, and P → 0 if mvaB ≫ ~.
2 Dirac equation (Relativistic QM)(Problem 1 in Sec. 3, 2011) The Dirac equation can be written in terms of two Pauli-type2-component spinors, ϕ and χ. For a free particle of mass m and momentum p, these areplane waves whose spinor coefficients satisfy
mϕ+ p · σχ = Eϕ,
p · σϕ−mχ = Eχ,
(13)(14)
where E is the energy eigenvalue and σ = (σ1, σ2, σ3) are the Pauli matrices. We work innatural units c = ~ = 1.
(a) Solve the two equations for E and give a physical interpretation of every solution fora given momentum p.
Answer: The Dirac equation is the relativistic version of the Schrödinger equation ofa spin-1/2 particle. In this problem, we solve for free-particle energy eigenstates of a givenmomentum p. The two coupled matrix equations can be rewritten as
(E −m)ϕ = p · σχ, (E +m)χ = p · σϕ. (15)
Since E ±m are c-numbers that commute with any 2× 2 matrix p · σ, we have
(E2 −m2)ϕ = (p · σ)2ϕ =
(pz px − ipy
px + ipy −pz
)2
ϕ = p2ϕ, (16)
where p = |p| =√p2x + p2y + p2z is the magnitude of the momentum p. The same eigen-
equation holds for χ. To get a nonzero ϕ (or χ), we have
E = ±√m2 + p2. (17)
Clearly, the E =√m2 + p2 solution gives the relativistic dispersion relation of a free particle
with mass m and momentum p. But what about E = −√m2 + p2? Do the negative-energy
states really exist? If so, why would particles occupy the positive-energy states instead ofthe negative ones that are lower in energy?
54
This also puzzled Dirac. To solve this problem, he made a wildly brilliant assumptionbased on the the Pauli exclusion principle. He assumed that the negative-energy states havealready been occupied, forming the so-called Dirac sea, so that a positive-energy particlecannot drop into one of these states by Pauli exclusion. We can’t feel the existence of theDirac sea because it is everywhere and we are used to it (like the atmospheric pressure).Dirac also deduced that if a negative-energy particle in the Dirac sea gets excited into apositive-energy state, it will leave behind a hole in the Dirac sea. This hole corresponds to a“deficit” of an electron with momentum p and energy −
√m2 + p2. The hole as an excitation
would thus appear to be an anti-particle with energy√m2 + p2 and momentum −p. The
antiparticle of an electron is a positron. Dirac predicted its existence in 1928, and four yearslater, in 1932, the positron was observed experimentally.
(b) The scattering by a spin-independent central potential V between the initial and finalmomenta p and p ′ is described in the Born approximation by the matrix element
V (p ′, p)(ϕ†p ′ϕp + χ†
p ′χp
),
where V (p ′, p) is the spatial matrix element of V between momentum eigenstates. Usingyour solution to part (a), show that
χ†p ′χp = ϕ†
p ′Σ(p′, p)ϕp, (18)
where Σ(p ′, p) = F1(p′, p) + σ · F2(p
′, p). Determine F1 and F2.
Answer: This problem wants us to find the relativistic correction to a spin-independentscattering problem. In non-relativistic quantum mechanics, we only have one 2-componentspinor ϕ, instead of ϕ and χ. From Eq. (15), we see that ϕ and χ are related by
χ†p ′χp = ϕ†
p ′(p ′ · σ)(p · σ)
(Ep +m)(Ep ′ +m)ϕp = ϕ†
p ′Σ(p′, p)ϕp. (19)
One may verify the formula (A · σ)(B · σ) = (A · B)I2×2 + i(A× B) · σ. Therefore,
Σ(p ′, p) =(p · p ′)I2×2 + i(p ′ × p) · σ
(Ep +m)(Ep ′ +m). (20)
Coefficients F1 and F2 can be read from this expression. The relativistically corrected matrixelement of the scattering potential is therefore given by
V (p ′, p)(ϕ†p ′ϕp + χ†
p ′χp
)= V (p ′, p)ϕ†
p ′
[I2×2 +
(p · p ′)I2×2 + i(p ′ × p) · σ(Ep +m)(Ep ′ +m)
]ϕp. (21)
(c) Suppose that the electron is initially moving in the x3 direction, with its spin alignedalso in the x3 direction, and that it is scattered by a very weak potential V into the x2
55
direction. Show that, in the non-relativistic limit, the probability of spin flip in the courseof such a scattering event is small. Quantify what is meant by “small” here.
Answer: In relativity, spin and spatial motion are related and mutually convertible.The initial state of the electron is given by ϕ = (1, 0)T , i.e., spin up in x3 direction. Giventhat its initial momentum p = pz is also in the x3 direction, we obtain
χ =p · σE +m
ϕ =p
E +m
[1 00 −1
](10
)=
p
E +m
(10
). (22)
We are interested in the transition probability into the final state with ϕ′ = (0, 1)T , i.e., spindown in x3 direction, and p ′ = py in the x2 direction. We are assuming an elastic scatteringEp = Ep ′ = E. The spinor χ′ corresponding to ϕ′ with momentum p ′ is given by
χ′ =p ′ · σE +m
ϕ′ =p
E +m
[0 −ii 0
](01
)=
p
E +m
(−i0
). (23)
The relativistically corrected matrix element in Part (b) is therefore
⟨p ′ ↓|V |p ↑⟩ = V (p ′, p)(ϕ′†ϕ+ χ′†χ
)=
ip2
(E +m)2V (p ′, p). (24)
One can obtain the same result using Eq. (21). The transition rate as in Fermi’s golden rulefrom |p ↑⟩ to |p ′ ↓⟩ is therefore proportional to
|⟨p ′ ↓|V |p ↑⟩|2 = p4
(E +m)4|V (p ′, p)|2. (25)
In order for this to be small, we require p ≪ E + m =√p2 +m2 + m, or equivalently
p ≪ 2mc in SI units. This problem tells us that the spin cannot be flipped by a spin-independent potential in non-relativistic quantum mechanics, but can be flipped when p iscomparable to mc, where m is the rest mass.
3 Spin-orbit interaction(Problem 2 in Sec. 4, 2010) Consider a hydrogen atom. The spin-orbit interaction
Hso = A(r)S · L, (26)
where S and L are the spin and orbital angular momenta of the electron. In Gaussian units,
A(r) =e2
2m2ec
2r3. (27)
56
(a) Describe in words the origin of the spin-orbit interaction.
Answer: In Griffith’s book § 6.3.2, the spin-orbit interaction is explained using theclassical picture of an electron orbiting a proton. In the co-moving frame of the electron, wesee the proton orbiting the electron, forming a circular current that generates a magneticfield along L at the electron. The magnetic dipole moment of the electron spin is along −S,so the coupling term in the Hamiltonian is proportional to S · L. The co-moving frame of theelectron is a non-inertial frame. When transforming back to the lab frame, the coefficient ofS · L gets corrected by a factor of 1
2due to the Thomas precession.
Griffith’s explanation implies that the type of force that binds the electron in the atommatters. The spin-orbit coupling constant would be different if the electrostatic potentialwere replaced by, e.g., a gravitational potential of the same value. This is because the spin-orbit coupling is a relativistic effect. Even though in the lab frame a gravitational potentialcan be made equal to an electrostatic potential, the difference emerges in the co-movingframe of the electron, because the electromagnetic potentials (ϕ, A) and the gravitationalcurvature tensor of spacetime would Lorentz transform differently.
(b) Construct the basis wave functions that diagonalize Hso.
Answer: Without Hso, electronic states are labeled by |nlmms⟩. Now that S and L arecoupled, we transform from the uncoupled basis to the coupled basis |nljmj⟩, where j andmj are the quantum numbers of the total angular momentum J = L+ S. The two bases arerelated by the Clebsch-Gordan coefficients. In the coupled basis, the spin-orbit interactionHso in Eq. (26) is angular diagonal. We have
⟨n′l′j′m′j|Hso|nljmj⟩ = ⟨n′l′|A(r)|nl⟩⟨l′j′m′
j|S · L|ljmj⟩. (28)
Since S · L = 12(J2 − L2 − S2) takes eigenvalues in the coupled basis, we have
⟨l′j′m′j|S · L|ljmj⟩ =
~2
2
[j(j + 1)− l(l + 1)− 3
4
]δll′δjj′δmjm′
j. (29)
The radial integral ⟨n′l′|A(r)|nl⟩ need not be orthogonal with respect to n and n′. But themixing of n = n′ states is only a small relativistic correction compared with the energysplitting due to the principal quantum numbers n, n′ before considering Hso. So we normallyneglect this effect and restrict ourselves to the same electron shell n = n′.
(c) Obtain the spin-orbit interaction energies for hydrogen in the state with principalquantum number n = 2.
Answer: For n = 2, we have the l = 0 and l = 1 states (2s and 2p). From Eq. (29),there is no spin-orbit coupling for l = 0, because j = 1/2 and the matrix element of S · L iszero. We are left with the l = 1 and j = 1/2 or 3/2 states.
57
The rest of this answer is not required for the exam problem. Since the radial wavefunction R21(r) of hydrogen is not given, you are not required to evaluate the radial integral⟨21|A(r)|21⟩ explicitly but only need to give a formal answer. From
R21(r) =r
2√6a
5/2B
e−r/2aB , (30)
we can evaluate the radial integral
⟨21|A(r)|21⟩ =∫ ∞
0
4πr2dr
(r
2√6a
5/2B
e−r/2aB
)2e2
2m2ec
2r3(31)
=πe2
12m2ec
2a3B=πα4mec
2
12~2, (32)
where α = e2/~c = 1/137 is the fine structure constant in Gaussian units, aB = ~/αmec =0.529 Å is the Bohr radius. From Eq. (29), we have for l = 1 that
⟨ljmj|S · L|ljmj⟩ =~2
2or − ~2 (33)
which corresponds to j = 3/2 or 1/2, respectively. From Eq. (28), the spin-orbit couplingenergy is thus correspondingly given by
⟨nljmj|Hso|nljmj⟩ =πα4mec
2
24or − πα4mec
2
12. (34)
The energy splitting due to spin-orbit coupling is O(α4mec2), smaller than the ground-state
energy O(α2mec2) by a factor of α2. Numerically we have mec
2 = 0.511 MeV, α2mec2 =
27.2 eV= 1 Hartree, and α4mec2 = 1.45 meV is the energy scale of spin-orbit effects.
4 Solid-state physics(Problem 4 in Sec. 4, 2013) Consider a two-dimensional square lattice in which noninter-acting electrons move. We will treat the motion of the electrons in the nearly free motionapproximation. We consider the effects of the lattice using a two-dimensional potential. Twopossibilities to describe the potential are:
Choice A,
V (x, y) = V0 cos2πx
acos
2πy
a. (35)
Choice B,
V (x, y) = V0
(cos
2πx
a+ cos
2πy
a
). (36)
58
Here a is the lattice constant and V0 is the strength of the potential. Let us ask whetherthese potentials will open a band gap in the electronic states at the point in momentumspace located at (π/a, π/a).
(a) In the limit where V0 goes to zero, what is the degeneracy of the lowest-energy stateat the point (kx, ky) = (π/a, π/a)?
Answer: This problem requires some understanding of solid-state physics and bandtheory. In the nearly free limit, the problem is also a good testing ground for degenerateperturbation theory. In a crystal with many noninteracting electrons, the single-particleSchrödinger equation is given by[
− ~2
2me
∇2 + V (x)
]ψ(x) = Eψ(x), (37)
where V (x) is the periodic lattice potential that can be expanded into a Fourier series
V (x) =∑G
VG eiG·x, (38)
where G sums over the whole reciprocal lattice. In the momentum space, the Schrödingerequation (37) becomes
p2
2me
ψ(p) +∑G
VGψ(p− ~G) = Eψ(p). (39)
As we can see from Eq. (39), ψ(p) only gets related with ψ(p − ~G) in the eigen-equation,where G is any lattice wave vector. In the limit of V0 → 0, we have a free electron withenergy E = ~2π2/ma2 when it is in momentum space located at k1 = (π/a, π/a). Thereare 3 other momenta in the first Brillouin zone k2 = (−π/a, π/a), k3 = (−π/a,−π/a), andk4 = (π/a,−π/a), which have the same energy and differ from k1 by a lattice wave vector.Therefore, we need to solve a 4-fold degenerate perturbation problem.
(b) In the presence of a nonzero value of V0, which one of the two potentials will open aband gap at the point (kx, ky) = (π/a, π/a)? Justify.
Answer: For choices A and B, we have
VA(x, y) =V04
[ei
2πa(x+y) + e−i 2π
a(x+y) + ei
2πa(x−y) + e−i 2π
a(x−y)
]. (40)
VB(x, y) =V02
(ei
2πax + e−i 2π
ax + ei
2πay + e−i 2π
ay). (41)
59
The matrices of VA(x, y) and VB(x, y) under plane wave basis k1, k3, k2, k4 are given by
VA =V04
0 1 0 01 0 0 00 0 0 10 0 1 0
, VB =V02
0 0 1 10 0 1 11 1 0 01 1 0 0
. (42)
We changed the order of the 4 wave vectors to give VA and VB better block structures.For small V0, we restrict ourselves to the 4-dimensional subspace spanned by k1, k2, k3, k4,because other momenta that have a different energy will contribute very little to the wavefunction. This is the idea of degenerate perturbation theory. We now find the eigenvalues ofVA and VB to get the energy corrections to the plane waves. We have
det (EI4×4 − VA) =
[E2 −
(V04
)2]2
= 0 ⇒ E = ±V04, (43)
det (EI4×4 − VB) = E2(E2 − V 20 ) = 0 ⇒ E = 0,±V0. (44)
Therefore, a band gap is opened at (π/a, π/a) by VA but not by VB, because there is anE = 0 solution that is unaffected by VB to first order. The 4-fold degeneracy split into 2+ 2for both choices A and B.
5 Josephson junction(Problem 2 in Sec. 4, 2010) Two superconductors separated by a thin insulator (called aJosephson junction) are connected to a DC voltage V , as shown in the figure below.
Let ψ1 be the wave function of the Bose-condensed superconducting electron pairs on oneside of the superconductor and ψ2 be the wave function on the other side. The two wave
60
functions are related by the time-dependent Schrödinger equationsi~dψ1
dt= eV ψ1 +Kψ2,
i~dψ2
dt= −eV ψ2 +Kψ1,
(45)
(46)
where the constant K is a real parameter of the junction related to the tunneling process ofthe electron pairs across the insulator, and V is the DC voltage applied. In this problem,the wave functions are expressed as ψ1 =
√n1 e
iθ1 and ψ2 =√n2 e
iθ2 , where n1 and n2 arethe condensation densities, and θ1 and θ2 are the phases of the condensate wave functionsof superconductors 1 and 2, respectively.
(a) Show that the current density of this junction is given by
J =dn1
dt= −dn2
dt= J0 sin δ, (47)
where δ = θ2 − θ1. Find the expression for J0 in terms of K, n1 and n2.
Answer: From the Schrödinger equations for ψ1 and ψ2, we can derivei~d
dt(ψ∗
1ψ1) = K(ψ∗1ψ2 − ψ∗
2ψ1),
i~d
dt(ψ∗
2ψ2) = K(ψ∗2ψ1 − ψ∗
1ψ2).
(48)
(49)
Using ψ1 =√n1 e
iθ1 and ψ2 =√n2 e
iθ2 , we obtain
J =dn1
dt= −dn2
dt=
2K
~Im (ψ∗
1ψ2) =2K
~√n1n2 sin (θ2 − θ1). (50)
Since δ = θ2 − θ1, our result agrees with Eq. (47), with J0 =2K~√n1n2.
(b) Assume that initially at t = 0 the condensation densities n1 and n2 are equal andlarge, and the tunneling probability K is small so we have n1(t) ≈ n2(t) afterwards. Showthat the current density J derived in Part (a) oscillates periodically over time. Find thefrequency of oscillation in terms of the applied DC voltage V .
Answer: From the Schrödinger equations of ψ1 and ψ2 again, we can derivei~d
dt(ψ∗
2ψ1) = 2eV ψ∗2ψ1 +K(n2 − n1),
i~d
dt(ψ∗
1ψ2) = −2eV ψ∗1ψ2 +K(n1 − n2).
(51)
(52)
61
Adding these two equations, we cancel the K(n2 − n1) term and obtain
d
dt(√n1n2 cos δ) = −2eV
~√n1n2 sin δ. (53)
Since the densities n1(t) ≈ n2(t) are approximately equal and n1(t) + n2(t) = const, theamplitude J0 ∝ √
n1n2 takes its extremum and its oscillation is negligible. We can thenassume √
n1n2 ≈ const and obtain
d
dtcos δ = − sin δ
dδ
dt= −2eV
~⇒ δ =
2eV
~t+ δ0, (54)
where δ0 is the initial phase difference of ψ1 and ψ2. So we have from Eq. (50) that
J =2K
~√n1n2 sin
(2eV
~t+ δ0
), (55)
which gives us the oscillating frequency ω = 2eV/~. Interestingly, we don’t get an equationof motion for J that resembles the simple-harmonic oscillator, but get a linear equation forδ which produces the oscillation. This is because the frequency is determined by the appliedvoltage V rather than some intrinsic parameters.
62
Lecture 6 Thermodynamics and Other Topics
Thermodynamics and lots of other topics such as statistical physics of noninteractingparticles, dynamics of ideal fluids, geometric optics and many more can appear in Secs. 5 – 6of the Quals. Aside from exact calculations, order of magnitude estimation and dimensionanalysis can be useful as well.
1 Statistical physics1.1 Boltzmann distribution(Problem 5 in Sec. 5, 2013) Consider a one-dimensional chain consisting of N molecules.Each molecule can exist in one of two configurations, A and B, with energies EA and EB
and lengths lA and lB, respectively.
(a) Write down the partition function for the system at temperature T , assuming theexternal force applied F = 0.
Answer: Let’s denote the state of the ith molecule as ni, which can be either A or B.The molecules are assumed distinguishable so that it is valid to label them, and sequencedso that they never swap positions. The partition function is the sum of the Boltzmann factorover all microstates
Z =∑ni
exp
(−β
N∑i=1
Eni
), (1)
where β = 1/kBT and ni = (n1, n2, . . . , nN). Some manipulation of Z yields that
Z =∑ni
N∏i=1
exp(−βEni) =
N∏i=1
∑ni=A,B
exp(−βEni) = ZN
1 , (2)
63
where Z1 = e−βEA + e−βEB is the partition function of a single molecule. For noninteractingand distinguishable particles, the partition function of the canonical system they form issimply the product of the individual partition functions, i.e., Z = ZN
1 .
(b) Assume that EA > EB and lA > lB (still with F = 0). Give an expression for theaverage length L of the molecule chain as a function of temperature at fixed large N . Sketchthe result, give the high and low-T limits and indicate the characteristic temperature atwhich the length changes from the high-T to the low-T limits.
Answer: The mean value of the total length L =∑
j lnjis given by
⟨L⟩ = 1
Z
∑ni
[(N∑j=1
lnj
)N∏i=1
exp(−βEni)
]=
1
ZN1
N∑j=1
∑ni
lnj
N∏i=1
exp(−βEni) (3)
=1
ZN1
N∑j=1
∑nj=A,B
lnjexp(−βEnj
)∑
ni\nj
∏i =j
exp(−βEni) (4)
=1
Z1
N∑j=1
∑nj=A,B
lnjexp(−βEnj
) = N · lA exp(−βEA) + lB exp(−βEB)
exp(−βEA) + exp(−βEB). (5)
The mean total length is the sum of the mean lengths of each molecule. In the low-T limit,β = 1/kBT → +∞, we have exp(−βEA) ≪ exp(−βEB) given EA > EB. All molecules wantto stay in the lower-energy state B. The length is then ⟨L⟩ ≈ NlB. In the high-T limit,β = 1/kBT → 0, we have exp(−βEA) ≈ exp(−βEB) ≈ 1. This makes the A and B statesnearly equally likely as their energy difference EA − EB is much smaller than the thermalfluctuation kBT . The length is then ⟨L⟩ ≈ N(lA + lB)/2. The transition temperature willbe of the scale T ∼ (EA − EB)/kB.
(c) Calculate the linear response function χ = ∂⟨L⟩/∂F showing how the length changesif a small tensile force is applied to the two ends of the chain as shown in the figure.
Answer: Since the force applied is a constant, we may introduce its potential energyand include it into the total energy of the system. The partition function becomes
Z =∑ni
exp
[−β
(N∑i=1
Eni− F
N∑i=1
lni
)]=
N∏i=1
∑ni=A,B
exp [−β(Eni− Flni
)] = ZN1 , (6)
where the partition function of a single molecule now becomes
Z1(β, F ) = exp [−β(EA − FlA)] + exp [−β(EB − FlB)] . (7)
64
Repeating the same derivation as in Part (b), we obtain
⟨L⟩ = N · lA exp [−β(EA − FlA)] + lB exp [−β(EB − FlB)]
exp [−β(EA − FlA)] + exp [−β(EB − FlB)]. (8)
Let’s define ∆(F ) = EA − EB − F (lA − lB), so the isothermal linear response function
χT =
(∂⟨L⟩∂F
)T
= N∂
∂F
(lA + lB eβ∆(F )
1 + eβ∆(F )
)β
= N∂
∂F
(lB +
lA − lB1 + eβ∆(F )
)β
(9)
= − N(lA − lB)
[1 + eβ∆(F )]2 e
β∆(F )βd∆(F )
dF=
N(lA − lB)2
4kBTsech2
[∆(F )
2kBT
]. (10)
The elasticity function χT = χT (T, F ) is the reciprocal of the spring constant (or stiffness).In case EA > EB and lA > lB, the elasticity χT increases as F increases if F < Fc =(EA − EB)/(lA − lB). Then χT decreases as F increases in the F > Fc regime. If EA − EB
and lA − lB have opposite signs, the elasticity χT decreases monotonically. This problemgives a model for the non-constant elasticity of rubber bands.
1.2 Bose-Einstein distribution(Problem 6 in Sec. 5, 2013) A large isolated box of volume V is filled with black bodyradiation in thermal equilibrium. The radiation has an initial total energy U0 (relative tovacuum in the same volume). A cyclic engine pulls heat from the boxed radiation, which issubsequently dumped into a large thermal reservoir whose temperature remains fixed at T .
(a) What is the minimum value of work W that must be expended on the cyclic engineto bring the radiation in the box arbitrarily close to being a vacuum if the boxed radiationtemperature was initially T?
Answer: The partition function of the boxed black body radiation is given by
Z =∏p,ζ
∞∑n=0
e−βnϵp =∏p,ζ
1
1− e−βcp, (11)
where p sums over all single-photon momenta allowed by the box V and ζ sums over the twopossible polarizations. The dispersion relation is ϵp = cp for photons. The Helmholtz freeenergy F = −kBT lnZ of the boxed black body radiation (photon gas) is given by
F = kBT∑p,ζ
ln(1− e−βcp) = kBT2V
(2π~)3
∫ ∞
0
4πp2 ln(1− e−βcp)dp (12)
=k4BT
4V
π2~3c3
∫ ∞
0
u2 ln(1− e−u)du = −π2k4BT
4V
45~3c3. (13)
65
Recall that the integral can be evaluated using the Taylor expansion of logarithm, i.e.,∫ ∞
0
u2 ln(1− e−u)du = −∞∑n=1
1
n
∫ ∞
0
u2e−nudu = −∞∑n=0
2
n4= −2ζ(4), (14)
where ζ(4) = π4/90. The Riemann zeta function ζ(2n) at even integers can be computedanalytically by applying Parseval’s identity to a constructed Fourier series. We will notreproduce the result here. The numbers ζ(2) = π2/6, ζ(3) = 1.202, ζ(4) = π4/90 can be puton your formula sheet if necessary. Now the total internal energy of the photon gas
U0 =1
Z
∏p,ζ
∞∑n=0
nϵp e−βnϵp = − 1
Z
(∂Z
∂β
)V
= −T 2 ∂
∂T
(F
T
)V
=π2k4
BT4V
15~3c3. (15)
Therefore, if we want the heat engine to dump all the energy U0 of the photon gas intothe heat reservoir at temperature T , the minimum amount of heat we must dump into thereservoir, in order not to violate the increase of entropy, is TS, where S is the entropy of thephoton gas. The minimum work that has to be done by the heat engine is therefore
W = TS − U0 = −F =π2k4
BT4V
45~3c3=
U0
3. (16)
(b) What is the average energy in the lowest-frequency mode ω0 = 2πf0 at arbitrarytemperature T?
Answer: The average energy of a single mode is equal to the energy ϵ0 = ~ω0 of thatmode times the average number of photons, which is given by Bose-Einstein distribution
⟨n0⟩ =∞∑n=0
ne−βnϵ0/ ∞∑
n=0
e−βnϵ0 =1
eβϵ0 − 1. (17)
Therefore, the average energy is given by
⟨E0⟩ = ϵ0⟨n0⟩ =~ω0
eβ~ω0 − 1. (18)
1.3 Fermi-Dirac distribution(Problem 5 in Sec. 5, 2019) Electron-positron equilibrium
(a) A box is filled with Planckian radiation of temperature kBT ≫ mec2, where me is the
electron mass. A population of e± pairs is maintained in equilibrium with radiation throughthe reaction e+ + e− ↔ γ + γ. Find the number density of positrons in the box.
Answer: In this problem, the box is a fixed-volume system with variable number ofparticles of 3 types. And it is assumed to be electrically neutral. Therefore, at equilibrium,
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all three species have zero chemical potential. The number of positrons (or electrons) in thebox volume V is therefore given by
⟨Ne⟩ =∑p,s
1
eβϵp + 1=
2V
(2π~)3
∫ ∞
0
4πp2dp
eβ√
m2ec
4+p2c2 + 1, (19)
where p sums over all single-positron (electron) momenta allowed by the boundary conditionsof the box V and s sums over the 2 possible spins. Let ϵ =
√m2
ec4 + p2c2 be the new
integration variable, so that the density
⟨ne⟩ =⟨Ne⟩V
=1
π2~3c3
∫ ∞
ϵ0
√ϵ2 − ϵ20
eβϵ + 1ϵdϵ, (20)
where ϵ0 = mec2 is the rest-energy of a positron (or electron). In case that kBT ≫ mec
2, wehave βϵ0 → 0. Therefore we set ϵ0 ≈ 0 as an approximation to obtain
⟨ne⟩ ≈1
π2~3c3
∫ ∞
0
ϵ2dϵ
eβϵ + 1=
3ζ(3)
2π2
k3BT
3
~3c3= 0.183
(kBT
~c
)3
, (21)
where we have used the formula∫ ∞
0
xndx
ex + 1=
(1− 1
2n
)n! ζ(n+ 1), (22)
and ζ(3) = 1.202, both of which were given in the original exam problem.
(b) Consider the same problem but now assume that the box is also filled with neutralelectron-proton matter at the equilibrium temperature T . The proton number density np isgiven. Find the electron chemical potential µ assuming µ ≪ kBT .
Answer: At temperature kBT ≫ mec2, the neutral electron-proton matter quickly
dissociates into free electrons and protons. We therefore have 4 noninteracting species:protons, positrons, electrons and photons. The charge neutrality condition n−
e = n+e + np
requires that there are more electrons than positrons. This makes the chemical potential µof the electrons higher than the chemical potential µ′ of the positrons. Another condition tosatisfy is that the reversible reaction
e+ + e− ↔ γ + γ (23)
is in phase equilibrium, so we have µ + µ′ = 0. Therefore, µ′ = −µ and from Eq. (20), thepositron and electron densities are given by
⟨n±e ⟩ =
1
π2~3c3
∫ ∞
ϵ0
√ϵ2 − ϵ20
eβ(ϵ±µ) + 1ϵdϵ. (24)
67
Assuming µ ≪ kBT , we have βµ ≪ 1 and can then use the Taylor expansion
1
eβ(ϵ±µ) + 1=
1
eβϵ + 1∓ βµeβϵ
(eβϵ + 1)2+ · · · (25)
in terms of βµ to obtain
np = ⟨n−e ⟩ − ⟨n+
e ⟩ ≈2
π2~3c3
∫ ∞
ϵ0
βµeβϵ
(eβϵ + 1)2
√ϵ2 − ϵ20 ϵdϵ. (26)
From the condition kBT ≫ mec2 given in Part (a), we have βϵ0 ≪ 1 and therefore
np ≈2
π2~3c3
∫ ∞
0
βµeβϵ
(eβϵ + 1)2ϵ2 dϵ = − 2βµ
π2~3c3∂
∂β
∫ ∞
0
ϵdϵ
eβϵ + 1(27)
=4k2
BT2µ
π2~3c3
∫ ∞
0
xdx
ex + 1=
2k2BT
2µ
π2~3c3ζ(2) =
k2BT
2µ
3~3c3, (28)
using Eq. (22) and ζ(2) = π2/6. Therefore the chemical potential of electrons
µ =3~3c3
k2BT
2np. (29)
This formula is for small np only such that µ ≪ kBT is satisfied.
2 Thermodynamics2.1 Equation of state(Problem 6 in Sec. 5, 2011) Suppose you have discovered a substance whose equation of stateis given by p = 5T 3/V . Note that the units for the constant “5” must be Nm/K3. Yourcolleagues have carried out another experiment where they have found that the internalenergy is a function of V and T given by U(V, T ) = BT x ln(V/V0) + f(T ), where B, x andV0 are constants, and f(T ) is an unmeasured function that only depends on T . What arethe values of B and x? (In terms of the constants you already know).
Answer: From the TdS equations of thermodynamics for a closed system, we have
dU = TdS − pdV = T
[(∂S
∂T
)V
dT +
(∂S
∂V
)T
dV
]− pdV. (30)
We can derive Maxwell’s relations using the Helmholtz free energy F = U −TS, which gives
dF = d(U − TS) = −SdT − pdV ⇒(∂S
∂V
)T
=
(∂p
∂T
)V
. (31)
68
Therefore, we can determine the internal energy from the equation of state. We have
dU = T
(∂S
∂T
)V
dT +
[T
(∂p
∂T
)V
− p
]dV = cV dT +
10T 3
VdV, (32)
where we have plugged in the specific heat at constant volume cV = T (∂S/∂T )V and theequation of state p = 5T 3/V . Now from the colleague’s expression for the internal energyU(V, T ) = BT x ln(V/V0) + f(T ), we have
dU =
[xBT x−1 ln
(V
V0
)+ f ′(T )
]dT +
BT x
VdV. (33)
Comparing Eqs. (32) & (33), we obtain
BT x
V=
10T 3
V⇒ B = 10Nm/K3, x = 3, (34)
so that the internal energy U(V, T ) = 10T 3 ln(V/V0) + f(T ) is determined. This problemgives an example that in thermodynamics, we can use the equation of state to determinethe internal energy from Maxwell’s relations.
2.2 Liquid-gas phase transition(Problem 6 in Sec. 5, 2009) Consider a liquid and its vapor, both at the same temperature inequilibrium with each other. Let the equilibrium vapor pressure be P g
0 . An inert, insolublegas is added to the closed container holding the liquid. Find an expression for the newequilibrium partial vapor pressure P g of the liquid in terms of V l
mol, the molar volume of theliquid, P , the partial pressure of the inert gas, P g
0 , the vapor pressure without the inert gas,and T , the temperature of the system. Assume the gas phase of the liquid obeys the idealgas law, and the liquid is incompressible.
Answer: The TdS equation for the Gibbs free energy of an open system is given by
dG = −SdT + V dp+ µdn, (35)
69
where n = N/NA is the amount of substance. And from G = µn, we have
dµ = −S
ndT +
V
ndp = −SmoldT + Vmoldp. (36)
This is a thermodynamic identity that applies to both the liquid and its vapor. At a constanttemperature, we have dT = 0, so we only need to look at the second term. The change ofthe chemical potential µl of the liquid is given by
∆µl = V lmol∆pl = V l
mol(P + P g − P g0 ), (37)
assuming V lmol is incompressible. The change of the chemical potential µg of the vapor is
∆µg =
∫ P g
P g0
V gmol(p)dp =
∫ P g
P g0
RT
pdp = RT ln
P g
P g0
, (38)
where R = kBNA = 8.314 J/(mol ·K) is the ideal gas constant. Therefore, by requiring∆µl = ∆µg, we obtain
V lmol(P + P g − P g
0 ) = RT lnP g
P g0
⇒ P g = P g0 exp
[V lmol
RT(P + P g − P g
0 )
]. (39)
This is a transcendental equation for P g, which may be solved iteratively using P g0 as an
initial guess. The final pressure P g > P g0 when some inert gas with partial pressure P is
added to the system. This result may be a bit counter-intuitive. When we add pressure, wenormally expect vapor to liquefy. But in this problem, it’s the opposite that happens.
The reason lies in the difference between applying pressure to the whole liquid-vaporsystem and introducing inert gas that only increases the pressure of the liquid while leavingthe partial pressure of the vapor unchanged. The first situation is what we normally call“adding pressure”. In this case, both µl and µg increase, but since V l
mol ≪ V gmol, we have
µg > µl so that some vapor has to liquefy to balance the two phases. In the second situation,only µl increases because of the extra partial pressure P from the inert gas while µg staysthe same, so that µl > µg. Therefore, the inert gas will cause some liquid to vaporize.
Based on the analysis above, V lmol/RT should be small because liquid is dense compared
with gas. This means good results of P g can be found by just a few iterations starting fromthe initial guess P g
0 . Therefore, we obtain the approximate answer
P g ≈ P g0 exp
(V lmol
RTP
), (40)
by doing only one such iteration. You may plug in data of water and find the approximateformula quite accurate compared with the iteration fixed point of Eq. (39) within a muchwider range of inert gas pressure P than linear response.
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3 Fluid dynamics3.1 Variables and equationsA classical fluid is described by its local density ρ(r, t), velocity field v(r, t) and pressuredistribution p(r, t). Fluid dynamics is often studied without considering its coupling withheat conduction. As a result, there is a simple constitutive relation of pressure and density
p = p(ρ). (41)
The mass of the fluid is conserved in a flow. Therefore, we have the continuity equation
∂ρ
∂t+∇ · (ρv) = 0. (42)
Newton’s laws of motion applied to a fluid is called Euler’s equation, which is given by
ρdv
dt≡ ρ
[∂v
∂t+ (v · ∇)v
]= −∇p− ρ∇ϕ, (43)
where the total time derivative d/dt is introduced in the sense that we follow a moving masspoint in the fluid and consider the changing rate of the velocity v of that mass point, i.e.,its acceleration. The quantity ϕ = ϕ(r) is the potential energy per unit mass of the fluid. Atypical example will be the gravitational potential ϕ = gz. No fluid resistance is included inthe equation, which implies a perfect fluid (frictionless).
3.2 Bernoulli’s equationIn the steady flow of a perfect (frictionless) incompressible fluid under gravity, at any pointalong the same streamline, we have Bernoulli’s equation (1752)
p+ ρgz +1
2ρv2 = const, (44)
where the mass density ρ = const no longer depends on the pressure p in an incompressiblefluid. We can derive Bernoulli’s equation from Euler’s equation and the continuity equation.Since ρ = const, the continuity equation (42) simplifies to ∇ · v = 0. From the steady flowcondition, we have ∂p/∂t = 0. And in a time-independent potential field ϕ(r), we have∂ϕ/∂t = 0. Therefore, from Euler’s equation (43), we have
ρv · dvdt
= −v · ∇p− ρv · ∇ϕ = − d
dt(p+ ρϕ), (45)
⇒ d
dt
(p+ ρϕ+
1
2ρv2)
= 0. (46)
From our definition of d/dt in Eq. (43), the quantity p+ ρϕ+ 12ρv2 is a constant along any
streamline. Plugging in ϕ = gz, we obtain Bernoulli’s equation.
71
3.3 Sound waves in airAs sound waves travel in air, the uniform pressure p0 and density ρ0 (neglecting gravity)get slightly perturbed. Since the air is a poor conductor of heat and sound waves oscillatecomparatively very quickly, the constitutive equation can be written as
p
p0=
(ρ
ρ0
)γ
, (47)
assuming adiabatic compression of ideal gas, where γ = 7/5 is the ratio of specific heats of di-atomic molecules because the essential constituents N2 and O2 are both diatomic. Neglectinggravity, we may simplify Euler’s equation (43) into
ρ
[∂v
∂t+ (v · ∇)v
]= −∇p. (48)
And we also have the continuity equation (42). Now we linearize all three equations in thevicinity of ρ = ρ0, p = p0 and v = 0. The advective acceleration (v · ∇)v is second-order inv and thus neglected. By keeping only first-order terms, we have
p0 + δp
p0=
(ρ0 + δρ
ρ0
)γ
⇒ δp = γp0ρ0
δρ, (49)
(ρ0 + δρ)dv
dt= −∇(p0 + δp) ⇒ ρ0
∂v
∂t= −∇δp, (50)
∂(ρ0 + δρ)
∂t+∇ · [(ρ0 + δρ) v] = 0 ⇒ ∂δρ
∂t+ ρ0∇ · v = 0, (51)
Plugging Eq. (49) into Eq. (50), applying ∂/∂t to Eq. (51) and the divergence to Eq. (50),we obtain a linear wave equation for the density fluctuation δρ given by
∂2δρ
∂t2= γ
p0ρ0
∇2δρ. (52)
Therefore, the speed of sound in air is found to be
c =
√γp0ρ0
=
√γRT
Mmolair
, (53)
where R = 8.314 J/(mol ·K) is the ideal gas constant and Mmolair = 29 g/mol is the molar
mass of air. At 15 C (59 F), T = 288 K, the speed of sound in air is u = 340 m/s. Eq. (53)is called the Newton-Laplace equation, which was first published by Newton (1686) and thencorrected by Laplace (1816) to include the γ factor by considering the compression of airdue to sound waves to be adiabatic rather than isothermal.
72
3.4 Velocity potential(Problem 3 in Sec. 5, 2009) The steady-state flow of a fluid is specified by the fluid velocityv(r) at position r. Consider an incompressible fluid (∇ · v = 0) undergoing irrotational flow(∇ × v = 0). An impenetrable solid sphere of radius R is fixed in position while the fluidflows around it. Far from the sphere, the fluid flows uniformly in the z direction (v = v0z
for r → +∞). For this azimuthally symmetric problem, find the r and θ components of thefluid velocity v(r, θ) for all r > R.
Answer: Since the velocity field of the flow satisfies ∇ × v = 0, we may introduce ascalar potential ϕ such that v = −∇ϕ. Then from ∇ · v = 0, we have
∇2ϕ = 0. (54)
Therefore, we get essentially an electrostatic problem. The Laplace equation (54) can besolved by separation of variables in spherical coordinates. Since we already had a lecture onelectrostatics (see Lecture 2), we will directly give the general solution here:
ϕ(r, θ) =∞∑l=0
(alr
l +blrl+1
)Pl(cos θ), r > R, (55)
where the Pl(·)’s are the Legendre polynomials. However, the boundary conditions hereare slightly different. Since the sphere is impenetrable, we have the normal component ofvelocity equals zero, i.e.,
∂ϕ
∂r
∣∣∣∣r=R+0
= 0. (56)
At r → +∞, we require that
ϕ ≈ −v0r cos θ. (57)
73
The first boundary condition Eq. (56) gives us
lalRl−1 − (l + 1)bl
Rl+2= 0, , l = 1, 2, 3, . . . (58)
The second boundary condition Eq. (57) gives us
al = −v0 δl1. (59)
Therefore, we obtain the specific solution
ϕ(r, θ) = −v0
(r +
R3
2r2
)cos θ, r > R. (60)
The velocity field for r > R is then given byvr = −∂ϕ
∂r= v0
(1− R3
r3
)cos θ,
vθ = −1
r
∂ϕ
∂θ= −v0
(1 +
R3
2r3
)sin θ.
(61)
(62)
A scalar field is mathematically easier to solve than a vector field. An irrotational vectorfield is equivalent to the gradient of a scalar potential. This technique is used extensively inelectrostatics (electrostatic potential), magnetostatics (magnetic scalar potential), and fluiddynamics (velocity potential), etc.
All these contemporary studies in various fields of 18th–19th-century physics were amaz-ingly related by the same set of PDEs we reviewed in Lecture 2. Modern fluid dynamicsinvolves a lot more richness and complexity, e.g., viscosity, turbulence, Reynolds number,nonlinearity, chaos, etc. These topics are beyond the scope of this book and the Quals, butthey do open a new era of complexity science and emergent phenomena.
Congratulations! You have finished my lecture notes of Quals review. May you pass theexams smoothly and be a successful researcher!
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