20201021 Quantum Mechanics II
Special Relativity Preparatory Course
Teaching Assistant: Oz Davidi
October 27-28, 2019
Disclaimer: These notes should not replace a course in special relativity, but should serve
as a reminder. If some of the topics here are unfamiliar, it is recommended to read one of the
references below or any other relevant literature.
Notations and Conventions
1. We use τ as a short for 2π.1
References
There exist lots of references about the subject. Many books about general relativity include
good explanations in their first chapters. Other sources are advanced books on mechanics and
electromagnetism. Here is a list of some examples which covers the subject from those different
points of view. In each of them, look for the relevant chapters.
1. Classical Mechanics, H. Goldstein.
2. Classical Electrodynamics, J. D. Jackson.
3. A First Course in General Relativity, B. F. Schutz.
4. Gravitation and Cosmology, S. Weinberg.
1See https://tauday.com/tau-manifesto for further reading.
1
2 INDEX NOTATION
1 Motivation
One of the main topics of the Quantum Mechanics II course is to develop a (special) relativistic
treatment of quantum mechanics, which is done in the framework of quantum field theory. We
will learn how to quantize (relativistic) scalar and fermionic fields, and about their interactions.
For this end, a basic knowledge in special relativity is needed.
2 Index Notation
We will find that index notation is the most convenient way to deal with vectors, matrices, and
tensors in general. Let us focus on tensors of rank 2 and below.
• For a vector ~v =(v1 v2 · · · vn
)T, we denote the i’s component by vi.
• For a matrix M =
m11 m12 · · ·m21 m22 · · ·
......
. . .
, we denote the [ij]’s entry by Mij.
Notice: In general, Mij 6= Mji, but Mji =[MT
]ij
.
• When we multiply a vector by a matrix from the left, we get a new vector ~u = M~v. The
i’s component of the new vector is given by ui = [M~v]i =∑
jMijvj.
From now on, we will use Einstein’s Summation Convention:
1. If an index appears twice, we sum over it
Mijvj ≡∑j
Mijvj . (2.1)
2. An index will NEVER appear more then twice!
• What about multiplying by a matrix from the right, ~vTM? Again, we get a new vector
~wT = ~vTM . In index notation[~wT]i
= [~w]i = wi =[~vTM
]i
=[~vT]jMji = vjMji.
Here is an example why this is so useful:
Example 2.1. Prove that ~u×(~v × ~w) = ~v (~u · ~w)− (~u · ~v) ~w.
2
3 FAST INTRODUCTION TO SPECIAL RELATIVITY
Proof. By using index notation
[~u×(~v × ~w)]i = εijkuj [~v × ~w]k
= εijkεklmujvlwm
= εijkεlmkujvlwm
= (δilδjm − δimδjl)ujvlwm= viujwj − ujvjwi= [~v (~u · ~w)− (~u · ~v) ~w]i .
3 Fast Introduction to Special Relativity
3.1 Defining Special Relativity (B. F. Schutz: 1.1, 1.2)
At first, Einstein’s theory of special relativity was understood algebraically, as a set of (Lorentz)
transformations that move us from one inertial observer’s system to another. Special relativity
can be deduced from two fundamental postulates:
1. Principle of Relativity (Galileo): No experiment can measure the absolute velocity of an
observer; the results of any experiment performed by an observer do not depend on his
speed relative to other observers who are not involved in the experiment.
2. Universality of the Speed of Light (Einstein): The speed of light relative to any unacceler-
ated (inertial) observer is 3×108 m/s, regardless of the motion of the light’s source relative
to the observer. Let us be quite clear about this postulate’s meaning: two different iner-
tial observers measuring the speed of the same photon will each find it to be moving at
c = 3× 108 m/s relative to themselves, regardless of their state of motion relative to each
other.
But what is an “inertial observer”? An inertial observer is simply a coordinate system
for spacetime, which makes an observation by recording the location(x y z
)and time
(t)
of any event. This coordinate system must satisfy the following three properties to be called
inertial :
1. The distance between point P1 =(x1 y1 z1
)and point P2 =
(x2 y2 z2
)is indepen-
dent of time.
3
3.2 Transformation Rules 3 FAST INTRODUCTION TO SPECIAL RELATIVITY
2. The clocks that sit at every point, ticking off the time coordinate t, are synchronized and
all run at the same rate.
3. The geometry of space at any constant time t is Euclidean.
3.2 Transformation Rules
Let us derive the transformation rules of special relativity in 1 + 1 dimensions (1 space and 1
time dimensions).
• Imagine two systems, O and O′, with respective velocity v between them.
• We choose x = x′ = 0 at t = t′ = 0.
• The position of a wave-front in system O is measured to be
x = ct . (3.1)
• We would like to see how this wave form is seen (parametrized) in system O′. We take the
transformation to be linear x′ = ax + bt, where a (which is dimensionless) and b (which
has dimensions of velocity) will be found below. The physical reason is that we want
O −→T1O′ −→
T2O′′ to be identical to O −−−−−→
T1“+”T2O′′ (you can compare it to rotations).2
• The origin of O′ in the O system is given by x = vt, hence
0 = (av + b) t =⇒ b = −av =⇒ x′ = a(x− vt) . (3.2)
• The inverse transformation is given by changing the sign of the velocity, namely
x = a(x′ + vt′) . (3.3)
• Plugging x′ into x gives
t′ = at+(1− a2)x
av. (3.4)
• Now, we demand that a wave-front in O, i.e. x = ct, is also a wave-front in O′, i.e.
x′ = ct′ (here, we demand that the speed of light is the same for all observers). By using
2We will make this statement more precise once we study group theory.
4
3.2 Transformation Rules 3 FAST INTRODUCTION TO SPECIAL RELATIVITY
the expression for x′, Eq. (3.2), and the expression for t′, Eq. (3.4), and substituting
x = ct, we get
a(c− v) = ca+(1− a2) c2
av. (3.5)
Solving for a, one gets
a ≡ γ =1√
1− v2
c2
. (3.6)
To summarize, the transformation rules are
x′ = γ(x− vt) , (3.7)
t′ = γ(t− v
c2x). (3.8)
As an important side note: Always check the dimensions of the quantities you look for.
Indeed, a turned out to be dimensionless, and b has the dimensions of velocity.
An immediate result is that the time coordinate is not universal! This is depicted in Fig. 1.
In classical mechanics, an event A =(tA xA yA zA
)shares the same time with an infinite
number of events B =(tA xB yB zB
). They all have the same time, meaning that events
that happen simultaneously at one inertial system, also happen at the same time in another.
On the other hand, the same event A, under special relativity, has a unique “now”. Other
events have their own “now”, hence different observers may not agree on the relative time
between events.
A trajectory x(t) of a particle for example, is called a world line. A world line must cross a
constant time slice once (and only once), but the crossing point can be at any point, depending
on the observer. The slope of a world line is the velocity reciprocal, v−1 = x−1. Because the
velocity is bounded from above by the same constant value c in all reference frames, at each
point of the trajectory x(t), one can draw a light-cone, and all inertial observers will agree that
the trajectory is within this light-cone.
We will sometime use β = v/c, and from now on, we set the speed of light to 1
c = 1 . (3.9)
Using the general 3 + 1 dimensional transformation rules, one can show that while different
5
4 THE METRIC
Figure 1: Spacetime structure in classical mechanics (top) and special relativity (bottom). In classical me-chanics, a universal time slice exists, while in special relativity, each event defines a light-cone. (B. F. Schutz:1.6)
inertial observers determine different world-lines for the same particle, they agree that
(∆t)2 − (∆x)2 = (∆t′)2 − (∆x′)
2. (3.10)
We take the infinitesimal limit, and define the interval ds2
ds2 ≡ dt2 − dx2 = dt′2 − dx′2 . (3.11)
The interval is a Lorentz scalar - it is invariant under Lorentz transformations (to be discussed
in Sec. 5).
4 The Metric
Minkowski pointed out that space (~x) and time (t) should be treated all as coordinates of a four-
dimensional space, which we now call spacetime. We thus define the spacetime four-vector
xµ =(t ~x
). The index µ ∈ { 0, 1, 2, 3 } is called the Lorentz index.
The Minkowski spacetime is not Euclidean. In order to measure distances, we define the
6
4 THE METRIC
metric as a symmetric function which maps two four-vectors to R
g(v1, v2) = g(v2, v1) ∈ R . (4.1)
Note that we did not define g to be positive definite. In special relativity, we parametrize
the metric by the rank-2 tensor ηµν = diag(1,−1,−1,−1).3 The metric with upper indices is
identical ηµν = diag(1,−1,−1,−1).
Einstein’s Summation Convention (Revisited):
1. If an index appears twice, once as a lower and once as an upper index, we sum over
it.
2. An index will NEVER appear twice as a lower/an upper index!
3. An index will NEVER appear more then twice!
We raise and lower indices using the metric
vµ = ηµνvν , vµ = ηµνvν . (4.2)
Example 4.1. What is xµ?
xµ = ηµνxν =
(t −~x
). (4.3)
Exercise 4.1. What is ηµν?
Exercise 4.2. What is ηµµ?
The interval, Eq. (3.11), can be now defined as
ds2 = ηµνdxµdxν . (4.4)
The modern way to define special relativity is the following. We start from the interval/metric,
and ask what are the symmetries of the system. Namely, what are the transformation rules
between different frames of reference that leave the interval invariant.
3This is the choice in Quantum Mechanics II course. Some authors use ηµν = diag(−1, 1, 1, 1), and you needto pay attention what is the convention of the book you read!
7
5 LORENTZ TRANSFORMATIONS
5 Lorentz Transformations
The allowed transformations, dxµ → dx′µ = Λµνdx
ν , are those that leave the interval ds2
invariant
ηµνdxµdxν → ηµνdx
′µdx′ν = ηµνΛµρdx
ρΛνσdx
σ =(ΛTηΛ
)ρσdxρdxσ
!= ηρσdx
ρdxσ . (5.1)
We see that Lorentz transformations are given by all transformations that preserve the metric.4
Comment: Note that Λµν 6=
(ΛT) µ
ν! Tensors do not behave as simple matrices. We use
the matrix notation to make expressions more familiar and intuitive, but the position of the
indices dictates the identity of the tensor! We will go back to this when we learn about Lorentz
transformations in the course. When dealing with tensors, pay special attention.
Let us count degrees of freedom:
• Λµν has 16 parameters (µ, ν ∈ { 0, 1, 2, 3 }).
• We have the condition ηµν = ηρσΛρµΛσ
ν −→
µ = ν gives 4 equations
µ 6= ν gives 6 equations.
• 16− 4− 6 = 6 −→ in general Λµν has 6 continuous parameters.
How do we interpret them?
• Three boosts, e.g. Λµν =
cosh(ηx) sinh(ηx) 0 0
sinh(ηx) cosh(ηx) 0 0
0 0 1 0
0 0 0 1
.
• Three rotations, e.g. Λµν =
1 0 0 0
0 1 0 0
0 0 cos(θx) sin(θx)
0 0 − sin(θx) cos(θx)
.
This gives Λ(ηi, θi). We will come back to that when we get to the subject of group theory.
Defining arctanh(ηx) = vxc
, we get the same transformation between two inertial observers,
that we developed in the old-fashion point of view in Eqs. (3.7,3.8).
4In the second part of the Quantum Mechanics II course, we will define the Lorentz transformations as thesymmetries of spacetime (with the flat metric ηµν), and using group theory, we will gain a lot of informationwhich is not evident when looking at Eqs. (3.7,3.8).
8
6 HYPARBOLIC STRUCTURE OF SPACETIME
Figure 2: Minkowski spacetime. The curves describe constant value of the interval ds2 = dt2 − d~x2.
6 Hyparbolic Structure of Spacetime
Recall that ds2 = dt2 − d~x2, and that all inertial observers measure the same value of ds2. We
can plot curves of constant values of ds2. A curve determines the same event as observed by
different inertial observers which have the same origin. We distinguish between three different
scenarios:
• Light-Like: For light, ds2 = 0 ⇒ dxdt
= ±1. This describes the dashed curves of Fig. 2.
All inertial observers with same origin will determine the same curve.
• Time-Like: ds2 > 0. This describes the top and bottom parts of Fig. 2. Here, causal
order is well defined, i.e. all observers will agree which event happened before another
event.
• Space-Like: ds2 < 0. This describes the right and left parts of Fig. 2. Here, causal order
is not well defined, namely events which are space-like cannot affect each other.
As an example, consider Rocket Raccoon and Groot, which are sitting at rest on the x axis:
Rocket at x = 1, and Groot at x = 2. They synchronized their clocks in advance, and decided
to push a button at t = 0.5 Drax the Destroyer and Mantis, sitting at rest at the origin, will
5Hopefully, Groot pushes the right button https://youtu.be/Hrimfgjf4k8.
9
7 THE POLE, THE BARN, AND SCHRODINGER’S CAT
say that Rocket and Groot pushed the button at the same time, and will mark the blue squares
of Fig. 2.
Star-Lord on the Milano, flying to the right direction (and passes through x = 0 at t = 0),
will say that Rocket (x = 1 in the rest frame) pushed the button after Groot (x = 2 in the rest
frame). He will mark the red squares of Fig. 2.
Finally, Gamora on the Benatar, flying to the left direction (and passes through x = 0 at
t = 0), will say that Rocket (x = 1 in the rest frame) pushed the button before Groot (x = 2
in the rest frame). She will mark the green squares of Fig. 2.
7 The Pole, the Barn, and Schrodinger’s Cat
The problem is presented pictorially in Fig. 3.
Einstein sits at rest inside his barn. His rest frame is denoted by O, and the length of his
barn in this frame is Lbarn = 10 m. Einstein’s barn has two doors. The left one is open, and
the right one is closed.
Meanwhile, Heisenberg and Schrodinger are sitting on a horizontal pole which arrives from
the left of the barn with a velocity β =√
3/2. Heisenberg sits on the tail of the pole (the
leftmost point of the pole), while Schrodinger sits on the head of the pole, and he holds a box
with a quantum cat inside. The rest frame of Heisenberg and Schrodinger is denoted by O′,and the length of the pole in this frame is Lpole = 10 m.
1. Einstein has a smart barn. When Schrodinger reaches the right door, it opens automat-
ically. Because he doesn’t like to keep all doors open, as soon as Heisenberg enters the
barn, the left door automatically closes.
2. Schrodinger told Einstein that when he will exit his barn, he is going to check whether
the cat is dead or alive. He doesn’t know that Heisenberg turned his box into a classical
system, by connecting it to a button. If Heisenberg pushes the button, the cat will
die, otherwise it lives. Heisenberg decides to push the death button when he enters the
barn. Schrodinger and Heisenberg do not know what will Schrodinger find inside the box.
Einstein, on the other hand, do knows.
The first question is: how do we measure the length of the pole in the barn rest frame O?
The way we measure length is by determining the x coordinate of the head and tail of the pole
at the same time.
10
7 THE POLE, THE BARN, AND SCHRODINGER’S CAT
Figure 3: The pole and the barn paradox - Schrodinger’s cat version. Heisenberg (H) and Schrodinger (S)sit on a pole that moves to the right with velocity β. Einstein (E) sits inside a barn. wµ and w′µ are thecoordinates of the event of measuring the pole’s tail position, and yµ and y′µ are the coordinates of the eventof measuring the pole’s head position, in the barn frame O and pole frame O′ respectively. zµ and z′µ are thecoordinates of the event of Heisenberg entering the barn.
• We choose the origin of O and O′ to determine the head of the pole and the entrance of
the barn at t = t′ = 0. This is the point y = y′ =(
0 0)
in Fig. 3.
• The point wµ (see figure) is the length of the pole in the barn rest frame. By definition,
its time component is set to zero, so wµ =(
0 −`pole)
. Here, `pole stands for the yet
unknown length of the pole in the barn rest frame.
• Now, we determine the same point in the pole rest frame. Since both systems share
the same origin, by definition, the space component of the tail of the pole is minus
its length, Lpole (recall that the pole does not move within its rest frame). Therefore,
w′µ =(T ′ −10
). Note that the time is yet unknown.
• β =√
3/2 ⇒ γ = (1− β2)−1/2
= 2. Using our transformation rules, Eqs. (3.7,3.8), and
11
7 THE POLE, THE BARN, AND SCHRODINGER’S CAT
their inverse, we getwµ =
(0 −`pole
)w′µ =
(T ′ −10
) =⇒
t = 0 = 2(T ′ −
√32
10)
= γ(t′ + βx′)
x = −`pole = 2(−10 +
√32T ′)
= γ(x′ + βt′). (7.1)
The solutions are T ′ = 5√
3 and `pole = 5.
We see that the pole is shorter in the barn system by a factor of two. It means that in the barn
rest frame, O, at time tH = `pole/β = 10/√
3, Heisenberg enters the barn, and Schrodinger is
still inside. Both of the doors are closed and the pole is locked inside the barn. Furthermore,
Heisenberg pushes the button, killing the cat before Schrodinger checks to see if it is alive or
not. When Schrodinger checks the box, the cat is dead.
The Classic Paradox - We can repeat the exercise in the pole rest frame, O′, and find that
the barn is twice shorter in this frame. How come that the pole can enter into a barn which is
half of its size?
Solution - In the pole frame, the time it takes for Schrodinger to arrive to the right door
of the barn is (one can use Lorentz transformations, but we can derive everything using simple
kinematics) t′S = 5/β = 10/√
3. The time it takes for Heisenberg to arrive to the left door
is given by t′H = 10/β = 20/√
3 = 2t′S. According to Heisenberg’s (and Schrodinger’s) clock,
Heisenberg enters the barn after Schrodinger had already exited, with a time difference of t′S.
The left door closes only after the right door opens.
The Quantum Paradox - In the pole frame, Schrodinger checks the cat’s status before
Heisenberg pushes the button. This means that the cat is alive! But it is dead in the barn
frame...
Solution - In order to be able to affect the cat’s condition, Heisenberg must send a signal
which is faster than light. The interval between the events (Heisenberg sending a signal and
Schrodinger checking the box) is space-like - the events are causally disconnected! Because
Heisenberg can’t control the fate of the cat, and because cats usually have more than one soul
anyway, the cat is alive, and can be found outside of the faculty building.
12
8 LORENTZ SCALARS AND MORE FOUR-VECTORS
8 Lorentz Scalars and More Four-Vectors
In order to get Lorentz scalars, i.e. objects that do not transform under Lorentz transforma-
tions, we contract all Lorentz indices using the metric.
A ·B ≡ ηµνAµBν = AµBµ , (8.1)
A ·B −→ A′ ·B′ = ηµνΛµρΛ
νσA
ρBσ = ηρσAρBσ = A ·B . (8.2)
Let us consider few important examples related to four-vectors.
8.1 Four-Momentum
The four momentum is given by
pµ =(E ~p
), (8.3)
where E is the energy, and ~p is the three momentum. The scalar which is obtained by taking
the four-momentum square, is the particle’s rest mass squared
p2 ≡ p · p = pµpµ = m2 . (8.4)
The four-momentum is related to the four-velocity by pµ = muµ, where uµ ≡ dxµ√ds2
= γ(
1 ~v)
.
Note that for light, γ →∞ but m→ 0, so the four-momentum is well defined, and E = |~p|.
8.2 Derivatives, Currents and Electromagnetism (J. D. Jackson:
11.6, 11.9)
Another important four-vector is the four-derivative ∂µ ≡ ∂∂xµ
=(∂t ~∇
). It transforms as
∂′µ = ∂∂x′µ
= Λ νµ ∂ν , as expected (see Jackson). Notice that while xµ or pµ were defined with an
upper index (contravariant vectors), the four-derivative is defined with a lower index (covariant
vector).
The four-dimensional Laplacian, the d’Alembertian � ≡ ∂µ∂µ = ∂2t − ~∇2, is a Lorentz
scalar, and hence also the wave equation.
Recall the continuity equation ∂ρ∂t
+ ~∇ · ~J = 0, which must hold at any frame of reference.
We can define the four-current Jµ =(ρ ~J
). Using this definition, the continuity equation is
clearly satisfied at all frames, and can be written as ∂µJµ = 0.
13
8.2 Derivatives, Currents and Electromagnetism (J. D. Jackson: 11.6, 11.9)8 LORENTZ SCALARS AND MORE FOUR-VECTORS
8.2.1 Electromagnetism
In Lorentz gauge, ∂tφ+ ~∇ · ~A = 0, the wave equations for the scalar and vector potentials are
∂2 ~A
∂t2− ~∇2 ~A = 2τ ~J , (8.5)
∂2φ
∂t2− ~∇2φ = 2τρ . (8.6)
It is just natural to define the four-potential Aµ =(φ ~A
). Then, the gauge can be written as
∂µAµ = 0, and the wave equations are
�Aµ = 2τJµ . (8.7)
You can check that the electromagnetic tensor is given by
Fµν = ∂µAν − ∂νAµ =
0 Ex Ey Ez
−Ex 0 −Bz By
−Ey Bz 0 −Bx
−Ez −By Bx 0
, (8.8)
where we used ~E = −∂ ~A∂t− ~∇φ and ~B = ~∇× ~A.
Exercise 8.1. Find F µν = ηµρηνσFρσ.
How do Fµν and F µν transform? Each index transforms as a vector, i.e.
F ′µν = Λ ρµ Λ σ
ν Fρσ , F ′µν = ΛµρΛ
νσF
ρσ . (8.9)
Exercise 8.2. Convince yourself that F µνFµν is a Lorentz scalar.
Exercise 8.3. Find the transformation rules of ~E and ~B by using Eq. (8.9).6
Most of the time in relativistic theories, we use Aµ and Fµν , and not ~E and ~B.
6In the tutorial on Lorentz transformations, we will see the “group theory way” of finding the transformationrules for the electromagnetic fields.
14
9 FOURIER TRANSFORM
9 Fourier transform
As a last comment, we will make use of the relativistic Fourier transform
f(p) =
∞∫−∞
d4x f(x) eix·p~ , f(x) =
∞∫−∞
d4p
h4f(p) e−i
x·p~ , (9.1)
where x · p is another useful Lorentz scalar.
We welcome aboard Planck’s constant. This is the first place where we see an integration
between special relativity and quantum mechanics.
15