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. T • For a vector ~v = v1 v2 ··· vn , we denote the i’s component by vi. m11 m12 ··· • For a matrix M =m21 m22 ···, we denote the [ij]’s entry by Mij. . . . . . .. T Notice: In general, Mij 6= Mji, but Mji = M ij.
• When we multiply a vector by a matrix from the left, we get a new vector ~u = M~v. The P i’s component of the new vector is given by ui = [M~v]i = j Mijvj.
From now on, we will use Einstein’s Summation Convention:
1. If an index appears twice, we sum over it
X Mijvj ≡ Mijvj . (2.1) j
2. An index will NEVER appear more then twice!
• What about multiplying by a matrix from the right, ~vT M? Again, we get a new vector T T T T T ~w = ~v M. In index notation ~w i = [~w]i = wi = ~v M i = ~v j Mji = 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 = ijkuj [~v × ~w]k
= ijkklmujvlwm
= ijklmkujvlwm
= (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- 8 ated (inertial) observer is 3×10 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 8 c = 3 × 10 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 O0, with respective velocity v between them.
• We choose x = x0 = 0 at t = t0 = 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 O0. We take the transformation to be linear x0 = 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 −→ O0 −→ O00 to be identical to O −−−−−→ O00 (you can compare it to rotations).2 T1 T2 T1“+”T2 • The origin of O0 in the O system is given by x = vt, hence
0 = (av + b) t =⇒ b = −av =⇒ x0 = a(x − vt) . (3.2)
• The inverse transformation is given by changing the sign of the velocity, namely
x = a(x0 + vt0) . (3.3)
• Plugging x0 into x gives (1 − a2) x t0 = at + . (3.4) av • Now, we demand that a wave-front in O, i.e. x = ct, is also a wave-front in O0, i.e. x0 = ct0 (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 x0, Eq. (3.2), and the expression for t0, Eq. (3.4), and substituting x = ct, we get (1 − a2) c2 a(c − v) = ca + . (3.5) av Solving for a, one gets 1 a ≡ γ = . (3.6) q v2 1 − c2
To summarize, the transformation rules are
x0 = γ(x − vt) , (3.7) v t0 = γ t − x . (3.8) c2
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 = (∆t0)2 − (∆x0)2 . (3.10)
We take the infinitesimal limit, and define the interval ds2
ds2 ≡ dt2 − dx2 = dt02 − dx02 . (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 3 the metric by the rank-2 tensor ηµν = diag(1, −1, −1, −1). 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
ν µ µν vµ = ηµνv , v = η vν . (4.2)
Example 4.1. What is xµ? ν xµ = ηµνx = t −~x . (4.3)
µ Exercise 4.1. What is η ν?
µ Exercise 4.2. What is η µ?
The interval, Eq. (3.11), can be now defined as
2 µ ν ds = ηµνdx dx . (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.
3 This is the choice in Quantum Mechanics II course. Some authors use ηµν = diag(−1, 1, 1, 1), and you need to pay attention what is the convention of the book you read!
7 5 LORENTZ TRANSFORMATIONS
5 Lorentz Transformations
µ 0µ µ ν 2 The allowed transformations, dx → dx = Λ νdx , are those that leave the interval ds invariant
µ ν 0µ 0ν µ ρ ν σ T ρ σ ! ρ σ ηµνdx dx → ηµνdx dx = ηµνΛ ρdx Λ σdx = Λ ηΛ ρσ dx dx = ηρσdx dx . (5.1)
We see that Lorentz transformations are given by all transformations that preserve the metric.4 µ T µ Comment: Note that Λ ν 6= Λ 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:
µ • Λ ν has 16 parameters (µ, ν ∈ { 0, 1, 2, 3 }). µ = ν gives 4 equations ρ σ • We have the condition ηµν = ηρσΛ µΛ ν −→ . µ 6= ν gives 6 equations
µ • 16 − 4 − 6 = 6 −→ in general Λ ν has 6 continuous parameters.
How do we interpret them? cosh(ηx) sinh(ηx) 0 0 sinh(ηx) cosh(ηx) 0 0 • Three boosts, e.g. Λµ = . ν 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 • Three rotations, e.g. Λµ = . ν 0 0 cos(θ ) sin(θ ) x x 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.
vx Defining arctanh(ηx) = c , 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 the symmetries of spacetime (with the flat metric ηµν ), and using group theory, we will gain a lot of information which 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:
2 dx • Light-Like: For light, ds = 0 ⇒ dt = ±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 Schr¨odinger’sCat
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 Schr¨odingerare 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 Schr¨odingersits on the head of the pole, and he holds a box with a quantum cat inside. The rest frame of Heisenberg and Schr¨odinger is denoted by O0, and the length of the pole in this frame is Lpole = 10 m.
1. Einstein has a smart barn. When Schr¨odingerreaches 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. Schr¨odingertold 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. Schr¨odingerand Heisenberg do not know what will Schr¨odingerfind 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 - Schr¨odinger’scat version. Heisenberg (H) and Schr¨odinger(S) sit on a pole that moves to the right with velocity β. Einstein (E) sits inside a barn. wµ and w0µ are the coordinates of the event of measuring the pole’s tail position, and yµ and y0µ are the coordinates of the event of measuring the pole’s head position, in the barn frame O and pole frame O0 respectively. zµ and z0µ are the coordinates of the event of Heisenberg entering the barn.
• We choose the origin of O and O0 to determine the head of the pole and the entrance of the barn at t = t0 = 0. This is the point y = y0 = 0 0 in Fig.3.
• The point wµ (see figure) is the length of the pole in the barn rest frame. By definition, µ its time component is set to zero, so w = 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, w0µ = T 0 −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 get
√ µ 0 3 0 0 w = 0 −`pole t = 0 = 2 T − 2 10 = γ(t + βx ) =⇒ √ . (7.1) 0µ 0 3 0 0 0 w = T −10 x = −`pole = 2 −10 + 2 T = γ(x + βt ) √ 0 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 Schr¨odingeris 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 Schr¨odinger checks to see if it is alive or not. When Schr¨odinger checks the box, the cat is dead. The Classic Paradox - We can repeat the exercise in the pole rest frame, O0, 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 Schr¨odingerto arrive to the right door of the barn is (one can use Lorentz transformations, but we can derive everything using simple √ kinematics) t0 = 5/β = 10/ 3. The time it takes for Heisenberg to arrive to the left door S √ 0 0 is given by tH = 10/β = 20/ 3 = 2tS. According to Heisenberg’s (and Schr¨odinger’s)clock, 0 Heisenberg enters the barn after Schr¨odingerhad already exited, with a time difference of tS. The left door closes only after the right door opens. The Quantum Paradox - In the pole frame, Schr¨odingerchecks 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 Schr¨odingerchecking 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.
µ ν µ A · B ≡ ηµνA B = A Bµ , (8.1) 0 0 µ ν ρ σ ρ σ A · B −→ A · B = ηµνΛ ρΛ σA B = ηρσA B = 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
2 µ 2 p ≡ p · p = p pµ = m . (8.4)
µ The four-momentum is related to the four-velocity by pµ = muµ, where uµ ≡ √dx = γ 1 ~v . ds2 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 0 ∂ ν µ µ ∂µ = ∂x0µ = Λµ ∂ν, as expected (see Jackson). Notice that while x or p were defined with an upper index (contravariant vectors), the four-derivative is defined with a lower index (covariant vector). µ 2 ~ 2 The four-dimensional Laplacian, the d’Alembertian ≡ ∂ ∂µ = ∂t − ∇ , is a Lorentz scalar, and hence also the wave equation. Recall the continuity equation ∂ρ + ∇~ · J~ = 0, which must hold at any frame of reference. ∂t We can define the four-current J µ = ρ J~ . Using this definition, the continuity equation is µ clearly satisfied at all frames, and can be written as ∂µJ = 0.
13 8.2 Derivatives, Currents and Electromagnetism SCALARS(J. D. Jackson: AND 11.6, MORE 11.9)8LORENTZ FOUR-VECTORS
8.2.1 Electromagnetism ~ ~ In Lorentz gauge, ∂tφ + ∇ · A = 0, the wave equations for the scalar and vector potentials are
∂2A~ − ∇~ 2A~ = 2τJ,~ (8.5) ∂t2 ∂2φ − ∇~ 2φ = 2τρ . (8.6) ∂t2 It is just natural to define the four-potential Aµ = φ A~ . Then, the gauge can be written as µ ∂µA = 0, and the wave equations are
µ µ A = 2τJ . (8.7)
You can check that the electromagnetic tensor is given by 0 Ex Ey Ez −Ex 0 −Bz By Fµν = ∂µAν − ∂νAµ = , (8.8) −E B 0 −B y z x −Ez −By Bx 0
~ ∂A~ ~ ~ ~ ~ where we used E = − ∂t − ∇φ and B = ∇ × A.
µν µρ νσ Exercise 8.1. Find F = η η Fρσ.
µν How do Fµν and F transform? Each index transforms as a vector, i.e.
0 ρ σ 0µν µ ν ρσ Fµν = Λµ Λν Fρσ ,F = Λ ρΛ σF . (8.9)
µν Exercise 8.2. Convince yourself that F Fµν is a Lorentz scalar.
Exercise 8.3. Find the transformation rules of E~ and B~ by using Eq. (8.9).6
µ ~ ~ Most of the time in relativistic theories, we use A and Fµν, and not E and B.
6In the tutorial on Lorentz transformations, we will see the “group theory way” of finding the transformation rules for the electromagnetic fields.
14 9 FOURIER TRANSFORM
9 Fourier transform
As a last comment, we will make use of the relativistic Fourier transform
∞ ∞ Z Z 4 4 i x·p d p −i x·p f˜(p) = d x f(x) e ~ , f(x) = f˜(p) e ~ , (9.1) h4 −∞ −∞ 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