# Introduction to Symmetry and Broken Symmetry in Physics

B.A. van Tiggelen

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B.A. van Tiggelen. Introduction to Symmetry and Broken Symmetry in Physics. Master. Introduction to Symmetry and Broken Symmetry in Physics, Ecole Normale Supérieure de Lyon, France. 2025, pp.176. hal-04989778v4

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Distributed under a Creative Commons Attribution 4.0 International License
Introduction to Symmetry and Broken Symmetry in Physics

Master 2 Lecture at ENS Lyon 2025

B.A. van Tiggelen*

Univ. Grenoble Alpes, CNRS, LPMMC, 38000 Grenoble, France

(Dated: version September, 2025)

"The Unreasonable Effectiveness of Mathematics in the Natural Sciences."

(Eugene Wigner, Nobel Prize in physics 1963)

"Our job in physics is to see things simply, to understand a great many complicated phenomena
in a unified way, in terms of a few simple principles."

(Steven Weinberg, Nobel Prize in physics 1979)

<figure><img src="image_2.png" /></figure>

<footnote>* bart.van_tiggelen@ens-lyon.fr</footnote>
CONTENTS

I. Introduction 5

II. Time-line of events 8

III. Lorentz Transformations 12

A. Miscellaneous Consequences of Lorentz invariance 14
→ Invariance of electromagnetic quantum vacuum 14
→ Cosmological constant problem 16
→ Unruh effect and black hole temperature 18
→ The Expanding Universe. 20

IV. Principle of Least Action & Equation of Motion 24

V. Symmetry Groups in Physics and their Representations 28

A. Properties of a Group 28
1. Representations of a group 34

B. Lie Groups and Lie Algebra 35
→ Adjoint representation of Lie algebra 40

C. The group $O(n)$ 41
→ Lévi-Civita as a third-order pseudo tensor in $O(3)$ 42
→ The group $L^2(S^2)$ and spherical harmonics. 45
→ The representation of the group $SO(2)$ in $\mathbb{C}^2$ is not irreducible. 47
→ The group $SO(3)$ is not simply connected 48

D. The group $U(n)$ 49
→ Equivalence of $U(n)$ and $SU(n) \times U(1)$ 50
→ $U(3)$ invariance of the isotropic 3D oscillator 51

E. The group $SL(2, \mathbb{C})$ 57

F. Dual Space and Dual Representation 59
1. Dual representation of Lie group and Lie algebra 61

G. The restricted Lorentz Group $SO^+(1, 3)$ and Spin 64
1. The proper Lorentz group and spin $\frac{1}{2}$ 69
→ The Lorentz group $SO^+(1, 3)$ is not simply connected 70

2. From $SU(2) \times SU(2)$ back to Minkowski space 72
<table>
  <tbody>
    <tr>
      <td>3. Spinor representation of SO<sup>+</sup>(1, 3): the Dirac equation</td>
      <td>73</td>
    </tr>
    <tr>
      <td>&rarr; C, P and T symmetry and CPT theorem</td>
      <td>76</td>
    </tr>
    <tr>
      <td>&rarr; Solutions of the free Dirac equation &amp; anti-particles</td>
      <td>79</td>
    </tr>
    <tr>
      <td>4. Gyromagnetic factor of electron</td>
      <td>82</td>
    </tr>
    <tr>
      <td>H. Irreducible Representations of SU(2) &times; SU(2)</td>
      <td>83</td>
    </tr>
    <tr>
      <td>1. Tensor representations</td>
      <td>85</td>
    </tr>
    <tr>
      <td>&rarr; Tensor representation of three spins 1</td>
      <td>87</td>
    </tr>
    <tr>
      <td>2. Tensor representations of the Lorentz group</td>
      <td>88</td>
    </tr>
    <tr>
      <td>&rarr; Tensor representation with spin 1/2 as building block</td>
      <td>89</td>
    </tr>
    <tr>
      <td>3. Two Tensor Representations of the Electromagnetic Field</td>
      <td>92</td>
    </tr>
    <tr>
      <td>VI. Symmetry &amp; Noether charges</td>
      <td>94</td>
    </tr>
    <tr>
      <td>A. Noether theorem after Second quantization</td>
      <td>99</td>
    </tr>
    <tr>
      <td>1. Second quantization of massive scalar real field</td>
      <td>100</td>
    </tr>
    <tr>
      <td>2. Second quantization of Dirac field</td>
      <td>102</td>
    </tr>
    <tr>
      <td>B. Wick's theorem for two particles</td>
      <td>104</td>
    </tr>
    <tr>
      <td>C. Electromagnetic Lagrangian</td>
      <td>108</td>
    </tr>
    <tr>
      <td>&rarr; Proca Lagrangian and photon mass</td>
      <td>111</td>
    </tr>
    <tr>
      <td>&rarr; Maxwell field as representation of the Lorentz group</td>
      <td>112</td>
    </tr>
    <tr>
      <td>&rarr; The Casimir effect</td>
      <td>116</td>
    </tr>
    <tr>
      <td>&rarr; The Lamb shift</td>
      <td>117</td>
    </tr>
    <tr>
      <td>&rarr; Angular momentum of the Maxwell field</td>
      <td>119</td>
    </tr>
    <tr>
      <td>D. Quantum Electrodynamics</td>
      <td>120</td>
    </tr>
    <tr>
      <td>&rarr; Electron self-energy</td>
      <td>123</td>
    </tr>
    <tr>
      <td>&rarr; Quantum Vacuum Polarization</td>
      <td>125</td>
    </tr>
    <tr>
      <td>E. Local symmetry</td>
      <td>129</td>
    </tr>
    <tr>
      <td>&rarr; Dirac-Maxwell Hamiltonian</td>
      <td>130</td>
    </tr>
    <tr>
      <td>VII. Spontaneous Symmetry Breaking</td>
      <td>133</td>
    </tr>
    <tr>
      <td>&rarr; Local and global symmetry &amp; tower of states</td>
      <td>134</td>
    </tr>
    <tr>
      <td>&rarr; Néel state in anti-ferromagnetism: the perfect SSB</td>
      <td>137</td>
    </tr>
    <tr>
      <td>A. Order parameter and cosets of broken symmetry states</td>
      <td>140</td>
    </tr>
    <tr>
      <td>&rarr; Interacting Bose-Einstein Condensate (superfluid)</td>
      <td>142</td>
    </tr>
    <tr>
      <td>&rarr; the Josephson effect</td>
      <td>146</td>
    </tr>
  </tbody>
</table>
→ Time crystals 149

B. Goldstone's Theorem 149

→ the superfluid 151

C. Long-range Correlations 152

→ Long-range order as a signature for SSB 153

→ No quantum time crystal in a many-body ground state 154

D. Hohenberg-Mermin-Wagner Theorem 156

→ The Peierls argument for the Ising model 160

→ The BKT transition 161

VIII. Symmetry breaking in the presence of gauge fields 163

A. Anderson-Higgs mechanism 163

→ The Meissner effect 164

B. The Electroweak Force 165

1. Charge, Hypercharge and Isospin 168

2. Masses of the gauge bosons 170

3. Fermion Masses 173

References 175

*Cover Image: Source CERN: a Historical Profile of the Higgs Boson.*
# I. INTRODUCTION

All known physical forces are described by mathematical laws that have undergone severe tests. The ultimate hope, considered by many to be quite illusionary, is that some day all physics can be described by one single law or principle. If that equation indeed exists, why then does it pop up in so many different disguises, from the Einstein equation for gravity, to the basic equations of quantum mechanics?

Contrary to the electromagnetic force and Newton gravity, other forces in Nature cannot directly be measured. For generalized gravity we must rely on observations of movements either on large scales in the Universe or in the proximity of fancy objects such as black holes. The weak and strong forces have ranges smaller than the atomic nucleus and can only be tested from scattering experiments at high energies. These observations are precious but we need additional tools to deduce the microscopic details of the underlying force. It has long been recognized that two elements play a fundamental key role in this process: the principle of least action and Poincaré-symmetry. All physical laws that we know so far obey these principles. The symmetry of the action, i.e. its observed invariance under a certain symmetry operation such as rotation determines conserved quantities. Poincaré symmetry, defined as the combined invariance to Lorentz transformations and space-time translations, results very generally in the conservation of momentum, angular momentum and energy, the invariance on phase factors in quantum mechanics is deeply related to charge conservation. Gauge freedom, originally inferred from Maxwell's equations, has been exploited to extend global symmetries to local symmetries. This “minimal coupling” principle identifies gauge fields as the carriers of the force, and the mathematical properties of their invariance determines in great detail the nature of the force. The link of gauge invariance to local symmetries rather than to global symmetries excludes an association with conservation laws. Nevertheless, they completely determine the interaction of matter.

A second major progress has been the discovery of broken symmetry. Observations have revealed states of matter with lower symmetry than the underlying equations of motion. Well-known examples exist in magnetism, and more recently in Bose-Einstein condensation where $U(1)$ phase symmetry is broken in the ground state, or even the chair that you are sitting on, clearly without translational symmetry. However such broken symmetry induces a major problem in field theory. A mathematical theorem in field theory demonstrates excitations in models with broken symmetry to be massless, and a way out has to be found to “create mass”. In particle physics
this is known as the Higgs mechanism, in condensed matter its non-relativistic version is known
as the Anderson-Higgs model. Anderson used this mechanism to explain why the electromagnetic
gauge field achieves mass in a superconductor where U(1) symmetry is broken. On this occasion
he wrote a legendary article on epistemology, i.e. the theory of scientific knowledge, that is an
absolute "obligation to read" for all students in physics. In this work, microscopic knowledge is
argued not be sufficient to understand the physical laws that we observe on macroscopic scales,
briefly that "More is Different" [1]. The concept of "Spontaneous Symmetry Breaking" in quantum
mechanics is put forward as a convincing example for the existence a fundamental principle that
was not rooted in one of its axioms, but that pops to understand observations. Of course, this
idea was greatly inspired by the introduction of entropy in statistical physics at the end of the 19th
century, where, despite the time-invariance $t \rightarrow -t$ of all microscopic wave equations, irreversibility
emerges on a macroscopic scale or equivalently spontaneously broken time- invariance. Similarly,
the Higgs mechanism was greatly inspired by the theory of phase transitions proposed first by Lev
Landau [2].

The unification of the electromagnetic force and the weak force has been a major achievement. It
uses the Higgs mechanism to give mass to both the gauge bosons and the interacting fermions, via
the intermediate action of a new particle, the Higgs boson. It took some time and effort (and a
lot of money) but this particle was finally observed indirectly (because of his rapid disintegration
and its large mass of 125 GeV) by the Large Hadron Collider at CERN in 2012. The electroweak
theory is currently described by a Lorentz-invariant action with $SU(2) \times U(1)$ symmetry that is
broken at low energies, with the massless, uncharged photon and the massive particles W and Z as
the carriers of the electroweak force. Exactly three gauge bosons exist as well as one Higgs boson
because the underlying original symmetry has 3 + 1 generators.

Advanced tools have been developed to cope with these issues: spontaneous symmetry breaking (SSB) and the tower of states. The first concept, borrowed from Landau's approach of phase transitions, leads to a so-called order parameter of the matter, an operator that does not commute with all symmetries of the Hamiltonian. When continuous symmetries are globally broken, quantum or thermal fluctuations generate so-called Goldstone modes. These modes are to be associated with observed excitations or particles, but turn out to be massless bosons (in relativistic terminology where $E^2 = m^2c_0^4 + p^2c_0^2$ they are said to be "gapless"). Because many observed excitations do have a mass, another tool is necessary: the celebrated Anderson-Higgs mechanism. Spontaneous symmetry breaking is hampered in low dimensions, because quantum and thermal fluctuations destroy
long-range order. The tower of states refers to the large number of microstates that accumulates
near the official ground state of a macroscopic body. The stable state chosen by the object is a
linear combination of all of them. This is an eigenstate of the order parameter, but not necessarily
of the Hamiltonian.

The intention of this course is to make the student familiar with the basic aspects and tools that have been developed concerning symmetry and broken symmetry. Several observed phenomena, mostly dating from the last 50 years, are discussed in this context, such as superfluidity, anti-ferromagnetism, the Meissner effect in superconductivity and recently, time crystals. The lecture intends to show that mathematical arguments, combined with observations, make it possible to advance our knowledge of physics. The above statement made by Eugene Wigner, "*the unreasonable effectiveness of mathematics in the natural science*", is worth contemplating. His discovery that elementary particles are unitary irreducible representations of the Lorentz-Poincaré group has become one of the most important pillars of theoretical physics.

Bart van Tiggelen, last updated version, October 2025
## II. TIME-LINE OF EVENTS

* 1746 Pierre-Louis Maupertuis formulates the principle of least action, inspired by the Fermat principle in optics, derived earlier.

* 1864 James Clerc Maxwell publishes the unification of electric and magnetic fields and predicts a velocity of light $c_0 = 310.740.000$ km/s.

* 1873 Sophus Lie lays the mathematical foundations of continuous “Lie” groups.

* 1887 Michelson-Morley experiments show $c_0$ to be a universal constant in moving frames suggesting the absence of aether.

* 1900 Hendrik Lorentz and Joseph Larmor reveal invariance of Maxwell equations under Lorentz transformations.

* 1905 Henri Poincaré establishes the Lorentz and Poincaré Group (= Lorentz transformations + space-time translations).

* 1905 Albert Einstein proposes Special Relativity.

* 1911 Heike Kamerling-Onnes discovers superconductivity of Mercury (no explanation).

* 1915 Albert Einstein publishes the theory of General Relativity and proposes Cosmological Constant (considered by him as his “biggest mistake ever”).

* 1915 Emmy Noether publishes the Noether theorem that relates continuous symmetries to conserved quantities.

* 1918 Hermann Weyl introduces the notion of gauge invariance (“Eichinvarianz”) in General Relativity and electromagnetism.

* 1922 Louis De Broglie and his student Alexandru Proca estimate the photon mass to be less than $10^{-18}$ eV, consistent with 0 and thus with gauge invariance associated with local $U(1)$ symmetry.

* 1924 Satyendra Bose introduces Bose statistics to explain Planck's law.

* 1925 Wilhelm Lenz and Ernst Ising introduce and solve the 1D classical Ising model with no magnetic phase transition.
• 1925 Albert Einstein predicts Bose-Einstein condensation.

* 1928 Paul Dirac publishes the relativistic Dirac equation for the electron and predicts anti-particles.

* 1931 Hermann Weyl and Eugene Wigner introduce group theory in quantum mechanics.

* 1933 Walter Meissner discovers the Meissner effect as the perfect paramagnetic behavior of a superconductor (no explanation).

* 1935 Fritz London and his brother Heinz explain the Meissner effect phenomenologically via the existence of the London current involving the electromagnetic vector potential, anticipating broken gauge invariance.

* 1936 Rudolf Peierls publishes an entropic argument in favor of a phase transition in the 2D Ising model.

• 1937 Piotr Kapitsa discovers superfluidity of Helium-4.

* 1937 Ettore Majorana predicts real-valued solutions of Dirac equation, where particle and anti-particle are equal.

* 1939 Eugene Wigner finds the irreducible unitary representations of the Poincaré Group [3], interpreted as elementary particles.

* 1944 Lars Onsager solves the classical Ising model in 2D, with spontaneous magnetization at low temperatures.

* 1948 Louis Néel discovers antiferromagnetism, introduced first theoretically by Lev Landau in 1933.

* 1952 Philip Anderson proposes Spontaneous Symmetry Breaking and Tower of States to explain the antiferromagnetic state [7].

* 1953 Chen Ning Yang and Robert Mills introduce gauge theory in particle physics with $SU(N)$ gauge invariance.

• 1954 Wolfgang Pauli proves the CPT theorem.

* 1960 Theoretical prediction of massless Nambu-Goldstone modes by Yoichiro Nambu (in particle physics) and Jeffrey Goldstone.
* 1961 Murray Gell-Mann presents QCD as a $SU(3)$ gauge theory for the strong interaction and predicts quarks.

* 1962 Steven Weinberg, Sheldon Glashow and Abdus Salam present the theory of electroweak interaction, with broken $SU(2)$ symmetry and broken chiral symmetry.

* 1962 Philip Anderson proposes the Higgs-Anderson model to explain the Meissner effect, with broken $U(1)$ symmetry but without Higgs boson.

* 1962 Brian Josephson predicts the Josephson effect (tunneling of supercurrent between two superconductors (observed and patented by Philip Anderson and John Rowell in 1963)).

* 1964 Robert Brout, François Englert and Peter Higgs propose the Higgs mechanism to give mass to particles and predict the Higgs boson.

* 1965 Arno Penzias and Robert Wilson measure “accidentally” the Cosmic Background Radiation at 4 GHz and estimate the temperature $T \approx 3.5$ K of the Universe.

* 1967 David Mermin, Herbert Wagner and Pierre Hohenberg demonstrate the HWM theorem that excludes spontaneous symmetry breaking in low dimensions.

* 1971 Martinus Veltman and his PhD student Gerard 't Hooft regularize the infinite quantum corrections in the theory of the weak interaction.

* 1972 Philip Anderson, More is Different [1].

* 1973 Prediction of the KTB phase transition in $2D$ by John Michael Kosterlitz and David Thouless after work by Vadim Berezinskii, showing that the HWM theorem does not exclude phase transitions of topological nature in low dimensions.

* 1973 Discovery of the $W^\pm$ and $Z$ bosons at CERN, carriers of the weak force.

* 1975 Stephen Hawking predicts Black Holes to radiate thermally and to have entropy.

* 1995 Observation of Bose-Einstein condensation by Eric Cornell and Carl Wieman (Rubidium atoms) and Wolfgang Ketterle (Sodium atoms), with broken $U(1)$ symmetry.

* 1998 Discovery of the acceleration of the expanding Universe by Saul Perlmutter and Adam Reiss, suggesting a positive Cosmological Constant attributed to dark energy.
* 2011 Franck Wilczek predicts time crystals defined by the spontaneous symmetry breaking of time translational symmetry.

* 2012 Discovery of the Higgs boson with mass 125 GeV by the ATLAS and CMS experiments at the Large Hadron Collider (LHC) at CERN, demonstrating the relevance of broken symmetry in particle physics

* 2015 Haruki Watanabe and Masaki Oshikawa demonstrate the absence of quantum time crystals in thermal equilibrium [9].
### III. LORENTZ TRANSFORMATIONS

Let $x = (c_0t, x)$ be a four-dimensional vector. The Lorentz boost $x \rightarrow \Lambda(v) \cdot x = x'$ to a reference $R'$ frame moving with speed $v$ in the original frame $R$ along the $x_1$-axis is defined by

$$ 
\begin{aligned}
c_0 t' &= \gamma(c_0 t - \beta x^1) \\
(x')^1 &= \gamma(-\beta c_0 t + x^1) \\
(x')^2, (x')^3 &= x^2, x^3
\end{aligned} 
\qquad (1) $$

and $\gamma = 1/\sqrt{1-\beta^2}$, with $\beta = v/c_0 \le 1$. The standard Einstein notation is $(x')^\nu = \Delta_\mu^\nu x^\mu$ with implicit summation over crossed indices and $c_0t = x^0$. We see that $\Delta_\mu^0 = \Delta_0^\mu$ for the boosts but this symmetry does not hold for the rotations that are anti-symmetric. With slight abuse of notation of exponents, Lorentz transformations can be defined as those that leave $(x^0)^2 - (x^1)^2 - (x^2)^2 - (x^3)^2 = x^\mu g_{\mu\nu} x^\nu$ invariant, with $g_{\mu\nu}$ the diagonal metric tensor $D(1, -1, -1, -1)$, or equivalently, that $\Lambda(v)^T \cdot g \cdot \Lambda(v) = g$, or:

$$ g_{\mu\nu} \Delta_{\alpha}^{\mu} \Delta_{\beta}^{\nu} = g_{\alpha\beta} \qquad (2) $$

For any Lorentz transformation $\Delta$ is $(\det \Lambda)^2 = 1$, in particular the volume element $d^4x' \equiv c_0 dt' d^3x' = |\det \Delta| c_0 dt d^3x$ in Minkowski space is a Lorentz scalar. Equation (2) also expresses that the metric tensor is a second-rank four-tensor invariant under Lorentz transformations. This notion will have crucial consequences in other representations of the Lorentz group that we will encounter later. It is customary to lower or lift indices using the metric tensor: $g_{\mu\nu} \Delta_\alpha^\mu = \Delta_{\nu\alpha}$. From the identity (2) we derive that $\Delta_\nu^\alpha \Delta_\beta^\nu = g_\beta^\alpha = \delta_\beta^\alpha$. If we compare this to the definition of inverse matrix, $(\Delta^{-1})_\nu^\alpha \Delta_\beta^\nu = \delta_\beta^\alpha$ we infer that $(\Delta^{-1})_\nu^\alpha = \Delta_\nu^\alpha$. A 4D vector $A^\mu$ is said to be a covariant four-vector if it transforms as $(A')^\mu = \Delta_\nu^\mu A^\nu$ under a Lorentz transformation. Its contra-variant equivalent $A_\mu = g_{\mu\nu} A^\nu$ transforms as $(A')_\mu = g_{\mu\nu} \Delta_\alpha^\nu g^{\alpha\beta} A_\beta = \Delta_\mu^\beta A_\beta$. The 4D gradient $\partial_\mu = \partial/\partial x^\mu$ is a contravariant four-vector since $\partial'_\mu = \partial x^\nu/\partial(x')^\mu \partial_\nu = (\Delta^{-1})_\mu^\nu \partial_\nu = \Delta_\mu^\nu \partial_\nu$. Hence the 4D divergence $\partial_\mu A^\mu$ is Lorentz-invariant.

**Important Remark:** The choice $g_{\mu\nu} = D(1, -1, -1, -1)$ is not unique. A different widely used choice, such as $g_{\mu\nu} = D(-1, 1, 1, 1)$ will have many consequences, such as when gauge fields are introduced and canonical (contravariant) momentum $p_\mu + qA_\mu$ is introduced. In the present convention, covariant position vectors correspond to $\{x^i\} = (x, y, z)$, and achieve a minus sign when lowered to $\{x_i\} = (-x, -y, -z)$. This is a drawback because often a logical choice to write a vector in $\mathbb{R}^3$ is $x_i$ when no special relativity is concerned, and thus a source for confusion. A common
habit is to write {$A^\mu$} = ($\phi$, $A_x$, $A_y$, $A_z$), i.e. only use covariant notation if the indices $\mu$ are numbers (0, 1, 2, 3) and not axes. No confusion exists if a vector is not part of a four-vector such as angular momentum $L$. In this course Greek indices $\mu$ always correspond to a four-vector, and normal indices $i$ always to Cartesian vectors.

Improper transformations have $\det \Lambda = -1$ and proper transformations have $\det \Lambda = +1$. We can see that $\Delta_0^0 \ge 1$ (orthochronous Lorentz transformation) or $\Delta_0^0 \le -1$ (non-orthochronous Lorentz transformation). Time-reversal $T = D(-1, 1, 1, 1)$ and space inversion $P = D(1, -1, -1, -1) = g$ can change proper, orthochronous Lorentz transformations into improper or non-orthochronous Lorentz transformations. They are not continuous operations, i.e. impossible to perform infinitely many times them "infinitesimally". The product of any a proper, orthochronous Lorentz transformation is again proper and orthochronous and 6 independent operations exist in the four-dimensional Minkowski space to generate them. This anticipates the statement that proper, orthochronous Lorentz transformations constitute a 6 dimensional group.

In four-dimensional Minkowski space it is not possible to introduce a vector product that defines the angular moment $L = \sum_n (\mathbf{x}^{(n)} \times \mathbf{p}^{(n)})$ of N particles with relativistic momentum $\mathbf{p}^{(n)}$. One can define the second rank tensor

$$M^{\alpha\beta} = \sum_{n=1}^{N} x^{(n),\alpha} p^{(n),\beta} - p^{(n),\alpha} x^{(n),\beta} \quad (3)$$

with the two four-vectors $x^{(n)} = (c_0 t^{(n)}, \mathbf{x}^{(n)})$ and $p^{(n)} = (E^{(n)}/c_0, \mathbf{p}^{(n)})$ in terms of energy $E^{(n)}$ and momentum $\mathbf{p}^{(n)}$ of each particle. By construction is $M^{00} = M^{ii} = 0$ and $M$ transforms as a second-rank Lorentz tensor,

$$(M')^{\alpha\beta} = \Lambda_{\mu}^{\alpha} \Lambda_{\nu}^{\beta} M^{\mu\nu}$$

<figure><img src="image_3.png" /><figcaption>Figure 1. Lorentz boost along $x_1$ axis, with $vt = (v/c_0)x_0$.</figcaption></figure>
At any observation at time t the tensor M can be written as the anti-symmetric 4 × 4 matrix with 6 independent entries,

$$ M^{\mu\nu} = \begin{pmatrix} 0 & N_x c_0 & N_y c_0 & N_z c_0 \\ -N_x c_0 & 0 & L_z & -L_y \\ -N_y c_0 & -L_z & 0 & L_x \\ -N_z c_0 & L_y & -L_x & 0 \end{pmatrix} \qquad (4) $$

with the total angular momentum $L = \sum_n \mathbf{x}^{(n)} \times \mathbf{p}^{(n)}$ and the vector $N = t\mathbf{P} - (E/c_0^2)\mathbf{X}$ in terms of total momentum $\mathbf{P}$, total energy $E$ and $\mathbf{X} = \sum_n E^{(n)}\mathbf{x}^{(n)}/E$. Note that $M_{ij} = \epsilon_{ijk}L^k$ with $\{L^k\} = (L_x, L_y, L_z)$ (see earlier remarks on index convention). This relation links the normal 3–tensor $M_{ij}$ to the pseudo-vector $L^k$ and we shall come back to this. In non-relativistic physics, translation symmetry imposes conservation of total momentum, and rotational symmetry imposes conservation of angular momentum (together 6 constraints). In a relativistic picture rotation and Galilean boots are replaced by a higher Lorentz symmetry that involves a 4-tensor imposing "center-of-mass" to be replaced by center of energy. The full symmetry is completed when also space-time translations are included, figuring the rest mass $m$ and establishing Lorentz-Poincaré symmetry.

## A. Miscellaneous Consequences of Lorentz invariance

### 1. Invariance of the electromagnetic quantum vacuum (Einstein, 1917).

Energy density of electromagnetic radiation density is measured per steradian $d\Omega = \sin\theta d\theta d\varphi$ and per frequency band $d\omega: dI(\theta, \varphi, \omega) = \rho(\theta, \varphi, \omega)d\cos\theta d\varphi d\omega$. According to Maxwell's equations is $dI = d(\frac{1}{2}\mathbf{E}^2 + \frac{1}{2}\mathbf{B}^2)$ in terms of electric and magnetic fields. The latter are obtained from the four-vector potential $A^\mu = (\Phi, \mathbf{A})$ according to $\mathbf{E} = -\nabla\Phi - c_0^{-1}\partial_t\mathbf{A}$ and $\mathbf{B} = \nabla \times \mathbf{A}$. From this it follows that they transform under a Lorentz-boost as

$$ \begin{aligned} \mathbf{E}' &= \gamma (\mathbf{E} + \beta \times \mathbf{B}) - (\gamma - 1)\hat{\beta}\hat{\beta} \cdot \mathbf{E} \\ \mathbf{B}' &= \gamma (\mathbf{B} - \beta \times \mathbf{E}) - (\gamma - 1)\hat{\beta}\hat{\beta} \cdot \mathbf{B} \end{aligned} \qquad (5) $$

In particular, an electric field transforms partly to a magnetic field and vice-versa. Note that these equations are invariant under $\mathbf{E} \to \mathbf{B}$ and $\mathbf{B} \to -\mathbf{E}$ (called the Hodge-dual transformation).
## Exercise III.1

* Show that the complex vector $F_{\pm} = E \pm iB$ transforms under a boost as $F'_{\pm} = M_{\pm} \cdot F_{\pm}$ with the hermitian matrix $M_{nm}(\pm) = \gamma(\delta_{nm} \pm i\epsilon_{nmk}\beta_k) - (\gamma - 1)\hat{\beta}_n\hat{\beta}_m$. Show that $M(\pm)^T \cdot M(\pm) = I_2$ and $\det M = 1$ so that $F_{\pm}^2 = (E^2 - B^2) \pm 2iE \cdot B$ identifies exactly two Lorentz scalars (evidently also invariant under rotation) as well as the complex electromagnetic field $E = \pm iB$ as a Lorentz-invariant configuration.

* Let us define the direction of $\beta$ along the z-axis and write for the wave vector $k = \hat{k}\omega/c_0$ of an electromagnetic wave,

$$ \hat{k} = \cos \theta \hat{z} + \sin \theta (\cos \varphi \hat{x} + \sin \varphi \hat{y}). $$

Show that the spectral energy density transforms as $dI' = \gamma^2(1 - \beta \cos \theta)^2 dI$

* The angle $\varphi$ is not affected by the Lorentz boost. Because $k^{\mu} = (\omega/c_0, k)$ is a four-vector, the volume-element transforms as $d^4k' = |\det \Delta|d^4k = d^4k$ and is Lorentz-invariant. Show that frequency and angle of propagation transform as

$$ \begin{aligned} \omega' &= \gamma\omega(1 - \beta \cos \theta) \\ \cos \theta' &= \frac{\cos \theta - \beta}{1 - \beta \cos \theta} \end{aligned} $$

This transformation has the Jacobian

$$ \frac{d(\omega', \cos \theta')}{d(\omega, \cos \theta)} = \frac{1}{\gamma} \frac{1}{1 - \beta \cos \theta} $$

* Show that the spectral radiation density transforms as

$$ \rho'(\omega', \cos \theta') = \rho\left(\frac{\omega'}{\gamma(1 - \beta \cos \theta)}, \cos \theta\right) \gamma^3(1 - \beta \cos \theta)^3 $$

Any *isotropic* radiation density is invariant under rotations but only the energy spectrum $\rho(\omega, \cos \theta) \sim \omega^3$ is also invariant under boosts. This spectrum corresponds to the electromagnetic quantum vacuum $\rho(\omega) = \frac{1}{2}\hbar\omega \times \omega^2/\pi^2 c_0^3$, the last factor being the well-known density of electromagnetic micro-states per unit volume. The quantum vacuum was not yet discovered but Einstein predicted it already to be Lorentz-invariant.
* Planck's law for the energy density of a perfect black body is given by

$$ \rho(\omega, \cos \theta) = \frac{\hbar \omega^3}{\pi^2 c_0^3} \frac{1}{e^{\hbar \omega / kT} - 1} \quad (6) $$

For the finite temperature $T_0 \approx 2.73$ K of the Universe the thermal energy density should add up to the one of the electromagnetic quantum vacuum. Show that for electromagnetic waves with frequency $f > 60$ GHz (in the microwave region of the electromagnetic spectrum), the vacuum energy dominates. For comparison, the Planck law peaks at 161 GHz and the Cosmic Background Radiation was first measured in 1965 at 4 GHz by Penzias and Wilson. These frequencies are all comparable. Does the quantum vacuum component really exist?

If the Earth moves with a velocity $v$ with respect to the Cosmic Background Radiation show that the thermal component emerges as a Doppler-induced radiation profile $T = T_0(1 - v \cos \theta/c_0)$ in the observed temperature, with $\theta$ the observation angle with respect to the velocity. The latest measurements fix $\theta$ and indicate $v = 369$ km/s, much larger than the rotation speed (30 km/s) of the Earth around the Sun and the rotation speed (100 km/s) of our Solar system around the Center of the Milky Way.

* Consider a radiation power spectrum of the form $\rho'(\omega', \cos \theta') \sim \omega'^4 \cos^2 \theta'$. This corresponds to the energy density generated by a radiating electric dipole at rest with its charge oscillating along $\theta' = \pi/2$. We might suppose that a boost perpendicular to the acceleration changes the frequency as $\omega' = \gamma\omega$, independent of angle. Show that in the frame $R$ that sees the dipole moving at nearly the speed of light ($\beta \approx 1 - 1/2\gamma^2 \approx 1$) the observed power spectrum is given by,

$$ \rho(\omega, \cos \theta) \sim \gamma \omega^4 \frac{(\cos \theta - \beta)^2}{(1 - \beta \cos \theta)^5} \rightarrow 8 \omega^4 \gamma^7 \frac{(1 - \gamma^2 \theta^2)^2}{(1 + \gamma^2 \theta^2)^5} $$

When $\gamma \gg 1$ the dipole radiation becomes strongly peaked in the direction along the line of motion. This relativistic effect is used to produce highly collimated synchrotron radiation.

## 2. Cosmological Constant

Conservation of energy and momentum of electromagnetic waves can be described by the energy stress four-tensor $T^{\mu\nu}$, which has components $T^{00} = \frac{1}{2}\boldsymbol{E}^2 + \frac{1}{2}\boldsymbol{B}^2$, $T^{0i} = T^{i0} = S^i = (\boldsymbol{E} \times \boldsymbol{B})^i$ and $T^{ij} = E^i E^j + B^i B^j - \delta_{ij}T^{00}$ the tensor describing electromagnetic stress (if it is isotropic we call it radiation pressure). Electromagnetic quantum fluctuations obey the fluctuation dissipation theorem that relates electromagnetic fluctuations $\langle E_i E_j \rangle$. No negative frequencies occur in second
quantization since terms $\exp(+i\omega t)$ are converted into the creation of a photon, equal to its anti-particle. Electromagnetic quantum fluctuations at zero temperature obey, as we shall see later,

$$ \langle T^{\mu\nu} \rangle = -2\hbar \operatorname{Im} \int_0^\infty \frac{d\omega}{2\pi} \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{1}{(\omega + i0)^2/c_0^2 - k^2} \begin{pmatrix} \omega^2/c_0^2 & \omega k^j/c_0 \\ \omega k^i/c_0 & k^i k^j \end{pmatrix} \quad (7) $$

### Exercise III.2

Verify that this tensor behaves as a traceless four-tensor with $T^\mu_\mu = 0$ under Lorentz transformation and that any upper cut-off in frequency to get rid of the divergence would break Lorentz-invariance. If we assume (and the use of distributions is far from evident in diverging integrals) that $\operatorname{Im} 1/(x + i0) = -\pi\delta(x)$, we would get

$$ \langle T^{\mu\nu} \rangle = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{3} \boldsymbol{I}_3 \end{pmatrix} \times \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \frac{1}{2} \hbar \omega_k $$

with $\omega_k = kc_0$, and featuring a zero energy current, zero momentum density and a diverging zero-point energy as Lorentz scalar. A tensor of this form has an energy density equal to 3 times the pressure, characteristic for ultra-relativistic dust and is still traceless $T^\mu_\mu = 0$.

The tensor $\langle T^{\mu\nu} \rangle$ obtained in Exercise III.2 is not proportional to the Lorentz-invariant metric tensor $g_{\mu\nu}$ which means that in this form the vacuum energy, even divergent, cannot be a candidate for the cosmological constant in Einstein's gravity theory, whose righthand side features $(8\pi G/c_0^4)T^{\mu\nu} - \Lambda g^{\mu\nu}$. If we would just take a regularized value of $T^{00}$ (the energy density), not only would we break Lorentz invariance, the cosmological constant would also be large and negative. The divergent vacuum energy is one of the most important unsolved problems in physics and is protected by Lorentz invariance and thus not easy to repair. What cancels this infinite energy, apparently with extreme fine-tuning since $\Lambda$ can be estimated from the acceleration of the Universe and is seen to be “slightly” positive? Let us try to “solve” the ultraviolet catastrophe in a Lorentz-invariant way by assuming the existence of some (maybe too..) speculative extra function $f(s^2 = k^2 - \omega^2/c_0^2)$ in the nominator of Eq. (7) with the “normal” value $f(s^2) \approx 1$ for small $s$, that converges rapidly to 0 for large arguments and that is analytic for $\operatorname{Re}\omega > 0, \operatorname{Im}\omega > 0$ for causality reasons.
### Exercise III.3

Show that by performing a “Wick rotation” in the complex plane ($\omega/c_0 \to ik'$) and by carrying out the remaining 2D integration over $dkdk'$, the energy-stress tensor takes the form

$$ \langle T^{\mu\nu} \rangle = -g^{\mu\nu} \times \frac{\hbar c_0}{32\pi^2} \int_0^{\infty} ds s^3 f(s^2) \equiv -\rho_0 g_{\mu\nu} $$

Being proportional to the metric tensor, this can indeed represent a cosmological constant $\Lambda = (8\pi G/c_0^4)\rho_0$. If $f(s^2) > 0$ the minus sign of $\langle T^{\mu\nu} \rangle$ would imply a positive value, which agrees with the observed acceleration of the universe. The current estimate $\Lambda = 1.10 \cdot 10^{-52} m^{-2}$ is equivalent to a cosmic energy density $\Lambda c_0^4/8\pi G = 5.35 \cdot 10^{-10} J/m^3$. This implies that $f(s^2)$ must differ from 1 at scales $\pi/s$ smaller than typically tens of microns where new gravitational laws would show up and where the Minkowski metric of Maxwell’s equations breaks down. In particular, Newton’s $1/r$ gravitation law should break down at this scale.

### 3. Unruh effect and black hole temperature.

If the quantum vacuum is the same for all observers linked by Lorentz-transformation, how does it emerge in a reference frame that sees itself accelerating with constant acceleration $a$? Let the speeds in our frame and the accelerating frame be $u$ and $u'$ all assumed co-linear. Acceleration transforms under a Lorentz transformation as

$$ \frac{du'}{dt'} = \frac{du}{dt} \frac{1}{\gamma^3} \frac{1}{(1 - uv/c_0^2)^3} $$

Since $du'/dt' = a$ and $u = v$ if we want to have the accelerator to be instantaneously co-moving, it follows that we see him with a velocity that grows with time as

$$ \frac{dv}{dt} = a \left( 1 - \frac{v^2}{c_0^2} \right)^{3/2} $$

with solution $v(t) = at/\sqrt{1 + a^2 t^2/c_0^2}$.

### Exercise III.4

* The proper time $\tau$ is defined by the time in the instantaneously co-moving frame. The length $ds = (c_0^2 dt^2 - dx^2)^{1/2}$ is Lorentz invariant. In the co-moving frame is $ds = c_0 d\tau$ whereas in rest frame is $ds = c_0 dt(1 - u(t)^2/c_0^2)^{1/2}$. Hence $d\tau = dt/\gamma$ (equivalently we can put $\tau = t'$
and $x = vt$ in Eq. (1)). Show that the time experienced by the accelerating observer relates to our time $t$ as $t = (c_0/a) \sinh a\tau/c_0$, and that $\boldsymbol{x}^2(t) - c_0^2 t^2 = -(4c_0^4/a^2) \sinh^2(a\tau/2c_0)$

* Since in vacuum we have $\nabla \cdot \boldsymbol{E} = 0$, we can use the gauge with $\Phi = 0$, $\nabla \cdot \boldsymbol{A} = 0$ and set $\boldsymbol{E} = -c_0^{-1} \partial_t \boldsymbol{A}$. Equation (7) is consistent with the correlation function for the vector potential in our rest frame at $T = 0$,

$$ \langle A_i(0,0)A_j(t, \boldsymbol{x}) \rangle = 2\pi\hbar\delta_{ij} \int_0^\infty \frac{d\omega}{2\pi} \int \frac{d^3\boldsymbol{k}}{(2\pi)^3} \delta\left(\frac{\omega^2}{c_0^2} - k^2\right) \exp(i\boldsymbol{k} \cdot \boldsymbol{x} - i\omega t) $$

Show that the correlation function measured by the moving observer, expressed in its own proper time, is

$$ \langle A_i(t = 0, \boldsymbol{x} = 0)A_j(t, \boldsymbol{x}(t)) \rangle = \frac{\hbar c_0 \delta_{ij}}{(2\pi)^2 \boldsymbol{x}^2(t) - c_0^2 t^2} = -\frac{\hbar \delta_{ij}}{(2\pi)^2 c_0 \sinh^2(a\tau/2c_0)} $$

* We can compare this to the time-correlation function in a rest frame at a finite temperature when also thermal photons exist. We will demonstrate in section VI.B that this changes the correlation function to,

$$ \begin{aligned} \langle A_i(0,0)A_j(t,0) \rangle &= 2\pi\hbar\delta_{ij} \int_0^\infty \frac{d\omega}{2\pi} \int \frac{d^3\boldsymbol{k}}{(2\pi)^3} \delta\left(\frac{\omega^2}{c_0^2} - k^2\right) (f_\omega + 1) \exp(-i\omega t) + f_\omega \exp(i\omega t)) \\ &= \frac{\hbar\delta_{ij}}{(2\pi)^2 c_0} \int_0^\infty d\omega \, \omega \, \cosh\left(\frac{\hbar\omega}{2kT}\right) \cos(\omega t) \end{aligned} \quad (8) $$

with the familiar Boltzmann factor $f_\omega = 1/[\exp(\hbar\omega/kT)-1]$ so that $2f_\omega+1 = \cosh\hbar\omega/2kT$. The real part has been taken because we expect the imaginary part to vanish for real-valued $A_i(\boldsymbol{x}, t)$. For $t > 0$, show that by moving the frequency integral to Matsubara frequencies (the poles on the positive imaginary axis),

$$ \langle A_i(t = 0,0)A_j(t,0) \rangle = -\frac{\hbar\delta_{ij}}{(2\pi)^2 c_0 \sinh^2(\pi k T t/\hbar)} $$

The correlation function of the vector potential at finite temperature has exactly the same form as the one for the vacuum fluctuations at $T = 0$ observed in the accelerated frame and we can identify $kT = \hbar a/2\pi c_0$. An accelerated body at zero temperature infers the quantum vacuum as blackbody radiation. This is the Unruh effect, and is still unobserved. The correlation function diverges universally as $-\hbar/c_0 t^2$ for small times due to the abundance of extremely high frequencies in the quantum vacuum.

Since acceleration and gravitation are related by Einstein's equivalence principle, this argument can be used to show that gravitating bodies radiate at a finite temperature. Hawking (1975) predicted
black holes to radiate thermally at their event horizon and argued that this is why they finally evaporate. The so-called Schwarzschild radius $r_S$ of a black hole is known to be $r_S = 2GM/c_0^2$ where total rest energy $Mc_0^2$ and gravitational energy $GM^2/r_S$ become comparable. The gravitational acceleration at this distance is $g = GM/r_S^2 = c_0^2/4GM$. Thus

$$kT = \frac{\hbar g}{2\pi c_0} = \frac{\hbar c_0^3}{8\pi GM}$$

which is exactly the Hawking temperature $T = 6.17 \cdot 10^{-8}(M_\odot/M)$ K. The finite temperature, however small, guarantees the existence of black hole entropy, predicted to be proportional to the surface $4\pi r_S^2$ of emission which violates by many orders of magnitude the result from GR that black holes are characterized only by their mass, size and angular momentum and thus have low entropy. Again a giant unsolved puzzle!

## 4. The Expanding Universe & Cosmological Principle.

In Einstein's General Relativity (GR), the metric tensor is no longer Minkowskian and space-time becomes curved. Its application on scales as large as our Universe relies on the Cosmological Principle that assumes a perfectly spherical, homogeneous and isotropic Universe seen by all observers on large scales. The Principle eliminates off-diagonal terms in the energy stress-tensor that should take the simple form

$$\langle T^{\mu\nu} \rangle = \begin{pmatrix} \rho(t) & 0 \\ 0 & I_3 p(t) \end{pmatrix}$$

involving energy density and pressure of the Universe. It was shown by Robertson and Walker in 1937 that the Cosmological Principle also restricts the metric to only three possibilities,

$$ds^2 = c_0^2 dt^2 - a^2(t) \left[ \frac{dr^2}{1 - kr^2} + r^2 (\sin^2 \theta d\phi^2 + d\theta^2) \right]$$

with $k = -1, 0, 1$ depending on the geometry (hyperbolic for $k = -1$, flat or Euclidian for $k = 0$ or elliptic for $k = +1$), and $r, \theta, \phi$ local dimensionless spherical coordinates; $a(t)$ is a unique characteristic length scale of the Universe that evolves with time and the local “co-moving” distance is defined as $dx = a(t)dr$. This means that all distances measured at the same universal time $t$ scale with $a$ and the volume of the universe $V(t)$ at time $t$ is proportional to $a(t)^3$. The space curvature is the same throughout the Universe, given by $k/a^2(t)$. Clearly, for $k < 0$ we require that $r < 1$ so that the Universe must be closed on scales of order $a$ at all times. Einstein's GR provides
two equations that describe the time-evolution of $a(t)$, known as the Friedman equations,

$$ \left(\frac{\dot{a}}{a}\right)^2 - \frac{8\pi G}{3c_0^2}\rho = -\frac{kc_0^2}{a^2} + \frac{\Lambda c_0^2}{3} \qquad (9) $$

$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3c_0^2}(\rho + 3p) + \frac{\Lambda c_0^2}{3} \qquad (10) $$

The first equation is the GR balance of kinetic energy + gravitational energy = curvature + energy from unknown matter. The second is an equation of motion for the Universe with the *unknown matter having a pressure that equals minus its energy density*.

Einstein noticed the possibility to add a Cosmological Constant $\Lambda$ to his GR equations and was satisfied to see that it facilitated a static solution with $\dot{a} = 0$. For a flat Universe this would yield the mass density $\rho_0/c_0^2 = -\Lambda c_0^2/8\pi G$ and requires a negative Cosmological Constant. A model with $\dot{a} = da/dt > 0$ describes an expanding Universe and was first predicted from GR in 1927 by Georges Lemaître. The expansion of the Universe was observed in 1929 by Edward Hubble, seven years after the Friedmann equations were published. Hubble established the phenomenological law $v = H_0D$ for the radial separation speed $v$ of a distant galaxy - measured from its observed redshift - at a distance $D$, obtained from another method. Today's official value is $H_0 = \dot{a}/a = 73.8$ km/s/Megaparsec. Einstein rapidly admitted his erroneous initial vision of a static Universe. The acceleration $\ddot{a} > 0$ of the cosmic expansion was observed much more recently in 1998 by Saul Perlmutter and Adam Reiss which established for the first time - via the second Friedmann equation, the undeniable need for a positive Cosmological Constant interpreted as *dark energy*, characterized by a negative pressure.

The introduction of $\Lambda$ in GR is generally recognized as the first theoretical indication for the existence of dark energy and his “mistake” has undoubtedly been one the greatest contributions of Einstein to physics. As argued above, a relation is suspected between the energy density $\rho_D = \Lambda c_0^4/8\pi G$ of the dark energy and some regularized version of the Lorentz-invariant quantum vacuum.

According to Eq. (6), the total number of photons at frequency $\omega$ in the Universe is proportional to $Vd\omega\omega^2[\exp(\hbar\omega/kT) - 1]$. The wavelength of a photon in an expanding universe red-shifts “adiabatically” as $\lambda(t) \sim a(t) \sim 1/\omega(t)$. It follows that the Planck spectral distribution is unaltered by the expansion although the temperature cools down as $T(t) \sim 1/a(t)$. The total radiative energy density on the other hand $\rho_R(t) = \sigma T(t)^4$ decays as $1/a(t)^4$. For the matter energy density we expect that $\rho_M(t) \sim 1/V(t) \sim 1/a(t)^3$ since it exerts no pressure when it expands so that $dU = d(\rho_M V) = TdS = 0$ for an (adiabatic) change that involves no heat exchange. As the Universe expands, the radiative energy density declines faster than the energy density of the matter
(because it exerts a positive pressure $p = \rho_R/3$ and $dU + pdV = 0$). If we extrapolate the present picture back, some time must have existed when radiation dominated the energy.

Dark energy is characterized by an equation of state $p = -\rho_D$. This implies that $d(\rho_D V) - \rho_D dV = d(\rho_D)V = 0$. The dark energy density does not decay with the expansion of the Universe and will for that reason eventually dominate.

### Exercise III.4: Time dependence and Age of the Universe

1. Show from both Friedmann equations that for a flat matter-dominated Universe ($k = 0$, $p = 0$ and $\Lambda = 0$), the expansion is algebraic and increases as $a(t) \sim t^{2/3}$.

2. Show that for a flat Universe dominated by dark-matter, the expansion is exponential and $a(t) \sim \exp(c_0\sqrt{\Lambda/3}t)$ so that the temperature cools down exponentially.

3. It is customary to define the energy densities $\Omega = \rho/\rho_c$ for radiation (photons and neutrinos), matter and dark matter as measured today relative to the “critical” density $\rho_c = 3c_0^2 H_0^2 / 8\pi G \approx 8.7 \cdot 10^{-27}$ kg/m$^3$. The Cosmic Background Explorer (COBE) satellite launched in 1989 revealed the first values $\Omega_R = 9 \cdot 10^{-5}$, $\Omega_M = 0.31$ and $\Omega_D = 0.69$, showing both that the Universe has entered his dark-matter-dominated stage and that the total energy density is equal (within error bars) to the critical energy density $\rho_c$. If we neglect the small period that radiation dominated and normalize $a \to a/a_0$ to its current value, show that the first Friedmann equation can be recast into,

$$ \frac{da}{dt} = H_0 \sqrt{\frac{\Omega_M(t_0)}{a} + \Omega_D a^2} $$

with $H_0 = 73.8$ km/s/Megaparsec $= 2.2 \cdot 10^{-18}/s$ the currently accepted value for the Hubble constant. The solution for the age of our Universe is

$$ t_0 = \frac{1}{H_0} \int_0^1 \frac{da}{\sqrt{\Omega_M(t_0)a^{-1} + \Omega_D a^2}} = \frac{0.95}{H_0} = 13.6 \text{ billion years} $$

which is close to the currently accepted age of 13.7 billion years.

4. Show that

$$ \frac{H(t)^2}{H_0^2} = \Omega(t) - \frac{kc_0^2}{a^2(t)H_0^2} $$

so that today,

$$ 1 - \Omega(t_0) = -\frac{kc_0^2}{a_0^2 H_0^2} \approx -\frac{k}{10} \frac{a_0(\text{obs})^2}{a_0^2} $$
in terms of the radius of the current *observable* Universe $a_0(\text{obs}) = 46.5$ billion light years $= 4.4 \cdot 10^{26}$ m. The Wilkinson Microwave Anisotropy Probe launched in 2001 reported the mass density $\rho(t_0) = 9.9 \cdot 10^{-27}$ kg/m$^3$ of the observable Universe which leads to $\Omega(t_0) = 1.14$ with some error bar. The possibility that $k = +1$ is excluded by the observed accelerated expansion, and the equality $k/10 = 0.14a_0(\text{obs})^2/a_0^2$ is generally accepted to imply a flat Universe with $k = 0$ and the obvious suspicion that the Universe is larger than the one we observe.
IV. PRINCIPLE OF LEAST ACTION & EQUATION OF MOTION

Given a real-valued scalar field $\Psi(t, x)$, we define the Lagrangian,

$$L[t, \Psi] = \int d^3x \mathcal{L}(\Psi, \partial_\mu \Psi) \qquad (11)$$

in terms of the Lagrangian density $\mathcal{L}$. The Lagrangian depends on time and is "functionally dependent" on the entire wave function $\Psi$ everywhere in space-time, as well as on its 4 derivatives $\partial_\mu \Psi$. If the field is complex-valued, rather than using real and imaginary parts of the wave function as independent degrees of freedom, we can use the wave function and its complex conjugate. The volume integral is performed over $\mathbb{R}^3$ and we assume sufficient decay for this integral to converge. The action during the time slot $t_B - t_A$ is

$$S[\Psi] = \int_{t_A}^{t_B} dt L[t, \Psi] = \int d^4x \mathcal{L}(\Psi, \partial_\mu \Psi) \qquad (12)$$

where we defined the four vector $x = (c_0t, x)$.

If the action $S$ is Lorentz-invariant, then so is $\mathcal{L}$. The *principle of least action* postulates that the equation of motion for $\Psi$ in space-time be determined by a local minimum for $S$. Hence for any small variation $\delta\Psi(x)$ at all points $x$ in space time around the physical solution $\Psi(x)$, given that $\delta\Psi(a) = \delta\Psi(b) = 0$, must obey $\delta S = 0$. This can be mathematically expressed more precisely using the concept of functional derivative. Consider the functional

$$S[f] = \int_{t_A}^{t_B} dt L(f, \partial_t f)$$

<figure><img src="image_4.png" /><figcaption>Figure 2. Principle of minimal action to go from A to B.</figcaption></figure>
For a small variation $\delta f(t)$ of the function $f$, it follows that

$$ 
\begin{align*}
\delta S[f] &= \int_{t_A}^{t_B} dt \left( \frac{\partial L}{\partial f} \delta f + \frac{\partial L}{\partial (\partial_t f)} \delta (\partial_t f) \right) \\
&= \int_{t_A}^{t_B} dt \left( \frac{\partial L}{\partial f} - \frac{\partial}{\partial t} \frac{\partial L}{\partial (\partial_t f)} \right) \delta f + \int_{t_A}^{t_B} dt \frac{\partial}{\partial t} \left( \frac{\partial L}{\partial (\partial_t f)} \delta f \right) \\
&= \int_{t_A}^{t_B} dt \left( \frac{\partial L}{\partial f} - \partial_t \frac{\partial L}{\partial (\partial_t f)} \right) \delta f + \frac{\partial L}{\partial (\partial_t f)} \delta f(t_B) - \frac{\partial L}{\partial (\partial_t f)} \delta f(t_A)
\end{align*}
$$

If we vary the function $f$ in the interval $(t_A, t_B)$ while keeping the values at the boundaries constant, it follows,

$$ \delta S[f] = \int_{t_A}^{t_B} dt \left( \frac{\partial L}{\partial f} - \partial_t \frac{\partial L}{\partial (\partial_t f)} \right) \delta f(t) \quad (13) $$

A minimum of $S[f]$ is characterized by $\delta S[f] = 0$ for arbitrary small $\delta f$. Hence,

$$ \frac{\partial L}{\partial f} - \partial_t \frac{\partial L}{\partial (\partial_t f)} = 0 \qquad (14) $$

With this in mind, we reconsider Eq. (12) and vary the field $\Psi$. We find,

$$ \delta S[\Psi] = \int d^4 x \left( \frac{\partial \mathcal{L}}{\partial \Psi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \Psi)} \right) \delta \Psi + \int d^4 x \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \Psi)} \delta \Psi \right) $$

We have defined the covariant derivative $\partial_{\mu} := \partial/\partial x^{\mu}$. It has the property that $g_{\mu\nu}\partial_{\nu}\partial_{\mu} = g^{\mu\nu}\partial_{\nu}\partial_{\mu} = \partial^{\mu}\partial_{\mu} = c_0^{-2}\partial_t^2 - \nabla^2$ is a Lorentz-invariant operator. The last integrand takes the form of a time derivative $\partial_t(\partial_{\mu}\mathcal{L})/\partial(\partial_t\Psi)\delta\Psi(t, \mathbf{x})$ plus the divergence of the vector field $v_i = \delta\Psi \times \partial_{\mu}\mathcal{L}/\partial(\partial_i\Psi)$. If we assume that $\delta\Psi(t_A, \mathbf{x}) = \delta\Psi(t_B, \mathbf{x}) = 0$ for all $\mathbf{x}$ the first term disappears as above. By Gauss' theorem, the second term transforms into a surface integral at infinity and vanishes if the fields at infinity vanish. Therefore, $\delta S = 0$ for arbitrary $\delta\Psi(t, \mathbf{x})$ yields

$$ \frac{\partial \mathcal{L}}{\partial \Psi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \Psi)} = 0 \qquad (15) $$

This equation is known as the Euler-Lagrange equation. It is Lorentz-invariant because the contraction $\partial_{\mu}\partial/\partial(\partial_{\mu}\Psi)$ is Lorentz-invariant. To see this recall that in flat Minkowskian space $\partial x^{\nu}/\partial x'^{\mu}$ is a constant so that

$$ \partial'_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial'_{\mu} \Psi)} \equiv \frac{\partial}{\partial x'^{\mu}} \frac{\partial \mathcal{L}}{\partial (\partial \Psi / \partial x'^{\mu})} = \frac{\partial x^{\nu}}{\partial x'^{\mu}} \frac{\partial}{\partial x^{\nu}} \frac{\partial x'^{\mu}}{\partial x^{\alpha}} \frac{\partial \mathcal{L}}{\partial (\partial \Psi / \partial x^{\alpha})} = \delta^{\nu}_{\alpha} \partial_{\nu} \frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \Psi)} = \partial_{\nu} \frac{\partial \mathcal{L}}{\partial (\partial_{\nu} \Psi)} $$

If the field is complex-valued and/or has more components, each one has its own Euler-Lagrange equation if we can vary them all independently.

Any transformation $\mathcal{L} \to \alpha\mathcal{L}$ with $\alpha \neq 0 \in \mathbb{R}$ leaves the Euler-Lagrange equations of motion invariant (although a maximum could change into a minimum for $\alpha < 0$).
Exercise IV.1 : Higgs Lagrangian

Given a complex scalar field $\Psi(t, x)$ in space time, consider the action

$$ S[\Psi, \Psi^*] = \frac{1}{2} \int dt \, d^3x \left( \frac{1}{c_0^2} (\partial_t \Psi)(\partial_t \Psi^*) - (\partial_n \Psi)(\partial_n \Psi^*) - \frac{m^2 c_0^2}{\hbar^2} \Psi \Psi^* - \frac{\lambda}{2} (\Psi \Psi^*)^2 \right) \\ = \frac{1}{2} \int d^4x \left( (\partial_\mu \Psi)(\partial^\mu \Psi^*) - \frac{m^2 c_0^2}{\hbar^2} \Psi \Psi^* - \frac{\lambda}{2} (\Psi \Psi^*)^2 \right) \qquad (16) $$

with $\partial^\mu = g^{\mu\nu}\partial_\mu = (c_0^{-1}\partial_t, -\nabla)$. The presence of $\hbar$ in the mass term anticipates second quantization with four-momentum $p_\mu = -i\hbar\partial_\mu$. Discuss the symmetries of the action and derive the equation of motion for $m^2, \lambda > 0$. If we interpret the integrand as $T-V$ (kinetic minus potential energy), then in the usual situation $m^2 > 0$ the potential has a stable minimum for $\Psi\Psi^* = 0$ (the “traditional vacuum state” since after second quantization $\Psi\Psi^*$ represents the particle number operator). If $m^2 < 0$ however, the potential $V$ the minimum occurs at $\Psi\Psi^* = -m^2c_0^2/\hbar^2\lambda \equiv v^2 > 0$. The maximum at $\Psi\Psi^* = 0$ is unstable since any small fluctuation lowers the potential energy and thus increases kinetic energy. Similar to second order phase transitions, the quantum field prefers to settle in a new stable vacuum associated with one such minimum with finite amplitude and well-defined phase (say equal to zero) that breaks the rotational symmetry. If we expand it around this new vacuum state,

$$ \Psi(x) = v + i\phi_{\perp}(x) + \phi_H(x) \qquad (17) $$

with $\phi_\perp(x)$ and $\phi_H$ two new real-valued fields along real and imaginary axis. Show that the new Lagrangian can be written as

$$ \mathcal{L}(\phi_\perp, \phi_H) = \partial_\mu \phi_H \partial^\mu \phi_H - \frac{1}{2} \lambda v^4 - 2 \lambda v^2 \phi_H^2 \\ + \partial_\mu \phi_\perp \partial^\mu \phi_\perp \\ + \mathcal{O}[\phi_H(\phi_H^2 + \phi_\perp^2), (\phi_H^2 + \phi_\perp^2)^2] \qquad (18) $$

This shows that $\phi_H$ represents a scalar field of mass $M_H^2 c_0^2/\hbar^2 = 2\lambda v^2 = (\text{hence } M_H^2 = -2m^2)$, and $\phi_\perp$ a scalar field of mass zero. This derivation anticipates the massless Goldstone mode $\phi_\perp$, as well as the massive Higgs field $\phi_H$ when a global symmetry is broken in a Lorentz-invariant Lagrangian.
Exercise IV.2

Given a four-vector $K^\mu$, show that any extra term of the form $\partial_\mu K^\mu$ in the Lagrangian density $\mathcal{L}$ leaves the equation of motion unchanged. Hint: use Gauss' law generalized to space-time:

$$ \int_{\Omega} d^4 x \partial_{\mu} K^{\mu} = \int_{\delta\Omega} d^3 x n_{\mu} K^{\mu} $$

with $\delta\Omega$ the surface (with normal vector $n_\mu$) surrounding the space-time volume $\Omega$.
# V. SYMMETRY GROUPS IN PHYSICS AND THEIR REPRESENTATIONS

Quantum physics is described by complex fields $\Psi(x)$ - that can have more than one component- that live in Hilbert space and operators act on fields in Hilbert space. Both depend on space-time. The *Euclidian Principle of Relativity* states that space and space-time are homogeneous and as isotropic as they can be. This means we can rotate the axes, translate them, invert them, boost them and expect physics to be invariant. *The physical laws cannot depend on the way we label space-time.* We will encounter several non-trivial consequences of this apparently innocent statement.

Symmetry operations in Hilbert space leave its scalar product unchanged, more precisely its absolute value. An important theorem, due to Eugene Wigner, is that such transformations are either unitary or anti-unitary. Recall that given a scalar product $\langle\Psi|\Phi\rangle$ in Hilbert space a unitary operation $U$ obeys $\langle U\Psi|U\Phi\rangle = \langle\Psi|\Phi\rangle$, often written as $UU^\dagger = 1$ in terms of its adjoint operator $U^\dagger$. An anti-unitary information $U_a$ on the other hand, obeys $\langle U_a\Psi|U_a\Phi\rangle = \overline{\langle\Psi|\Phi\rangle} = \langle\Phi|\Psi\rangle$. In particular, for any $\lambda \in \mathbb{C}$, $U|\lambda\Psi\rangle = \lambda U|\Psi\rangle$ is a linear operator whereas $U_a|\lambda\Psi\rangle = \lambda U_a|\Psi\rangle$ is an anti-linear operator. It follows that $U_a^2 = U$ is a unitary operator. In this lecture we will mainly be concerned by continuous symmetries that are unitary. However, anti-unitary transformations exist, such as time-reversal and charge conjugation. They give rise to the most fundamental theorem in quantum field theory: the CPT theorem.

## A. Properties of a Group

It is convenient to disconnect the Hilbert space of physical states from the mathematical structure that governs the symmetry. The relation between symmetry operations and its actions in Hilbert space are discussed in the next section.

A group $G$ is a set of elements $\{g\}$ characterized by 4 conditions:

1. An operation $G \times G \to G$, $g_1 \cdot g_2 = g$, is defined.

2. The operation is associative, i.e. $(g_1 \cdot g_2) \cdot g_3 = g_1 \cdot (g_2 \cdot g_3)$. If $g_1 \cdot g_2 = g_2 \cdot g_1$ the group is said to be Abelian.

3. The identity operator $g_0 = e$ such that $g \cdot e = e \cdot g$ for all $g \in G$ is part of the group. It is necessarily unique. If the operation refers to multiplication then $e = 1_G$, if it refers to addition then $e = 0_G$.
4. Every operator $g$ must have an inverse $g^{-1}$ in the group such that $g \cdot g^{-1} = e$.

One can distinguish discrete groups (with finite or countable number of elements) from continuous groups. Discrete groups can represent symmetries corresponding to rotations over 90 degrees in all directions (cubic symmetry) or inversion symmetry $P$ where $r \to -r$. Continuous groups are characterized by a continuous parameter that can be small. For instance the group $U(1)$ with elements $\{g(\theta)\}, \theta \in \mathbb{R}$ with operation $g(\theta_1) \cdot g(\theta_2) = g(\theta_1+\theta_2)$ has a straightforward representation on the numbers $\exp(i\theta)$ on the unit circle of the complex plane. Only one real-valued parameter is needed to span this group, the group is continuous with dimension equal to 1. A group $G$ may have one or more subgroups that obey themselves the group conditions.

For each element $g$ in a group $G$, we can define left ($L_n$) and right ($R_n$) cosets of the subgroup $H$ as all elements ($gH$) or ($Hg$). Different elements $g_1$ and $g_2$ inside the same coset are called equivalent: $g_1 \sim g_2$. If $gH = Hg$ for all $g \in G$, i.e. all left and right cosets are equal, the subgroup $H$ is said to be normal. The set of cosets of a normal subgroup constitutes the quotient group $G/H$ ("G modulo $H$"). This group splits the group $G$ up in different parts that individually preserve the symmetry of $H$. When a symmetry $G$ is broken, leaving behind a lower symmetry described by its subgroup $H$, cosets collect operators that generate "equivalently" broken symmetry states whereas different cosets generate states with non-equivalently broken symmetry. The set $G/H$ generating non-equivalently broken states only has group structure when the subgroup $H$ is normal.

**Exercise V.1: Cosets**

(https://en.wikipedia.org/wiki/Coset)

Let $H$ be a subgroup of the group $G$.

1. Show that $g_1 \sim g_2$ implies that $g_1 = g_2 h$ for some element $h \in H$.

2. Show that cosets are either identical or non-overlapping and that $\bigcup_n L_n = \bigcup_n R_n = G$.

3. Show that the number of different left cosets and right cosets of $H$ and $G$ is equal. It is called the *index* or *order* $[G : H]$ of $H$ in $G$.

4. Show that the equivalence relation satisfies convexity ($g \sim g$), symmetry (if $g_1 \sim g_2$ then $g_2 \sim g_1$) and transitivity (if $g_1 \sim g_2$ and $g_2 \sim g_3$ then $g_1 \sim g_3$).
5. Show that any group $G$ is always normal subgroup of itself, in particular $gG = Gg = G$ for all $g \in G$. (Hint: choose $g_1 \in gG$, thus $g_1 = g \cdot g_0$ for some $g_0 \in G$. Choose $g_2 = g_1 \cdot g_0 \cdot g_1^{-1} \in G$ and show that $g_2 \cdot g = g_1$ hence $g_1 \in Gg$. If a $g_1 \in G$ would exist such that $g_1 \neq g \cdot g_0$ for any $g_0 \in G$ then $g_1 \cdot g^{-1} \notin G$. This violates group properties, hence $gG = Gg = G$). Show that $[G : G] = 1$.

6. *Lagrange theorem*. Let $|G|$ be the number of elements of the group $G$, assumed finite for simplicity, and $|H|$ the number of elements of the subgroup $H$. Show that $|G|/|H| = [G : H]$.

7. The group operation $(aH) \cdot (bH) := (a \cdot b)H$ can be defined between cosets of the quotient set but $(ah_1) \cdot (bh_2) \neq (ab)h_3$ in general so that the product would depend on the choice of $a$ and $b$ in the coset. Show that only if the subgroup $H$ is normal in $G$, the quotient set $G/H$ with this operation constitutes a group. (https://en.wikipedia.org/wiki/Quotient_group).

Several useful aspects can be identified with group structure.

* A cyclic group $C$ is generated by one single element $g_0$, i.e. all group elements, including its inverse $g_0^{-1}$, are products of $g_0$.

* A homomorphism between two groups $G$ and $H$ is a map $f : G \to H$ that preserves group operation: $f(g_1 \cdot g_2) = f(g_1) \cdot f(g_2)$. The map $f$ is an isomorphism if it is also bijective (one-to-one correspondence in two directions), and one writes $G \cong H$. The kernel $K(f)$ of the homomorphism $f$ is the subset $K$ of $G$ such that $f(g \in K) = e_H$. The image $f(G)$ of $f$ is the subset $I$ in $H$ that obeys $f(g) = h$ for some $g \in G$.

* The kernel $K(f)$ of a homomorphism $f : G \to H$ is a normal subgroup of $G$. We first check that $K$ is a group. Clearly, $e_G \in K$ because $f(e_G) = e_H$. For any $k_{1,2} \in K$, $f(k_1 \cdot k_2) = f(k_1) \cdot f(k_2) = 1$. Hence $k_1 \cdot k_2 \in K$. Finally $k^{-1} \in K$ because $f(k^{-1}) \cdot f(k) = f(k^{-1}) = f(k \cdot k^{-1}) = e_H$. The subgroup $K$ is normal: for any $g \in G$ we have $g \cdot k \cdot g^{-1} \in K$ since $f(g \cdot k \cdot g^{-1}) = e_H$. Hence $gK = Kg$ for all $g \in G$.

* Given two groups $G$ and $H$, its direct product $G \times H$ is defined as the Cartesian product $(g \in G, h \in H)$ with operation $(g_1, h_1) \cdot (g_2, h_2) = (g_1 \cdot g_2, h_1 \cdot h_2)$, and constitutes a group (and can be generalized straightforwardly if $G$ and $H$ have different operations). This operation is well-defined for groups and is not to be confused with the tensor product, only defined for representations of $G$ and $H$ in a vector space $V$ and often written as $G \otimes H$.
<figure><img src="image_5.png" /><figcaption>Figure 3. Homomorphism $f: G \to H$.</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>Figure 4. Kernel of $f$ containing the elements in $G$ that map to the unit element in $H$.</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Figure 5. Image of $f$ in $H$. Note that $f: G/K \to f(G)$ is an isomorphism</figcaption></figure>

* The general definition of direct sum is somewhat more subtle, especially because for infinite groups sums may not converge. We restrict to a (not necessarily Abelian) group $L$ with two normal subgroups $G$ and $H$. The group $L$ is said to be the direct sum $L = G + H$ if $G$ and $H$ are normal subgroups with trivial overlap (i.e. share only $e$) and if for all $\ell \in L$ we have $\ell = g \cdot h$. Because the subgroups only share $e$ as common element, the decomposition $\ell = g \cdot h$ is unique (Hence the reference direct: the proof is straightforward: $g_1 \cdot h_1 = g_2 \cdot h_2 = \ell$ implies that $g_2^{-1} \cdot g_1 = h_2 \cdot h_1^{-1}$ is part of both groups and thus equal to $e$.) Because both groups are normal in $L$, for all $g_1$ some $h_2 \in H$ exists such that $g_1 \cdot h_1 = h_2 \cdot g_1$ and for all $h_1$ some $g_2$ exists such that $g_1 \cdot h_1 = h_1 \cdot g_2$. We conclude that $h_1 = h_2$ and $g_1 = g_2$ and $\ell = g \cdot h = h \cdot g$ decomposes in a unique way, and which we could write as $g + h$.

We can define the map $f: G \times H \to G + H$ as $f(g, h) = g \cdot h$. It defines an isomorphism
between direct product and direct sum, so that $G \times H \cong G + H$. In physics and chemistry we like to use matrix representations $M_G$ of groups $G$ in vector spaces endowed with a basis. The direct sum of two groups is then represented by a block-diagonal matrix,

$$M_{G \oplus H} = \begin{pmatrix} M_G & 0 \\ 0 & M_H \end{pmatrix}$$

This is for instance the case for the infinitely many representations of $SO(3)$ in the Hilbert space $L^2(S^2)$ of square-integrable functions on the unit sphere, discussed below. The direct sum will also be encountered when we search for the irreducible representations of the Lorentz group, isomorphic to $SU(2) \times SU(2)$.

A few consequences are highly useful and are not difficult to check:

* A cyclic group is Abelian.

* For any $n \in \mathbb{N}$, the infinite set $n\mathbb{Z} = \{\dots, -2n, -n, 0, n, 2n, \dots\}$ forms a group under summation with $e_N = 0$ as the unit operator. It is a subgroup of $\mathbb{Z}$ of all integers and Abelian because summation commutes. $n\mathbb{Z}$ has $n$ cosets in $\mathbb{Z}$ of infinite order $(0n\mathbb{Z}) = \{\dots, -2n, -n, 0, n, 2n, \dots\}$, $(1n\mathbb{Z}) = \{-2n+1, -n+1, 1, n+1, 2n+1, \dots\}$, $\dots$, $((n-1)n\mathbb{Z}) = \{\dots, -n-1, 1, n-1, 2n-1, 3n-1, \dots\}$ with forms a group with $(in\mathbb{Z})(jn\mathbb{Z}) = ((i+j)n\mathbb{Z})$ as group operation and $(0n\mathbb{Z})$ as unit operator. This group is cyclic because $(jn\mathbb{Z}) = (1n\mathbb{Z})^j$. This quotient group is isomorphic to $\mathbb{Z}_n = \{0, 1, \dots, n-1\}$ under addition modulus $n$: $\mathbb{Z}_n \cong \mathbb{Z}/n\mathbb{Z}$ as well as to the group $C_n$ of discrete rotations $(\exp(2\pi ik/n) = r^k$ on the unit circle in the complex plane, $k = 0, 1, \dots, n-1$.

* If $f: G \to H$ is an homomorphism, $f(G)$ is subgroup of $H$ (and surjective if $f(G) = H$) with $e_H = f(e_G)$ as unit operation, and $K(f)$ is a normal subgroup of $G$ (and injective if $K(f) = \{e_G\}$). The quotient group $G/K$ is a group because $K$ was seen earlier to be a normal subgroup of $G$, and isomorphic to $f(G)$ (this is the first isomorphism theorem). If $f$ is an isomorphism, then $f(G) = H$, $K(f) = \{e_G\}$ and $G/K$ is just equal to $G$ and obviously isomorphic to $H$, as demonstrated by the very existence of $f$ itself.

* $G \times H \cong H \times G$;

* The group $G \times H$ contains $G$ and $H$ as distinct and normal subgroups in the sense that $G \times e_H \cong G$ and $e_G \times H \cong H$ are subgroups of $G \times H$, with $(G \times e_H) \cap (e_G \times H) = (e_G, e_H)$.
The subgroups are normal: for any element $(g, h) \in G \times H$, the left coset $(g, h)G \times e_H$ is equal to the right coset $G \times e_H(g, h)$ because the sets $gG$ and $Gg$ are equal.

* If $H_1$ and $H_2$ are both subgroups of the group $G$ and the map $f: H_1 \times H_2 \to G$ is defined by $f(h_1, h_2) = h_1h_2$ then the map is injective if and only if $H_1 \cap H_2 = \{e\}$. In particular, if $H_1 = H_2$ this map is not injective unless $H_1 = H_2 = \{e\}$. Proof: Let $h \in H_1 \cap H_2$. If $f$ is injective then $f(e, h) = f(h, e)$ implies $(h, e) = (e, h)$ thus $h = e$. On the other hand, if $H_1 \cap H_2 = \{e\}$, $f(h_1, h_2) = f(k_1, k_2)$ implies $k_1^{-1}h_1 = h_2^{-1}k_2^{-1}$, thus both part of both $H_1$ and $H_2$ and necessary equal to $e$. Hence $h_1 = k_1$ and $h_2 = k_2$ so that $f$ is injective.

* $(G+H)/H \cong G$ for normal subgroups $G$ and $H$.

* $\mathbb{R}/\mathbb{Z} \cong U(1)$: This quotient set is equal to all cosets of infinite discrete collection of numbers $\{x \in [0, 1] | x + \mathbb{Z}\}$ labeled by $x$. Every coset in this set maps on one point of the unit circle $\exp(2\pi ix)$ which is an Abelian group under complex number multiplication itself and also isomorphic to rotations over the angle $2\pi x$ in $\mathbb{R}^2$. Both are representations of the group $U(1)$. The identity can also be understood as a manifestation of the first isomorphism theorem: The map $x \to \exp(2\pi ix)$ from $\mathbb{R}$ to $U(1)$ is surjective but has kernel $K = \mathbb{Z}$.

## Exercise V.2: Dihedral group

The Dihedral group $D_n$ describes the rotations $(\exp(2\pi ik/n) = r^k$ in the complex plane with $n$ even and $k = 0, 1, \dots, n-1$, as well as complex conjugation $s$ and all their combinations.

1. Show that the group $D_2$ is Abelian and cylic.

2. The group $D_4$ has 8 elements $(1, i, -1, -i, s, si, -s, -si)$. With $si = -is$ check the multipli-
cation table,

<table><thead><tr><th></th><th>1</th><th>i</th><th>-1</th><th>-i</th><th>s</th><th>si</th><th>-s</th><th>-si</th></tr></thead><tbody><tr><th>1</th><td>1</td><td>i</td><td>-1</td><td>-i</td><td>s</td><td>si</td><td>-s</td><td>-si</td></tr><tr><th>i</th><td>i</td><td>-1</td><td>-i</td><td>1</td><td>-si</td><td>s</td><td>si</td><td>-s</td></tr><tr><th>-1</th><td>-1</td><td>-i</td><td>1</td><td>i</td><td>-s</td><td>-si</td><td>s</td><td>si</td></tr><tr><th>-i</th><td>-i</td><td>1</td><td>i</td><td>-1</td><td>si</td><td>s</td><td>-is</td><td>-s</td></tr><tr><th>s</th><td>s</td><td>si</td><td>-s</td><td>-si</td><td>1</td><td>si</td><td>-1</td><td>-i</td></tr><tr><th>si</th><td>si</td><td>s</td><td>-si</td><td>-s</td><td>i</td><td>1</td><td>-i</td><td>-1</td></tr><tr><th>-s</th><td>-s</td><td>si</td><td>s</td><td>-si</td><td>-1</td><td>-i</td><td>1</td><td>i</td></tr><tr><th>-si</th><td>-si</td><td>-s</td><td>si</td><td>s</td><td>-1</td><td>-i</td><td>i</td><td>1</td></tr></tbody></table>

Use this Table to argue that complex conjugation $s$ and rotations $r^k$ do not always commute so that $D_4$ is a non-Abelian, with cyclic subgroups $C = (1, s)$ and $I = (1, -1)$. The latter is also normal because the inversion $-1$ commutes with all elements. Its 4 cosets are given by $D_4/I = \{(1, -1), (i, -i), (s, -s), (-si, si)\}$ and constitutes a group. The first element $e = \{(1, -1)\}$ is the identity operation in the quotient group.

3. Show in general that for $n > 2$, $D_n$ is a non-Abelian group of size $2n$ and isomorphic to a discrete subgroup of $O(2)$ (see below). Its normal cyclic subgroup $\{r^k\}$ is isomorphic to the group $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$ of integers with addition modulo $n$ (see below). The normal subgroup $H = (1, -1)$ has $n$ cosets so that $|D_n| = 2n$, $|H| = 2$ and index $[D_n : H] = n$ and Lagrange's theorem applies.

1. Representations of a group

(https://en.wikipedia.org/wiki/Irreducible_representat

A *representation* of a group $G$ is a homomorphism $D : G \to V$ to a vector space $V$ (i.e. endowed with operations $av_1 + bv_2$ with $a, b \in \mathbb{R}, \mathbb{C}, \dots$). The representation maps group operations $g \in G$ to operations on elements $v \in V$ in this vector space. In physics it is customary to chose a basis $\{e_i\}$ of the vector space. The dimension of the representation is defined as the dimension of the vector space, i.e. the number of basis vectors. The operation on each basis vector must again be a superposition of basis vectors, i.e. $D(g)e_i = \sum_j e_j D(g)_{ji}$ so that an arbitrary vector $v = \sum_i v_i e_i$
is mapped to $v' = \sum_{ij} e_j D(g)_{ji} v_i$. Hence the coordinates $v'_j$ of the transformed vector can be obtained by matrix operation, $v'_j = \sum D_{ji} v_i$. We obtain a matrix representation of the group $G$ if $D(g_1)D(g_2) = D(g_1g_2)$.

A linear subspace $W \subset V$ is called *G-invariant* if $D(g)w \in W$ for all $g \in G$ and $w \in W$ and the restriction of $D(G)$ to $W$ is called a sub-representation of $G$ in $V$. In the matrix representation sub-representations appear as upper-triangular blocks on the diagonal for a suitable chosen basis in $V$, i.e.

$$D(g) = \begin{pmatrix} D(g)_{11} & D(g)_{12} \\ 0 & D(g)_{22} \end{pmatrix}$$

where all elements are again matrices of lower dimensionality. In this form the representation acts on vectors $(v_1 \in W, v_2 \in V/W)$ and all vectors $(v_1, 0)$ map on their own space $W$. The off-diagonal block implies that elements outside the subspace $W$ can also map to $W$. If the space $W$ does not contain itself a $G$-invariant subspace, i.e. cannot be further put in upper-triangular sub-blocks, the representation $D(g)$ restricted to $W$ is said to be *irreducible*. If a basis can be found that makes the matrix entirely block-diagonal associated with subspaces $W_n$ with $\cup_n W_n = V$, the representation is said to be *decomposable* and can be expressed as a direct sum of the representations of $G$ in the subspaces $W_n$. It follows that an irreducible representation is not further decomposable. Note that a decomposable representation may appear non-diagonal in a different basis. The matrix representation also makes it clear that any one-dimensional representation, such as $U(1)$ on the ring $S^1 = \{\exp(i\theta)\}$ in the complex plane, must be irreducible although it has many discrete subgroups that are however not $U(1)$-invariant. The representation is said to be *faithful* if the homomorphism $D$ is injective so that every image of $D$ in $V$ has a different original $g \in G$.

In mathematics one insists on the difference between the group structure with multiplication algebra and the representation in some vector space. In physics and chemistry, matrix representations are popular, with the unfortunate consequence that a representation that is often confused with the group itself. For instance the group $SU(2)$ is often directly associated with the group of $2 \times 2$ matrices with unit determinant but this is in fact the fundamental representation and others do exist! An important theorem in group theory is that the number of fundamental representations

## B. Lie Groups and Lie Algebra

Lie groups were first introduced by Sophus Lie in 1873, and have been studied in great detail in the mathematical literature ever since, especially because they have become the building block of
quantum field theory. We will give here a short explanation. The two most important features of a Lie group are its continuity and its differentiability. They allow to perform differential calculus and for instance to look for local variations of a group, mostly near the unit element $e = 1$, and henceforth try to construct global group structure from their local behavior. They are important in physics because only continuous symmetries lead to conserved quantities and local variations define conserved currents via Noether's theorem, as we shall see later. The discussion of the group $SO(N)$ in the next section will make clear that "infinitesimal" variations of a group characterized by a continuous parameters determine many properties of the group and sometimes even "generate" the entire group. In fact, the local properties of a Lie group $G$ define a vector space $\mathfrak{g}$ called the Lie algebra, and are characterized by a Lie bracket. Its formal definition is the following. A Lie algebra over a field $F$ (that can be $\mathbb{R}$, $\mathbb{C}$, the unit circle etc.) is a vector space $\mathfrak{g}$ (where addition and multiplication $a\mathfrak{g}_1 + b\mathfrak{g}_2$ of its elements are defined with $a, b \in F$), endowed with a binary operation $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ called the Lie bracket $[\mathfrak{g}_1, \mathfrak{g}_2]$, satisfying,

1. bilinearity: $[a\mathfrak{g}_1 + b\mathfrak{g}_2, \mathfrak{g}_3] = a[\mathfrak{g}_1, \mathfrak{g}_3] + b[\mathfrak{g}_2, \mathfrak{g}_3]$ for all $a, b \in F$, and idem for righthand side of bracket.

2. the alternating property : $[\mathfrak{g}_i, \mathfrak{g}_i] = 0$ for all $\mathfrak{g}_i \in \mathfrak{g}$.

3. the Jacobi identity

$$[\mathfrak{g}_1, [\mathfrak{g}_2, \mathfrak{g}_3]] + [\mathfrak{g}_3, [\mathfrak{g}_1, \mathfrak{g}_2]] + [\mathfrak{g}_2, [\mathfrak{g}_3, \mathfrak{g}_1]] = 0$$

From the first two conditions we can easily check that $[\mathfrak{g}_1, \mathfrak{g}_2] = -[\mathfrak{g}_2, \mathfrak{g}_1]$. It is straightforward to show that the Lie algebra $\mathfrak{g}$ constitutes a group with the addition as operation and $\mathfrak{g} = 0$ as null element. For instance, the vector product in $\mathbb{R}^3$, $[\mathbf{x}, \mathbf{y}] = \mathbf{x} \times \mathbf{y}$ defines a Lie algebra over $\mathbb{R}$. Similarly, the commutator $[Q_a, Q_b] = Q_a Q_b - Q_b Q_a$ for matrices in some vector space such as $\mathbb{C}^n$ constitutes a Lie algebra. In most applications in physics the vector space is $\mathbb{R}^n$ or $\mathbb{C}^n$ and with respect to a chosen basis, all elements of $\mathfrak{g}$ reduce to $n \times n$ matrices with some specific properties such as vanishing trace, antisymmetry or hermiticity that respect the axioms of a vector space as well as a commutator specific for the underlying group.
Intermezzo: exponential of a matrix

(https://en.wikipedia.org/wiki/Matrix_exponential)

Given a matrix A, one defines

$$e^A := \sum_{k=0}^{\infty} \frac{A^k}{k!} \qquad (19)$$

Elementary properties can be found in literature

* $(\exp(A)]^T = \exp(A^T)$

* $\det \exp(A) = \exp(\mathrm{Tr} A)$.

* $\exp(A) \exp(B) = \exp(C)$ with $C = A + B + \frac{1}{2}[A, B] + \frac{1}{12}[A, [A, B]] - \frac{1}{12}[B, [A, B]] + \dots$ (Baker-Campbell-Hausdorff formula).

* $e^A = \lim_{n \to \infty} (1 + \frac{A}{n})^n$: The De Moivre law for matrices: it shows that the exponential operation can be seen as repeating the operation infinitesimally small infinitely many times.

From this a few useful identities follow,

* $\exp(A) \exp(-A) = 1$, hence $\exp A$ is always invertible.

* if $[A, B] = 0$ then $\exp(A) \exp(B) = \exp(A + B)$

* $\det \exp(A) = 1$ if $\mathrm{Tr} A = 0$.

* $\exp(iA)$ is unitary if $A$ is hermitian.

* If $\mathrm{Tr} A = \mathrm{Tr} B = 0$ then $\exp(A) \exp(B) = \exp(C)$ with $\mathrm{Tr} C = 0$

From the Lie algebra we can construct the exponential map from $\mathfrak{g}$ to a group $G$,

$$G = \{e^X | X \in \mathfrak{g}\} \qquad (20)$$

Because the commutators in $\mathfrak{g}$ are closed, the Baker-Campbell-Hausdorff formula guarantees that the product of two elements in $G$ is again in $G$. We want $G$ to be associative, i.e. $(g_1g_2)g_3 = g_1(g_2g_3)$, we can check this to be guaranteed by the Baker-Campbell-Hausdorff formula, to this end just collect all commutators $[[X, Y], Z]$ in the exponent of $(\exp(X) \exp(Y)) \exp(Z) =$
$\exp(X)(\exp(Y)\exp(Z))$. Of course an inverse $\exp(-X)$ of $\exp(X)$ always exists and $-X \in \mathfrak{g}$. We conclude that $G$ is indeed a group. If we choose a basis $\pi_k$, $k = 1, \dots, D$ for the Lie algebra $\mathfrak{g}$ we see that

$$G = e^{\sum_n \alpha_n \pi_n} \qquad (21)$$

where the $\alpha_n$ can be real or complex-valued. They can be varied continuously and facilitate differential calculus: The group $G$ is a Lie group, and is near $g = e$ locally determined by the generators $\pi_n$ that span the vector space of $\mathfrak{g}$. When the basis is identified, the Lie bracket becomes the matrix commutator in $\mathbb{R}^n$ or $\mathbb{C}^n$. Because $\mathfrak{g}$ is a group it can have different representations in different vector spaces. The Lie group it generates will also be different. If $\pi(\mathfrak{g})$ is a representation of the Lie algebra $\mathfrak{g}$ in some vector space it must obey $\pi(\mathfrak{g}_1 + \mathfrak{g}_2) = a\pi(\mathfrak{g}_1) + b\pi(\mathfrak{g}_2)$. We can define its Lie bracket as $[\pi(\mathfrak{g}_1), \pi(\mathfrak{g}_2)] = \pi([\mathfrak{g}_1, \mathfrak{g}_2])$. From this we can verify that the three Lie axioms are satisfied.

The Lie group $\Pi(G)$ that is generated as such arguably becomes

$$e^{\pi(X)} \stackrel{?}{=} \Pi(e^X) \qquad (22)$$

with $X \in \mathfrak{g}$ and $\exp(X) \in G$. This would imply that the representation of the Lie algebra uniquely determines the group, but unfortunately that is not true in general. In a *connected group* each group member is not necessarily equal to an exponential but can still be expressed as a product of exponentials, i.e. for each $g \in G$,

$$g = e^{X_1} e^{X_2} \dots e^{X_n} \qquad (23)$$

with $X_i$ a linear span of Lie generators of the group. The representation of this element can then be defined as

$$\Pi[g] = \Pi[g_1]\Pi[g_2]\dots\Pi[g_n] \qquad (24)$$

and $\Pi[g_i] = \exp(i\pi[X_i])$. A well-known counter example of Eq. (22) is given by the two Lie groups $SO(3)$ and $SU(2)$ that will be introduced in the next sections. The fundamental representation of $SO(3)$ is the Lie group of rotations of vectors in $\mathbb{R}^3$, for $SU(2)$ this is the rotation of spinors in $\mathbb{C}^2$. The 3 generators $(M_n)_{ij} = \epsilon_{ijn}$ of $SO(3)$ span a Lie algebra group $\mathfrak{so}(3)$ over $\mathbb{R}$ with Lie bracket the commutator in the matrix space $\mathbb{R}^{3 \times 3}$. For $SU(2)$ the 3 generators are the three $2 \times 2$ Pauli matrices $\sigma_n$. Because the $M_n$ and the $\sigma_n$ are both traceless, they generate matrices with unit
determinant. Because the $M_n$ are anti-symmetric, $\exp(\alpha_n M_n)$ is always unitary for $\alpha_n \in \mathbb{R}$. In physics it is customary to write $\exp(-i\alpha_n M_n)$ and to redefine the generators $(M_n)_{ij} = -i\epsilon_{ijn}$ as complex hermitian matrices. The resulting Lie algebra $[M_n, M_m] = i\epsilon_{nmk}M_k$ is equal to the one for the Pauli matrices in $\mathfrak{su}(2)$ and both are vector spaces over $\mathbb{R}$: The two Lie algebras are isomorphic. Nevertheless, the groups $SO(3)$ and $SU(2)$ are not isomorphic, the latter being the double cover of the first (Exercise V.4) with also spin $\frac{1}{2}$ included. This gives rise to a subtle topological difference discussed later that even knowing that both groups are connected and with surjective exponential map, the group $SO(3)$ is not simply connected, i.e. not every loop of subsequent infinitesimal group actions can be continuously shrunk into a point by a continuous change in parameter space. For the group $SL(2, \mathbb{C})$ we will find that not all group elements are covered by the exponential map but that the group is (singly) connected so that the Lie algebra still reaches the entire group via Eq. (24).

We conclude that not every Lie group can be written as $\exp\{\pi(\mathfrak{g})\}$ as was questioned by Eq. (22). The local version of Eq. (20) remains valid, and for $t \approx 0$ and any $X \in \mathfrak{g}$,

$$e^{\pi(tX)} = \Pi(e^{tX})$$

or, because $\pi(tX) = t\pi(X)$, the representation being a homomorphism between two linear vector spaces,

$$\pi(X) = \frac{d}{dt} \Pi(e^{tX})_{t=0} \qquad (25)$$

In physics we mostly concentrate on unitary Lie groups (an important exception will be the non-unitary Lorentz group) and use a $D$-dimensional basis for the Lie algebra, $X = -i\alpha_a Q_a$ so that

$$U = e^{-i \sum_a \alpha_a Q_a} \qquad (26)$$

with $D$ real-valued parameters $\alpha_a$ and $Q_a$ hermitian matrices. This implies that $UU^\dagger = 1$ so that the transformation $U$ is indeed unitary. If we also want that $\det U = 1$, as is the case for the special unitary group $SU(N)$, the $Q_a$ have to be traceless. The Lie bracket can be defined for the generators of the group,

$$[Q_a, Q_b] = i \sum_c f_{abc} Q_c \qquad (27)$$

and via the first axiom of the Lie algebra defines the Lie bracket for the entire group. If indeed the group is unitary and all $Q_a$ are hermitian, the structure constants $f_{abc}$ must be real-valued. When
<table><thead><tr><th>Group</th><th>Properties</th><th>Dimension D</th><th>Lie Generators Q<sub>a</sub></th></tr></thead><tbody><tr><td>SO(n) &isin; &#8477;<sup>n&times;n</sup></td><td>O<sup>T</sup> &sdot; O = 1, det SO = 1</td><td>n(n &minus; 1)/2</td><td>i&times; anti-symmetric real matrices</td></tr><tr><td>SU(n) &isin; &#8477;<sup>n&times;n</sup></td><td>U<sup>&dagger;</sup> &sdot; U = 1, det U = 1</td><td>n<sup>2</sup> &minus; 1</td><td>traceless hermitian matrices</td></tr><tr><td>SL(n, &#8477;)</td><td>det SL = 1</td><td>n<sup>2</sup> &minus; 1</td><td>i&times; traceless real matrices</td></tr><tr><td>SL(n, &#8477;)</td><td>det SL = 1</td><td>2(n<sup>2</sup> &minus; 1)</td><td>traceless matrices</td></tr><tr><td>SO<sup>+</sup>(1, 3)</td><td>&Delta;<sup>T</sup> &sdot; g &sdot; &Delta; = g, det &Delta; = 1, &Delta;<sub>00</sub> &gt; 0</td><td>6</td><td>Pauli matrices (double cover)</td></tr></tbody></table>

the generators are anti-hermitian, a case we will encounter for the boosts in the Lorentz group, we can check from Eq. (27) that the structure constant are purely imaginary. It can be verified that Jacobi's identity is obeyed if $f_{abc} = f_{cab} = f_{bca}$ (cyclic symmetry) and $f_{abc} = -f_{bac}$ (anti-symmetry upon permutation). This still allows some of them to be equal to zero even when $a \neq b \neq c$. implying that some generators may actually commute. The *rank* of a Lie group can be defined as the maximum number of its mutually commuting generators. The set $Q_a$ with addition defines a Lie algebra that is itself a group and closed under the action of the Lie bracket. Clearly, an Abelian symmetry is characterized either by one generator only or by vanishing structure constants and has rank one. Table 1 shows different Lie groups with matrix representations in terms of traceless matrices. For $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ algebras we have $f_{abc} = \epsilon_{abc}$. Only 1 independent constant $\epsilon_{123} = 1$ exists and ant two generators do not commute. Both Lie groups thus have rank 1. The $\mathfrak{su}(3)$ algebra has 8 independent non-vanishing generators and in the usual Gellmann matrix basis,

$$f_{123} = 1, f_{147} = -f_{156} = f_{246} = f_{257} = f_{345} = -f_{367} = \frac{1}{2}, f_{458} = f_{678} = \frac{\sqrt{3}}{2}.$$

The group $SU(3)$ has rank 2.

### Example: Adjoint representation of Lie algebra

Consider $N$ generators $Q_a$ of a Lie algebra with bracket given by Eq. (27), featuring the real-valued structure constants $f_{abc}$. We can define $N$ matrices of size $N \times N$ by $(T_a)_{bc} = -if_{abc}$. These matrices are traceless and hermitian. It is straightforward to verify that they obey the same Lie bracket,

$$[T_a, T_b] = i \sum_c f_{abc} T_c$$

and as such constitute a representation of the Lie algebra on $\mathbb{C}^{N \times N}$, called the adjoint representation. In particular the Jacobi identity is satisfied. A straightforward example is the Lie algebra
shared by SU(2) and SO(3), with $f_{abc} = \epsilon_{abc}$. The adjoint representation is just the one of the rotations SO(3) in $\mathbb{R}^3$ with $(R_n)_{ij} = -i\epsilon_{ijn}$.

### C. The group $O(n)$

The group $O(n)$ can be represented by the set of $n \times n$ real-valued orthogonal matrices, i.e. $O^T O = 1$ with $T$ denoting transpose: $(O^T)_{ij} = O_{ji}$. These matrices preserve the scalar product $(x_1 \cdot x_2)$ of vectors in $\mathbb{R}^n$. They have $n^2$ real-valued entries like any matrix in the General Linear group $GL(n, \mathbb{R})$, but the constraint of orthogonality lowers the number of independent parameters. Since the matrix $O^T O$ is symmetric, it has $n$ independent parameters to be put equal to 1 on the diagonal, and $n(n-1)/2$ off-diagonal elements to be put equal to 0. This results in $n+n(n-1)/2 = n(n+1)/2$ constraints. The dimension of the group $O(n)$ is thus $D = n^2 - n(n+1)/2 = n(n-1)/2$.

Clearly $(\det O)^2 = 1$, so that $\det O = \pm 1$. The subgroup $SO(n)$ restricts to matrices with determinant $+1$ but has the same dimension. For $n = 2$ for instance, $SO(2)$ restricts to pure rotations over any angle $\theta$ and excludes the mirror operations in $x$- or $y$-axis and $xy$-permutations which all have determinant equal to $-1$. The product of two consecutive reflections in both axes is equal to the parity operation $P: r \to -r$, equal to a rotation over $\pi$ and thus part of $SO(2)$. Contrary to $SO(2)$, $O(2)$ is non-Abelian, for instance the mirror-image operation in the $x$-axis does not commute with ordinary rotation, but instead commutes with a rotation over $-\theta$

$$ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} \cos -\theta & \sin -\theta \\ -\sin -\theta & -\cos -\theta \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \neq \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad (28) $$

The same property can be checked for the $xy$-permutation $P_{xy}$. For all matrices $g$ in $O(2)$, we can verify that $g \cdot SO(2) = SO(2) \cdot g$, so that $SO(2)$ is identified as a normal, Abelian subgroup of $O(2)$. The quotient set $O(2)/SO(2)$ is a group that contains only two elements, $SO(2)$ itself as the unit element and $P_{xy} \cdot SO(2)$ that generates all matrices with determinant $-1$. We can equivalently define the normal subgroup $\mathcal{I} = \{I_2, P_{xy}\}$ and write $\mathcal{I} \times SO(2) \cong O(2)$. The group $SO(2)$ has one-dimensional representations such as the one by complex numbers $z = \exp(in\theta)$ for any $n \in \mathbb{Z}$, or using real numbers with as operation the addition modulo $2\pi$. They are all irreducible but only faithful for $n = \pm 1$.
For $n > 2$ the group $SO(n)$ is no longer Abelian either. In general we can write for any element $M_n \in SO(n)$

$$M_n = \exp \left( \sum_i \alpha_i R_i \right) \qquad (29)$$

where $R_i$ are the $D$ generators of the group $SO(n)$. Because $\det M_n = e^{\text{Tr} \sum_i \alpha_i R_i} = 1$ and $(M_n)^T(M_n) = 1$, the generators must be real-valued anti-(or skew-)symmetric, $R_i^T = -R_i$. It can be checked that this leaves us again with $D = n(n-1)/2$ independent choices. For $n = 2$ only one generator $R$ exists and we can verify that

$$M_2 = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} = e^{\theta R} = \lim_{n \to \infty} \left(1 + \frac{\theta}{n} R\right)^n, \quad R = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

In view of the second *De Moivre* formula for matrices, the matrix $R$ can be identified as the generator for infinitesimal rotations $\delta\theta$ along the axis perpendicular to the plane. The Lie algebra is thus one-dimensional and Abelian. We could use here the notation $iR = L$ for the hermitian operator $L$ and write $SO(2) = e^{-i\theta L}$ but for $SO(n)$ no complex numbers need to come in. The inversion operation $(x, y, z) \to (-x, -y, -z)$ with $\det = -1$ is part of $O(3)$ but cannot be equal to any number of rotations (necessarily with $\det = +1$) and is thus not part of $SO(3)$, like it was in $SO(2)$. This time, 3 generators exist in the Lie algebra, to be identified with infinitesimal rotations along the 3 axes $x, y$ and $z$ that do not commute.

In general, the direct product $SO(n) \times \{1_n, P_n\} \cong O(n)$ applies only when $n$ is odd, where $1_n$ is the $n \times n$ identity matrix and $P_n$ the inversion in the origin. When $n$ is even, $\mathcal{I} = \{1_n, P_n\}$ is a normal subgroup of $SO(n)$. Therefore, the map $f: SO(n) \times \mathcal{I} \to SO(n) \cdot \mathcal{I}$ to $O(n)$ cannot be injective (see above) and is clearly not surjective either since it never maps on matrices in $O(n)$ with negative determinant. For $n$ odd on the other hand, $f$ is an isomorphism, with $SO(n) \cdot I_n$ mapping on all matrices with positive determinants and $SO \cdot P_n$ mapping on all matrices with negative determinants. Note that in all cases the parity operation $P_n$ commutes with rotations in $SO(n)$. This is again a manifestation of the Euclidean principle of relativity.

### Example : Lévi-Civita as a third-order pseudo tensor in $O(3)$

The Lévi-Civita tensor $\epsilon_{ijk}$ is defined as $\epsilon_{123} = 1$ and $\epsilon_{ijk} = 1$ for all even permutations of 123 such as $\epsilon_{231} = 1$, $\epsilon_{ijk} = -1$ for all odd permutations and zero in all other cases, such as $\epsilon_{112} = 0$. This definition only depends on indices. To be a proper tensor, like for instance $D_{ijk} = x_i x_j x_k$, the transformation under $O(3)$ should be given by $\epsilon'_{ijk} = \epsilon_{i'j'k'} R_i^{i'} R_j^{j'} R_k^{k'}$ for any matrix $R$ in $O(3)$.
As we can check explicitly for different combinations $ijk$, the following identity holds

$$ \epsilon_{i'j'k'} R_i^{i'} R_j^{j'} R_k^{k'} = \epsilon_{ijk} \det \mathbf{R} \qquad (30) $$

For any rotation in $SO(3)$, we have $\det \mathbf{R} = 1$ and we conclude that the Lévi-Civita tensor transforms indeed as a "proper" third-order tensor in $SO(3)$ to itself. For the parity operation $R = -I_3$ in $\mathbb{R}^3$ however, part of $O(3)$, $\det \mathbf{R} = -1$, and the Lévi-Civita tensor should change sign if it were a real third-order tensor, whereas by definition it does not. The Lévi-Civita tensor is a third-order pseudo-tensor in $O(3)$, and is a number independent on how we rotate or invert axes: $\epsilon'_{ijk} = \epsilon_{ijk}$

### Exercise V.3: Generation of pseudo-vector from anti-symmetric proper tensor

A proper vector transforms as $A'_n = R_n^i A_i$ in $O(3)$, both for rotations and inversions. A pseudo vector $B_i$ is a vector that transforms "properly" under $SO(3)$ rotations, but whose coordinates do not change sign upon inversion. Let $F^{ij}$ be a proper second-order tensor, i.e. transforming as $(F')^{i'j'} = R_i^{i'} R_j^{j'} F^{ij}$ and thus parity-even under inversions. We can define the vector $B_i = \epsilon_{ijk} F^{jk}$. Well-known pseudo-vectors constructed as such are the magnetic field or the angular momentum. We can assume the tensor $F^{ij}$ to be anti-symmetric, since any symmetric part would not contribute to $B_i$. Since the left-handed side does not change sign under inversion and the Lévi-Civita tensor $\epsilon_{ijk}$ is just a number independent of reference frame, the vector $B_i$ must be a pseudo-vector. Show from Eq. (30) that the vector $B_i$ transforms in $O(3)$ as

$$ R_{i'}^i B_i = \epsilon_{i'j'k'} (F')^{j'k'} \det \mathbf{R} $$

Use that $R_{i'}^i = (R^{-1})^{i'k}$. Conclude that for rotations in $SO(3)$, the vector $B_i$ behaves as a proper vector, whereas under space-inversion, $B_i$ does not change sign. The relativistic convention is used here for indices but is redundant in $\mathbb{R}^3$ since the metric tensor is just $\delta_{ij}$ so that contra-variant and covariant notation coincide. It will be discussed in more detail for the relativistic generalization $SO^+(1,3)$ of the rotation group $SO(3)$.

### Exercise V.4: eigenfunctions from Lie Algebra (taken from Merzbacher [11])

Consider a group $G$ having the 3 hermitian generators $J_x, J_y$ and $J_z$ satisfying the Lie algebra $[J_x, J_y] = iJ_z$ and the 2 cyclic equivalents. We define $J_{\pm} = J_x \pm iJ_y$. This group can be $SO(3)$ but also $SU(2)$ discussed below, that share the same Lie algebra (more precisely, the Lie algebra's are isomorphic).
* Show that $[J_z, J_+] = J_+, [J_-, J_z] = J_-, [J_\pm J_\mp, J_z] = 0$ and $[J_+, J_-] = 2J_z$

* Show that $\mathbf{J}^2 = J_\pm J_\mp \mp J_z + J_z^2 = J_\mp^\dagger J_\mp \mp J_z + J_z^2$ so that $[\mathbf{J}^2, J_z] = 0$ and in general $[\mathbf{J}^2, \mathbf{J}] = 0$ since the $z$ axis was chosen arbitrarily. The operator $\mathbf{J}^2$ is a *Casimir operator* that commutes with all generators. For rank-one groups $SO(3)$ and $SU(2)$ there is only one such operator, in general there can be more, and *their eigenvalues, or "good quantum numbers", label different irreducible representations of the group*.

* Let $|\lambda m\rangle$ be a simultaneous eigenvector of both $\mathbf{J}^2$ and $J_z$ in some vector space $V$,

$$J_z|\lambda m\rangle = m|\lambda m\rangle; \quad \mathbf{J}^2|\lambda m\rangle = \lambda|\lambda m\rangle$$

Show that $\lambda \ge m^2$ (expressing that “total norm must exceed or equal any component”).

* Show that $J_z J_+ |\lambda m\rangle = (m+1) J_+ |\lambda m\rangle$ and $\mathbf{J}^2 J_+ |\lambda m\rangle = \lambda J_+ |\lambda m\rangle$ so that $J_+ |\lambda m\rangle = C_{\lambda m}^+ |\lambda(m+1)\rangle$ with some normalization constant $C_{\lambda m}^+$. Similarly, we find $J_- |\lambda m\rangle = C_{\lambda m}^- |\lambda(m-1)\rangle$.

* A rotation over the angle $2\pi k$ along the axis $\hat{n}$ is defined by $U_k = \exp(-2\pi k i \hat{n} \cdot \mathbf{J})$. If we choose the $z$-axis along $\hat{n}$, we find $U_k = (-1)^{2mk} = \pm 1$ for any odd $k$. Given the operator $\mathbf{J}^2$ (the value for $j$), the *Euclidean principle of relativity imposes rotations to be equal for all eigenvectors of $J_z$*, if not, the rotation would depend on our label $m$. Hence, all $m$ are either *integer* or *half-integer* values for $j$, but cannot contain both.

* Show that some real-valued $j > 0$ must exist such that $J_+ |\lambda(m=j)\rangle = 0$. Since $\mathbf{J}^2 = J_- J_+ + J_z + J_z^2$ it follows that $\lambda = j(j+1)$. Similarly, a $j' < 0$ must exist such that $J_- |\lambda(m=-j')\rangle = 0$ and $\lambda = j'(j'-1)$. Thus $j = j'$ and $m = -j, -j+1, \dots, j-1, j$ can take $2j+1$ values, in steps of 1. Conclude that $j$ must be *integer* or *half-integer*.

* Evaluate $\langle \lambda m | J_+^\dagger J_+ | \lambda m \rangle$ to find that $C_{\lambda m}^+ = \sqrt{(j-m)(j+m+1)}$. Similarly, $C_{\lambda m}^- = \sqrt{(j+m)(j-m+1)} = C_{\lambda(-m)}^+$.

* The operator $U(a) = \exp(-i n_a \cdot \mathbf{J})$ acts on wave functions in the vector space $V$. The subspaces $V_\lambda$ in $V$ of eigenfunctions $|\lambda\{m\}\rangle$ are all generated by eigenfunctions of $\mathbf{J}^2$. Hence,

$$U(a)|\lambda m\rangle = \sum_{m'} D_{mm'}^\lambda(a)|\lambda m'\rangle$$
with $D_{mm'}^{\lambda}(a) = \langle \lambda m | U_a | \lambda m' \rangle$ since eigenfunctions are orthonormal. We can perform a second rotation $U_b$,

$$U(b)U(a)|\lambda m\rangle = \sum_{m''m'} D_{mm''}^{\lambda}(b)D_{m''m'}^{\lambda}(a)|\lambda m\rangle$$

because of the group structure the joint operation $ab$ stands for an operator $U(ba)$ with matrix element $D_{mm'}^{\lambda}(ba)$ satisfying

$$D_{mm'}^{\lambda}(ba) = \sum_{m''} D_{mm''}^{\lambda}(b) D_{m''m'}^{\lambda}(a) = (D^{\lambda}(b) \cdot D^{\lambda}(a))_{mm'}$$

This implies an isomorphism between $G$ and the subgroup $V_{\lambda}$. For any allowed $\lambda$, the group $G$ has a matrix-representation in the $2j+1$ dimensional subspace of $V$ spanned by the eigenfunctions $|\lambda m\rangle$.

### Example: Representation of $SO(3)$ in $S^2$ and $L^2(S^2)$ and spherical harmonics.

In physics the group $SO(3)$ has the well known “fundamental” representation in terms of rotations on the unit sphere $S^2$ in $\mathbb{R}^3$. In $S^2$ the rotations are given by $\exp(-i\alpha_n M_n)$ with generators $(M_n)_{ij} = -i\epsilon_{ijn}$. This group is a faithful matrix representation of $SO(3)$, in physics often even identified with $SO(3)$ itself.

* Using the contraction rule for the Lévi-Civita tensor, $\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ show that $[M_x, M_y] = iM_z$ with all cyclic permutations.

* Show that $(M^2)_{ij} = 2\delta_{ij}$ so that this representation is three dimensional and has $j = 1$ (spin 1), $m = -1, 0, 1$ in the notation above. The rotations in $\mathbb{R}^3$ have Abelian subgroups that correspond to rotations over any fixed axis and are representations of $U(1)$. Nevertheless, the representation of the group $SO(3)$ in $S^2$ is irreducible since if we take any axis and call it the $z$-axis, rotations over $x-$ or $y$-axis do not conserve $m$ and do not have an invariant subspace.

Another widely used representation of $SO(3)$ is on the Hilbert space $L^2(S^2)$ of square-integrable scalar complex wave functions defined on the two-dimensional unit sphere $S^2$ and endowed with the standard scalar product $\langle \Psi | \Phi \rangle = \int d^2\hat{x} \bar{\Psi}(\hat{x})\Phi(\hat{x})$. In Dirac notation, widely used in physics, wave functions are spanned by the complete set $|\hat{x}\rangle$ of eigenfunctions of the position operator $\hat{r}$, with $\langle \hat{x} | \hat{y} \rangle = \delta(\hat{x} - \hat{y})$, and $\langle \hat{x} | \Psi \rangle \equiv \Psi(\hat{x})$. The bra notation used here is actually a linear form
that maps elements in Hilbert space to $\mathbb{C}$, as will be discussed later. Rotation operators $U(\alpha)$ in $L^2(S^2)$ are related to rotations on the sphere $S^2$ by

$$U(\alpha)|\hat{x}\rangle = |\mathbf{R}(\alpha) \cdot \hat{x}\rangle.$$

with $\mathbf{R}$ the rotation of the vector $\hat{x}$ on $S^2$. Because

$$\langle \hat{y} | \hat{x} \rangle = \langle \mathbf{R} \cdot \hat{y} | \mathbf{R} \cdot \hat{x} \rangle = \langle \hat{y} | U^{\dagger} U | \hat{x} \rangle$$

we can check that $U^\dagger = U^{-1}$, so that the operators $U$ take over the unitary properties of $\mathbf{R}$ although they operate in different spaces. For an arbitrary wave function in $S^2$ we find

$$\Psi'(\hat{x}') = (U\Psi)(\hat{x}') = \Psi(\hat{x}) = \Psi(\mathbf{R}^{-1} \cdot \hat{x}') \quad (31)$$

In the following we show that the set of operators $U$ on $L^2(S^2)$ constitute an infinite, reducible representation of the group $SO(3)$. From quantum mechanics we know that rotations in $L^2(S^2)$ are given by the unitary transformations $U(\alpha) = \exp(-i\alpha \cdot L)$, generated by the 3 components of the orbital angular momentum operator $L = r \times p$. Because of the commutation rule $[r_n, p_m] = i\delta_{nm}$, this operator satisfies the Lie algebra $[L_x, L_y] = iL_z$ with all cyclic permutations. Thus, we have a physical and faithful representation of the group $SO(3)$ in the infinite-dimensional Hilbert space $L^2(S^2)$.

* Verify the infinitesimal transformation $[1 - i\alpha \cdot (r \times p)]\Psi(x) = \Psi(x_i + \epsilon_{ijn}\alpha_n x_j)$ that connects rotations on $S^2$ to rotations on $L^2(S^2)$.

* In view of exercise V.4 we can write $L^2 = \ell(\ell + 1)$ with $\ell$ integer or half-integer. Argue that this representation of the group $SO(3)$ does not cover the operator $-I_3$, so that half-integer representations are excluded and $\ell = 0, 1, 2, \dots$.

The Lie algebra and the number of generators are the same for rotations in both $S^2$ and $L^2(S^2)$, and are both representations of the group $SO(3)$. For rotations on $S^2$ the representation of $SO(3)$ is faithful and irreducible. However, many sub-representations exist for the representation in $L^2(S^2)$ because unlike $M^2 = 2$, $J^2 = \ell(\ell + 1)$ can take infinitely many values.

Irreducible representations of $SO(3)$. Each value for $\ell$ generates an irreducible, indecomposable representation of physical dimension $2\ell + 1$ of the group $SO(3)$. For ordinary rotations in $S^2$ only $\ell = 1$ exists and there is no (non-trivial) sub-representation. If we use the basis $\{|\ell, m\rangle\}$ of $L^2(S^2)$, different $\ell$ appear as different blocks in the matrix representation without "cross-talk" and this
subspace $L_{\ell}^2(S^2)$ has become a $SO(3)$-invariant sub-representation. A matrix representation of $SO(3)$ in the finite vector space $\mathbb{C}^{2\ell+1}$ can be obtained by calculating the three $(2\ell+1) \times (2\ell+1)$ matrices $R(i)_{mm'}^{\ell} = \langle \ell, m | \mathbf{J}_i | \ell, m' \rangle$. Of course, in physics this decomposition corresponds to different states of orbital angular momentum that are conserved in time. Here it emerges entirely from mathematical group structure.

This non-Abelian sub-representation is irreducible but its implementation on $L^2(S^2)$ cannot cover all rotations on the unit sphere because any wave function is necessarily a linear superposition of $|\ell, m\rangle$ involving not only all allowed $m$ but also all $\ell$. We define the infinite set of spherical harmonics $Y_m^{\ell}(\hat{x}) \equiv \langle \hat{x} | \ell, m \rangle$ that constitutes an orthonormal complete set on the Hilbert space $L_{\ell}^2(S^2)$ equipped with the standard scalar product above. Because this is an invariant subspace we have

$$U|\ell, m\rangle = \sum_{m'} D_{mm'}^{\ell}|\ell, m'\rangle$$

with $D_{mm'}^{\ell}$ the Wigner $d$-matrix involving 3 Euler angles. Hence,

$$U|\Psi\rangle = \sum_{\ell,m} \psi_{\ell m} U|\ell, m\rangle = \sum_{\ell,m,m'} \psi_{\ell m} D_{mm'}^{\ell}|\ell, m'\rangle$$

and which defines the rotation $U$ for arbitrary wave functions in $L^2(S^2)$. This is in block-diagonal form. The direct sum of all irreducible sub-representations constitutes an infinite-dimensional representation $\Pi$ of $U \in SO(3)$ on the Hilbert space $L^2(S^2)$, and

$$\Pi(U) \cong \bigoplus_{\ell=0}^{\infty} \Pi_{\ell}(U) = (0) \oplus (1) \oplus (2) \oplus (3) \oplus \dots \qquad (32)$$

with each sub-representation being an irreducible representation of $SO(3)$. The second equality is a popular notation in physics. The symbol $\oplus$ has been used for the direct sum because here we consider matrix representations of the group, rather than the group itself. A group representation of $SO(3)$ using only the irreps $(\ell)$ involves putting the unity anywhere else in the matrix $D_{mm'}^{\ell}$. Clearly this fails to be a faithful representation of the entire group $SO(3)$ because its kernel is large.

### Example: The representation of the group $SO(2)$ in $\mathbb{C}^2$ is not irreducible.

The fundamental representation of $SO(2)$ in $S^1$ (the unit sphere in $\mathbb{R}^2$) is

$$R(\theta) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$
which can be seen to be unitary equivalent to

$$R' = U \cdot R \cdot U^{\dagger} = \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix}$$

and is as such a representation of $SO(2)$ in $\mathbb{C}^2$ with clearly two one-dimensional, necessarily irreducible subspaces. All maps $\Pi_k : R(\theta) \to \exp(ik\theta)$ for $k \in \mathbb{Z}$ are irreducible representations of $SO(2)$ in $\mathbb{C}^1$, the unit sphere $|z| = 1$ in $\mathbb{C}$. For $|k| > 1$ the representation is not faithful, e.g. for $k = 2$, $\theta$ and $\theta \pm \pi$ map on the same complex number.

### Example: The group $SO(3)$ is not simply connected

A group is called connected if two arbitrary elements can be reached by subsequent operations $g_1g_2g_3\dots$, controlled by a continuous parameter. A group is simply connected if all loops of operations leaving and returning to a given group element can be continuously shrunk into one point. We can illustrate these concepts for the group $SO(3)$. Every element $R(\phi, n)$ in this group is characterized by the orientation of the rotation axis $n$ and one angle $\phi \in [-\pi, \pi)$ along which the rotation is performed. For $\phi = 0$, $R = 1$ and the rotation axis is arbitrary. Furthermore $R(-\phi, n) = R(\phi, -n)$ and $R(\pi, n) = R(\pi, -n)$. This implies that we can restrict to $0 \le \phi \le \pi$ and map the group on a sphere in $\mathbb{R}^3$ with radius $\pi$: the orientation defines the unit vector $n$ and the distance from the source equals $\phi$. The condition for $\phi = \pi$ implies that the antipodal surface points $\pm \pi n$ must be declared equal. It is obvious in this map that all points in this sphere can be connected by a continuous line describing subsequent small rotations: the group $SO(3)$ is connected.

Any small loop connecting a single point $\pi n$ via a path inside the sphere can clearly be deformed continuously to the point itself (yellow line in Fig. 6). Consider the path that goes straight through the origin from one antipode $\pi n$ to the other at $-\pi n$ (blue dashed). According to the above it must be considered as a loop, corresponding to a rotation over $2\pi$ along the axis $\pm n$. However, it is impossible to deform the loop continuously and shrink it to a point, since it will always be stuck at both anti-nodes: the group $SO(3)$ is not simply connected. If we would consider the same path, jump back from one antipode to the other and take the same path again (a loop trajectory that corresponds to a rotation over $4\pi$), this actually can be done. We can move both paths continuously to ones over the surface (solid blue line). This creates a great circle loop over the sphere surface through both antipodes that can be shrunk into the center (the identity rotation)) so that $SO(3)$ is doubly connected. We will see later that the group $SU(2)$, with similar Lie algebra is simply
<figure><img src="image_8.png" /><figcaption>Figure 6. The representation of $SO(3)$ rotations on the sphere. Distance from the origin is equal to the rotation angle $0 \le \phi \le \pi$. The direction indicates the unit vector of the rotation axis. The yellow roundtrip can be shrunk continuously into the point $\pi n$, the blue loop is stuck at the antipodes and can only be deformed to the surface.</figcaption></figure>

connected with the rotation over $2\pi$ being equal to *minus* the identity. The double connectivity of rotations in $\mathbb{R}^3$ implies that rotations over $2\pi$ are topologically “non-trivial” in the sense that they cannot be transformed smoothly to the identity, which is what we would intuitively expect. It is a remarkable property, sometimes referred to as the “memory of $2\pi$ rotations in $\mathbb{R}^3$”, in whose physical consequences have been debated.

D. The group $U(n)$

The group $U(n)$ can be defined as the set of unitary, complex $n \times n$ matrices in $\mathbb{C}^{n \times n}$ acting on vectors in $\mathbb{C}^n$: $U^\dagger U = 1$. Any such matrix has $n^2$ complex-valued entries, hence $2n^2$ real-valued parameters. The unitarian condition imposes, like for $SO(3)$, $n$ for the real-valued diagonal constraints for $U^\dagger U$ but $2 \times n(n-1)/2$ real-valued constraints for its off-diagonal elements. Hence the dimension of $U(n)$ is equal to $2n^2 - n - n(n-1) = n^2$. The determinant must be on the unit
circle, $\det U = \exp(i\phi)$. If we restrict to unitary matrices with determinant equal to 1, we find the subgroup $SU(n)$. Because we restrict one more parameter $\phi = 0$, the dimension of $SU(n)$ is equal to $n^2 - 1$. The dimension of traceless matrices in $\mathbb{C}^d$ is equal to $d^2 - 1$. The Lie group of $SU(n)$ has faithful representations in the vector space $\mathbb{C}^d$ when its dimension is larger or equal to the number of generators, hence if $d \ge n$. The rank of $SU(n)$ can be seen to be equal to $n - 1$ by the following argument. The generators must be traceless and commuting generators can be simultaneously diagonalized using an appropriate basis. In dimension $d \ge n$ we have $n$ possible entries but the constraint of trace zero (which is independent on choice of basis) leaves only $n - 1$ possibilities exist to diagonalize. For or $n = 2$ there is only one, corresponding to spin along one axis.

**Example:** We demonstrate that

$$U(n) \cong (SU(n) \times U(1))/\mathbb{Z}_n \quad \text{or} \quad U(n)/SU(n) \cong U(1) \qquad (33)$$

We can propose the map $f: SU(n) \times U(1) \to U(n)$, $f(S, z) = Sz$. This map is a homomorphism to $U(n)$ if $|z| = 1$. It is however not injective because $f(S, z) = I_n$ does not have a unique solution. To have $Sz = I_n$ it follows that $\det S = \det(I_n/z) = z^{-n} = 1$. Hence $K(f) = (S_k, z_k)$ with $S_k = I_n e^{-2\pi i k/n}$ and $z_k = e^{2\pi i k/n}$ for $k = 0, 1, \dots, n - 1$. We can check that this group is isomorphic to the group $\mathbb{Z}_n$ e.g. by mapping the set $(S_k, z_k)$ to $k \in \mathbb{Z}_n$ that has as operation the addition modula $n$. The map $f$ of the group $SU(n) \times U(1))/\mathbb{Z}_n$ into $U(n)$ with the same image as $(SU(n) \times U(1))$.

Remains to show that it is also surjective and that all $U \in U(n)$ are actually imaged by $f$. Clearly for any $Sz$ we find $|\det Sz|^2 = 1$ so $f[(SU(n) \times U(1))] \subseteq U(n)$. Can any $U \in U(n)$ always be written as $Sz$? Set $\det U = \xi$ with $|\xi| = 1$. Write $X = Uz$ and $z = \xi^{1/n}$. We see that $z \in U(1)$ and $\det U = z^{-n} \det X = 1$. Hence $U \in SU(n)$, and $f[(SU(n) \times U(1))] \supseteq U(n) f$. We conclude that $f$ is an isomorphism.

The second very similar identity follows from the map $g: U(n) \to U(1)$, $g(U) = \det U$. This map defines a homomorphism between the groups $U(n)$ and $U(1)$ and is surjective since any $\exp(i\theta)$ is covered. It is not injective and its kernel $K$ is exactly $SU(n)$ with $\det SU(n) = 1$. From the first isomorphism theorem is follows that $U(n)/SU(n)$ is isomorphic to $U(1)$.
### Example: $U(3)$ invariance of the isotropic 3D oscillator

The 3D harmonic oscillator has the equation of motion with Hamiltonian that is in appropriate units,

$$H = \frac{1}{2} \boldsymbol{p}^2 + \frac{1}{2} \boldsymbol{q}^2$$

The high isotropy of this Hamiltonian in 6D phase space ($\boldsymbol{p}, \boldsymbol{q}$) looks like $O(6)$ with dimension $D = 15$, but we must also make sure that the quantum canonical relations $[q_i, p_j] = i\delta_{ij}$ and $[p_i, p_j] = [q_i, q_j] = 0$ are kept invariant. We can propose the six-dimensional real-valued transformation

$$\begin{pmatrix} \boldsymbol{p}' \\ \boldsymbol{q}' \end{pmatrix} = \begin{pmatrix} \boldsymbol{U}' & \boldsymbol{U}'' \\ -\boldsymbol{U}'' & \boldsymbol{U}' \end{pmatrix} \begin{pmatrix} \boldsymbol{p} \\ \boldsymbol{q} \end{pmatrix}$$

and verify that both Hamiltonian and canonical relations are invariant if $\boldsymbol{U} \equiv \boldsymbol{U}' + i\boldsymbol{U}''$ is a complex unitary $3 \times 3$ matrix. These transformations cover a small subgroup of $O(6)$, isomorphic to $U(3)$ with much lower dimension $D = 9$ (i.e. $\det \boldsymbol{U} = \exp(i\phi)$ is allowed). If we introduce the 3 non-hermitian operators $b_i = (p_i - iq_i)/\sqrt{2}$ we find,

$$H = \sum_{i=1}^{3} b_i^\dagger b_i + \frac{3}{2}$$

with $[b_i, b_j^\dagger] = \delta_{ij}$. These operators transform as $b_i^\dagger = (U^\dagger)_{ij} b_j$ and $(b_i^\dagger)^\dagger = b_j^\dagger U_{ji}$. This invariance is not a real symmetry but rather a degree of freedom to define canonical operators or, equivalently, the operators $b_i$, facilitated by the high isotropy. The real symmetries in this model are $U(1) \times SO(3)$ (dimension $D = 1+3=4$, with 4 associated conserved quantities- norm and angular momentum - as well as time-translation symmetry that conserves total energy (this will be made explicit in Exercise VI.3 of the next section)).

### Exercise V.5 Representations of the group $SU(2)$

The group $SU(2)$ is defined as the set of complex unitary operators $U$ in 2 dimensions and with determinant equal to 1. Then, 3 operators $J_i$ can be identified subject to the Lie algebra $[J_x, J_y] = iJ_z$ with all cyclic permutations, or equivalently $[J_n, J_m] = i\epsilon_{nml}J_l$. We can apply the same procedure as we did for $SO(3)$ since the Lie algebra for $J_i$ is the same. Groups like $SU(2)$ have faithful representations in $\mathbb{C}^n$ even for large $n$.

Contrary to $SO(3)$, the operator $U = -I_2$ is part of $SU(2)$. This means that we can allow both integer and half-integer representations. Below we establish that $SU(2)/\mathbb{Z}_2 \cong SO(3)$. A rotation
over $2\pi$ along the z-axis is given by $U = \exp(-2\pi im) = -1$ for half-integer spin. For $j = 0$ the generator of $SU(2)$ in any vector space $\mathbb{C}^n$ is just the identity and the representation reduces to the trivial identity $U = 1$. It is clearly not faithful since all group members in $SU(2)$ map to the identity in $\mathbb{C}^n$.

1. The fundamental faithful representation (used as definition) of the group $SU(2)$ for $j = 1/2$ is obtained in $\mathbb{C}^2$ (hence the name). We introduce the basis vectors $(1, 0) = |j = \frac{1}{2}, m = \frac{1}{2}\rangle$, $(0, 1) = |j = \frac{1}{2}, m = -\frac{1}{2}\rangle$. Show that with this choice, the generators of $SU(2)$ become $J_i = \frac{1}{2}\sigma_i$, where the Pauli matrices are given by (hint: use the operators $J_\mp$ introduced above to find the images of $J_x$ and $J_y$),

$$ \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad (34) $$

(a) Show that $\det U(\alpha_a) = \exp(i\mathrm{Tr} \alpha_a \sigma_a) = 1$.

(b) Verify that these generators obey the Lie algebra $[\sigma_a, \sigma_b] = 2i\epsilon_{abc}\sigma_c$ and anti-commute, i.e. $\sigma_a\sigma_b + \sigma_b\sigma_a = 2\delta_{ab}$

(c) Show that $(\hat{n} \cdot \sigma)^2 = 1$ and from this that the symmetry transformations of $SU(2)$ on 2D spinors can be written as a generalized Euler equation

$$ U(\hat{n}, \theta) = \exp \left( -\frac{i}{2} \theta \hat{n} \cdot \sigma \right) = 1 \cos \frac{\theta}{2} - i \hat{n} \cdot \sigma \sin \frac{\theta}{2} $$

Contrary to rotations in $\mathbb{R}^3$, they change sign upon rotation over $2\pi$.

(d) Show that $U(\hat{n}, \theta_1)U(\hat{n}, \theta_2) = U(\hat{n}, \theta_1 + \theta_2)$, i.e. subsequent rotations over identical axes commute and $U(\hat{n}, \theta)^{-1} = U(\hat{n}, -\theta)$. The set of rotations over any fixed axis constitutes a subgroup of $SU(2)$, isomorphic to $U(1)$.

(e) Using $(\hat{n} \cdot \sigma)(\hat{m} \cdot \sigma) = \hat{n} \cdot \hat{m} + i(\hat{n} \times \hat{m}) \cdot \sigma$ show that $U(\hat{n}, \theta_1)U(\hat{m}, \theta_2)$ is again of the form $C + i(\hat{n}' \cdot \sigma)S$ (tough) but that both rotations do not commute when $\hat{m} \neq \hat{n}$ (easy).

(f) For a reflection $P_{zy}$ in the $zy$ plane we have $x \to -x, y \to y, z \to z$. We expect it to commute only with rotations along the $x$-axis, but not with those along the two other axes $y$ and $z$. Since $P_{zy}^2 = 1$, we expect, like in Eq. (28), that

$$ P_{zy}U(n_x, n_y, n_z, \theta)P_{zy} = U(n_x, -n_y, -n_z, \theta) $$
so that $P_{zy}\sigma_x P_{zy} = \sigma_x$ but $P_{zy}\sigma_{y,z}P_{zy} = -\sigma_{y,z}$. Show that one can choose $P_{zy} = \sigma_x$, and similarly $P_{zx} = \sigma_y$ and $P_{xy} = \sigma_z$. The parity operator is, up to a phase factor, equal to 3 subsequent reflections in 3 orthogonal planes, for SU(2) this implies $P = P_{zy}P_{zx}P_{xy} = \sigma_x\sigma_y\sigma_z = i\mathbf{1}$. Hence the Pauli-matrices are invariant under parity, reminiscent of angular momentum operators. In general any operation in SU(2) commutes with parity: $PU(\hat{n}, \theta)P = U(\hat{n}, \theta)$. This is again a manifestation of the Euclidian principle of relativity: rotations and inversion around a given point should commute.

(g) If $\mathbf{L} = (L_x, L_y, L_z)$ are the 3 generators in $SO(3)$ associated with orbital angular momentum, and $\mathbf{S} = \frac{1}{2}\sigma$ denotes the spin degree of freedom, argue that the group $SO(3) \times SU(2)$ has generators $(\mathbf{L} \times e_S) + (e_L \times \mathbf{S})$, often written as loosely $\mathbf{L} + \mathbf{S}$, that obey again the spin Lie Algebra. Show that $\mathbf{L} \times e_S$ and $e_L \times \mathbf{S}$ commute (hint: use that operators acting on different vector spaces necessarily commute).

2. Other representations of SU(2) may be obtained. Choose $j = 1, m = 0, \pm 1$ and introduce the 3 basis vectors $(1, 0, 0) = |1, 1\rangle$, $(0, 1, 0) = |1, 0\rangle$ and $(0, 0, 1) = |1, -1\rangle$. Show that a representation of the Lie algebra in $\mathbb{C}^3$ is,

$$ J_x = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \quad J_y = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \quad J_z = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \quad (35) $$

and which is used to describe a massive particle with spin 1 in quantum mechanics. This matrix representation of the $su(2)$ Lie algebra (with respect to a Cartesian basis) is isomorphic to the Lie algebra $so(3)$ in $\mathbb{R}^3$ with basis $(R_n)_{ij} = -i\epsilon_{ijn}$ and a unitary transformation $U$ exists such that $U^\dagger R_n U = J_n$, given by

$$ U = \sqrt{\frac{i}{2}} \begin{pmatrix} 1 & 0 & -1 \\ i & 0 & i \\ 0 & -\sqrt{2} & 0 \end{pmatrix} $$

The three-dimensional representations of the Lie algebra's of SU(2) and SO(3) for spin 1 are isomorphic.

3. The fundamental representations SU(2) in $\mathbb{C}^2$ and SO(3) in $\mathbb{R}^3$ have 3 generators with identical Lie algebras. A connection is likely to exist but the Lie groups are not isomorphic.
For any $U \in SU(2)$, we can propose a transformation in $SO(3)$, $x' = R \cdot x$ such that

$$ (\boldsymbol{\sigma} \cdot \boldsymbol{x}') = U^{\dagger} \cdot (\boldsymbol{\sigma} \cdot \boldsymbol{x}) \cdot U $$

Show that, equivalently,

$$ R_{ij} = \frac{1}{2} \mathrm{Tr} U^{\dagger} \cdot \boldsymbol{\sigma}_{i} \cdot U \cdot \boldsymbol{\sigma}_{j} \quad ; \quad U^{\dagger} \cdot \boldsymbol{\sigma}_{i} \cdot U = R_{ji} \sigma_{j} = (R^{-1})_{ij} \sigma_{j} \qquad (36) $$

(Hint: use that $\sigma_n \cdot \sigma_i + \sigma_i \cdot \sigma_n = 2\delta_{ni}$). Show (using that $\det \boldsymbol{\sigma} \cdot \boldsymbol{x} = -|\boldsymbol{x}|^2$) that this map defines a homomorphism (i.e. the operation $U_2U_1$ in $SU(2)$ is mapped to $R_2R_1$ in $SO(3)$). Establish that indeed $R \in SO(3)$, i.e. unitary and directional (that is, not equal to $-I_n$: the vector $x \sim n$ always maps to itself and never to $-n$). However, both $U$ and $-U$ transform to the same $R$ and the map is not injective. Similar to the isomorphism (33) we find $SU(2)/\mathbb{Z}_2 \cong SO(3)$, i.e. the map is a "two-to-one" homomorphism. Show that for infinitesimal rotations in $SU(2)$ over $\theta$ along axis $\hat{n}$, $U(\hat{n}) = \mathbf{1} - i(\hat{n} \cdot \boldsymbol{\sigma})\theta/2$, the corresponding rotation in $SO(3)$ is given by

$$ R_{ij} = \delta_{ij} - \theta \epsilon_{ijk} \hat{n}_k = \delta_{ij} - i \theta (-i \epsilon_{ijk}) \hat{n}_k $$

i.e. rotation in $\mathbb{R}^3$ over twice the angle along the same axis. This relation shows that the Lie algebra's are isomorphic, but suggests that $SU(2)$ has "twice as many" elements. In the local expansion of $SU(2)$ near the unit operation $U = 1$, the double cover due to $-U$ is ruled out: *Unlike the Lie groups themselves, the Lie algebras of $SU(2)$ and $SO(3)$ are isomorphic.*

4. The $2D \times 2D$ (tensor) representation of the group $SU(2) \times SU(2)$, written as $(1/2) \otimes (1/2)$, describes two particles with spin $1/2$. It acts on product states $|\frac{1}{2}, \pm\frac{1}{2}\rangle_1 |\frac{1}{2}, \pm\frac{1}{2}\rangle_2$. The generators for rotations in the direct product space $C_1^2 \times C_2^2$ are $S = S_1 \times 1 + 1 \times S_2$. These generators obey the Lie algebra $[S_n, S_m] = i\epsilon_{nml}S_l$ and thus have their own representation for each (half-) integer $j$, with $S_z = -j + 1, \dots, j$. Because $S_z = S_{1,z} + S_{2,z}$ can only take the values 0 and 1 it follows that $j = 0$ or $j = 1$. As was seen above, $j = 0$ leads to a single asymmetric singlet state $|0, 0\rangle$ with total spin 0 and a one-dimensional representation (0) of $SU(2)$, whereas the triplet $|1, m = 0, \pm 1\rangle$ leads to a 3D representation (1) of $SU(2)$ with total spin 1 and symmetric wave functions. This is formally expressed as

$$ (1/2) \times (1/2) \cong (0)_A \oplus (1)_S \qquad (37) $$

expressing that the direct product of the representation $(1/2)$ with itself is isomorphic to the direct sum of the two irreducible representations $(0)_A$ and $(1)_S$ of $SU(2)$. The indices $A$
and $S$ are important since a spin 0 state can also be symmetric, as is the case for real spin 0 or for the direct product of two spins 1. The dimension of the group (total number of Lie generators) is $2 \times 2$ on the lefthand side of Eq. (37), and $1+3$ on the righthand side.

5. A vector operator $A(\mathbf{x}) = (A_x(\mathbf{x}), A_y(\mathbf{x}), A_z(\mathbf{x}))$ acting in some Hilbert space can be defined as an operator whose quantum expectation values transform as a vector in $\mathbb{R}^3$. This means that $\langle U\Psi|A_i(\mathbf{x}')|U\Phi\rangle = R_{ij}\langle\Psi|A_j(\mathbf{x})|\Phi\rangle$ for all wave functions $\Psi, \Phi$ in Hilbert space. If $U = \exp(-i\alpha \cdot \mathbf{J})$ describes rotations in Hilbert space, it follows that we require

$$U^\dagger \cdot A_i(\mathbf{x}) \cdot U = R_{ij} A_j(\mathbf{R}^{-1} \cdot \mathbf{x}) \quad (38)$$

for $A$ to be a vector operator, to be compared to the transformation of a scalar field in Eq. (31). For instance, the position operator itself is indeed a vector operator because it commutes with angular momentum operator as $[J_n, x_i] = i\epsilon_{nij}x_j$. For a small rotation along the axis $\alpha$ in Hilbert space this confirms that

$$(1 + i\alpha_n J_n)x_i(1 - i\alpha_n J_n) = x_i + i\alpha_n [J_n, x_i] = x_i - \epsilon_{ijn}\alpha_n x_j = R_{ij}x_j$$

Any external vector, such as an external electric or magnetic field, does not transform under $U$ and does not obey this identity: the momentum vector $p$ of a particle can be checked to a vector operator but the canonical momentum $p - qA(r)$ in the presence of an external magnetic field is not.

**Exercise V.6:** $SU(2)/U(1) \sim S^2$ and $(SU(2) \times SU(2))/\mathbb{Z}_2 \cong SO(4)$.

We can imbed $U(1)$ inside $SU(2)$ using the isomorphism $\exp(i\phi) \to U(1)(\phi) = D[\exp(i\phi), \exp(-i\phi)] = \cos\phi I + i\sin\phi\sigma_z$. This rotation along the $z$-axis constitutes a subgroup of $SU(2)$ with the Abelian algebra of $U(1)$. The subgroup is not normal because a rotation $U(1)$ followed by a given $SU(2)$ transformation is not necessarily equal to first a $U(1)$ transformation (not necessarily the same one) followed by the same $SU(2)$ transformation, i.e. $gU(1) \neq U(1)g$ for any $g \in SU(2)$). We show that the quotient set is the $2D$ unit sphere $S^2$ in $\mathbb{R}^3$ and is for that reason not necessarily a group. The right coset of a matrix $SU$ in $SU(2)$ is the collection of operators $\{SU \cdot U(1)(\phi)|\phi \in [0, 2\pi]\}$. We have $SU(\alpha) = \exp(-i\alpha \cdot \sigma)$ for $\alpha \in \mathbb{R}^3$.

* Show from the transformation between $SO(3)$ and $SU(2)$ discussed in the previous exercise that $\exp(-i\alpha \cdot \sigma)(\hat{n} \cdot \sigma) \exp(+i\alpha \cdot \sigma) = (R \cdot \hat{n}) \cdot \sigma$ with $R$ a 3D rotation operator that
depends on the direction of $\alpha$. Show from this that

$$SU(\alpha)U(\phi) = e^{-i\alpha \cdot \sigma} e^{i\phi\sigma_z} = e^{i\phi(R \cdot \hat{z}) \cdot \sigma} e^{-i\alpha \cdot \sigma}$$

For any given $\alpha$ the right coset $SU(\alpha)U(\phi)$ is the set of operations $SU(\alpha)$ followed by an arbitrary rotation over an axis given by the unit vector $R \cdot \hat{z} \in S^2$. The latter is not always part of $U(1)$ so that $U(1)$ is not normal in $SU(2)$. The quotient set is not a group because the product of two elements in this set is not necessarily equal to another coset. Nevertheless, each element in the (right) quotient set $SU(2)/U(1)$ that is obtained as we vary $\alpha \in \mathbb{R}^3$ can be associated with an element on the sphere $S^2$.

* **Hopf Fibration.** Alternatively, any $SU(2)$ matrix can uniquely be written as the quaternion $SU = b_0I + ib_j\sigma_j$ with implicit summation over $j = x, y, z$, and real-valued coefficients satisfying $b_0^2 + b_x^2 + b_y^2 + b_z^2 = 1$. Thus the four-vector $(b_0, b_j)$ is on the 3-sphere $S^3$ in $\mathbb{R}^4$. The uniqueness implies that multiplication of two $SU(2)$ matrices defines a quaternion product on $S^3$ that we will not specify here, and that guarantees group structure of $S^3$. It follows that $SU(2) \cong S^3$ and in particular that $SU(2)$ is simply connected, unlike $SO(3)$. Define the two complex numbers $z_1 = b_0 + ib_3$ and $z_2 = b_1 + ib_2$ with $|z_1|^2 + |z_2|^2 = 1$. Show that both transform under $U(1)(\phi)$ as $z_i \to z_i \exp(i\phi)$.

* The set $S^2$ is equivalent to $(z \in \mathbb{C}, x \in \mathbb{R})$ with $|z|^2 + x^2 = 1$. Thus $(2z_1z_2^*, |z_1|^2 - |z_2|^2) \cong S^2$. This element is unaffected by all operations $U(1)(\phi)$ and corresponds to exactly one point on $S^2$ (although $-b$ projects to the same point). The entire group $SU(2)$ spans the whole 2-sphere (and even doubly since $b$ and $-b$ project to the same point in $S^2$): We can write $SU(2)/U(1) \sim S^2$, but with clear abuse of notation because of the double cover, and explicitly not a real group equivalence.

This discussion reveals a remarkable result in group theory: $S^1 \cong U(1)$ because the group operation $x_1 \cdot x_2$ for 2 elements in $S^1$ can be defined as complex-number multiplication, and in $S^3 \cong SU(2)$ group operation can be defined by quaternion multiplication. Clearly $S^2/S^1 \cong \{+1, -1\}$ is a real group equivalence, but $S^2 \sim S^3/S^1$ is not, in lack of a group operation. Do not confuse $S^2$ with $SO(2)$ that is a group performing rotations on elements of $S^2$ and whose multiplication is clearly defined as the Abelian 2D matrix product.

* The dimension of $SO(4)$ is 6, equal to the one of $SU(2) \times SU(2)$. From the previous argument we can see that the group $SU(2) \times SU(2)$ corresponds to 2 points $(b_0, b_j)$ on the sphere $S^3$
among which we can associate a rotation $SO(4)$. It is straightforward to show that the homomorphism conditions are met and that the map has a double cover because two points $(-b_0, -b_j)$ map to the same rotation. Hence we have the group equivalence

$$(SU(2) \times SU(2))/\mathbb{Z}_2 \cong SO(4)$$

The quotient set $SU(2)$ modulo $U(1)$ written $SU(2)/U(1)$ of righthand cosets can be identified as the set of states of a system after $SU(2)$ symmetry has been broken to the lower symmetry $U(1)$. This is for instance the case for the Néel state in antiferromagnetism, where spins point along or opposite to a direction of $S^2$, with remaining rotation symmetry along this axis. Recall that elements inside a coset were called “equivalent”. Different cosets have “non-equivalent” broken symmetry. One coset is associated with a set of states with “equivalent” broken symmetry since the remaining symmetry can transform one into the other.

### E. The group $SL(2, \mathbb{C})$

The group $SL(2, \mathbb{C})$ has as fundamental representation the set of all complex $2 \times 2$ matrices with unit determinant, but not necessarily unitary. It plays a fundamental role in the spin $1/2$ representation of the Lorentz group. It has $SU(2)$ as a unitary subgroup. Its Lie algebra must be the vector space of traceless complex matrices $C$ which we can always write as $C = C' + iC''$ with both $C'$ and $C''$ hermitian, traceless matrices in $\mathbb{C}^2$. Each of them generates the $SU(2)$ group with $SU(2) = \exp(iC') = \exp(iC'')$ and we know that its Lie algebra has 3 generators for which we can take the hermitian Pauli matrices. We expect therefore that the Lie algebra of $SL(2, \mathbb{C})$ is closely related to the direct sum of two real-valued $SU(2)$ Lie algebras $\mathfrak{su}(2)_{\mathbb{R}}$ were it not for the extra $i$ in front of $C''$, or alternatively one complex $SU(2)$ Lie algebra $\mathfrak{su}(2)_{\mathbb{C}}$ for which multiplication with complex numbers is allowed. The complexification of Lie algebras was investigated by Hermann Weyl. A rigorous mathematical statement, made only plausible here, is that

$$\mathfrak{sl}(2, \mathbb{C}) \cong \mathfrak{su}(2)_{\mathbb{C}} \cong \mathfrak{su}(2)_{\mathbb{R}} \oplus \mathfrak{su}(2)_{\mathbb{R}} \qquad (39)$$

The righthand side is the Lie algebra of the group $SU(2) \times SU(2)$.

**Intermezzo: $SL(2, \mathbb{C})$ as a connected Lie group.**
Consider the matrix

$$g = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$$

which is part of $SL(2, \mathbb{C})$. However, it can not be written as $g = \exp(iQ)$ with $Q$ some complex linear combination $Q = z_n\sigma_n$ of Pauli matrices: The matrix $Q$ can be always be diagonalized and has distinct eigenvalues $\lambda = \pm\sqrt{z_1^2 + z_2^2 + z_3^2}$ because its trace vanishes and thus $Q$ different eigenvectors. Yet, $g$ has a doubly degenerated eigenvalue $-1$ associated with a single eigenvector $(1, 0)$, and cannot be diagonalized. The connectivity of a Lie group was defined in Eq. (23). The group $SL(2, \mathbb{C})$ is indeed connected and for instance for the specific example here,

$$g = g_1 g_2 = \exp \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix} \exp \begin{bmatrix} i\pi & 0 \\ 0 & -1 \end{bmatrix}$$

The second operation is just a rotation $2\pi$ over the $z$-axis, in $SU(2)$ equivalent to $g_2 = -1$. The first transformation is a joint combination of boost along $x$ and rotation along $y$ and only its first order survives, $g_1 = -g$.

Unlike for $SU(2)$, not each element in the group $SL(2, \mathbb{C})$ is covered by its Lie algebra via $\exp(\mathfrak{sl}(2, \mathbb{C}))$ and we have to define how the representation of the Lie algebra extends to such elements. For a $P \in SL(2, \mathbb{C})$, we would like to write $P = \exp(i\alpha_n\sigma_n + \beta_n\sigma_n)$ but it was already argued in the section on Lie groups that not all elements of a group are necessarily covered by the exponential map of its Lie algebra. It can be shown that any $P \in SL(2, \mathbb{C})$ can be written uniquely as $P = U \exp(H)$, with $U \in SU(2)$ and $H$ a traceless hermitian complex matrix. This implies that for all $P \in SL(2, \mathbb{C})$,

$$P = e^{i\alpha_n\sigma_n} e^{\beta_n\sigma_n} \qquad (40)$$

It is instructive to write the general expressions,

$$e^H = \begin{pmatrix} x_0 + x_3 & x_1 + ix_2 \\ x_1 - ix_2 & x_0 - x_3 \end{pmatrix} ; \quad U = \begin{pmatrix} y_1 + iy_4 & y_2 + iy_3 \\ -y_2 + iy_3 & y_1 - iy_4 \end{pmatrix}$$

in terms of the four-vector $x_\mu$ on the surface $M^3$, i.e. the set of four-vectors with $x_\mu x^\mu = 1$ in $4D$ Minkowski space, and the Euclidean vector $y_i$ on the surface $S^3$ with $y^2 = 1$ in $\mathbb{R}^4$. The uniqueness of $U(P)$ and $H(P)$ implies that products in the group $SL(2, \mathbb{C})$ generate product operations for both vectors $x_\mu$ and $y_i$ so that we can identify a continuous and smooth homomorphism from
$SL(2, \mathbb{C})$ to $M^3 \times S^3$ that is connected. For $H = 0$, i.e. in $SU(2)$, this reduces to the quaternion product defined on $S^3$ discussed previously with the conclusion that $SU(2) \cong S^3$. The surface $S^3$ is well-known to be simply connected and therefore so is $SU(2)$. For $U = 1$ this operation defines the set of boosts in a given direction.

Because $H$ must have two real-valued eigenvalues $\pm\lambda$ with opposite sign, $\exp(H)$ has eigenvalues $\exp(\pm\lambda)$, both positive. From the expression above we infer that these eigenvalues are $\exp(\pm\lambda) = x_0 \pm \sqrt{x_0^2-1}$, so that $x_0 = \cosh \lambda > 1$. The surface $x_\mu x^\mu = 1$ disconnects into two surfaces for $x_0 > 1$ and $0 < x_0 < 1$ that touch at $x_0 = 1$, and each of them is simply connected. We conclude that $SL(2, \mathbb{C})$ is simply connected. In the next section we will see that the group $SL(2, \mathbb{C})$ is a double cover of the restricted Lorentz group $SO^+(3, 1)$ with also 6 generators, 3 for the unitary rotations ($C'$) and 3 for the non-unitary boosts ($C''$).

## F. Dual Space and Dual Representation

A linear form $\ell$ is defined as a linear map to $\mathbb{R}$ or $\mathbb{C}$ of a vector space $V$ over real or complex numbers $a, b$. As such it must satisfy $\ell(aw_1 + bv_2) = a\ell(v_1) + b\ell(v_2)$. For instance, in the vector space $\mathbb{R}^n$, all linear maps of vectors $\mathbf{x} = (x_1, x_2, \cdots, x_n)$ to $\mathbb{R}$ can be written as $\ell[\mathbf{w}](\mathbf{x}) = \sum_{i=1}^n w_i x_i = \mathbf{w}^T \mathbf{x}$ with $w_i \in \mathbb{R}$. Every linear form is associated with a new vector $\mathbf{w} = (w_1, w_2, \cdots, w_n) \in \mathbb{R}^n$. The dual space $\tilde{V}$ of $V$ (the symbol $V^*$ is widely used in literature but stands here already for complex conjugation) is defined as the set of all linear forms of the vector space $V$. With the rule $a\ell[\mathbf{w}_1] + b\ell[\mathbf{w}_2] = \ell[aw_1 + bw_2]$ the dual space of $\mathbb{R}^n$ is a vector space as well and just equal to itself. If the linear map is from $\mathbb{C}^n$ to $\mathbb{C}$ these statements change *mutatdis mutandis*, but the bilinear form is not equal to the scalar product in $\mathbb{C}^n$. The map $\ell[\mathbf{w}](\mathbf{x}) = \mathbf{w}^\dagger \cdot \mathbf{x}$ is still a linear form on $\mathbb{C}^n$ but the dual space has become anti-linear: $(a\ell[\mathbf{w}_1] + b\ell[\mathbf{w}_2]) \equiv \ell[aw_1 + bw_2] = a^* \times \ell[\mathbf{w}_1] + b^* \times \ell[\mathbf{w}_2]$. In physics the Dirac notation $\langle \mathbf{w} | \mathbf{v} \rangle$ is widely used to define the bilinear form $\langle \mathbf{w} | \mathbf{v} \rangle$ and its anti-linearity is adopted.

**Exercise V.7: Dual space of $\mathbb{R}^\infty$**.

https://en.wikipedia.org/wiki/Dual_space

The vector space $\mathbb{R}^\infty$ is defined as all infinite sets $\mathbf{x} = (x_1, x_2, x_3, \cdots)$ with only a finite number of non-zero entries and endowed with the usual multiplication and addition rules. The linear form $\mathbf{x}^T \cdot \mathbf{y}$ involves a finite sum and is thus always well defined.
1. Argue that $\mathbb{R}^\infty$ has a basis that is infinite but countable, i.e. has a basis that can be indexed by $\mathbb{N}$, with for each $i \in \mathbb{N}$ the $i^{\text{th}}$ basis element given by $e_i = (0, \dots, x_i = 1, 0, \dots)$.

2. Show that the dual space is $\mathbb{R}^\mathbb{N}$, i.e. all infinite sets $(x_1, x_2, x_3, \dots)$ of real numbers, with no countable basis, and with no scalar product properly defined. The linear form $e_i^T \cdot y$ samples the vector $y$ at the $i^{\text{th}}$ component $y_i$ and can be written in Dirac notation as $y_i = \langle i | y \rangle$.

Let us assume that a bilinear form $(v_1, v_2)$ exists in the vector space $V$ so that all linear forms on $V$ can be written as $\ell[w](v) = (w, v)$ for $w \in V$. With this definition, the dual space is again a linear vector space, with $\ell[w_1](\cdot) + \ell[w_2](\cdot) = \ell[w_1 + w_2](\cdot)$. Let $A$ be a linear operator in $V$. If the vector space $V$ has a basis $\{\hat{v}_i\}$ (not necessarily orthonormal), the image of $A$ of any basis vector is necessarily a superposition of all basis vectors, i.e. $A(\hat{v}_j) = \sum_i \hat{v}_i A_{ij}$ so that a general vector maps as $(Av)_i = A(\sum_j v_j \hat{v}_j)_i = \sum_j A_{ij} v_j$. The dual operator $\tilde{A}$ in the dual space $\tilde{V}$ can be defined formally as $\ell(Av) = (\tilde{A}\ell)(v)$ for all $v \in V$ and $\ell \in \tilde{V}$. If the linear form is given by $\ell[w](v) = w^T v = \sum_i w_i v_i$, we immediately see that

$$ \ell[w](Av) = \sum_{ij} w_i A_{ij} v_j = \sum_{ij} (A^T)_{ji} w_i v_j = \ell[A^T w](v) $$

so that $(\tilde{A}\ell)[w] = \ell[A^T w]$ in terms of the usual transpose $A^T$ of a matrix.

Given the dual vector spaces $V$ and $\tilde{V}$, the notion of duality can be extended to representations of groups in vector spaces. Let $\Pi[G]$ be a matrix representation $\Pi: G \to V$ of the group $G$ in the vector space $V$. This means that given the basis $\{\hat{v}_i\}$ of $V$, for each $g \in G$, $\Pi[g]\hat{v}_j = \sum_i \hat{v}_i G_{ij}[g]$ a matrix $G_{ij}[g]$ is defined satisfying $G[g_1 \cdot g_2] = G[g_1] \cdot G[g_2]$) that acts on column vectors with the coordinates of the vectors $v \in V$. The dual representation $\tilde{\Pi}[G]$ in $\tilde{V}$ is a matrix representation of the group $G$ in the dual space $\tilde{V}$ acting on column vectors that are the coordinates with respect to a basis $(\hat{\ell}_j)$ in $\tilde{V}$. We assert,

$$ \tilde{\Pi}[g] = \Pi[g^{-1}]^T \quad \text{for all } g \in G \qquad (41) $$

We can verify that the group property is obeyed for $\tilde{\Pi}$ if it is obeyed for $\Pi$. Because the group is not necessarily Abelian, one needs the presence of the inverse. For instance, from Eq. (41) it follows that $\tilde{\Pi}(g_1) \cdot \tilde{\Pi}(g_2) = \tilde{\Pi}(g_1 \cdot g_2)$. Dual representations may not be unique but what is convenient is that the representations in Eq. (41) together preserve the linear form, i.e. $(\tilde{\Pi}[g]\ell)(\Pi[g]v) = \ell(v)$ for all $v \in V$ and $\ell \in \tilde{V}$. For the matrix representation above is $\ell(v) = w^T \cdot v$, so that

$$ (\tilde{\Pi}[g]w)(\Pi[g]v) = (\Pi[g^{-1}]^T w)^T \cdot (\Pi[g]v) = w^T \Pi[g^{-1}] \cdot \Pi[g]v = w^T \Pi[g^{-1}]g \cdot v = w^T \cdot v $$
Note that $\Pi[g^{-1}] = (\Pi[g])^{-1}$ since the representation must preserve the group structure of $G$. A unitary representation has $\Pi^{-1} = \Pi^{\dagger}$ must satisfy $\tilde{\Pi}[g] = \Pi[g]^{*}$, i.e. the dual group representation is just the complex conjugate.

### Example: Dual representations of $U(1)$.

An illustrative example is the group $U(1)$ of rotations on the vector space $S^1$ composed of all numbers $z_v = \exp(i\theta)$ for $\theta \in [0, 2\pi)$ on the unit circle $S^1$ in $\mathbb{C}$. We use $z_g$, $z_v$ and $z_w$ to discriminate, respectively, group $z_g \in G$, vector space $z_v \in V$ and dual space $z_w \in \tilde{V}$ that here all coincide. Representations of $U(1)$ in $S^1$ are $\Pi_n[z_g] = \exp(i\theta_g n) = z_g^n$. Their action on any $z_v$ on $S^1$ is $\Pi_n[z_g](z_v) = z_g^n z_v$ and which amounts to a rotation in the complex plane. They all satisfy $\Pi_n[z_1z_2] = \Pi_n[z_1]\Pi_n[z_2]$ provided that $n \in \mathbb{Z}$. For instance, $n = \frac{1}{2}$ is excluded because we would get $\Pi_{1/2}[1] = 1$ with $\theta = 0$. But $\Pi_{1/2}[-1] = i$ as $\theta = \pi$. Hence $1 = \Pi_{1/2}[-1] \neq \Pi_{1/2}[-1] \times \Pi_{1/2}[-1] = -1$. All representations $\Pi_n$ are seen to be Abelian, unitary, one-dimensional and thus necessarily irreducible. Many discrete subgroups exist but they are not invariant under the continuous $U(1)$ action. The “irreps” are faithful for $n = \pm 1$ only. The $n$ group elements $z_g(m) = z_0 \exp(2\pi mi/n)$ for $m = 0, 2, \dots, n-1$, all map under $\Pi_n$ on the same rotation $z_0^n$.

The bilinear forms to the unit circle are given by $\ell[z_w](z_v) = (z_w, z_v) = z_w \times z_v$ with $z_w \in S^1$, so that the dual space $\tilde{S}^1$ is again the unit circle. The dual representations are $\tilde{\Pi}_n[z_g] = \Pi_n[z_g^{-1}] = \exp(-i\theta_g n) = \Pi_{-n}[z_g]$, the transpose being redundant in one dimension. The linear form $(\tilde{\Pi}_n[z_g]z_w)(\Pi_n[z]z_v) = z_w z_g^{-n} z_g^n z_v = z_w z_v$ maps on $S^1$ and is the same for all representations.

#### 1. Dual representation of Lie group and Lie algebra

For a Lie group $G$ we have identified earlier the Lie algebra $\mathfrak{g}$ as a group acting in a vector space with vector elements $X$. Representations of the Lie group $\Pi$ and the Lie algebra $\pi[X]$ are related by Eq. (25). Using the definition (41) for the dual representation of $\Pi$, the dual representation of the Lie algebra becomes,

$$ \tilde{\pi}[X] = \frac{d}{dt} \tilde{\Pi} [e^{tX}]_{t=0} = \frac{d}{dt} \Pi [e^{-tX}]_{t=0}^{T} = -\pi[X]^{T} \quad (42) $$

In physics one prefers to write the Lie algebra as $X = -iQ$ and we get,

$$ \tilde{\pi}[Q] = -\pi[Q]^{T} \quad (43) $$
The Lie algebra for the representation $\pi[Q]$ is defined as $[\pi[Q_1], \pi[Q_2]] = \pi[[Q_1, Q_2]]$ and satisfies the Lie axioms imposed for a representation of the Lie group $G$. The dual bracket satisfies

$$[\tilde{\pi}[Q_1], \tilde{\pi}[Q_2]] = -[\pi[Q_1], \pi[Q_2]]^T = -\pi[[Q_1, Q_2]]^T = \tilde{\pi}[[Q_1, Q_2]] \quad (44)$$

and thus also satisfies the Lie axioms. Unitary representations have hermitian generators so that $\tilde{\pi}[Q] = -\pi[Q]^*$. If $\langle w | v \rangle$ is a scalar product on a complex Hilbert space $H$, then $\langle w^* | v \rangle$ is a bilinear form. The representation $\Pi[Q]$ in $V$ of the group element $Q$ transforms to the vector $\Pi[Q]v$, written as the $ket \Pi[Q] |v\rangle$ in Hilbert space. In dual space, the corresponding transformation is given by $\tilde{\Pi}[Q]^* w^*$, written $\langle \tilde{\Pi}[Q]^* w^* | = \langle w^* | \Pi[Q^{-1}]$ in dual space, leaving the scalar product $\langle \tilde{\Pi}^* w^* | \Pi v \rangle$ unchanged. In physics, where scalar products are far more common than bilinear forms, one often considers the set of bra's $\langle w | H|$ to be the dual space of linear forms, and defines $\tilde{\Pi}[Q]^* = \Pi[Q^{-1}]^\dagger$ as the complex-dual representation of $G$ with transformed bra $\langle \tilde{\Pi}[Q] w | = \langle w | \Pi[Q^{-1}]$. However, with this definition, the dual space becomes anti-linear, as was mentioned earlier, and its generators are given $\tilde{\pi}[Q] = \pi[Q]^* = \pi[Q]^\dagger$. The minus sign disappears because the complex conjugation acts on the extra factor $i$ that was removed in Eq. (43).

### Example: Dual representation of SU(2) in $\mathbb{C}^2$

The generators of the fundamental irreducible, unitary representation of $SU(2)$ in $\mathbb{C}^2$ are the 3 Pauli-matrices $\sigma_n$ given by Eq. (34), acting on a spinor $\eta$. Since the representation is unitary, we have the dual representation

$$\tilde{\sigma}_n = -\sigma_n^* = -\sigma_n^T \quad (45)$$

We infer that the dual matrices $\tilde{\sigma}_n$ satisfy the same commutator relations as the Pauli matrices $\sigma_n$. We can verify that $\tilde{\sigma}_n = \sigma_2\sigma_n\sigma_2 = \sigma_2\sigma_n\sigma_2^\dagger$. The group $SU(2)$ is self-dual in the sense that the dual elements are obtained by a unitary transformation. We also infer that the representations $\exp(-i\alpha_n\sigma_n)$ and $\exp(+i\alpha_n\sigma_n^*)$, mutually dual and also each others complex conjugates, are related by the unitary transformation $\sigma_2$. The group $SU(2)$ is said to be pseudo-real. The bilinear form of two spinors is $(\xi, \eta) = \xi^T\eta$ and is not altered by the double transformation $\exp(-i\alpha_n\sigma_n)\eta$ and $\exp(-i\alpha_n\tilde{\sigma}_n)\xi$. If on the other hand we consider the bilinear form $(\xi, \eta) = \xi^\dagger\eta$, the complex-dual representation of $SU(2)$ has generators $\tilde{\sigma}_n = \sigma_n^\dagger = \sigma_n$, and the self-duality of $SU(2)$ is explicit.

### Example: Dual representation of SL(2, $\mathbb{C}$)
With $P = \exp(i\alpha_n\sigma_n)\exp(\beta_m\sigma_m)$ for any $P \in SL(2, \mathbb{C})$, the complex-dual representation of $SL(\mathbb{C}, 2)$ in $\mathbb{C}^2$ is the group $\tilde{P} = (P^{-1})^{\dagger} = \exp(i\alpha_n\sigma_n)\exp(-\beta_m\sigma_m)$. Unlike what was seen for $SU(2)$, its dual representation is not the same because the boosts are not unitary. We can verify that the $P$ and its complex-dual are related by the transformation $\sigma_2 P^* \sigma_2^\dagger = (P^\dagger)^{-1}$. The extra complex-conjugation prevents the group from being self-dual. It is customary to introduce the anti-symmetric real-valued matrix $\epsilon = -i\sigma_2$ and write this as $(P^\dagger)^{-1} = \epsilon P^* \epsilon^T$. With this notation it follows that when the spinor $\eta \in \mathbb{C}^2$ transforms as $P\eta$, the spinor $\xi = \epsilon\eta^*$ transforms as $(P^{-1})^\dagger\xi$, i.e. as a dual spinor, leaving invariant the bilinear form $\xi^\dagger\eta$. The fact that the Lorentz group has two different spinor representations in $\mathbb{C}^2$ associated with spin $\frac{1}{2}$ has an important consequence,

*The spin $\frac{1}{2}$ representation of the Lorentz group in $\mathbb{C}^2$ is accompanied by its dual and different representation that accounts for the anti-particle, with different properties. This follows from a pure group-theoretical argument and is associated with the non-unitarity of the boosts.*

The fundamental role of the anti-symmetric tensor $\epsilon_{ab}$ as the invariant metric tensor for the spin $\frac{1}{2}$ representation of the Lorentz group in $\mathbb{C}^2$ will become clear in the next section.
G. The restricted Lorentz Group $SO^{+}(1,3)$ and Spin

https://fr.wikipedia.org/wiki/Groupe_de_Lorentz

The Lorentz transformation was already defined in section III, and denotes transformations $\Lambda$ in four-dimensional Minkowski space that preserve the metric defined by the tensor $g_{\mu\nu}$: $x'^{\mu}g_{\mu\nu}x'^{\nu} = x^{\mu}x_{\mu}$. This defines the group $O(1,3)$ which is the equivalent in Minkowski space of the group $O(3)$ in $\mathbb{R}^3$. Since $\Lambda^T \cdot g \cdot \Lambda = g$, any $\Lambda \in O(1,3)$ has determinant equal to $\pm 1$. Like for $O(3)$ we wish to exclude some discrete operations such as spatial inversions $P = D(1, -1, -1, -1)$ and focus on operations with $\det \Lambda = 1$ that define $SO(1,3)$. The additional complication for $SO(1,3)$ is that time-reversal $T = D(-1, 1, 1, 1)$ also conserves the metric and is, like $P$, part of $O(1,3)$ but not of $SO(1,3)$. Nevertheless, their combined effect $PT = -1$ is again part of $SO(1,3)$ although non-orthochronous. We want to restrict to orthochronous boosts that have $\Delta_0^0 > 1$ but for mathematical reasons, we do not want to exclude sets that are not subgroups. Fortunately, the operator $PT = -1$ stands on its own and no Lorentz boosts or spatial rotations exist such that $\Lambda_1\Lambda_2 = -1$. Therefore the restricted Lorentz group $SO(1,3)^+ \equiv SO(1,3)/(1,P,T,-1)$ is still a subgroup and even a normal subgroup in $O(3,1)$. The group $(1,P,T,-1)$ is known as the Klein group, the smallest non-cyclic group.

The existence of an orthochronous normal subgroup of the Lorentz group $SO(1,3)$ means we can apply group theoretical arguments to phenomena that are observed to be causal.

Every orthochronous Lorentz transformation is known to be equal to the product of a boost and a rotation. The group of spatial rotations $SO(3)$ in $\mathbb{R}^3$ is a subgroup of $SO^+(1,3)$ but is not a normal subgroup because the product of first a rotation and then a boost is not necessarily equal to the product of first (another) boost and then the same rotation: $\Delta SO(3) \neq SO(3)\Delta$ for arbitrary $\Delta \in SO^+(1,3)$. Nevertheless, the 3 generators $J_n$ in $SO(3)$ are also generators of $SO^+(1,3)$,

$$ (J_n)^\mu_\nu = \begin{pmatrix} 0 & \mathbf{0} \\ \mathbf{0} & -i\epsilon_{n\mu\nu} \end{pmatrix} \qquad (46) $$

The boosts, on the other hand, do not even constitute a group on their own because products of boosts in different directions are not equal to another boost but also involve a rotation. For the boosts (1) we can choose the direction of the velocity in 3 different directions and for $v_j/c_0 = \beta_j \ll 1$
we obtain $\Lambda_{\nu}^{\mu} = g_{\nu}^{\mu} - i\beta_{n}(K_{n})^{\mu}_{\nu}$ with $g_{\nu}^{\mu} = I_{4} = D(1, 1, 1, 1)$ and the generators

$$ (K_x)^\mu_\nu = i \begin{pmatrix} 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad (K_y)^\mu_\nu = i \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad (K_z)^\mu_\nu = i \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \quad (47) $$

In this form, the $J_{n}$ are hermitian, but the boost generators $K_{n}$ are anti-hermitian. Let us write an infinitesimal Lorentz transformation in the double contra-variant form, using the anti-symmetric tensor $\omega_{\mu\nu}$ as $\omega_{0n} = -\omega_{n0} = \beta_{n}$ and $\omega_{lm} = -\omega_{ml} = \epsilon_{lmn}\alpha_{n}$.

$$ \Delta_{\mu\nu} = g_{\mu\nu} + \omega_{\mu\nu} \quad (48) $$

## Exercise V.8

Unlike $g_{\mu\nu} = D(1, -1, -1, -1)$, the matrix $\Delta_{\mu\nu}$ is not symmetric. Use $\Delta_{\nu\alpha}\Delta_{\beta}^{\nu} = g_{\alpha\beta}$ to show that $\omega_{\mu\nu} = -\omega_{\nu\mu}$ i.e. anti-symmetric for both rotations and boosts. Note that $(J_{n})_{\mu\nu}$ and $(K_{n})_{\mu\nu}$ are both hermitian matrices, although for the boosts this is a deceptive trick since the physical action of two consecutive boosts is described by $(K_{n})_{\mu\lambda}g^{\lambda\nu'}(K'_{n})_{\nu'\nu}$ and not by their direct matrix product!

In the following we want to find the 6 generators of the Lorentz group and a non-perturbational expression for all Lorentz transformations $\Delta$ which are 4 × 4 matrices. We define the second-order antisymmetric four-tensors $(M^{\mu\nu})_{\alpha\beta}$ as $(M^{0n})_{\alpha\beta} = -(M^{n0})_{\alpha\beta} = (K_{n})_{\alpha\beta}$ and $(M^{lm})_{ij} = \epsilon_{lmn}(J_{n})_{ij} = \epsilon_{lmn}i\epsilon_{nij} = i(\delta_{li}\delta_{mj} - \delta_{lj}\delta_{mi})$. Any infinitesimal Lorentz transformation can be expressed as,

$$ \Delta_{\alpha\beta} = g_{\alpha\beta} - \frac{i}{2} \omega_{\mu\nu} (M^{\mu\nu})_{\alpha\beta} + \mathcal{O}(\omega_{\mu\nu})^2 \quad (49) $$

The factor $\frac{1}{2}$ compensates for the double contraction. Because $M^{\mu\nu} = -M^{\nu\mu}$ again only six non-zero matrices exist, equal to the dimension of the Lorentz group. We can write equivalently,

$$ (M^{\mu\nu})_{\alpha\beta} = i (g_{\alpha}^{\mu} g_{\beta}^{\nu} - g_{\beta}^{\mu} g_{\alpha}^{\nu}) \quad (50) $$

with $g_{\alpha}^{\mu} = g^{\mu\nu}g_{\nu\alpha} = \delta_{\mu}^{\alpha}$. Via Eq. (49) they generate the Lorentz group, in the same way that the rotation group $SO(3)$ is generated as by the 3 generators $(J_{n})_{ij} = i\epsilon_{ijn}$ (note that the double
covariant notation generates a minus sign to $(J_n)_j^i = -i\epsilon_{ijn}$ in the vector space $\mathbb{R}^3$. The generators $M^{\mu\nu}$ satisfy their own Lie algebra that we shall omit. The algebra guarantees that finite Lorentz transformations can be written in the desired matrix form,

$$ \Delta_{\beta}^{\alpha} = \exp \left( -\frac{i}{2} \omega_{\mu\nu} M^{\mu\nu} \right)_{\beta}^{\alpha} \qquad (51) $$

In particular, products of two transformations will have again this form. We have raised the index $\alpha$ to make sure that unity is obtained on both sides when $\omega_{\mu\nu} = 0$. Of course it has to be verified if this exponential map is surjective, i.e. every element in $SO^+(1, 3)$ is covered by this map. This is indeed the case.

## Intermezzo : fully anti-symmetric pseudo four-tensor

In the group $SO(3)$ the Lévi-Cività tensor $\epsilon_{ijk}$ is fully antisymmetric with $\epsilon_{123} = 1$ and as such unique. It shows up as structure factor of the $SO(3)$ Lie group and in addition, for $k = x, y, z$ acts as generator of the rotation group. The pseudo-tensor $\epsilon_{ijk}$ defines a third-order tensor invariant under $SO(3)$ but also under inversion, as was made explicit in Eq. (30). In the Lorentz group one can define the tensor $\epsilon^{\alpha\beta\gamma\delta}$ in the same way though with respect to Minkowski metric. We define $\epsilon^{0123} = 1$ and impose antisymmetry for any permutation so that in particular $\epsilon^{ijk} = \epsilon^{ikj}$. We can lower the indices as usual with the metric tensor: $\epsilon_{\alpha\beta}^{\ \ \gamma\delta} = g_{\alpha\alpha'}g_{\beta\beta'}\epsilon^{\alpha'\beta'\gamma\delta}$. We can check that $\epsilon_{0123} = -1$ and $\epsilon^{1230} = -1$ since 3 permutations are involved, hence for the anti-symmetric 4-tensor no invariance exists under cyclic permutation. The tensor has many interesting properties, the most important being its *invariance under proper Lorentz transformations*. This can be expressed as a generalization of the transformation of $\epsilon_{ijk}$ in $O(3)$. A real four-tensor must transform

$$ \epsilon_{\alpha'\beta'\mu'\nu'} \Delta_{\alpha}^{\alpha'} \Delta_{\beta}^{\beta'} \Delta_{\mu}^{\mu'} \Delta_{\nu}^{\nu'} = \epsilon_{\alpha\beta\mu\nu} \det \Delta \qquad (52) $$

For proper Lorentz transformations $\det \Delta = 1$ so that the fully anti-symmetric 4-tensor transforms indeed as a genuine four-tensor. Concerning time-reversal $T$ and space inversion $P$ however, $\epsilon_{\alpha\beta\mu\nu}$ should transform with an extra minus sign if it were a proper four-tensor, but it does not. Therefore, the tensor is a pseudo 4-tensor invariant under proper Lorentz transformations but also under parity and time-reversal. Other useful properties are,

$$ \epsilon^{\mu\nu\gamma\delta} \epsilon_{\gamma\delta\alpha\beta} = -2 (g_{\alpha}^{\mu} g_{\beta}^{\nu} - g_{\beta}^{\mu} g_{\alpha}^{\nu}) $$

so that the Lorentz group generators can be written as

$$ (M^{\mu\nu})_{\alpha\beta} = -\frac{i}{2} \epsilon^{\mu\nu\gamma\delta} \epsilon_{\gamma\delta\alpha\beta} $$
i.e. a contraction of two fully anti-symmetric tensors unlike SO(3) that has only one, and manifestly hermitian for all 6 choices of $\mu\nu$. We can also check that

$$ \epsilon^{\mu\nu\gamma\delta} \epsilon_{\gamma\delta\mu\beta} = -6g_{\beta}^{\nu} $$

and

$$ \epsilon^{\mu\nu\gamma\delta} \epsilon_{\gamma\delta\mu\nu} = -24 $$

The anti-symmetric pseudo four-tensor makes it possible to transform tensors to pseudo-tensors. For instance given a second-order tensor $F_{\mu\nu}$ we can define a new tensor,

$$ G_{\alpha\beta} = \epsilon_{\alpha\beta}^{\mu\nu} F_{\mu\nu} \qquad (53) $$

that transforms as a real tensor under proper Lorentz transformations. However, under parity and time-reversal both tensors differ. For instance if we consider the electromagnetic tensor $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ we can check that $F_{0i}$ is parity-odd but even under time reversal, whereas $F_{ij}$ is parity-even and odd under time reversal. Since $G_{0i} = \epsilon_{0i}^{nm}F_{nm\neq i}$ we see that $G_{0i}$ is parity-even and odd under time-reversal. Similarly $G_{ij}$ is seen to be parity-odd and even under time-reversal. This will lead later to the important conclusion that two different dual group representations exist for the electromagnetic field.

**Exercise V.9** With $\omega_{0n} = -\omega_{n0} = \beta_n$ and $\omega_{ij} = \epsilon_{ijn}\alpha_n$ verify that we recover rotations and boosts from the generators $M^{\mu\nu}$ of the Lorentz group

$$ \frac{1}{2}\omega_{\mu\nu}M^{\mu\nu} = \beta_n K_n + \alpha_n J_n \qquad (54) $$

**Exercise V.10: Finite Boosts**

An infinitesimal boost in the $z$ direction is described by $\Delta(\beta)^\alpha_\gamma = \delta^\alpha_\gamma + \beta B^\alpha_\gamma$ with, restricting to the two coordinates $\alpha, \gamma = (t, z)$.

$$ (B^\alpha_\gamma) = -i(K^\alpha_\gamma) = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = -\sigma_x $$

• Show that

$$ (B_{\alpha\gamma}) = -i(K_{\alpha\gamma}) = \begin{pmatrix} 0 & -1 \\ +1 & 0 \end{pmatrix} = -i\sigma_y $$

and thus making $(K_{\alpha\gamma})$ a traceless, antisymmetric hermitian $2 \times 2$ matrix.
* Show from its definition that a finite boost to a reference frame with velocity $\beta = v/c_0$ can be written as

$$ \Delta(\beta)^{\alpha}_{\gamma} = \begin{pmatrix} \cosh \xi & -\sinh \xi \\ -\sinh \xi & \cosh \xi \end{pmatrix} = (e^{-\xi \sigma_x})^{\alpha}_{\gamma} $$

with $\tanh \xi = \beta$ and $-\infty < \xi < \infty$. These transformations constitute an Abelian group $K(\beta \in [-1, 1])$ and are not unitary since $\Delta\Delta^{\dagger} = e^{-2\xi\sigma_x} \neq I_2$, but hermitian and with unit determinant and as such a subgroup of $SL(2, \mathbb{C})$ of $2 \times 2$ representing matrices with unit determinant. The parameter $\xi$ is additive in two consecutive boosts in the same direction and takes over the role of the velocity $\beta$ in Galilean transformations. We could redefine velocity addition in consecutive, rectilinear boosts as $\beta_1 \oplus \beta_2 = \tanh(\arctanh\beta_1 + \arctanh\beta_2) = (\beta_1 + \beta_2)/(1 + \beta_1\beta_2)$, that also covers the asymptotical values $\beta = \mp 1$ and check that this group is an Abelian representation of the group of real numbers on the interval $\beta \in [-1, 1]$ with addition $\oplus$ and unit element $\beta = 0$: $\Delta(\beta_1)\Delta(\beta_2) = \Delta(\beta_1 \oplus \beta_2)$.

* Argue that the map $K \to \mathbb{C}^2$: $\Delta(\beta) \to U(\beta) = \exp(-i\pi\beta\sigma_x)$ to rotations along the x-axis in $\mathbb{C}^2$ is both surjective and injective and establishes an isomorphism between boosts and the subgroup of $SU(2)$ generated by $\sigma_x$.

Let us work in the “normal” matrix representation with $K_n$ the matrix $(K_n)^\mu_\nu$. It can be checked that in addition to $[J_x, J_y] = iJ_z$ cyclic, we have $[K_x, K_y] = -J_z$ cyclic, $[J_x, K_y] = iK_z$ cyclic (and thus in particular $[J_n, K_n] = 0$, all follow directly from the notion that $iK_n$ is a rotation), so that commutators mix up boots and rotations. The combination of two boosts in different directions is not equal to another boost, but rather to a combined boost and rotation, a notion that leads to an effect known as Thomas precession of a spin. Thus, unlike the set $J_n$, the set $K_n$ has no group structure of its own. It is convenient to define complex matrices $A_n = (J_n + iK_n)/2$ and $B_n = (J_n - iK_n)/2$. It can easily be checked that both are traceless hermitian and

$$ [A_n, A_m] = i\epsilon_{nml}A_l, \quad [B_n, B_m] = i\epsilon_{nml}B_l, \quad [A_n, B_m] = 0 \qquad (55) $$

Hence with this complexification the matrices $A$ and $B$ obey mutually independent $SU(2)$ Lie-algebras that generate unitary transformations. This anticipates the important conclusion that $SO^+(1,3) \cong SU(2) \times SU(2)$, i.e. we can represent the restrictive Lorentz group by a double
presentation of SU(2). In the Exercise below we reveal this correspondence using the spin $\frac{1}{2}$ representation of SU(2). The equivalence of both groups establishes the fundamental relation between special relativity and particle spin, or to put it even stronger: *the existence of spin, and half-integer spin in particular, is the group-theoretical consequence of Lorentz symmetry*. And there is more: *the link between SO+(1,3) and SU(2) having two different representations (for spin $\frac{1}{2}$) predicts the anti-particle*. This is important although we should not forget that SU(2) has representations for higher angular momenta, and therefore so does SO+(1,3).

1. *The proper Lorentz group and spin* $rac{1}{2}$

**Exercise V.11: Spin $\frac{1}{2}$ representation of SO+(1,3) with SL(2, C) and SU(2) × SU(2)**

Spin $\frac{1}{2}$ is associated with the representation of SU(2) describing rotations in $\mathbb{C}^2$, with the Pauli matrices as generators, and acting on spinors. Equation (36) revealed the double cover of SU(2) over the group of rotations in $\mathbb{R}^3$. To cover the entire Lorentz group SO+(1,3) we need the much larger group SL(2, C) of matrices in $\mathbb{C}^2$ with unit determinant but not necessarily unitary; SL(2, C) is a group of dimension $2 \times 2^2 - 2 = 6$, equal to the one of SO(1,3).

* Define $\sigma^0 = I_2$ and define the Pauli 4-vector $\sigma^\mu = (I_2, \boldsymbol{\sigma})$ satisfying $\sigma_\nu\sigma^\nu = g_{\alpha\nu}\sigma^\alpha\sigma^\nu = -2I_2$. Show that 1) $\det \sigma^\mu x_\mu = x_\mu x^\mu$ is a Lorentz-invariant scalar, and that 2) $\text{Tr} \sigma^\mu \sigma^\nu = 2\delta_{\mu\nu}$. One sometimes writes $\bar{\sigma}^\mu = \sigma_\mu$ with the dual choice for the Pauli-matrices (see the discussion on duality in previous section), with only the space components achieving opposite sign, $\bar{\sigma}^\mu = (I_2, -\boldsymbol{\sigma})$. The introduction of $\bar{\sigma}_\mu$ makes it possible to respect the Lorentz contraction rule $\text{Tr} \bar{\sigma}_\mu \sigma_\nu = 2g_{\mu\nu}$. Then: $\text{Tr} \sigma_\mu \bar{\sigma}^\nu = 2g_\mu^\nu$. The dual form is often expressed in terms of the fully antisymmetric real-valued operator, in SU(2) given by $\epsilon \equiv -i\sigma^2$. Then $\bar{\sigma}^\mu = \sigma^2(\sigma^\mu)^*\sigma^{2,\dagger} = \epsilon(\sigma^\mu)^*\epsilon^T$.

* We can define a covariant four-vector $x_\mu = (x_0, -\boldsymbol{x})$ from the tensor product of a spinor $\eta$ and its complex-dual spinor $\xi = \epsilon \cdot \eta^*$ using $(x_\mu \sigma^\mu)_{\alpha\beta} = i\eta_\alpha \xi_\beta$. Show that $x_0 = \text{Im} \eta_1 \eta_2^*$ and $x_\mu x^\mu = 0$.

* Define the map $\rho : \text{SL}(2, \mathbb{C}) \to \text{SO}(1, 3) : \sigma^\mu x'_\mu = P \cdot \sigma^\mu x_\mu \cdot P^\dagger$, with $P \in \text{SL}(2, \mathbb{C})$. For all $P \in \text{SL}(2, \mathbb{C})$, $\rho$ defines a transformation in Minkowski space: $(x')_\nu = \Delta_\nu^\mu \cdot x_\mu$. Show that
$\Delta$ conserves the Minkowski norm and can be written as

$$ \Delta^{\nu\mu} = \frac{1}{2} \mathrm{Tr} \sigma_{\nu} P \sigma^{\mu} P^{\dagger} \quad \text{or} \quad \Delta_{\mu}^{\nu} = \frac{1}{2} \mathrm{Tr} \bar{\sigma}^{\nu} P \sigma_{\mu} P^{\dagger} \quad \text{or} \quad P \sigma^{\mu} P^{\dagger} = \Delta_{\nu}^{\mu} \sigma^{\nu} = (\Delta^{-1})_{\nu}^{\mu} \sigma^{\nu} \quad (56) $$

The map $\rho$, similar to the one discussed in Exercise IV.5-3 that defines $SU(2)$ to cover $SO(3)$, defines a two-to-one homomorphism from $SL(2, \mathbb{C})$ to $SO(1, 3)$, i.e. “doubly covers” $SO(1, 3)$. The operator $-I_2$ is part of $SL(2, \mathbb{C})$ but is the only element except the identity $I_2$ to have no action on the transformation and is thus part of the Kernel $K(\rho)$ of $\rho$. Since $\Delta_0^0 = \frac{1}{2} \mathrm{Tr} PP^\dagger \ge 1$ for any complex $2 \times 2$ matrix with unit determinant, the map is orthochronous and part of $SO^+(1, 3)$. If we would restrict $P$ to its special unitary subgroup $SU(2)$ we would map to rotations only, recalling Eq. (36). It was shown by Eugene Wigner that it is impossible to get a unitary representation of the Lorentz group in a finite dimensional vector space, so here $P$ cannot be unitary. If we would extend $P$ to complex matrices with arbitrary determinant and replace $P^\dagger$ by $P^{-1}$ in Eq. (56), $\rho$ would still conserve the Minkowski norm but many matrices $P$ would map on the unit matrix, and the map would not be just two-to-one injective.

### Example: The Lorentz group $SO^+(1, 3)$ is not simply connected

The group $SL(2, \mathbb{C})$ was previously seen to be simply connected and isomorphic to the group $M^3 \times S^3$, with $M^3$ the orthochronous unit surface in Minkowski space defined by $x_\mu x^\mu = 1$ and $x_0 > 1$, and $S^3$ the spherical unit surface in $\mathbb{R}^4$. From Eq. (56) we conclude that it doubly covers $SO^+(1, 3)$. Since $P = U \exp(H)$ for $P \in SL(2, \mathbb{C})$ both $P = \pm U \exp(H)$ map on the same element in $SO^+(1, 3)$ which can be expressed as

$$ SO^+(1, 3) \cong M^3 \times (S^3 / \{1, -1\}) \quad (57) $$

implying that opposite points on the surface $S^3$ should be considered equal. This notion does not change the connectivity but the group $SO^+(1, 3)$ is no longer simply connected. Any path along the surface $S^3$ linking two anti-nodes now has become a loop that cannot be shrunk into a single point.

We can define a second “dual” map $\rho' : SL(2, \mathbb{C}) \to SO(1, 3) : \bar{\sigma}^\mu x'_\mu = Q \cdot \bar{\sigma}^\mu x_\mu \cdot Q^\dagger$, with again $Q \in SL(2, \mathbb{C})$. Since this changes the sign of the spatial components $x_\mu$ and $x'_\mu$ with respect to
the operation involving $\sigma_n$ and $P$, this is equivalent to a parity-inverted picture of the former map. Space inversion was eliminated to construct the group $SO^+(1,3)$ so this map must be considered as fundamentally different from the map $\rho$. To get the same Lorentz transformation it follows that $P \neq Q$. Since $\bar{\sigma}^\mu = \sigma^2(\sigma^\mu)^*\sigma^2$ and $\sigma^2 B^T \sigma^2 = B^{-1}$ for a matrix with unit determinant, we infer $Q = (P^\dagger)^{-1}$, a relation that confirms the (complex-)dual nature of the second map. We combine both maps and introduce the $C^{4 \times 4}$ matrix

$$S(P) = \begin{pmatrix} P & 0 \\ 0 & (P^\dagger)^{-1} \end{pmatrix} \qquad (58)$$

working on a bispinor $(\eta, \xi)$. Note that $S(P_1)S(P_2) = S(P_1P_2)$. We introduce the Dirac matrices $\gamma^\mu$ from

$$\gamma(x) = \begin{pmatrix} 0 & \sigma^\mu x_\mu \\ \bar{\sigma}^\mu x_\mu & 0 \end{pmatrix} \equiv \gamma^\mu x_\mu \qquad (59)$$

Using Eq. (56) we can verify that

$$S(P)\gamma^\mu S(P)^{-1}x_\mu = \begin{pmatrix} 0 & P\sigma^\mu x_\mu P^\dagger \\ (P^\dagger)^{-1}\bar{\sigma}^\mu x_\mu P^{-1} & 0 \end{pmatrix} = \gamma^\nu \Delta_\nu^\mu x_\mu = (\Delta^{-1})^\mu_\nu \gamma^\nu x_\mu$$

Since this is valid for arbitrary four-vectors $x$ we conclude that the transformation $S(P)$ actually generates an inverse Lorentz transformation of the Dirac matrices,

$$S(P)\gamma^\mu S(P)^{-1} = (\Delta^{-1})^\mu_\nu \gamma^\nu \quad \text{or} \quad S(P)^{-1}\gamma^\mu S(P) = \Delta^\mu_\nu \gamma^\nu \qquad (60)$$

A second important relation is,

$$\gamma^0 S(P)^\dagger \gamma^0 = S(P)^{-1}$$

Because $S$ is not unitary, the matrix $\gamma^0$ conveniently converts the adjoint into its inverse. For any bi-spinor $\Psi(x) = (\eta(x), \xi(x))$, let us define the scalar

$$\mathcal{L}(x) = \Psi(x)^{\dagger}\gamma^{0}\gamma^{\mu}i\partial_{\mu}\Psi(x) \qquad (61)$$

that anticipates the Dirac Lagrangian. It can be verified that the transformation $\Psi'(x') \equiv L(\Delta)\Psi(x) = S(P)\Psi(\Delta^{-1}x')$ leaves the action $\int d^4x \mathcal{L}(x)$ invariant. With $(x')^\mu = \Delta^\mu_\nu x^\nu$, we have $\partial'_\mu = \Delta^\alpha_\mu \partial_\alpha$ and $x = \Delta^{-1}x'$. Hence,

$$\begin{aligned} \mathcal{L}'(x') &= \Psi'(x')^\dagger \gamma^0 \gamma^\mu i \partial'_\mu \Psi'(x') = (S\Psi(x))^\dagger \gamma^0 \gamma^\mu i \partial'_\mu S\Psi(x) \\ &= \Psi(x)^\dagger \gamma^0 S^{-1} \gamma^\mu S i \partial'_\mu \Psi(x) = \Psi(x)^\dagger \gamma^0 \Delta^\mu_\nu \gamma^\nu i \Delta^\alpha_\mu \partial_\alpha \Psi(x) \\ &= \Psi(x)^\dagger \gamma^0 \gamma^\nu i \partial_\nu \Psi(x) = \mathcal{L}(x) \end{aligned}$$
With the transformation $x' \to x$ in the integral and $d^4x = d^4x'$, we conclude that the transformation leaves the action invariant. This is a representation of the Lorentz group since

$$ 
\begin{aligned}
L(\Delta_1)L(\Delta_2)\Psi(x) &= L(\Delta_1)S(P_2)\Psi((\Delta_2)^{-1}\cdot x) = S(P_1)S(P_2)\Psi((\Delta_2)^{-1}\cdot (\Delta_1)^{-1}x) \\
&= S(P_1\cdot P_2)\Psi(\Delta_1\cdot \Delta_2)^{-1}\cdot x) = L(\Delta_1\cdot \Delta_2)\Psi(x)
\end{aligned} 
\quad (62) $$

In this transformation the spinor $\eta$ transforms under $P$ whereas its complex-dual spinor $\xi$ transforms under $(P^\dagger)^{-1}$, as was established in section V.E, and can be mapped to the spinor as $\xi = -i\sigma^2\eta^*$. We can also check that $\gamma^0$ acts as parity operator since $\gamma^0\gamma(x_0, x)\gamma^0 = \gamma(x_0, -x)$. Recall that the Pauli matrices themselves are parity invariant and therefore do not change sign.

## 2. From SU(2) × SU(2) back to Minkowski space

The 6 generators of $SL(2, \mathbb{C})$ are traceless $2 \times 2$ complex matrices which we can write as $C = C' + iC''$ in terms of two independent sets of 3 hermitian $2 \times 2$ traceless matrices. For both sets we can choose the Pauli-matrices that generate a representation of $SU(2)$ in $\mathbb{C}^2$. Hence we can choose the 6 generators $C'_n = \frac{1}{2}\sigma_n$ and $C''_n = \frac{1}{2}\sigma_n$ for $n = x, y, z$, with real-valued Lie parameters $\alpha_n$ and $\beta_n$ respectively. Upon expanding the map $\rho$ into infinitesimal transformations of $P$, we find from Eq. (56),

$$ \Delta_{\mu}^{\nu} = g_{\mu}^{\nu} + \frac{1}{2}i \sum_{n=1}^{3} \alpha_{n} \mathrm{Tr}[\sigma_{\mu}, \sigma_{\nu}] C'_{n} + \frac{1}{2}i \sum_{n=1}^{3} \beta_{n} \mathrm{Tr}\{\sigma_{\mu}, \sigma_{\nu}\} i C''_{n} + \mathcal{O}(\alpha_{n}^{2}) $$

We have $[\sigma^\mu, \sigma^\nu] = 2i\epsilon_{\mu\nu k}\sigma_k$ for $\mu, \nu \neq 0$ and zero otherwise. The anti-commutator has $\{\sigma_0, \sigma_\nu\} = 2\sigma_\nu$, $\{\sigma_\mu, \sigma_0\} = 2\sigma_\mu$, and $\{\sigma_i, \sigma_j\} = 2\delta_{ij}$ for $i, j = 1, 2, 3$ (which thus gives $\mathrm{Tr}\,\sigma_n = 0$ in the trace above). It follows that the terms involving $C'_n$ reproduce the hermitian generators $J_n$ of the rotations in $SO(3)$ and that the terms with $C''_n$ generate the anti-hermitian boosts $K_n$, given in Eqs. (46, 47):

$$ 
\begin{aligned}
\Delta_{\mu}^{\nu} &= g_{\mu}^{\nu} + \sum_{n=1}^{3} (-i\alpha_n J_n - i\beta_n K_n)^{\nu}_{\mu} \\
&= g_{\mu}^{\nu} - i \sum_{n=1}^{3} (\alpha_n + i\beta_n)(B_n)^{\nu}_{\mu} - i \sum_{n=1}^{3} (\alpha_n - i\beta_n)(A_n)^{\nu}_{\mu}
\end{aligned}
 $$

where $A_n$ and $B_n$ have been defined above and obey independent $SU(2)$ Lie algebra. The fact that the Lorentz group $SO^+(1, 3)$ is generated by a double $SU(2)$ cover agrees with the above. If we finally perform the map $i\beta_n \to \beta'_n \in \mathbb{R}$ so that boosts in imaginary time become unitary, this expression is just the infinitesimal expansion of the group $SU(2) \times SU(2)$ and hence its Lie algebra
$\mathfrak{su}(2) \times \mathfrak{su}(2)$. This implies that (apart from some mathematical objections on compactness that are not specified here), the Lie algebras are isomorphic,

$$ \mathfrak{so}^{+}(1,3) \cong \mathfrak{sl}(2, \mathbb{C}) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2) \text{ after complexification} \quad (63) $$

We have already seen that $SU(2)$ has representations in terms of integer or half-integer spin. We can consider different representations of its direct product. If we choose spin $\frac{1}{2}$ for one and spin 0 for the other ($\alpha_n = \pm \beta'_n$), we find the representation $(0, \frac{1}{2}) \oplus (\frac{1}{2}, 0)$ for $SO^+(1,3)$. The representations $(0, \frac{1}{2})$ and $(\frac{1}{2}, 0)$ are called the right- and left-chiral representations, for reasons that become clear later. They act on 2D left-handed spinors $\xi_\alpha$ or right-handed spinors $\eta_\alpha$, with $\alpha = 1, 2$ and the dot reminding that dual spinors transform differently: Under $SL(2, \mathbb{C})$, the spinor $\eta$ transforms to $P\eta$ and its dual spinor $\xi$ to $(P^\dagger)^{-1}\xi$. The group-theoretical nature of spin 1/2 particles emerges here naturally even before writing down the equation of motion.

### 3. Spinor representation of $SO^+(1,3)$: the Dirac equation

In the spin-0 representation of $SO^+(1,3)$ the field $\Psi(x)$ is scalar and symmetry transformations simplify to a change of global phase. Under a Lorentz transformation, $\Psi'(x') = \Psi(\Delta^{-1}x')$. This leads to an equation of the kind (16). For spin 1/2 we must use the spinor field and more precisely the representations $(0, \frac{1}{2})$ and $(\frac{1}{2}, 0)$ of $SU(2) \times SU(2)$. We will come back later to this tensor notation.

In Eqs. (61), (58) and (62) we obtained the transformation $\Psi'(x') = S(\Delta)\Psi(x) = S(\Delta)\Psi(\Delta^{-1}x')$ that leaves the Lagrangian density

$$ \mathcal{L} = \Psi^{\dagger}(x)\gamma^{0}\gamma^{\mu}i\partial_{\mu}\Psi(x) $$

invariant. Because $S(\Delta_1\Delta_2) = S(\Delta_1)S(\Delta_2)$ the transformation preserves the group structure of the Lorentz group and is as such a representation of the Lorenz group $SO^+(1,3)$ in the bispinor space $\mathbb{C}^4$. We recall the $4 \times 4$ (non-hermitian !) Dirac matrices as earlier announced in Eq. (59),

$$ \gamma^{\mu} = \begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma}^{\mu} & 0 \end{pmatrix} \quad (64) $$

From the algebra of the Pauli matrices we find the anti-commutator,

$$ \{\gamma^{\mu}, \gamma^{\nu}\} = 2g^{\mu\nu} = 2g_{\mu\nu} \quad (65) $$
It is customary to introduce the Dirac conjugate bi-spinor $\overline{\Psi} \equiv \Psi^\dagger \gamma^0$ to compensate for the non-unitarity of the Lorentz group. We see that the scalar $\overline{\Psi} \Psi$ is Lorentz-invariant, i.e. a Lorentz scalar. We can use it to can add a Lorentz-invariant mass-term $m \overline{\Psi} \Psi$ to the Dirac Lagrangian density, to get

$$ \mathcal{L} = \overline{\Psi} (i\gamma^{\mu} \partial_{\mu} - m) \Psi \qquad (66) $$

The Lorentz-invariance of this Lagrangian is truly remarkable since no new vectors are necessary to compensate for the transformation of $\partial_\mu$. The great contribution of Dirac was to find 4 × 4 matrices $\gamma^\mu$ that actually transform as four-vectors. From the Euler-Lagrange equation (15) with the spinors $\overline{\Psi}$ and $\Psi$ as independent (4D) parameters, we find the Dirac equation

$$ i\hbar\gamma^{\mu}\Psi\partial_{\mu} - mc_{0}\Psi = 0 \quad \text{or} \quad i\hbar\partial_{\mu}\overline{\Psi}\gamma^{\mu} + mc_{0}\overline{\Psi} = 0 \qquad (67) $$

where we have switched from natural to physical units. This can equivalently be written in Hamilton form

$$ i\hbar\partial_t \Psi = H\Psi \quad \text{with} \quad H = c_0\alpha \cdot p + \gamma^0 mc_0^2 \qquad (68) $$

with the three tensors $\alpha^i = \gamma^0 \gamma^i$ and $p = -i\hbar \nabla$ the usual momentum operator. From the properties of the Dirac matrices it is straightforward to verify that

$$ H^2 = [p^2 c_0^2 + (mc_0^2)^2] \times I_4 $$

Hence the Dirac equation can be considered as the “square root” of the Klein-Gordon equation, but to take the square root and keep Lorentz-invariance, we need 4 dimensions!

It is useful to define a fifth Dirac matrix,

$$ \gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3 \qquad (69) $$

This matrix has 4 particular properties that can easily be checked:

1. $\{\gamma^5, \gamma^\mu\} = 0$

2. $(\gamma^5)^2 = 1$

3. $(\gamma^5)^\dagger = \gamma^5$

4. $\gamma^5 = -\frac{i}{4!} \epsilon_{\mu\nu\alpha\beta} \gamma^\mu \gamma^\nu \gamma^\alpha \gamma^\beta$
The last identity follows from the fact that all different Dirac matrixes anti-commute, with 4! different orders possible and with the extra minus sign due to convention $\epsilon_{0123} = -1$. It brings $\gamma^5$ in a manifestly Lorentz-invariant form and implies that, contrary to the $\gamma^\mu$, the matrix $\gamma^5$ transforms as a scalar under Lorentz transformations: $S^{-1}\gamma^5 S = \gamma^5$. This in turn implies that $\overline{\Psi'}\gamma^5\Phi' = \overline{\Psi}\gamma^5\Phi$ for all spinors. In the *chiral* or *Weyl representation*, $\gamma^5 = D(-\sigma^0, \sigma^0)$ is block-diagonal, as is the transformation $S(\Delta)$.

### Exercise V.12: Chiral representation of Dirac Lagrangian

Chirality can be defined as the expectation value of the operators $L = (1-\gamma^5)/2$ and $R = (1+\gamma^5)/2$ that transform as Lorentz-invariant scalars. The wave functions $\Psi_L = L\Psi$ and $\Psi_R = R\Psi$ are eigenfunctions of $\gamma^5$ with eigenvalues $\pm 1$. Show that the Dirac Lagrangian can be written as

$$ \mathcal{L} = \overline{\Psi_L} i \gamma^\mu \partial_\mu \Psi_L + \overline{\Psi_R} i \gamma^\mu \partial_\mu \Psi_R - m (\overline{\Psi_L} \Psi_R + \overline{\Psi_R} \Psi_L) \quad (70) $$

Hint: show first that $(\overline{\Psi_L}) = \overline{\Psi} R$ which we could write as $(\overline{\Psi})_R$, and use that $R+L = 1 = R^2+L^2$.

The chiral, Weyl representation of the Dirac matrices is useful but is not unique. For instance we can obtain a different spinor representation $\Psi' = U\Psi$ straightforwardly and transform the Dirac matrices as $(\gamma^\mu)' = U\gamma^\mu U^{-1}$ accordingly. In this global transformation $\partial_\mu$ is unaffected. If the transformation is unitary ($U^\dagger U = 1$) the Dirac Lagrangian is invariant, as well as all properties above of the Dirac matrices, such as the anti-commutator of $\gamma^\mu$ in Eq. (65). In particular, any such new representation must be also a Lorentz-invariant representation and the new Dirac matrices transform again as four-vectors $(S')^{-1}(\gamma')^\mu S' = \Delta_\nu^\mu(\gamma')^\nu$. Physically, different representations may correspond to different particles.

### Exercise V.13: Dirac and Majorana representation

* Dirac representation associated with electron-positron: Show that the transformation

$$ U = \frac{1}{\sqrt{2}} [1 - \gamma^5(\text{Weyl})\gamma^0(\text{Weyl})] $$

is unitary. Show that this transformation leaves $\gamma^i$ unaffected but that $U\gamma^0(\text{Weyl})U^\dagger =$
$-\gamma^5(\text{Weyl})$ and $U\gamma^5(\text{Weyl})U^\dagger = \gamma^0(\text{Weyl})$. Thus, the Dirac representation has

$$ \gamma^0 = \begin{pmatrix} \sigma^0 & 0 \\ 0 & -\sigma^0 \end{pmatrix}; \quad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ \bar{\sigma}^i & 0 \end{pmatrix}; \quad \gamma^5 = \begin{pmatrix} 0 & \sigma^0 \\ \sigma^0 & 0 \end{pmatrix} \qquad (71) $$

* Marjorana representation with particle equal to its anti-particle : Consider

$$ U = -\frac{1}{\sqrt{2}}\gamma^5(\text{Weyl})[1 + \gamma^2(\text{Weyl})] $$

Show that $U = U^\dagger = U^{-1}$ and verify that $U$ transforms the Weyl representation into

$$ \gamma^0 = \begin{pmatrix} 0 & \sigma^2 \\ \sigma^2 & 0 \end{pmatrix}, \quad \gamma^1 = \begin{pmatrix} i\sigma^3 & 0 \\ 0 & i\sigma^3 \end{pmatrix}, $$

$$ \gamma^2 = \begin{pmatrix} 0 & -\sigma^2 \\ \sigma^2 & 0 \end{pmatrix}, \quad \gamma^3 = \begin{pmatrix} -i\sigma^1 & 0 \\ 0 & -i\sigma^1 \end{pmatrix} \qquad (72) $$

In this representation all $i\gamma^\mu$ are all real-valued and therefore so is Dirac's wave equation, as is the Klein-Gordon equation and the Maxwell equations, with real-valued solutions: $\Psi = \Psi^*$. By the built-in Lorentz-invariance of the Dirac equation, the real-valued representation must also be Lorentz-invariant and cannot be an artifact of a special frame. This will not be shown here.

### Example/Exercise V.14: C, P and T symmetry and CPT theorem

* Show that the relations $(\gamma^0)^\dagger = \gamma^0$, $(\gamma^i)^\dagger = -\gamma^i$ and $\gamma^0\gamma^\mu\gamma^0 = (\gamma^\mu)^\dagger$ hold in the Weyl representation and are valid for all representations. Show that in the Weyl representation $\gamma^2\gamma^\mu\gamma^2 = (\gamma^\mu)^*$ or equivalently, $\gamma^\mu\gamma^2 = -\gamma^2(\gamma^\mu)^*$ because $(\gamma^2)^2 = -1$. Similarly, show that with $T = \gamma^1\gamma^3$, $T\gamma^\mu T = -(\gamma^\mu)^T$ or $\gamma^\mu T = T(\gamma^\mu)^T$ because $T^2 = -1$. These last two last identities are not valid for arbitrary representations.

* Given a scalar product $(\Psi, \Phi)$ defined on a Hilbert space, the adjoint operator $A^\dagger$ of an operator $A$ is defined as $(A\Psi, \Phi) = (\Psi, A^\dagger\Phi)$. An anti-unitary operator is defined as $(U_a\Psi, U_a\Phi) = (\Phi, \Psi) = (\Psi, \Phi)^*$. It can always be written as $U_a = UK$ with $U$ unitary $(U^\dagger U = 1)$ and $K$ complex conjugation: $KAK = A^*$ hence in particular $K^2 = 1$ and $KiK = -i$. Show that $U_a^2$ is a unitary operator and that,

$$ (U_a\Psi, AU_a\Phi) = (\Phi, (U^\dagger A^\dagger U)^*\Psi) $$
* It is convenient to redefine the scalar product for spinors as $(\Psi, \Phi) := \bar{\Psi}\Phi = \Psi^\dagger\gamma^0\Phi$. Adjoint operators are denoted as $A^\dagger$, to be distinguished from the traditional adjoint $A^\dagger$ with respect to 4D scalar product of spinors. Show that in general $A^\dagger = \gamma^0 A^\dagger \gamma^0$. Show that the new scalar product is not positive definite. With $\mathcal{L} = \bar{\Psi}D\Psi$, $D = i\gamma^\mu\partial_\mu - m$, verify that the Dirac operator $D = D^\dagger$ is hermitian and that the Lorentz generator $S(\Delta)^{-1} = S(\Delta)^\dagger$ is unitary. In particular the Dirac matrices $\gamma^\mu$ are hermitian with respect to the new scalar product in any representation whereas the Dirac matrix $\gamma^5$ is anti-hermitian: $(\gamma^5)^\dagger = -\gamma^5$. This implies in particular that the Dirac norm $\bar{\Psi}\Psi$ is not positive definite which poses a problem to interpret the wave function as a particle.

* Let us anticipate the coupling of the Dirac field to the electromagnetic gauge field via the minimal coupling $i\partial_\mu \rightarrow i\partial_\mu - qA_\mu$ with $A_\mu = (\phi, -\mathbf{A})$. Thus, $\Psi(t, \mathbf{x})$ obeys

$$[\gamma^\mu (i\partial_\mu - qA_\mu(t, \mathbf{x})) - m] \Psi(t, \mathbf{x}) = 0 \quad (73)$$

Under time-reversal $\phi(t, \mathbf{x}) \rightarrow \phi(-t, \mathbf{x})$, $A(t, \mathbf{x}) \rightarrow -A(-t, \mathbf{x})$ and under parity $\phi(t, \mathbf{x}) \rightarrow \phi(t, -\mathbf{x})$, $A(t, \mathbf{x}) \rightarrow -A(t, -\mathbf{x})$ hence $PTA_\mu(x) = A_\mu(-x) = A_\mu(x)$. Under charge conjugation we have $q \rightarrow -q$ and $A_\mu(t, \mathbf{x}) \rightarrow -A_\mu(t, -\mathbf{x})$, so that $(CPT)qA_\mu = (C)qA_\mu = qA_\mu$. But the change of sign in $A_\mu$ is only true if we would conjugate all charges, not only the fermion in the Dirac equation. It is also essential to recall that $A_\mu$ is a real-valued vector field.

Show that in any representation the Dirac matrix $\gamma^0$ acts as parity operator $P$, with the operator $P = \gamma^0$ satisfying $P^2 = 1$, $P^\dagger = P^\dagger = P$), i.e. the wave function $\Psi'(t, \mathbf{x}) = P\Psi(t, -\mathbf{x})$ satisfies the Dirac equation. Hint: show that this would lead to

$$[\gamma^\mu (i\partial_\mu^P - qA_\mu(t, -\mathbf{x}) - m)] \Psi(t, -\mathbf{x}) = 0$$

and equal to zero because $\Psi$ obeys the Dirac equation with the parity transformation $\partial_\mu^P = (\partial_0, -\partial/\partial x^i)$ and $A_\mu(t, -\mathbf{x}) = (A_0, +\mathbf{A})$. Show that $\bar{\Psi}\gamma^5\Psi$ is a Lorentz-invariant pseudo-scalar and transforms oddly under parity.

* We define the operator $C = \gamma^2$ with the properties $C^2 = -1$, $C = -C^*$, $CC^\dagger = 1$ and $C^\dagger = C$ so that $CC^\dagger = -1$. Show that the transformed Weyl field $\Psi'(x) = CK\Psi(x)$ satisfies the Dirac equation with opposite charge $-q$.

$$[\gamma^\mu (i\partial_\mu - (-q)A_\mu(t, \mathbf{x}) - m)] \Psi'(t, \mathbf{x}) = 0$$
This implies that CK performs charge conjugation of the particle. Hint: insert $\Psi'(x, q)$ into the Dirac equation with opposite charge and use $\gamma^\mu\gamma^2 K = -\gamma^2 K\gamma^\mu$. If on the other hand we conjugate all charges then $A_\mu \to -A_\mu$ which corresponds to $\Psi'(x) = \Psi(x)$.

* We define the operator $T = i\gamma^3\gamma^1$ with properties $T^2 = 1$, $T = T^*$, $T^\dagger = T^*\gamma^1 = T$, $TT^*\gamma^1 = 1$. Show that the transformed Weyl field $\Psi'(t, \mathbf{x}) = TK\Psi(-t, \mathbf{x})$ satisfies the Dirac equation

$$[\gamma^\mu (i\partial_\mu - qA_\mu(t, \mathbf{x})) - m]\Psi'(t, \mathbf{x}) = 0$$

The operator TK performs time-reversal conjugation of the particle. Hint: insert $\Psi'(x)$ into the Dirac equation and use $\gamma^\mu TK = TK(\gamma^\mu)^\dagger$, as well as $A_\mu(-t, \mathbf{x}) = (A_0, +A^i)$.

* Show that the joint action of the 3 symmetries C, P and T is $CPT = \gamma^5$ and thus a Lorentz-invariant property. We have $(CPT)\Psi(x, q) = \gamma^5\Psi(-x, -q)$. We can investigate the Lagrangian is under CPT, using $(\gamma^5)^\dagger = \gamma^5$ and $(\gamma^5)^2 = 1$,

$$ \begin{aligned} \mathcal{L}'(x, q) &= \overline{\gamma^5 \Psi(-x)} [\gamma^\mu (i\partial_\mu - (-q)A_\mu(x)) - m] \gamma^5 \Psi(-x) \\ &= \overline{\Psi(-x)} (-\gamma^5)^\dagger [\gamma^\mu (i\partial_\mu - (-q)A_\mu(x)) - m] \gamma^5 \Psi(-x) \\ &= -\overline{\Psi(-x)} [-\gamma^\mu (i\partial_\mu - (-q)A_\mu(x)) - m] (\gamma^5)^2 \Psi(-x) \\ &= -\overline{\Psi(-x)} [\gamma^\mu (i\partial_\mu - qA_\mu(-x)) - m] \Psi(-x) \\ &= -\mathcal{L}(-x, q) \end{aligned} \qquad (74) $$

In the action we can change $-x \to x$. The minus sign is curious but does not affect the Euler-Lagrange equations. In general, the CPT theorem states that any wave equation with Lorentz symmetry is invariant under simultaneous C, P and T operations.

* Show that $\int d^4x \overline{\Psi}\gamma^5\Psi$ is parity-odd, i.e. is a pseudo-scalar and that $\int d^4x \overline{\Psi}\gamma^\mu\partial_\mu\Psi$ is parity even, and thus a scalar. Show that $\int d^4x \overline{\Psi}\gamma^\mu\gamma^5\Psi$ is parity-odd for $\mu = 0$ and parity-even for $\mu = 1, 2, 4$, i.e. is an axial four-vector. Show that $\int d^4x \overline{\Psi_L}\gamma^\mu\partial_\mu\Psi_L$ transforms under parity to $\int d^4x \overline{\Psi_R}\gamma^\mu\partial_\mu\Psi_R$ (use that $\gamma^0(L, R) = (R, L)\gamma^0$). Since L, R are Lorentz-invariant operators we conclude that the restriction to a single chirality preserves Lorentz-invariance but violates parity. This is used in the gauge theory of the weak nuclear force, that is known to break parity. However, a finite mass would couple the two chiralities by Lorentz-symmetry and destroy chiral symmetry. This is solved by the Higgs mechanism.
### Exercise V.15: $U(1)$ vector and axial symmetry

* In the chiral representation the kinetic Lagrangian (i.e. mass $m = 0$) is not only invariant under the global (“vector”) $U(1)$ transformation $\Psi \to \exp(-i\theta)\Psi$ but also under separate $U(1)$ transformations of both chiralities: $\Psi_R \to \exp(-i\theta_R)\Psi_R$ and $\Psi_L \to \exp(-i\theta_L)\Psi_L$. Using the anti-commutations of $\gamma^5$, show that this is equivalent to the two transformations

$$ \text{vector } U_V: \Psi \to e^{-i(\theta_L + \theta_R)/2}\Psi \qquad \bar{\Psi} \to \bar{\Psi}e^{i(\theta_L + \theta_R)/2} $$

$$ \text{axial } U_A: \Psi \to e^{-i\gamma^5(\theta_L - \theta_R)/2}\Psi \qquad \bar{\Psi} \to \bar{\Psi}e^{-i\gamma^5(\theta_L - \theta_R)/2} $$

* Show that the mass term obeys vector symmetry but violates axial symmetry.

### Exercise V.16/Example : Solutions of the free Dirac equation & anti-particles

* For the Free Dirac equation we expect plane wave solutions of the form $\Psi(k, x) = u_k \exp(-ik_\mu x^\mu)$ with $u_k = (\xi_k, \eta_k)$ and $k_\mu = (E/c_0, -\mathbf{k})$. Show that in the Dirac representation this solution must obey

$$ \begin{pmatrix} E/c_0 - mc_0 & -\boldsymbol{\sigma} \cdot \mathbf{k} \\ \boldsymbol{\sigma} \cdot \mathbf{k} & -E/c_0 - mc_0 \end{pmatrix} \begin{pmatrix} \xi_k \\ \eta_k \end{pmatrix} = 0 $$

so that $E = \pm E_k = \pm \sqrt{k^2 c_0^2 + (mc_0^2)^2}$.

* Show that each energy is doubly generated and that the eigenfunctions are given by the spinors,

$$ u_{ks} = N_k \begin{pmatrix} \hat{e}_{ks} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{k}}{E/c_0 + mc_0} \hat{e}_{ks} \end{pmatrix} = N_k \begin{pmatrix} \hat{e}_{ks} \\ \frac{sk\hat{e}_{ks}}{E/c_0 + mc_0} \end{pmatrix} $$

for $E > 0$, $s = \pm 1$ and

$$ u_{ks} = N_k \begin{pmatrix} \frac{\boldsymbol{\sigma} \cdot \mathbf{k}}{E/c_0 - mc_0} \hat{e}_{ks} \\ \hat{e}_{ks} \end{pmatrix} $$

for $E < 0$, $s = \pm 1$. The vectors $\hat{e}_{ks}$ are the two orthogonal 2D eigenvectors of the helicity operator $\boldsymbol{\sigma} \cdot \hat{\mathbf{k}}$ with eigenvalue $s$.
* The solutions for negative energy, when also considered at opposite momentum $-k_i$ can alternatively be considered as (the complex-conjugated) solutions $\exp(i(-E/c_0)t + i(-\mathbf{k}) \cdot \mathbf{x})$ at positive energy traveling backwards. Show that these can be written as $\Psi(k, x) = v_{ks} \exp(+ik_\mu x^\mu)$ with

$$v_{ks} = N_k \left( \frac{\frac{\sigma \cdot \mathbf{k}}{|E|/c_0 + mc_0} \hat{e}(-\mathbf{k})s}{\hat{e}(-\mathbf{k})s} \right) = N_k \left( \frac{-sk\hat{e}_{\mathbf{k}(-s)}}{|E|/c_0 + mc_0} \hat{e}_{\mathbf{k}(-s)} \right)$$

Negative energy particle solutions, going backward in time are equivalent to positive energy anti-particle solutions propagating forward in time with opposite helicity. It makes therefore sense to convert the negative energy solutions into anti-particles with positive energy and to regroup them with same helicity meaning that the vectors $\hat{e}_{ks}$ in $u_{ks}$ and $v_{ks}$ must be orthogonal.

* Show that $\bar{u}_{ks}u_{ks'} = 2mc_0\delta_{ss'}$, $\bar{v}_{ks}v_{ks'} = -2mc_0\delta_{ss'}$ and $\bar{u}_{ks}v_{ks'} = 0$ if we choose $N_k = \sqrt{E_k/c_0 + mc_0}$. Show that with this choice the hermitian relations are $u_{ks}^\dagger u_{ks'} = v_{ks}^\dagger v_{ks'} = 2E_k/c_0\delta_{ss'}$. However, their cross-term only vanishes for opposite momenta: $u_{ks}^\dagger v_{-ks'} = 0$ (with $\hat{e}_{-ks}\hat{e}_{\mathbf{k}(-s)}$).

* Using that $\sum_s \hat{e}_{ks}\hat{e}_{ks}^\dagger = I_2$ and $\sum_s s \times \hat{e}_{ks}\hat{e}_{ks}^\dagger = \sigma \cdot \hat{\mathbf{k}}$ show that

$$\sum_s u_{ks} \bar{u}_{ks} = \hbar \not{k} + mc_0 \quad ; \quad \sum_s v_{ks} \bar{v}_{ks} = \hbar \not{k} - mc_0 \qquad (75)$$

where we recall that $\bar{u}_{ks} = u_{ks}^\dagger \gamma^0$ and we used the popular Feynman slash notation $\gamma^\mu k_\mu := \not{k}$. We can check that these completeness relations are compatible with $(\hbar \not{k} - mc_0)u_{ks} = 0$ and $(\hbar \not{k} + mc_0)v_{ks} = 0$ because $(\hbar \not{k} - mc_0)(\hbar \not{k} + mc_0) = \hbar^2 k^2 - m^2 c_0^2 = 0$.

This analysis implies that the wave function

$$\Psi_{ks} = \alpha_{ks} u_{ks} e^{-ik_{\mu} x^{\mu}} + \beta_{ks}^{*} v_{ks} e^{+ik_{\mu} x^{\mu}}$$

is the general solution of the Dirac equation for given $k^\mu$, helicity $s$ and energy $E_k > 0$. Thus in real space,

$$\Psi(x) = \sum_{ks} \frac{1}{\sqrt{2E_k}} \left[ \alpha_{ks} u_{ks} e^{-ik_{\mu} x^{\mu}} + \beta_{ks}^{*} v_{ks} e^{+ik_{\mu} x^{\mu}} \right] \qquad (76)$$

The Fourier coefficient $\beta_{ks}^*$ is taken complex-conjugated because so is the plane wave. The extra factor $1/\sqrt{2E_k}$ is historically included to guarantee a Lorentz-invariant volume element $d^3\mathbf{k}/(2E_k)$
<figure><img src="image_9.png" /><figcaption>Figure 7. The solutions of the Dirac equation with negative energy are interpreted as anti-particles with positive energy and opposite momentum and spin.</figcaption></figure>

later when we calculate charge densities and compensates the hermitian norm $\sqrt{2E_k}$ given earlier to $u_k$ and $v_k$.

### Exercise V.17: Zitterbewegung

The equation of motion (67) for a Dirac field can be cast into a time-evolution equation

$$i\hbar\partial_t\Psi = H\Psi \qquad (77)$$

with $H = \gamma^0mc_0^2 + \alpha_i p_i c_0$ with $\alpha^i = \gamma^0\gamma^i$.

* Show that the expectation value of the velocity is given by

$$\frac{d}{dt}\langle\Psi|x_k|\Psi\rangle = \langle\Psi|\alpha^k c_0|\Psi\rangle$$

* Show that $[H, \alpha^k] = 2p_k - \alpha_k H$.

* Argue or calculate that the free electron oscillates at a frequency $\omega$ given by $\hbar\omega = 2mc_0^2$. This corresponds to a wavelength of 0.012 Angström, i.e. in the hard X-ray regime. This oscillation is a manifestation of the fact that boots and rotations do not commute in the Lorentz group. Zitterbewegung was predicted by Schrödinger himself in 1930. It can also be regarded as a Rabi oscillation known for two-level systems and defined here by the positive and negative energy solutions, separated by a gap $2mc_0^2$ at $k = 0$. Its existence was recently claimed in Bose-Einstein condensates in which the Dirac Hamiltonian was simulated using hyperfine atomic transitions coupled to their linear momentum. [10]