# Quantum field theory 1, lecture 17

### 9 Lorentz symmetry and the Dirac equation

Symmetries are basic concepts for the construction of a model. Particle physics in flat Minkowski space is invariant under Lorentz transformations. Even though the cosmological solutions are not Lorentz invariant, Lorentz invariance holds to a very good approximation on length and time scales that are small compared to the “size” (inverse Hubble parameter) of the universe. The functional integral formulation makes the implementation of symmetries easy. One imposes that the action $$S$$ is invariant under the symmetry transformations. This is sufficient if the functional measure is also invariant. All symmetry properties follow from the invariance of $$S$$ and the functional measure.

A given model is specified by the action $$S$$ that appears in the functional integral. Symmetries restrict the possible form that the action can take. Lorentz symmetry is therefore an important guiding principle for establishing possible models for particle physics. Together with other symmetries as gauge symmetries and the requirement that only a given finite number of derivatives of the fields can appear – this is related to renormalizability – the form of the action is often uniquely determined once a given set of fields is chosen. Because of this crucial importance we will discuss the Lorentz transformations in some detail.

The discussion of Lorentz transformations also introduces in a practical way some elements of the theory of Lie groups that are useful for the present lecture. A more profound knowledge of Lie groups is useful, but not required for the present lecture.

#### 9.1 Lorentz transformations and invariant tensors

Lorentz metric. The cartesian coordinates of space and time are $$t$$ and $$\bf x$$. They are denoted as the contravariant vector \begin{equation*} x^\mu = (t,\textbf{x}), \quad \quad \quad t= x^0. \end{equation*} The corresponding covariant vector is \begin{equation*} x_\mu = (-t,\textbf{x}) = (-x^0,\textbf{x}). \end{equation*} We can lower and raise indices with the metric tensor $$\eta _{\mu \nu }$$ and its inverse $$\eta ^{\mu \nu }$$, which have the same matrix representation, \begin{equation*} \eta _{\mu \nu } = \eta ^{\mu \nu } = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \\ \end{pmatrix}. \end{equation*} As raising and lowering are inverse operations, the multiplication of both matrices is the identity, \begin{equation*} \eta ^{\mu \nu }\eta _{\nu \rho } = \delta ^\mu _\rho . \end{equation*} The sign of the eigenvalues of a metric are called the ”signature of the metric“. In our case the signature is $$(-,+,+,+)$$. The explicit transformation rules are given by \begin{equation*} x_\mu = \eta _{\mu \nu }x^\nu \quad \quad \quad \text{and} \quad \quad \quad x^\mu = \eta ^{\mu \nu }x_\nu . \end{equation*}

Lorentz transformations. From the contravariant and covariant four-vectors one can form the scalar product $$x^\mu x_\mu = -t^2+\vec{x}^2. \label{eq:NN1}$$ The Lorentz transformations can be defined as those linear transformations of $$x^\mu$$ that leave $$x^\mu x_\mu$$ invariant. In other words, for the transformation $$x^\mu \to x'^\mu = \Lambda ^\mu _{\;\;\nu }x^\nu \label{eq:NN2}$$ we want to find the matrices $$\Lambda ^\mu _{\;\;\nu }$$ that leave $$x^\mu x_\mu$$ invariant. We require that the expression \begin{equation*} x'^\mu x'_\mu = x'^\mu x'^\nu \eta _{\mu \nu } = \Lambda ^\mu _{\,\,\rho } x^\rho \Lambda ^\nu _{\;\;\sigma }x^\sigma \eta _{\mu \nu } \end{equation*} is equal to $$x^\mu x_\mu$$. This results in the condition $$\Lambda ^\mu _{\;\;\rho }\Lambda ^\nu _{\;\;\sigma }\eta _{\mu \nu } = \eta _{\rho \sigma }. \label{eq:defLorentzTransfo}$$

Equation \eqref{eq:defLorentzTransfo} is the defining equation for $$\Lambda$$. All transformations that fulfill \eqref{eq:defLorentzTransfo} are called Lorentz transformations. A sequence of two Lorentz transformations leaves $$x^\mu x_\mu$$ invariant and is therefore again a Lorentz transformation. The inverse Lorentz transformation as well as the identity transformation exists. The Lorentz transformations form a group, the ”Lorentz group“, which is often denoted by SO(1,3). The group elements are the matrices $$\Lambda ^\mu _{\;\;\nu }$$.

So-called proper, orthochronous Lorentz transformations can be obtained as a sequence of infinitesimal transformations. Particle physics is invariant under the proper orthochronous Lorentz transformations for which we often use the shorthand “Lorentz transformations”. The general transformations \eqref{eq:NN2}, which we will often call “extended Lorentz transformations”, comprise discrete transformations like parity and time reversal. Particle physics is not invariant under those discrete transformations.

Transformation of tensors. Let us consider the contravariant and covariant four-momenta \begin{equation*} p^\mu = (E,\textbf{p}) \end{equation*} \begin{equation*} p_\mu = (-E,\textbf{p}) \end{equation*} As we already discussed, we can raise and lower indices of vectors with the metric tensor $$\eta _{\mu \nu }$$ and the inverse $$\eta ^{\mu \nu }$$. The Lorentz transformation of all contravariant vectors is defined as \begin{equation*} p'^\mu = \Lambda ^\mu _{\;\;\nu }p^\nu . \end{equation*} The corresponding transformation rule for covariant vectors can be found by expressing both sides in terms of covariant vectors \begin{equation*} \eta ^{\mu \rho }p'_\rho = \Lambda ^\mu _{\;\;\nu }\eta ^{\nu \sigma }p_\sigma . \end{equation*} Multiplication with an inverse metric yields \begin{equation*} p'_\kappa = \eta _{\kappa \mu }\Lambda ^\mu _{\;\;\nu }\eta ^{\nu \sigma }p_\sigma , \end{equation*} where we recognize a Lorentz matrix with different index positions, $$\Lambda ^{\;\;\sigma }_\kappa = \eta _{\kappa \mu }\Lambda ^\mu _{\;\;\nu } \eta ^{\nu \sigma }. \label{eq:9.2}$$ In consequence, covariant vectors transform as \begin{equation*} p'_\mu = \Lambda _\mu ^{\;\;\nu } p_\nu . \end{equation*}
A contravariant second rank tensor $$T^{\mu \nu }$$ is an object with two upper indices. It has the same transformation property as a product of two contravariant vectors $$T'^{\mu \nu }=\Lambda ^\mu _{\;\;\rho }\Lambda ^\nu _{\;\;\sigma }T^{\mu \nu }. \label{eq:NN3}$$ An example is the energy-momentum tensor that plays an important role in gravity. The relation \eqref{eq:NN3} ensures that product relations as $$T^{\mu \nu }=a^\mu b^\nu \label{eq:NN4}$$ are ”covariant“. This means that the same product relation holds after the Lorentz transformation $$T'^{\mu \nu }=a'^\mu b'^\nu . \label{eq:NN5}$$ This generalises to tensors with an arbitrary number of upper and lower indices. An example for the Lorentz transformation of a more complicated tensor is \begin{equation*} A'^{\mu \nu \rho }_{\;\;\;\;\;\;\sigma \tau } = \Lambda ^\mu _{\;\;\mu '}\Lambda ^\nu _{\;\;\nu '}\Lambda ^\rho _{\;\;\rho '}\Lambda ^{\;\;\sigma '}_\sigma \Lambda ^{\;\;\tau '}_\tau A^{\mu '\nu '\rho '}_{\;\;\;\;\;\;\;\;\sigma '\tau '}. \end{equation*}

Contractions. The product of a covariant and a contravariant vector is a scalar: It is invariant under Lorentz transformations, \begin{equation*} \begin{split} s &= a^\mu b_\mu ,\\ \Rightarrow ~s' &= \Lambda ^\mu _{~\rho } a^\rho \Lambda _\mu ^{~\sigma }b_\sigma = a^\rho \underbrace{\Lambda ^\mu _{~\rho } \eta _{\mu \nu } \Lambda _{~\tau }^{\nu }}_{\eta _{\rho \tau }} \eta ^{\tau \sigma }b_\sigma = a^\rho b_\rho = s. \end{split} \end{equation*} The summation over an upper and lower index is called an ”index contraction“. Contracted Lorentz indices do not contribute to transformations: \begin{equation*} \begin{split} (A')_{\mu \rho } (B')^{\rho \nu } &= \Lambda _\mu ^{\;\;\sigma } \Lambda ^\nu _{\;\;\tau } A_{\sigma \rho } B^{\rho \tau }, \end{split} \end{equation*} such that $$C_\mu ^{\;\;\nu } = A_{\mu \rho } B^{\rho \nu }$$ transforms as a 2-tensor. Similar to matrices, the order of indices matters for all tensors. For example, one has $$C^\nu _{\;\;\mu }=\eta ^{\nu \rho } C_\rho ^{\;\;\sigma }\eta _{\sigma \mu }, \label{eq:NN7}$$ which differs from $$C_\mu ^{\;\;\nu }$$.

Invariant tensors. The metric $$\eta _{\mu \nu }$$ is a symmetric invariant tensor under Lorentz transformations $$\eta '_{\mu \nu }=\Lambda _\mu ^{\;\;\rho }\Lambda _\nu ^{\;\;\sigma }\eta _{\rho \sigma }=\eta _{\mu \nu }. \label{eq:NN8}$$ This follows from the defining relation \eqref{eq:defLorentzTransfo} for the Lorentz transformations, that we can write in the form $$\Lambda _{\nu \rho }\Lambda ^{\nu \sigma }=(\Lambda ^T)_{\rho \nu }\Lambda ^{\nu \sigma }=(\Lambda ^T)_{\rho }^{\;\;\nu }\Lambda _{\nu }^{\;\;\sigma }=\delta _\rho ^\sigma , \label{eq:NN9}$$ where the transposed Lorentz matrix $$\Lambda ^T$$ obeys $$(\Lambda ^T)_{\rho \nu }=\Lambda _{\nu \rho }, \quad (\Lambda ^T)^{\rho \nu }=\Lambda ^{\nu \rho }. \label{eq:NN10}$$ We can identify $$(\Lambda ^T)_\rho ^{\;\;\sigma }$$ as the inverse matrix of the matrix $$\Lambda _\rho ^{\;\;\sigma }$$. With $$\Lambda ^T \Lambda =1$$ implying $$\Lambda \Lambda ^T=1$$ we also have the relation $$\Lambda _\rho ^{\;\;\nu }(\Lambda ^T)_{\nu }^{\;\;\sigma }=\delta _\rho ^\sigma , \label{eq:NN11}$$ which is equivalent to eq. \eqref{eq:NN8}. We can equivalently use eq. \eqref{eq:NN8} as the defining relation for the Lorentz transformations. From the standpoint of group theory this is a more natural definition since the group elements are defined as transformation matrices that leave the particular tensor $$\eta _{\mu \nu }$$ invariant. The naming SO(1,3) refers to the signature of $$\eta _{\mu \nu }$$. It is straightforward to see that $$\eta ^{\mu \nu }$$ and $$\delta _\mu ^\nu$$ are also invariant tensors.

There is only one more tensor that is invariant under Lorentz transformations. This is the totally antisymmetric tensor $$\epsilon _{\mu \nu \rho \sigma }$$, the relativistic generalization of the Levi-Civita tensor $$\epsilon _{ijk}$$. The Levi-Civita symbol with four indices $$\epsilon _{\mu \nu \rho \sigma }$$ is defined by total antisymmetry and \begin{equation*} \epsilon _{0123}= 1. \end{equation*} It equals $$1$$ for all cyclic permutations of $$(0,1,2,3)$$, and $$-1$$ for all anti-cyclic permutations. The $$\epsilon$$-tensor with raised indices, $$\epsilon ^{\mu \nu \rho \sigma }$$ has the opposite signs, e. g. $$\epsilon ^{0123} = -1$$.

Let us prove that $$\epsilon _{\mu \nu \rho \sigma }$$ is invariant under Lorentz transformations. In the following lines we will use the short hand notation $$\Lambda _\mu ^{\;\;\nu } \to \Lambda ,$$ and $$\eta _{\mu \nu } \to \eta$$. With this notation, the defining relation (??) reads $$\Lambda \eta \Lambda ^T = \eta , \label{eq:9.3}$$ If we compute the determinant on both sides, we find, using $$\text{det}(\Lambda )\text{det}(\Lambda ^T)=(\text{det}(\Lambda ))^2$$, \begin{equation*} \text{det}(\Lambda ) = \pm 1. \end{equation*} The determinant of $$\Lambda$$ can also be calculated by \begin{equation*} \text{det}(\Lambda ) = \frac{1}{4!} \Lambda _{\mu _1}^{\;\;\nu _1}\Lambda _{\mu _2}^{\;\;\nu _2}\Lambda _{\mu _3}^{\;\;\nu _3}\Lambda _{\mu _4}^{\;\;\nu _4} \epsilon _{\nu _1\nu _2\nu _3\nu _4}\epsilon ^{\mu _1\mu _2\mu _3\mu _4} =\frac{1}{4!}\epsilon '_{\mu _1\mu _2\mu _3\mu _4}\epsilon ^{\mu _1\mu _2\mu _3\mu _4}. \end{equation*} Here $$\epsilon '$$ is the Lorentz transformed tensor. We can verify that $$\epsilon '_{\mu \nu \rho \sigma }$$ is totally antisymmetric, thus $$\epsilon '_{\mu \nu \rho \sigma } = c \, \epsilon _{\mu \nu \rho \sigma }$$ with constant $$c$$. Using $$\epsilon _{\mu \nu \rho \sigma }\epsilon ^{\mu \nu \rho \sigma }$$ = 4! we obtain $$\det (\Lambda ) = c$$ or \begin{equation*} \epsilon '_{\mu _1\mu _2\mu _3\mu _4}= \text{det}(\Lambda )\epsilon _{\mu \nu \rho \sigma } = \pm \epsilon _{\mu _1\mu _2\mu _3\mu _4}. \end{equation*} Only Lorentz transformations with $$\text{det}(\Lambda ) = +1$$ will leave the $$\epsilon$$-tensor invariant (they are called proper Lorentz transformations). The Lorentz transformations which are continuously related to the unit transformation obey $$\det (\Lambda ) = 1$$.

Analogy to Rotations. Equation \eqref{eq:9.3} looks very similar to orthogonal transformations $$O_{\mu }^{\;\;\nu }$$ with \begin{equation*} O \mathbb{1} O^T = O O^T = \mathbb{1},\qquad \mathbb{1}_{\mu \nu } = \delta _{\mu \nu }, \end{equation*} where $$\mathbb{1}$$ is the unit matrix. The invariant tensor of the orthogonal transformations is the ”euclidean metric” $$\delta _{\mu \nu }$$. In eq. \eqref{eq:9.3} the euclidean metric is replaced by the metric tensor $$\eta _{\mu \nu }$$ for Minkowski space. In short,

• Orthogonal transformations : $$\delta _{\mu \nu }$$ invariant.
• Lorentz transformation: $$\eta _{\mu \nu }$$ invariant.
• Analytic continuation: $$\delta _{\mu \nu } \to \eta _{\mu \nu }$$.

The group of orthogonal transformations in four dimensions is denoted O$$(4)$$. The analogy that we just discussed motivates the name Pseudo orthogonal transformations O$$(1,3)$$ where the separated $$1$$ indicates the special role of time in special relativity. The ”special orthogonal transformations“ SO(4) have in addition the property $$\text{det}(\Lambda )=1$$. Similarly, for SO(1,3) one has $$\text{det}(\Lambda )=1$$.

Derivatives. The derivative with respect to a contravariant vector is a covariant vector, \begin{equation*} \partial _\mu = \frac{\partial }{\partial x^\mu }. \end{equation*} We will discuss later the Lorentz - transformations of fields more carefully. For the moment we just generalize what we know from three-dimensional rotations, namely that the divergence of a vector field transforms as a scalar field. With $$\partial _\mu a^\mu (x)$$ transforming as a scalar field, $$\partial _\mu$$ has to compensate the transformation acting on the indices $$a^\mu$$. For example one has \begin{equation*} \partial _\mu x^\mu = 4. \end{equation*} The momentum operator is \begin{equation*} \hat{p}_\mu = -i\partial _\mu . \end{equation*} It transforms as a covariant vector.

Four-dimensional Fourier transformation. The four-dimensional Fourier transformation of a function $$\psi (x)$$ is defined as \begin{equation*} \psi (x) = \int _p e^{i p_\mu x^\mu } \psi (p). \end{equation*} With $$p_\mu = (-\omega , \vec{p})$$ and $$p_\mu x^\mu =-\omega t + \vec{p}\vec{x}$$ this reads \begin{equation*} \psi (t,\vec{x}) = \int _\omega \int _{\vec{p}} e^{-i\omega t + i \vec{p}\vec{x}} \psi (\omega ,\vec{p}). \end{equation*} Note that $$p_\mu x^\mu$$ is Lorentz invariant.

Covariant equations. For a covariant equation the left hand side and right hand side have the same transformation properties. An example is \begin{equation*} \partial _\mu F^{\mu \nu } = J^\nu . \end{equation*} These are two of the four Maxwell equations. The other two are $$\partial _\mu \epsilon ^{\mu \nu \rho \sigma }F_{\rho \sigma }=0. \label{eq:NN12}$$ Thus the Maxwell equations are Lorentz-covariant. We will later derive them from a Lorentz-invariant action.

#### 9.2 Lorentz group

Group structure. If we have two elements $$g_1,g_2$$ that are elements of a group $$\mathscr{G}$$, the product of these two elements will still be an element of the group \begin{equation*} g_3=g_2 g_1 \in \mathscr{G}. \end{equation*} In particular, we can write for matrices \begin{equation*} (\Lambda _3)^\mu _{\;\;\nu } = (\Lambda _2)^\mu _{\;\;\rho }(\Lambda _1)^\rho _{\;\;\nu }. \end{equation*} A group contains always a unit element $$e$$ such that \begin{equation*} ge=eg=g \end{equation*} for every group element $$g$$. For matrices, this unit element is $$\delta ^\mu _{\;\;\nu }$$. Furthermore the inverse element $$g^{-1}$$ exists. Every matrix $$\Lambda ^\mu _{\;\nu }$$ has an inverse matrix because the determinant of $$\Lambda$$ is $$\pm 1$$. Finally, for a group the multiplication law has to be associative, which is trivial for matrix multiplications.

Discrete symmetries. The (extended) Lorentz transformations contain some discrete symmetries that we discuss next.

Space reflection (parity). The space reflection transformation changes the sign of all space coordinates, $$x^j \to - x^j$$ for $$j \in \{1,2,3\}$$ while time stays invariant $$t \to t$$. The corresponding matrix is \begin{equation*} P= \begin{pmatrix} +1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}. \end{equation*} The determinant is $$\det (P) = -1$$. The metric tensor $$\eta _{\mu \nu }$$ is kept invariant under a space reflection, $$P\eta P^T = \eta$$.

Time reflection. The time reflection transformation is $$x^j \to x^j$$ for $$j \in \{1,2,3\}$$ and $$t \to -t.$$ The corresponding matrix is \begin{equation*} T = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \\ \end{pmatrix}. \end{equation*} The determinant of $$T$$ is the same as for $$P$$, $$\det (T) = \det (P) = -1$$. Both transformations change the sign of the $$\epsilon$$-tensor and are therefore improper Lorentz transformations. Again, the metric tensor is invariant under $$T\eta T^T =\eta$$.

Space-time reflection. The combination of both space and time reflection is \begin{equation*} PT = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}. \end{equation*} This time the determinant is $$+1$$.

Continuous Lorentz Transformations. A continuous Lorentz transformation can be obtained as a product of infinitesimal transformations. We use Lorentz transformation for the continuous Lorentz transformations. Since no jumps are possible, the continuous Lorentz transformations have a determinant $$+1$$, so we can immediately conclude that the discrete transformations $$P$$ and $$T$$ can’t be described by continuous Lorentz transformations. As the product $$PT$$ has a determinant $$+1$$, one could first think that this may be obtained by continuous transformations, but this is not the case. The reason is that infinitesimal transformations will never change the sign in front of time variable, but actually, $$PT$$ does exactly this. However, a discrete transformation that can be obtained by infinitesimal ones is the reflection of $$x$$ and $$y$$. The product $$P_1 P_2$$ with \begin{equation*} P_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, \quad \quad \quad P_2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, \quad \quad \quad P_1 P_2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, \end{equation*} can be obtained as a continuous transformation, as familiar from rotations in two-dimensional space.

#### 9.3 Generators and Lorentz Algebra

Infinitesimal Lorentz Transformations. Let us consider the difference $$\delta p_\mu$$ between a four-momentum and the transformed four-momentum, \begin{equation*} \delta p_\mu = p'_\mu - p_\mu = (\Lambda _\mu ^{\;\;\nu } - \delta _\mu ^{\nu })p_\nu = \delta \Lambda _\mu ^{\;\;\nu } p_\nu , \end{equation*} with \begin{equation*} \Lambda _\mu ^{\;\;\nu }= \delta _\mu ^{\nu } + \delta \Lambda _\mu ^{\;\;\nu }. \end{equation*} In a matrix representation, the infinitesimal Lorentz transformation is given by $$\Lambda = \mathbb{1} + \delta \Lambda$$. The defining relation of a Lorentz transformation ($$\Lambda \eta \Lambda ^T = \eta$$) then leads to constraints for $$\delta \Lambda$$ as follows: \begin{equation*} \begin{split} &\Lambda \eta \Lambda ^T = \eta \\ &\Leftrightarrow (1+\delta \Lambda ) \eta (1+\delta \Lambda )^T = \eta \\ &\Leftrightarrow \delta \Lambda \; \eta + \eta \; \delta \Lambda ^T = 0. \end{split} \end{equation*} In this last line we neglected the 2nd order term in $$\delta \Lambda$$. If we write down this equation in the index notation of eq. (?? ), we have \begin{equation*} \begin{split} \delta \Lambda _{\rho }^{\;\;\mu }\eta _{\mu \sigma } + \delta \Lambda _\sigma ^{\;\;\nu }\eta _{\rho \nu } &= 0,\\ \delta \Lambda _{\rho \sigma } + \delta \Lambda _{\sigma \rho } &= 0. \end{split} \end{equation*} This equation tells us that $$\delta \Lambda _{\mu \nu }$$ is antisymmetric, while $$\delta \Lambda _\mu ^{\;\;\nu }$$ is not antisymmetric. The matrices have six independent elements, what is obvious for $$\delta \Lambda _{\mu \nu } = -\delta \Lambda _{\nu \mu }$$. The number of independent elements in an antisymmetric matrix equals the number of linearly independent antisymmetric matrices. The six matrices represent three infinitesimal rotations and three infinitesimal boosts.

Generators. Let us write the infinitesimal transformation of the momentum vector in the following way, $$\delta p_\mu = i\epsilon _z(T_z)_\mu ^{\;\;\nu } p_\nu , \quad \quad \quad z=1 \ldots 6, \label{eq:infTrafoMomentum}$$ where a sum over $$z$$ is implied. Any infinitesimal Lorentz transformation can be represented as a linear combination in this form $$\delta \Lambda _\mu ^{~ ~\nu } = i\epsilon _z (T_z)_\mu ^{~ ~\nu }. \label{eq:infLorTrafo}$$

For the six independent generators we choose $$\text{rotations}:\quad \quad (T_1)_{\mu \nu } = (T_1)_\mu ^{\;\nu } = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ \end{pmatrix}, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \label{eq:rotx}$$ \begin{equation*} (T_2)_{\mu \nu } = (T_2)_\mu ^{\;\nu } = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \\ \end{pmatrix},\quad (T_3)_{\mu \nu }=(T_3)_\mu ^{\;\nu } = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \label{eq:rotyz} \end{equation*} \begin{equation*} \text{boosts}: \quad \quad (T_4)_{\mu \nu } = \begin{pmatrix} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \quad (T_4)_\mu ^{\;\nu } = \begin{pmatrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \label{eq:boostx} \end{equation*} \begin{equation*} (T_5)_\mu ^{\;\nu } = \begin{pmatrix} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}, \quad (T_6)_\mu ^{\;\nu } = \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -i & 0 & 0 & 0 \\ \end{pmatrix}. \label{eq:boostyz} \end{equation*}

Some remarks may be useful:

• $$T_1$$is a rotation around the $$x$$-axis (only $$y$$ and $$z$$ components change). Similarly $$T_2$$ is a rotation around the $$y$$-axis and $$T_3$$ a rotation around the $$z$$-axis.
• For the rotation matrices, raising and lowering of indices doesn’t change anything. The reason is that the metric tensor has a -1 only in the zero component and the rotation matrices are zero in the first row and column.
• For the boost matrices, raising of the first index changes the sign of the first row of the matrix (see $$T_4$$). After raising the index, the boost matrices are not any longer antisymmetric. Explicitly, one has \begin{equation*} (T_4)^\mu _{\;\;\nu } = \eta ^{\mu \rho }(T_4)_{\rho \nu } = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} =-(T_4)_\mu ^{\;\nu }. \end{equation*}
• In order to see that $$T_1$$, $$T_2$$ and $$T_3$$ are the generators of rotations, we may compare them to the infinitesimal rotations in two dimensions. A general rotation matrix, \begin{equation*} R = \begin{pmatrix} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \\ \end{pmatrix}, \end{equation*} reads for an infinitesimal angle $$\phi = \epsilon$$ \begin{equation*} R= \begin{pmatrix} 1 & \epsilon \\ -\epsilon & 1 \\ \end{pmatrix}. \end{equation*} The difference to the identity is \begin{equation*} \delta R = \begin{pmatrix} 0 & \epsilon \\ -\epsilon & 0 \\ \end{pmatrix}, \end{equation*} This is equivalent to eq. \eqref{eq:infLorTrafo} if we multiply the generator \eqref{eq:rotx} with $$i\epsilon$$. The $$i$$ in the definition of the generators is chosen such that $$T_1, T_2, T_3$$ are hermitian matrices.
• The matrices $$T_4$$, $$T_5$$ and $$T_6$$ generate boosts in $$x$$, $$y$$ and $$z$$ direction.

Categories:

Updated: