User:Maschen/Symmetries in quantum mechanics

(Considering how painfully vague and unclear generators, groups, representations etc. are explained in the context of QM symmetries, as well as some important content for RQM which couldn't fit into that article, this page will be created soon, once all errors are fixed and more content added).

Symmetries in quantum mechanics are theoretically important for the mathematical structure of quantum mechanics, relativistic quantum mechanics and quantum field theory, because symmetries, invariance, and conservation laws are fundamentally important constraints for formulating physical theories and models. In practice they are also powerful methods for solving problems and predicting what could happen. While conservation laws do not always give the answer to the problem directly and alone, they form the correct constraints and the first steps to solving the problem.

This article outlines the connection between the classical form of continuous symmetries as well as their quantum operators, and relates them to the Lie groups, Lorentz group, and Poincaré group, with relativistic generalizations in the Lorentz transformation.

Overview of Lie group theory
If G is a group parametrized by real scalars ζ1, ζ2, ..., the group elements g in G are a function of the parameters:


 * $$g = G(\zeta_1, \zeta_2, \cdots )$$

All parameters set to zero returns the identity element:


 * $$I = G(0,0 \cdots )$$

A Lie group has a finite set of continuously varying parameters.

Taking derivatives of the group element with respect to the group parameters, then evaluating the result when the parameter is zero, yields the generators of the group:


 * $$X_j = \left. \frac{\partial g}{\partial \zeta_j} \right|_{\zeta_j = 0} $$

In quantum theory, the generators must be Hermitian for a unitary representation of the group, i.e. X = X†, which requires a factor of −i:


 * $$X_j = -i \left. \frac{\partial g}{\partial \zeta_j} \right|_{\zeta_j = 0} $$

The generators of the group form a vector space, i.e. linear combinations of generators also form a generator. The generators of a Lie algebra satisfy the commutator (Lie bracket?):


 * $$\left[X_a,X_b\right] = i f_{abc}X_c$$

where fabc are the structure constants of the group. Very often, this turns out to be an antisymmetric entity like the three dimensional Levi-Civita symbol εijk.

The representation of the group is denoted using a capital D and defined by:


 * $$D(\zeta_j) = e^{ i \zeta_j X_j}$$

without summation on the j index. Examples are given throughout the article.

Symmetry transformations on the wavefunction in non-relativistic quantum mechanics
Generally, the correspondence between continuous symmetries and conservation laws is given by Noether's theorem.

The form of the fundamental quantum operators becomes clear when one considers the initial state, then changes one parameter of it slightly. This can be done for displacements (lengths), durations (time), and angles (rotations). Additionally, the invariance of certain quantities can be seen by making such changes in lengths and angles, which illustrates conservation of these quantities.

In what follows, transformations on only one-particle wavefunctions in the form:


 * $$ \hat{\Omega}\psi(\mathbf{r},t) = \psi(\mathbf{r}',t) $$

where Ω denotes a Hermitian and unitary operator, the inverse is the Hermitian conjugate Ω−1 = Ω†, are considered. The results can be extended to many-particle wavefunctions. Written in Dirac notation as standard, the transformations on quantum state vectors are:


 * $$ \hat{\Omega}\left|\mathbf{r}(t)\right\rangle = \left|\mathbf{r}'(t')\right\rangle $$

Considering the action of Ω changes ψ(r, t) to ψ(r&prime;, t&prime;), the inverse Ω−1 = Ω† changes ψ(r&prime;, t&prime;) back to ψ(r, t), so an operator A invariant under Ω satisfies:


 * $$ \hat{A}\psi = \hat{\Omega}^\dagger\hat{A}\hat{\Omega}\psi \quad \Rightarrow \quad \hat{\Omega}\hat{A}\psi = \hat{A}\hat{\Omega}\psi $$

taking the Hermitian conjugate of both sides:


 * $$ (\hat{\Omega}\hat{A})^\dagger = (\hat{A}\hat{\Omega})^\dagger \Leftrightarrow \hat{A}^\dagger\hat{\Omega}^\dagger = \hat{\Omega}^\dagger\hat{A}^\dagger $$

Quantum operators must also be Hermitian so their eigenvalues are real, the operator equals it's Hermitian conjugate,A = A†, which returns the previous equation:


 * $$ \hat{\Omega}\hat{A} = \hat{A}\hat{\Omega} $$

Momentum as the generator of spatial translations
The translation operator acts on a wavefunction to shift the coordinates of all the particles by a constant displacement Δr:


 * $$\hat{X}(\Delta\mathbf{r})\psi(\mathbf{r},t) = \psi(\mathbf{r} + \Delta\mathbf{r},t)$$

To determine what X is, expand the right hand side in a Taylor series about r:


 * $$\psi(\mathbf{r} + \Delta\mathbf{r},t) = \sum_{n=0}^\infty \frac{1}{n!}\left.\left(\Delta\mathbf{r}\cdot\frac{\partial}{\partial \mathbf{r}}\right)^n\psi(\mathbf{r},t)\right |_{\mathbf{r}=\Delta\mathbf{r}} = \left[1 + \Delta\mathbf{r}\cdot \nabla + \cdots \right] \psi(\Delta\mathbf{r},t) $$

where the expression


 * $$\Delta\mathbf{r}\cdot\frac{\partial}{\partial \mathbf{r}} = \Delta\mathbf{r}\cdot \nabla = \Delta x \frac{\partial}{\partial x} + \Delta y \frac{\partial}{\partial y} + \Delta z \frac{\partial}{\partial z}$$

is understood to be an operator, and the partial derivatives are written in this way to clarify the derivatives are take with respect to the position coordinates and the derivatives taken together form a vector (see matrix calculus for more on the notation "&part;/&part;r = &nabla;"), this notation also parallels with other variables presented next. To first order in Δr, namely the first power n = 1 with terms n ≥ 2 neglected:


 * $$\hat{X}(\Delta\mathbf{r}) = 1 + \Delta\mathbf{r}\cdot \nabla $$

which can be rewritten using the momentum operator:


 * $$\hat{X}(\Delta\mathbf{r}) = 1 + \frac{i}{\hbar}\Delta\mathbf{r} \cdot \hat{\mathbf{p}} \,,\quad \hat{\mathbf{p}} = -i \hbar \nabla$$

The previous operator is only true for small displacements. A net translation can be composed as a sequence of smaller translations. To obtain the translation operator by a finite displacement, define an infinitesimal displacement by Δr = a/N where N is a positive non-zero integer and a a small displacement vector, then as N increases the magnitude of a becomes even smaller while leaving the direction unchanged. Acting the translation operator N times and taking the limit as N tends to infinity gives the translation by a finite amount:


 * $$ \lim_{N \rightarrow \infty} \left(1 + \frac{i}{\hbar}\frac{\mathbf{a}}{N} \cdot \hat{\mathbf{p}}\right)^N = \exp\left(\frac{i}{\hbar}\mathbf{a} \cdot \hat{\mathbf{p}}\right) = \hat{X}(\mathbf{a})$$

and the exponential function arises by it's definition as this limit, due to Euler. This is the translation operator reconstructed in terms of the momentum operator.

Spatial translations commute, which is physically intuitive,


 * $$\left[\hat{X}(\mathbf{a}_1), \hat{X}(\mathbf{a}_2) \right] = 0 $$

Energy as the generator of time translations
Similarly, the time translation operator acts on a wavefunction to shift the time coordinate by a constant duration into the future by Δt:


 * $$\hat{U}(\Delta t)\psi(\mathbf{r},t) = \psi(\mathbf{r},t + \Delta t)$$

To determine what T is, expand the right hand side in a Taylor series about Δt:


 * $$\psi(\mathbf{r},t+\Delta t) = \sum_{n=0}^\infty \frac{1}{n!}\left.\left(\Delta t \frac{\partial}{\partial t}\right)^n\psi(\mathbf{r},t)\right |_{t = \Delta t} = \left[ 1 + \Delta t \frac{\partial}{\partial t} + \cdots \right] \psi(\mathbf{r}, \Delta t) $$

To first order in Δt:


 * $$ \hat{U}(\Delta t) = 1 + \Delta t \frac{\partial}{\partial t}$$

in terms of the energy operator:


 * $$ \hat{U}(\Delta t) = 1 - \frac{i}{\hbar}\Delta t \hat{E} \,,\quad \hat{E}=i\hbar\frac{\partial}{\partial t}$$

Again, the previous operator is only true for small time increments. A net time translation can be composed by making lots of small time translations. Defining an infinitesimal time increment by Δt = t/N where N is an integer and t a small time duration, taking the limit as before:


 * $$ \lim_{N \rightarrow \infty} \left(1 - \frac{i}{\hbar}\frac{t}{N} \hat{E}\right)^N = \exp\left(-\frac{i}{\hbar}t\hat{E}\right) = \hat{U} $$

gives the exponential. This is the unitary time evolution operator, in reconstructed in terms of the energy operator:


 * $$\hat{U}(t) = \exp\left(- \frac{it\hat{E}}{\hbar}\right) $$

For a time-independent Hamiltonian, energy is conserved in time and and quantum states are stationary states: the eigenstates of the Hamiltonian are the energy eigenvalues E:


 * $$\hat{U}(t) = \exp\left( - \frac{i t E}{\hbar}\right) $$

All stationary states have the form


 * $$\psi(\mathbf{r},t + t_0) = \hat{U}(t) \psi(\mathbf{r},t_0) $$

where t0 is the initial time, usually set to zero since there is no loss of continuity where the initial time is set.

Time translations commute:


 * $$\left[\hat{U}(\Delta t_1), \hat{U}(\Delta t_2) \right] = 0 $$

Orbital angular momentum
The rotation operator acts on a wavefunction to rotate the spatial coordinates by a constant angle Δθ:


 * $$\hat{R}(\Delta\theta,\mathbf{n})\psi(\mathbf{r},t) = \psi(\mathbf{r}',t)$$

where r&prime; are the rotated coordinates about an axis defined by a unit vector n, given by a rotation matrix R:


 * $$\mathbf{r}' = R(\Delta\theta,\mathbf{n})\mathbf{r}\,.$$

It is not as obvious how to determine the rotational operator compared to space and time translations, so we may consider a special case and infer the general result. Considering the z-axis, the rotation matrix about this axis is:



R(\Delta\theta,\mathbf{e}_z) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

so that



\mathbf{r}' = \begin{pmatrix} x' \\ y' \\ z' \\ \end{pmatrix} = R(\Delta\theta,\mathbf{e}_z) \mathbf{r} = \begin{pmatrix} \cos\Delta\theta & -\sin\Delta\theta & 0 \\ \sin\Delta\theta & \cos\Delta\theta & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} $$

For small Δθ, we have the small angle approximations sin(Δθ) ≈ Δθ and cos(Δθ) ≈ 1, which yields



\begin{pmatrix} x' \\ y' \\ z' \\ \end{pmatrix} = \begin{pmatrix} x - \Delta \theta y \\ \Delta\theta x + y \\ z \\ \end{pmatrix} $$

and then


 * $$\hat{R}(\Delta\theta,\mathbf{e}_z)\psi(x,y,z,t) = \psi(x -\theta y, \theta x + y, z, t)$$

Taylor expanding about Δθ:


 * $$\psi(x - \Delta \theta y, \Delta\theta x + y, z, t)= \sum_{n=1}^\infty \frac{1}{n!} \left.\left[\Delta\theta \left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)\right]^n\psi(\mathbf{r},t)\right|_{\mathbf{r}=\mathbf{r}'} = \left[1 + \Delta\theta \left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right) + \cdots \right] \psi(\mathbf{r}',t) $$

so to first order in Δθ:


 * $$ \hat{R}(\theta,\mathbf{e}_z) = 1 + \theta \left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)$$

or in terms of thez-component of the angular momentum operator, in both Cartesian and spherical polar coordinates:


 * $$ \hat{R}_z = 1 - \frac{i}{\hbar}\theta \hat{L}_z \,,\quad \hat{L}_z = i\hbar\frac{\partial}{\partial \theta}$$

This can be done similarly for rotations about the y or x axes through angle $Δθ$:



R(\theta,\mathbf{e}_y) = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \\ \end{pmatrix}

\,,\quad

R(\theta,\mathbf{e}_x) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \\ \end{pmatrix} $$

There is nothing special about the particular x, y, or z axes, and in general for rotations about an axis defined by a unit vector n = (n1, n2, n3), the rotation matrix has components:


 * $$[R(\theta, \mathbf{n})]_{ij} = \cos\theta \delta_{ij} + (1-\cos\theta)\frac{n_i n_j}{n^2} - \frac{\sin\theta}{\theta}\varepsilon_{ijk}\theta_k $$

In terms of the pseudovector angular momentum operator L:


 * $$ \hat{R} = 1 - \frac{i}{\hbar}\theta \mathbf{n} \cdot \hat{\mathbf{L}} \,,\quad \hat{\mathbf{L}} = i\hbar \mathbf{n}\frac{\partial}{\partial \theta}$$

Again, a finite rotation can be made from lots of small rotations, letting θ = α/N and taking the limit:


 * $$ \lim_{N \rightarrow \infty} \left(1 - \frac{i}{\hbar}\frac{\alpha}{N} \mathbf{n} \cdot \hat{\mathbf{L}} \right)^N = \exp\left( - \frac{i}{\hbar}\alpha \mathbf{n} \cdot \hat{\mathbf{L}}\right) = \hat{R} $$

which is the rotation operator reconstructed in terms of the angular momentum operator.

Rotations about different axes do not commute:


 * $$\left[ R(\theta_1 \mathbf{n}_i), R(\theta_2 \mathbf{n}_j) \right] \neq 0$$

although, of course, rotations about the same axis does commute:


 * $$\left[ R(\theta_1 \mathbf{n}), R(\theta_2 \mathbf{n}) \right]=0$$

In this sense, orbital angular momentum has the common sense properties of rotations. Each of the above commutators can be easily demonstrated by holding an everyday object and rotating it through the same angle about any two different axes in both possible orderings; the final configurations are different.

In quantum mechanics, there is another form of rotation which mathematically appears similar to the orbital case, but has different properties, as described next.

The generators of rotations are given by the angular momentum matrices:


 * $$ J_z = i\left.\frac{\partial R(\theta,\mathbf{e}_x)}{\partial \theta}\right|_{\theta = 0} = i\begin{pmatrix}

0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$


 * $$ J_y = i\left.\frac{\partial R(\theta,\mathbf{e}_y)}{\partial \theta}\right|_{\theta = 0} = i\begin{pmatrix}

0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{pmatrix}$$


 * $$ J_x = i\left.\frac{\partial R(\theta,\mathbf{e}_z)}{\partial \theta}\right|_{\theta = 0} = i\begin{pmatrix}

0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix} $$

which have the commutator:


 * $$ [ J_a, J_b ] = i \hbar \varepsilon_{abc} J_c $$

Exponentiating these gives their representations.

Spin angular momentum
All previous quantities have classical definitions. Spin is a quantum property without any classical analogue. The spin operator is S. Rotations about an axis n through angle θ can also be described by:


 * $$ \hat{S} = \exp\left( - \frac{i}{\hbar}\theta \mathbf{n} \cdot \hat{\mathbf{S}}\right) $$

However, unlike orbital angular momentum in which the z-projection quantum number can only take positive or negative integer values (including zero), spin can take half-integer values also. There are rotational matrices for each spin quantum number.

Evaluating the exponential for a given z-projection spin quantum number s gives a spin matrix. This can be used to define a spinor as a column vector of 2s + 1 components which transforms to a rotated coordinate system according to the spin matrix. For the simple case of s = 1/2, the spin operator is given by the Pauli matrices:


 * $$ \mathbf{S} = \frac{\hbar}{2} \boldsymbol{\sigma} $$

so evaluating the matrix exponential gives:



\hat{S} = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & -e^{-i\phi}\sin\left(\frac{\theta}{2}\right) \\ e^{i\phi}\sin\left(\frac{\theta}{2}\right) & \cos\left(\frac{\theta}{2}\right) \\ \end{pmatrix} $$

The Pauli matrices in the standard representation are:



\sigma_1 = \sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \,,\quad \sigma_2 = \sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix} \,,\quad \sigma_3 = \sigma_z = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} $$

and they are the generators of the SU(2) Lie group. Their commutation relation is the same as for orbital angular momentum, aside from a factor of 2:


 * $$ [ \sigma_a, \sigma_b ] = 2i \hbar \varepsilon_{abc} \sigma_c $$

Total angular momentum
The total angular momentum operator is the sum of the orbital and spin


 * $$ \mathbf{J} = \mathbf{L} + \mathbf{S} $$

and is an important quantity for multi-particle systems, especially in nuclear physics and the quantum chemistry of multi-electron atoms and molecules.

We have a similar rotation matrix:


 * $$ \hat{J} = \exp\left( - \frac{i}{\hbar}\theta \mathbf{n} \cdot \hat{\mathbf{J}}\right) $$

Lorentz group in relativistic quantum mechanics
In relativistic quantum mechanics, wavefunctions are no longer single-component scalar fields, but now 2(2s + 1) component spinor fields, where s is the spin of the particle. The transformations of these functions in spacetime are given below.

Transformations of spinor wavefunctions in relativistic quantum mechanics
Under a proper orthochronous Lorentz transformation $(r, t) → Λ(r, t)$ in Minkowski space, all one-particle quantum states $ψ_{σ}$ locally transform under some representation $D$ of the Lorentz group:


 * $$\psi(\mathbf{r}, t) \rightarrow D(\Lambda) \psi(\Lambda^{-1}(\mathbf{r}, t)) $$

where $D(Λ)$ is a finite-dimensional representation, in other words a $(2s + 1)×(2s + 1)$ dimensional square matrix. Again, $ψ$ is thought of as a column vector containing components with the $(2s + 1)$ allowed values of $σ$. The quantum numbers $s$ and $σ$ as well as other labels, continuous or discrete, representing other quantum numbers are suppressed. One value of $σ$ may occur more than once depending on the representation.

Following is an overview of the Lorentz group; a treatment of boosts and rotations in spacetime. Throughout this section, see (for example) and E. Abers (2004).

Transformations of spacetime coordinates and spinor fields
Lorentz transformations can be parametrized by rapidity $φ$ for a boost in the direction of a three dimensional unit vector $a$, and a rotation angle $θ$ about a three dimensional unit vector $n$ defining an axis.

The Lorentz group has six generators; three for Lorentz boosts usually denoted $K = (K_{1}, K_{2}, K_{3})$, and three for rotations usually denoted $J = (J_{1}, J_{2}, J_{3})$. One aspect of generators in theoretical physics is they are operators corresponding to symmetries, and since operators can be written as matrices, so can generators. This is hand-in-hand with the idea that groups are powerful abstract objects for analyzing symmetries and invariance. In this case the generator of rotations is the angular momentum operator. However the term "boost" refers to the relative velocity between two frames and shouldn't be conflated with momentum as the generator of translations. The generators for time translations and space translations, taken together with the boost and rotation generators, constitute the Poincaré group.

The boost and rotation generators have (reducible) representations denoted $D(K)$ and $D(J)$ respectively, the capital $D$ in this context indicates a group representation.

For the Lorentz group, generators $K$ and $J$ and their representations $D(K)$ and $D(J)$ fulfill the following commutation rules.


 * {|class="wikitable"

!Pure boost !Pure rotation !Lorentz transformation !Generators !Representations
 * +Commutators
 * $$\left[J_a ,J_b\right] = i\varepsilon_{abc}J_c$$
 * $$\left[K_a ,K_b\right] = -i\varepsilon_{abc}J_c$$
 * $$\left[J_a ,K_b\right] = i\varepsilon_{abc}K_c$$
 * $$\left[{D(J_a)} ,{D(J_b)}\right] = i\varepsilon_{abc}{D(J_c)}$$
 * $$\left[{D(K_a)} ,{D(K_b)}\right] = -i\varepsilon_{abc}{D(J_c)}$$
 * $$\left[{D(J_a)} ,{D(K_b)}\right] = i\varepsilon_{abc}{D(K_c)}$$
 * }
 * }

Exponentiating the generators gives the boost and rotation operators which combine into the general Lorentz transformation, under which the spacetime coordinates transform from one rest frame to another boosted and/or rotating frame. Likewise, exponentiating the representations of the generators gives the representations of the boost and rotation operators, under which a particle's spinor field transforms.


 * {|class="wikitable"

!Pure boost !Pure rotation !Lorentz transformation !Transformations !Representations
 * +Transformation laws
 * $$B(\varphi,\hat{\mathbf{a}}) = \exp\left(-\frac{i}{\hbar} \varphi\hat{\mathbf{a}} \cdot \mathbf{K}\right)$$
 * $$R(\theta,\hat{\mathbf{n}}) = \exp\left(-\frac{i}{\hbar} \theta \hat{\mathbf{n}} \cdot \mathbf{J}\right) $$
 * $$\Lambda(\varphi,\hat{\mathbf{a}},\theta,\hat{\mathbf{n}}) = \exp\left[-\frac{i}{\hbar} \left(\varphi\hat{\mathbf{a}} \cdot \mathbf{K} + \theta\hat{\mathbf{n}} \cdot \mathbf{J}\right)\right] $$
 * $$D[B(\varphi,\hat{\mathbf{a}})] = \exp\left(-\frac{i}{\hbar} \varphi \hat{\mathbf{a}} \cdot D(\mathbf{K})\right)$$
 * $$D[R(\theta,\hat{\mathbf{n}})] = \exp\left(-\frac{i}{\hbar}\theta \hat{\mathbf{n}} \cdot D(\mathbf{J})\right) $$
 * $$D[\Lambda(\theta,\hat{\mathbf{n}},\varphi,\hat{\mathbf{a}})] = \exp\left[-\frac{i}{\hbar}\left( \varphi \hat{\mathbf{a}} \cdot D(\mathbf{K}) + \theta \hat{\mathbf{n}} \cdot D(\mathbf{J})\right)\right] $$
 * }
 * }

For example, a boost with velocity $ctanhφ$ in the x direction given by the standard Cartesian basis vector $e_{x}$, is the simplest Lorentz transformation:



B(\varphi,\mathbf{e}_x) = \begin{pmatrix} \cosh\varphi & \sinh\varphi & 0 & 0 \\ \sinh\varphi & \cosh\varphi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

similarly for boosts with velocity $ctanhφ$ the y or z directions:



B(\varphi,\mathbf{e}_y) = \begin{pmatrix} \cosh\varphi & 0 & \sinh\varphi & 0 \\ 0 & 1 & 0 & 0 \\ \sinh\varphi & 0 & \cosh\varphi & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}

\,,\quad

B(\varphi,\mathbf{e}_z) = \begin{pmatrix} \cosh\varphi & 0 & 0 & \sinh\varphi \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh\varphi & 0 & 0 & \cosh\varphi \\ \end{pmatrix} $$

Products of boosts give another boost, and products of rotations give another rotation (a frequent exemplification of a group), while products of rotations and boosts gives a mixture of translational and rotational motion. For more background see (for example) B.R. Durney (2011) and H.L. Berk et al and references therein.

The generators for the boosts (not translation) are given by:


 * $$ K_x = i\left.\frac{\partial B(\varphi,\mathbf{e}_x)}{\partial \varphi}\right|_{\varphi=0} = i \begin{pmatrix}

0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$


 * $$ K_y = i \left.\frac{\partial B(\varphi,\mathbf{e}_y)}{\partial \varphi}\right|_{\varphi=0} = i \begin{pmatrix}

0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$


 * $$ K_z = i\left.\frac{\partial B(\varphi,\mathbf{e}_z)}{\partial \varphi}\right|_{\varphi=0} = i \begin{pmatrix}

0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{pmatrix} $$

Exponentiating these gives their representations.

Combining boosts and rotations in relativistic quantum mechanics
The previous boost generators $K$ and rotation generators $J$ can be combined into one generator $M$, an antisymmetric four-dimensional matrix:


 * $$M^{0a} = -M^{a0} = K_a \,,\quad M^{ab} = \varepsilon_{abc} J_c \,.$$

or explicitly in the standard representation:


 * $$\begin{align}

J_1 &= J^{23} = -J^{32} = i\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&-1\\ 0&0&1&0\\ \end{pmatrix}, J_2 = J^{31} = -J^{13} = i\begin{pmatrix} 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&-1&0&0\\ \end{pmatrix}, J_3 = J^{12} = -J^{21} = i\begin{pmatrix} 0&0&0&0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{pmatrix},\\ K_1 &= J^{01} = J^{10} = i\begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix}, K_2 = J^{02} = J^{20} = i\begin{pmatrix} 0&0&1&0\\ 0&0&0&0\\ 1&0&0&0\\ 0&0&0&0\\ \end{pmatrix}, K_3 = J^{03} = J^{30} = i\begin{pmatrix} 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ \end{pmatrix}. \end{align}$$

The representation of this is denoted $D(M)$, with the corresponding representations of the boost generator $D(K)$ and rotation generators $D(J)$.

Exponentiating the generator gives the Lorentz transformation,


 * $$\Lambda = \exp\left(-\frac{i}{2}\Xi_{\alpha\beta}M^{\alpha\beta}\right)$$

under which spacetime coordinates, or more generally any four vector, transforms according to from some rest frame to a boosted and/or rotated frame. $Ξ_{αβ}$ is another antisymmetric four-dimensional matrix containing the boosts in the $a$ directions, and and rotations about the $c$ axes, which are respectively:


 * $$\Xi_{0a} = - \Xi_{a0} = \varphi \hat{\mathbf{a}}\,,\quad \Xi_{ab} = \theta \varepsilon_{abc} \hat{n}_c \,,$$

The corresponding representation is

under which the one-particle spinor field transforms according to.

The commutation relations the generators must satisfy are:

Real irreducible representations and spin
The irreducible representations of $D(K)$ and $D(J)$, in short "irreps", can be used to build to spin representations of the Lorentz group. Defining new generators:


 * $$\mathbf{A} = \frac{\mathbf{J} + i \mathbf{K}}{2}\,,\quad \mathbf{B} = \frac{\mathbf{J} - i \mathbf{K}}{2}\,,$$

so $A$ and $B$ are simply complex conjugates of each other, it follows they satisfy the symmetrically formed commutators:


 * $$\left[A_i ,A_j\right] = \varepsilon_{ijk}A_k\,,\quad \left[B_i ,B_j\right] = \varepsilon_{ijk}B_k\,,\quad \left[A_i ,B_j\right] = 0\,,$$

and these are essentially the commutators the orbital and spin angular momentum operators satisfy. Therefore $A$ and $B$ form operator algebras analogous to angular momentum; same ladder operators, z-projections, etc., independently of each other as each of their components mutually commute. By the analogy to the spin quantum number, we can introduce positive integers or half integers, $a, b$, with corresponding sets of values $a, a − 1, ... −a + 1, −a$ and $b, b − 1, ... −b + 1, −b$, to numerate all the irreducible representations. Adding and subtracting the generators back, the general representation of the boost and rotation generators are respectively denoted, with reference to the numbers $a, b$, by:


 * $$ D^{(a,b)}(\mathbf{K}) = -i\left(\mathbf{A} - \mathbf{B}\right)\,,\quad D^{(a,b)}(\mathbf{J}) = \mathbf{A} + \mathbf{B}\,,$$

Some authors write $(a, b)$ for $D^{(a, b)}$, etc. Applying this to particles with spin $s$;
 * left-handed $(2s + 1)$-component spinors transform under the real irreps $D^{(s, 0)}$,
 * right-handed $(2s + 1)$-component spinors transform under the real irreps $D^{(0, s)}$,
 * taking direct sums symbolized by $&oplus;$ (see direct sum of matrices for the simpler matrix concept), one obtains the representations under which $2(2s + 1)$-component spinors transform: $D^{(a, b)} &oplus; D^{(b, a)}$ where $a + b = s$. These are also real irreps, but as shown above, they split into complex conjugates.

In these cases the $D$ refers to any of $D(J)$, $D(K)$, or a full Lorentz transformation $D(Λ)$.

Relativistic wave equations
In the context of the Dirac equation and Weyl equation, the Weyl spinors satisfying the Weyl equation transform under the simplest irreducible representations of the Lorentz group, since the spin quantum number in this case is the smallest non-zero number allowed: 1/2. The 2-component left-handed Weyl spinor transforms under $D^{(1/2, 0)}$ and the 2-component right-handed Weyl spinor transforms under $D^{(0, 1/2)}$. Dirac spinors satisfying the Dirac equation transform under the representation $D^{(1/2, 0)} &oplus; D^{(0, 1/2)}$, the direct sum of the irreps for the Weyl spinors.

Colour
The Gell-Mann matrices λn are the generators for the SU(3) group, and important for quantum chromodynamics. They have the commutator:


 * $$\left[\lambda_a, \lambda_b \right] = 2i f){abc}\lambda_c$$

where the indices a, b, c take the values 1, 2, 3... 8 for the eight gluon color charges.