User:Maschen/sandbox

=Wikitasks=

Main space

 * Lorentz transformation
 * Wigner rotation
 * Thomas precession
 * Hamiltonian field theory
 * Lagrangian mechanics
 * Relativistic Lagrangian mechanics
 * Lagrangian (field theory)

User space

 * User:Maschen/Bargmann–Michel–Telegdi equation
 * User:Maschen/Bargmann-Wigner equations
 * User:Maschen/Dirac–Fierz–Pauli equations


 * User:Maschen/Relativistic Hamiltonian mechanics
 * user:Maschen/Lagrangian mechanics (damped and dissipative forces)


 * User:Maschen/Electromagnetic displacement tensor
 * User:Maschen/Relativistic electromagnetic dipole moments


 * User:Maschen/wave function
 * User:Maschen/Spin wave function

=Rapidity=

Definitions
Start with two Lorentz boosts B(ζ1) and B(ζ2) with rapidities


 * $$\boldsymbol{\zeta}_1 = \mathbf{n}_1\tanh^{-1}\beta_1\,, \quad \boldsymbol{\zeta}_2 = \mathbf{n}_2\tanh^{-1}\beta_2 $$

where the unit vectors are in the direction of the boosts,


 * $$\mathbf{n}_1 = \frac{\boldsymbol{\beta}_1}{\beta_1} \,, \quad \mathbf{n}_2 = \frac{\boldsymbol{\beta}_2}{\beta_2} $$

and in turn the betas are simply the relative velocities between frames defined by


 * $$\boldsymbol{\beta}_1 = \frac{\mathbf{v}_1}{c} \,, \quad \boldsymbol{\beta}_2 = \frac{\mathbf{v}_2}{c} $$

Composite formulae
There are two inequivalent composite Lorentz transformations B(ζ1)B(ζ2) and B(ζ2)B(ζ1). There are two composite boosts


 * $$\boldsymbol{\beta}_1 \oplus \boldsymbol{\beta}_2 =\frac{1}{1 + \boldsymbol{\beta}_1 \cdot \boldsymbol{\beta}_2 }\left[\boldsymbol{\beta}_1 + \frac{\boldsymbol{\beta}_2}{\gamma_1} + \frac{\gamma_1}{1+\gamma_1}(\boldsymbol{\beta}_1 \cdot \boldsymbol{\beta}_2)\boldsymbol{\beta}_1\right]$$


 * $$\boldsymbol{\beta}_2 \oplus \boldsymbol{\beta}_1 =\frac{1}{1 + \boldsymbol{\beta}_2 \cdot \boldsymbol{\beta}_2 }\left[\boldsymbol{\beta}_2 + \frac{\boldsymbol{\beta}_1}{\gamma_2} + \frac{\gamma_2}{1+\gamma_2}(\boldsymbol{\beta}_2 \cdot \boldsymbol{\beta}_1)\boldsymbol{\beta}_2\right]$$

where &oplus; is the relativistic velocity-addition, defined by the first expression above (second one follows immediately by switching betas).


 * $$\boldsymbol{\beta}_1 \oplus \boldsymbol{\beta}_2 = \frac{1}{1 + \boldsymbol{\beta}_1 \cdot \boldsymbol{\beta}_2}\left[\boldsymbol{\beta}_1 + \boldsymbol{\beta}_2 +

\frac{\gamma_1}{1+\gamma_1} \boldsymbol{\beta}_1 \times(\boldsymbol{\beta}_1 \times \boldsymbol{\beta}_2)\right]$$

The composite rapidities are


 * $$\boldsymbol{\zeta} = \mathbf{n}\tanh^{-1}\beta \,,\quad \overline{\boldsymbol{\zeta}} = \overline{\mathbf{n}}\tanh^{-1}\overline{\beta} $$

The unit vectors


 * $$ \mathbf{n} = \frac{\boldsymbol{\beta}_1\oplus\boldsymbol{\beta}_2}{|\boldsymbol{\beta}_1\oplus\boldsymbol{\beta}_2|} \,, \quad \overline{\mathbf{n}}= \frac{\boldsymbol{\beta}_2\oplus\boldsymbol{\beta}_1}{|\boldsymbol{\beta}_2\oplus\boldsymbol{\beta}_1|} $$

are separated by an angle (Wigner rotation). The magnitudes are equal,


 * $$ \beta = |\boldsymbol{\beta}_1\oplus\boldsymbol{\beta}_2| = |\boldsymbol{\beta}_2\oplus\boldsymbol{\beta}_1| = \overline{\beta} = \frac{\sqrt{(\boldsymbol{\beta}_1 + \boldsymbol{\beta}_2)^2 - (\boldsymbol{\beta}_1 \times \boldsymbol{\beta}_2)^2}}{1+\boldsymbol{\beta}_1\cdot\boldsymbol{\beta}_2}$$

In summary


 * $$\boldsymbol{\zeta} = \frac{\boldsymbol{\beta}_1\oplus\boldsymbol{\beta}_2}{\beta}\tanh^{-1}\beta \,,\quad \overline{\boldsymbol{\zeta}} = \frac{\boldsymbol{\beta}_2\oplus\boldsymbol{\beta}_1}{\beta}\tanh^{-1}\beta $$

so these vectors have the same magnitude


 * $$\zeta = \overline{\zeta} = \tanh^{-1}\beta$$

which is to be expected since composite boost velocities have the same magnitudes, but different directions.

In terms of rapidity


 * $$\boldsymbol{\beta}_1 \oplus \boldsymbol{\beta}_2 = \frac{1}{1 + \mathbf{n}_1 \cdot \mathbf{n}_2 \tanh \zeta_1\tanh \zeta_2 }\left[\mathbf{n}_1\tanh\zeta_1 + \frac{\mathbf{n}_2\tanh\zeta_2}{\cosh\zeta_1} + \frac{\cosh\zeta_1}{1+\cosh\zeta_1}(\mathbf{n}_1 \cdot \mathbf{n}_2)\mathbf{n}_1 \tanh^2\zeta_1\tanh\zeta_2\right]$$


 * $$\boldsymbol{\beta}_2 \oplus \boldsymbol{\beta}_1 = \frac{1}{1 + \mathbf{n}_2 \cdot \mathbf{n}_1 \tanh \zeta_2\tanh \zeta_1 }\left[\mathbf{n}_2\tanh\zeta_2 + \frac{\mathbf{n}_1\tanh\zeta_1}{\cosh\zeta_2} + \frac{\cosh\zeta_2}{1+\cosh\zeta_2}(\mathbf{n}_2 \cdot \mathbf{n}_1)\mathbf{n}_2 \tanh^2\zeta_2\tanh\zeta_1\right]$$


 * $$ \beta = \frac{\sqrt{(\mathbf{n}_1 \tanh\zeta_1 + \mathbf{n}_2 \tanh\zeta_2 )^2 - (\mathbf{n}_1 \times \mathbf{n}_2)^2 \tanh^2\zeta_1\tanh^2\zeta_2}}{1+\mathbf{n}_1\cdot\mathbf{n}_2 \tanh\zeta_1\tanh\zeta_2}$$

so in full,


 * $$\boldsymbol{\zeta} = \frac{\boldsymbol{\beta}_1\oplus\boldsymbol{\beta}_2}{\beta}\tanh^{-1}\beta \,,\quad \overline{\boldsymbol{\zeta}} = \frac{\boldsymbol{\beta}_2\oplus\boldsymbol{\beta}_1}{\beta}\tanh^{-1}\beta $$

Hyperbolic triangles
If ζ1 and ζ2 are separated by an angle, and have magnitudes ζ1 and ζ2, these are related to ζ by (add formula later...)


 * $$\gamma_1 = \cosh\zeta_1 \,,\quad \gamma_2 = \cosh\zeta_2 $$
 * $$\boldsymbol{\beta}_1 = \mathbf{n}_1 \tanh\zeta_1 \,,\quad \boldsymbol{\beta}_2 = \mathbf{n}_2 \tanh\zeta_2 $$
 * $$\gamma_1\boldsymbol{\beta}_1 = \mathbf{n}_1 \sinh\zeta_1 \,,\quad \gamma_2\boldsymbol{\beta}_2 = \mathbf{n}_2 \sinh\zeta_2 $$

=Relativistic wave equations=

General considerations

 * Line element and energy-momentum relation

In the following, the four spacetime coordinates of the particle are x = (ct, x) (a.k.a. "four position vector"), and its four momentum is p = (E/c, p). The Minkowski metric η throughout is (+, −, −, −). Natural units c = ħ = 1 will be used as standard in RQM.

Using the relativistic invariance of the infinitesimal line element


 * $$dx_\mu dx^\mu = d\tau^2$$

and the definition of four momentum


 * $$p_\mu = m \frac{d x_\mu}{d\tau} $$

for a massive particle obtains the energy-momentum relation in its covariant form


 * $$ p_\mu p^\mu = m^2$$

or explicitly in terms of energy and 3-momentum


 * $$E^2 - \mathbf{p}^2 = m^2 $$

This must also hold in relativistic quantum mechanics. There are a number of possibilities resulting from the equation which must be considered separately:


 * Massless particles (m = 0): $$ E = |\mathbf{p}| $$
 * Massive particles (m > 0): $$ E^2 - \mathbf{p}^2 = m^2 $$
 * "Positive energy" (E > 0): $$ E = +\sqrt{\mathbf{p}^2 + m^2} $$
 * "Negative energy" (E < 0): $$ E = -\sqrt{\mathbf{p}^2 + m^2} $$


 * Wigner's classification

Here is Wigner's classification [Ohlsson 2011, p.15] for the irreducible representations of the Poincaré group:


 * $$p^2=m^2 > 0 \,, \quad p^0 > 0 $$
 * $$p^2= 0 \,, \quad p^0 < 0 $$
 * $$p^2= 0 \,, \quad p \neq 0 \,,\quad p^0 > 0 $$
 * $$p^2= 0 \,, \quad p \neq 0 \,,\quad p^0 < 0 $$
 * $$p^2= p = 0 $$
 * $$p^2 = m^2 = < 0 $$


 * Postulates of QM

All RWEs must also be consistent with the Schrödinger equation, a linear partial differential equation in the wave function, because this is one of the fundamental postulates of QM. Even if the RWE does not directly take the form of the SE, it should be possible to reproduce the SE from it somehow.

In non-relativistic quantum mechanics, the energy and momentum observables are differential operators, and their eigenvalues are the results of measurement. In the index notation and four vector formalism here, these derivatives collect into a four momentum operator in terms of the four gradient with components


 * $$ \hat{p}_\mu = i \partial_\mu $$
 * $$\hat{p} = (\hat{E},\hat{\mathbf{p}})= i \left(\frac{\partial}{\partial t}, - \nabla \right)$$


 * Spin and angular momentum

Further, in non-relativistic quantum mechanics a particle with spin s is described by a wavefunction ψ with 2s + 1 components ψσ indexed by the spin projection quantum number σ = −s, −s + 1, ..., s − 1, s, often these are arranged into a column vector


 * $$\psi = \begin{pmatrix} \psi_s \\ \psi_{s-1} \\ \vdots \\ \psi_{-s+1} \\ \psi_{-s} \end{pmatrix}$$

Still in non-relativistic quantum mechanics, the spin operator for a particle of spin s is a vector s of matrices sx, sy, sz, each are (2s + 1)×(2s + 1) matrices.

There is also the relativistic angular momentum tensor M, and the Pauli-lubanski pseudovector is defined from M and p. The spacelike component of the PL four pseudovector are related to spin.


 * Poincaré group

The four momentum and angular momentum relations satisfy the commutation relations of the Poincaré group.


 * $$[p_\mu, p_\nu] = 0\,$$
 * $$\frac{ 1 }{ i }[M_{\mu\nu}, p_\rho] = \eta_{\mu\rho} p_\nu - \eta_{\nu\rho} p_\mu\,$$
 * $$\frac{ 1 }{ i }[M_{\mu\nu}, M_{\rho\sigma}] = \eta_{\mu\rho} M_{\nu\sigma} - \eta_{\mu\sigma} M_{\nu\rho} - \eta_{\nu\rho} M_{\mu\sigma} + \eta_{\nu\sigma} M_{\mu\rho}\, ,$$

For any representation of the Poincare group, the infinitesimal unitary operators corresponding to boosts U(B), rotations U(R), and translations U(a) on the Hilbert space for the system are


 * $$U(B) \approx I - i\boldsymbol{\zeta}\cdot\mathbf{K}$$
 * $$U(R) \approx I - i\boldsymbol{\theta}\cdot\mathbf{J}$$
 * $$U(a) \approx I + i a \cdot p$$

or


 * $$U(\Lambda) \approx I - \frac{i}{2}\omega_{\alpha\beta}M^{\alpha\beta} $$
 * $$U(a) \approx I + i a^\mu p_\mu $$

where ω is an antisymmetric parameter matrix for boosts and rotations.

In the finite case, one obtains the matrix exponentials. The generators J, K, and four displacement (not four position) a are all operators satisfying the commutation relations. They are not unique and can take different forms depending on the representation, provided the commutation relations are satisfied.

Heuristic construction of the KG and Dirac equations
Combining the four momentum operator and energy momentum relation leads to the KG eqn


 * $$ \hat{p}_\mu \hat{p}^\mu \psi = m^2 \psi $$

which applies for spin-0 particles, massive or massless. This can be factorized by difference of two squares into


 * $$ (\gamma^\mu\hat{p}_\mu + m)(\gamma^\nu\hat{p}_\nu - m) \psi = 0 $$

for suitable quantities γμ. Following LL vol 4 (at top of my head), expand the brackets


 * $$ \gamma^\mu{\hat {p}}_\mu\gamma^\nu{\hat {p}}_\nu\psi - m\gamma^\mu{\hat {p}}_\mu\psi + m\gamma^\nu{\hat {p}}_\nu \psi - m^2\psi =0 $$

and split the product γμγν into symmetric and antisymmetric products


 * $$ \frac{1}{2}(\gamma^\mu\gamma^\nu + \gamma^\mu\gamma^\nu) \hat {p}_\mu \hat {p}_\nu \psi + \frac{1}{2}(\gamma^\mu\gamma^\nu - \gamma^\mu\gamma^\nu) \hat {p}_\mu \hat {p}_\nu \psi - m^2\psi =0 $$

The summation of the antisymmetric tensor γμγν − γνγμ with the symmetric tensor pμpν is zero (this applies generally for the summation of any antisymmetric and symmetric tensors). Since the KG eqn must be recovered, the quantities γμ must satisfy the anticommutation relation


 * $$ \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2\eta^{\mu\nu} \,. $$

One of the factors


 * $$ (\gamma^\mu\hat{p}_\mu - m) \psi = 0 $$

is the Dirac equation, which applies to any spin-1/2 particle, massive or massless. If ψ satisfies this equation, it will also satisfy the KG equation, but the converse is not true because it can be shown the γμ, known as the gamma matrices, are related to the spin of the particle.

The KG equation is a second order linear PDE and cannot be put into the form of a Schrödinger equation, which is always first order in time. However this is not a problem. The Dirac equation can be by separating the (first order) time derivative then choosing a suitable Hamiltonian.

RWEs for particles of spin higher than 1/2 can be constructed heuristically from the Dirac equation. One approach is to form symmetric products of Dirac spinors to form a multicomponent spinor describing a particle of the appropriate spin. A set of Dirac equations apply to each index of the multicomponent spinor. Additional constraints are required. The results are the Bargmann-Wigner equations. Another approach is to form a homogenous polynomial of derivative operators involving a contraction with a symmetric (matrix-valued) multi-index coefficient. The degree of the polynomial is related to the spin particle spin. The entire resulting operator acts on the wavefunction in a single equation. This is the Joos–Weinberg equation.

Lagrangian densities
Using the field theoretic EL equations, there are Lagrangian densities for the KG and Dirac equations. For higher spin RWEs, finding a Lagrangian is not trivial.

Construction of any RWE using Lorentz group theory
Jaroszewicz and Kurzepa (1992), Sexl and Urbantke (1992), Weinberg's QFT (vol 1), Ryder's QFT, Tung's Group theory in Physics...

Tung's paper

Aims...

For the transformations of coordinates and wavefunctions/fields:


 * Summarize LTs using the J, K and A, B generators. Done
 * Link the A, B generators to representations of the Lorentz group (not just the labels (A,B) for classifying representations, but actually connect the generators A and B themselves to the D matrices)
 * Plainly write down how spinors corresponding to a particle of spin j transform according to the relevant D matrices.
 * Derive the transformation of Pauli spinors under boosts and rotations. Provide explicit formulae for the D(1/2,0) and D(0,1/2) for boosts and rotations (in terms of the complex parameters α = θ + i ζ and α* = θ − i ζ and spin-1/2 operator, rather than Pauli matrices).
 * Explain how to construct explicit formulae for D(j,0) and D(0,j) under boosts and rotations (at least in principle, ideally actually provide the explicit formulae). Any connection to the spin-j or spin-1/2 operator?
 * Discrete transformations (CPT, complex/Hermitian conjugation)

For the construction of RWEs:


 * establish the general conditions a RWE must satisfy (write the RWE in operator form, illustrate transformation properties)
 * Fourier transform of RWE
 * RWE in rest frame, then any frame
 * use the established representations to write down the RWEs

Extra background (LL vol 4, SU, Barut, Carmeli, description of spinors):


 * Connect spinors to vectors (and generally tensors)
 * spinor indices
 * relate quantities transforming under a given D representation to spinors or tensors

Summary of Lorentz transformations using the (J, K) generators
For every Lorentz transformation Λ on the coordinates in Minkowski spacetime M, there is a unitary operator D(Λ) on the Hilbert space H of allowed quantum states ψ of the physical system. The representations D depend on the system.

Any Λ can be decomposed into pure boosts B and pure rotations R. Boosts have parameters ζ and generators K. Rotations have parameters θ and generators J. The generators J and K can be expressed as the following 4×4 matrices


 * $$K_x = \begin{bmatrix}

0 & 1 & 0 & 0 \\   1 & 0 & 0 & 0 \\    0 & 0 & 0 & 0 \\    0 & 0 & 0 & 0 \\  \end{bmatrix}\,,\quad K_y = \begin{bmatrix} 0 & 0 & 1 & 0\\   0 & 0 & 0 & 0\\    1 & 0 & 0 & 0\\    0 & 0 & 0 & 0  \end{bmatrix}\,,\quad K_z = \begin{bmatrix} 0 & 0 & 0 & 1\\   0 & 0 & 0 & 0\\    0 & 0 & 0 & 0\\    1 & 0 & 0 & 0  \end{bmatrix} $$
 * $$J_x = \begin{bmatrix}

0 & 0 & 0 & 0 \\    0 & 0 & 0 &  0 \\    0 & 0 & 0 & -1 \\    0 & 0 & 1 &  0 \\  \end{bmatrix}\,,\quad J_y = \begin{bmatrix} 0 & 0 & 0 & 0 \\    0 &  0 & 0 & 1 \\    0 &  0 & 0 & 0 \\    0 & -1 & 0 & 0  \end{bmatrix}\,,\quad J_z =  \begin{bmatrix} 0 & 0 & 0 & 0 \\    0 & 0 & -1 & 0 \\    0 & 1 &  0 & 0 \\    0 & 0 &  0 & 0  \end{bmatrix} $$

The Λ form a group of transformations on spacetime coordinates leaving the line element invariant. The full set of Lorentz transformations (including parity and time reversal) is the Lie group O(1,3). Boosts and rotations are the Lie group SO(1,3).

The finite LT is


 * $$\Lambda (\boldsymbol{\zeta}, \boldsymbol{\theta}) = e^{-\boldsymbol{\zeta} \cdot\mathbf{K} + \boldsymbol{\theta} \cdot\mathbf{J} }$$

Individual boosts and rotations are


 * $$R(\boldsymbol{\theta}) = e^{ \boldsymbol{\theta} \cdot\mathbf{J} }$$
 * $$B(\boldsymbol{\zeta}) = e^{-\boldsymbol{\zeta} \cdot\mathbf{K} }$$

Explicit formulae for the representations for an n component object are


 * $$D(\Lambda(\boldsymbol{\zeta},\boldsymbol{\theta})) = $$
 * $$D(R(\boldsymbol{\theta})) = $$
 * $$D(B(\boldsymbol{\zeta})) = $$

Summary of Lorentz transformations using the (A, B) generators
Another basis for the Lorentz generators takes complex conjugates of the J and K generators: A = (J + i K)/2 and B = (J − i K)/2. The corresponding parameters are α = θ + i ζ and α* = θ − i ζ.

Using the above 4×4 matrices for J and K, A and B take the form


 * $$A_x = \frac{1}{2}\begin{bmatrix}

0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\   0 & 0 & 1 &  0 \\  \end{bmatrix}\,,\quad A_y = \frac{1}{2}\begin{bmatrix} 0 & 0 & i & 0\\ 0 & 0 & 0 & 1 \\    i &  0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}\,,\quad A_z = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0 & i\\ 0 & 0 & -1 & 0\\   0 & 1 & 0 & 0\\    i & 0 & 0 & 0 \end{bmatrix} $$
 * $$B_x = \frac{1}{2}\begin{bmatrix}

0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\   0 & 0 & 1 &  0 \\  \end{bmatrix}\,,\quad B_y = \frac{1}{2}\begin{bmatrix} 0 & 0 & -i & 0\\ 0 & 0 & 0 & 1 \\    -i &  0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}\,,\quad B_z = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0 & -i\\ 0 & 0 & -1 & 0\\   0 & 1 & 0 & 0\\    -i & 0 & 0 & 0 \end{bmatrix} $$

which satisfy the commutation relations


 * $$[A_x,A_y]=iA_z\,,\quad [B_x,B_y]=iB_z$$

where the rest can be found from cyclic permutations of x-, y-, and z-components of A and B. Each component of A commutes with each component of B. All the commutation relations in index notation are


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

The components of A and B each satisfy the commutation relations of the Lie algebra su(2). Taken together, A and B satisfy the commutation relations of the direct sum of their Lie algebras, su(2) &oplus; su(2). The corresponding Lie group is the tensor product SU(2) &otimes; SU(2).

The finite Lorentz transformation is


 * $$\Lambda (\boldsymbol{\alpha},\boldsymbol{\alpha}^{*}) = e^{\boldsymbol{\alpha}\cdot \mathbf{A}+\boldsymbol{\alpha}^{*} \cdot \mathbf{B} } $$

The generators A and B do not have to be the 4×4 matrices matrices above. The commutation relations they satisfy are exactly those of the spin operator, indicating they can be spin matrices. Let S(j) be the spin operator corresponding to spin j. Then A = S(A), a vector of three (2A + 1)×(2A + 1) matrices, and B = S(B), a vector of three (2B + 1)×(2B + 1) matrices. The z-component spin projection quantum numbers for Az are a = −A, −A + 1, ...,  A − 1, A, likewise for Bz they are b = −B,  −B + 1, ...,  B − 1, B. When these are exponentiated (with appropriate parameters included), the results will also be matrices of sizes (2A + 1)×(2A + 1) and (2B + 1)×(2B + 1). These matrices have the correct size for transformation matrices of (2A + 1)-component and (2B + 1)-component wave functions, respectively. Representations of the Lorentz group can be labelled and classified by this pair of angular momenta (A, B), each integer or half integer.

The inverse formulae are J = A + B and iK = A − B. In the extreme cases,
 * A = 0 (hence A = 0) while B arbitrary: B = −i K = J, in other words B = J = S(B) while K = i S(B).
 * B = 0 (hence B = 0) while A arbitrary: A = i K = J, in other words A = J = S(A) while K = −i S(A).

Notice J is both the generator of spatial rotations in spacetime, and an angular momentum operator as the sum of two angular momenta. The operator for their total angular momentum J is more accurately written using the direct or tensor product &otimes; as follows


 * $$\mathbf{J} = \mathbf{A}\otimes I_{2B+1} + I_{2A+1} \otimes \mathbf{B} $$

which has allowed quantum numbers J = A + B, A + B − 1..., |A − B| + 1, |A − B|, and In is the n-dimensional identity operator. As a matrix, J has the size ((2A + 1)(2B + 1))×((2A + 1)(2B + 1)), and In is the n×n identity matrix.

In index notation, the above operator is [Weinberg vol 1 somewhere]


 * $$[\mathbf{J}]_{a'b'ab} = [\mathbf{A}]_{a'a} \delta_{b'b} + \delta_{a'a}[\mathbf{B}]_{b'b} $$

where the multiple indices select the components of spinors the operator J acts on. Explicitly, if ξa is a 2A + 1 component spin wave function and ηb a 2B + 1 component spin wave function, J acts on the their tensor product as follows:


 * $$[\mathbf{J}]_{a'b'ab}\xi_a \eta_b = ([\mathbf{A}]_{a'a} \xi_a ) \eta_{b'} + \xi_{a'} ( [\mathbf{B}]_{b'b}\eta_b ) $$

useful link

another

another

Transformations of spinors according to the (A, B) representations (massive particles)
The tensor product representation is denoted and defined by [SU Particles and Fields p.231 eq 8.1.8 (in notation used here)]


 * $$D^{(A,B)}(\boldsymbol{\alpha}, \boldsymbol{\alpha}^*) = D^{(A)}(\boldsymbol{\alpha}^*) \otimes D^{(B)}(\boldsymbol{\alpha})$$

See also Clebsch-Gordan decomposition (c.f. addition of quantum angular momentum)


 * $$D^{(A,B)} = D^{(A)}\otimes D^{(B)} = D^{(A+B)}\oplus D^{(A+B-1)}\oplus\cdots D^{(|A-B|)}$$

Need to find connections between representations and generators, for boosts and rotations:


 * $$D^{(A,0)}(\boldsymbol{\alpha}^*) {\color{grey} \overset{?}{=} e^{i\boldsymbol{\alpha}^*\cdot\mathbf{A}} } \leftrightarrow \mathbf{A} = \mathbf{S}^{(A)}, \boldsymbol{\alpha}^* $$
 * $$D^{(0,B)}(\boldsymbol{\alpha}) {\color{grey} \overset{?}{=} e^{i\boldsymbol{\alpha}\cdot\mathbf{B}} } \leftrightarrow \mathbf{B} = \mathbf{S}^{(B)}, \boldsymbol{\alpha} $$
 * $$D^{(j)}(\boldsymbol{\alpha}) {\color{grey} \overset{?}{=} e^{i\boldsymbol{\alpha}\cdot\mathbf{S}^{(j)}} } \leftrightarrow \mathbf{S}^{(j)}, \boldsymbol{\alpha} $$

One can find D(1/2,0) and D(0,1/2) (to be derived soon).


 * General transformation of 2-component spinors

LL vol 4 "Fermions" chapter

A 2 component left handed spinor ζ and right handed spinor η generally transform according to


 * $$\xi' = M \xi \,,\quad \eta' = M^* \eta $$

in matrix notation


 * $$\begin{pmatrix} {\xi'}_1 \\ {\xi'}_2 \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix} \,,\quad \begin{pmatrix} {\eta'}_1 \\ {\eta'}_2 \end{pmatrix} = \begin{pmatrix} \alpha^* & \beta^* \\ \gamma^* & \delta^* \end{pmatrix} \begin{pmatrix} \eta_1 \\ \eta_2 \end{pmatrix}$$

in spinor index notation


 * $${\xi'}_\lambda = M_\lambda{}^\mu \xi_\mu \,,\quad {\eta'}_\dot{\lambda} = {M^*}_\dot{\lambda}{}^\dot{\mu} \eta_\dot{\mu} $$

where det(M) = αδ − γβ = 1. Spinor indices take the values 1 and 2.


 * D(1/2,0) and D(0,1/2)

Using Ryder QFT (p.38?), under boosts and rotations, a left handed Pauli spinor ζ transforms according to the (1/2,0) representation, while the right handed spinor η according to the (0,1/2) representation, explicitly


 * $$\xi' = D^{(1/2,0)} \xi \,,\quad \eta' = D^{(0,1/2)} \eta $$

where


 * $$D^{(1/2,0)} = e^{i\mathbf{S}^{(1/2)}\cdot \boldsymbol{\alpha}^*} \,, \quad D^{(0,1/2)} = e^{i\mathbf{S}^{(1/2)}\cdot \boldsymbol{\alpha}} $$

and


 * $$\mathbf{S}^{(1/2)} = \frac{1}{2}\boldsymbol{\sigma}$$

is the spin-1/2 operator, directly proportional to the Pauli matrices σ.


 * D(1/2,0) &oplus; D(0,1/2)

Dirac spinors transform as


 * $$\psi' = D^{(1/2,0)} \oplus D^{(0,1/2)} \psi $$

explicitly in matrix form


 * $$\psi=\begin{pmatrix} \xi \\ \eta \end{pmatrix} \,, \quad D^{(1/2,0)} \oplus D^{(0,1/2)} = \begin{pmatrix} D^{(1/2,0)} & 0 \\ 0 & D^{(0,1/2)} \end{pmatrix}$$


 * (D(1/2,0) &oplus; D(0,1/2))&otimes;n

For a tensor product of n Dirac spinors


 * $$ \begin{align} {\psi'}_{\rho_1} {\psi'}_{\rho_2} \cdots {\psi'}_{\rho_n} & = D_{\rho_1\sigma_1} \psi_{\sigma_1} D_{\rho_2\sigma_2} \psi_{\sigma_2} \cdots D_{\rho_n\sigma_n} \psi_{\sigma_n} \\

& = D_{\rho_1\sigma_1} D_{\rho_2\sigma_2}\cdots D_{\rho_n\sigma_n} \psi_{\sigma_1} \psi_{\sigma_2} \cdots \psi_{\sigma_n} \end{align}$$

in which D = D(1/2,0) &oplus; D(0,1/2), the product transforms as the n "tensor power" (tensor product of representation with itself n times)


 * $$(D^{(1/2,0)} \oplus D^{(0,1/2)} )^{\otimes n}$$

Likewise, a multicomponent spinor transforms in the same way


 * $$ {\psi'}_{\rho_1 \rho_2 \cdots \rho_n} = D_{\rho_1\sigma_1} D_{\rho_2\sigma_2}\cdots D_{\rho_n\sigma_n} \psi_{\sigma_1 \sigma_2 \cdots \sigma_n}$$

Bargmann–Wigner spinors transform like this.


 * D(j,0) and D(0,j)

Under boosts and rotations, a (2j + 1)-component left handed spinor ζ transforms according to the (j,0) representation, while a right handed spinor η transforms according to the (0,j) representation. Explicitly


 * $$\xi' = D^{(j,0)} \xi \,,\quad \eta' = D^{(0,j)} \eta $$

Need to find correspondence


 * $$D^{(j,0)}, D^{(0,j)} \leftrightarrow \mathbf{S}^{(j)} , \boldsymbol{\alpha}  , \boldsymbol{\alpha}^* $$


 * D(j,0) &oplus; D(0,j)

The 2(2j + 1)-component spinor


 * $$\psi=\begin{pmatrix} \xi \\ \eta \end{pmatrix} $$

transforms as


 * $$\psi' = D^{(j,0)} \oplus D^{(0,j)} \psi $$

where


 * $$D^{(j,0)} \oplus D^{(0,j)} = \begin{pmatrix} D^{(j,0)} & 0 \\ 0 & D^{(0,j)} \end{pmatrix}$$

The Joos-Weinberg wave function transforms like this.

Massive particles with spin s
Start by writing the RWE in general form


 * $$ \Pi(\hat{p},m) \psi(x) = 0 $$

where Π is a linear differential operator, and depends on the mass m of the particle. The linearity is required for consistency with the SE. (If the RWE equation was a nonlinear PDE, Π and ψ would not be separable from each other, because Π would be a function of ψ). For generality let ψ have n components, and Π be an n×n matrix to act on all the components of ψ. The relation between n and the spin quantum number s will be found later.

The four spacetime coordinates x change under a Lorentz transformation Λ, while the n components of the wave function collectively transform according to a n×n transformation matrix D(Λ),


 * $$x' = \Lambda x \,, \quad \psi'(x') = D(\Lambda)\psi(x) \,, $$

explicitly for all the components


 * $${x'}^\alpha = \Lambda^\alpha{}_\beta x^\beta \,, \quad \psi'(x')_\rho = D(\Lambda)_{\rho\sigma}\psi(x)_\sigma \,. $$

Notice the four spacetime coordinates x have spacetime (or "Lorentz") indices α and β which take values 0, 1, 2, 3. The wave functions ψ and transformation matrix D(Λ) simply have matrix indices ρ and σ which take the values 1, 2, ... n. The transformed wave function in terms of the original wave function, in the transformed coordinates throughout, is


 * $$ \psi'(x') = D(\Lambda)\psi(\Lambda^{-1}x') \,, $$

The transformation matrices D(Λ) must preserve the group composition properties of Lorentz transformations Λ, because for each coordinate change, the wave functions must also change correspondingly. In particular,


 * Composition: $$D(\Lambda_1\Lambda_2) = D(\Lambda_1)D(\Lambda_2)$$
 * Identity element: $$D(I) = I$$
 * Inverse element: $$D(\Lambda)^{-1} = D(\Lambda^{-1})$$

which indicates the D(Λ) must be appropriate representations of the Lorentz group.

The transformed wave equation is


 * $$\Pi'(\hat{p}',m) \psi'(x') = 0 \,. $$

Projecting the Hermitian conjugate wave functions ψ† onto their own wave equations (temporarily suppressing arguments for clarity),


 * $$\psi^\dagger \Pi \psi = 0 \,, $$
 * $${\psi'}^\dagger \Pi' \psi' = {\psi}^\dagger D^\dagger \Pi' D \psi = 0 \,, $$

If D(Λ) is unitary (D(Λ)† = D(Λ)−1), comparing these equations leads to the similarity transformation


 * $$ \Pi = D^{-1} \Pi' D \,, $$

in full


 * $$ \Pi(\hat{p},m) = D(\Lambda)^{-1} \Pi'(\hat{p}',m) D(\Lambda) \,, $$

Now the transformed operator Π&prime; can be expressed in in terms of the original Π, and in terms of the original momentum operators:


 * $$ D(\Lambda) \Pi(\hat{p},m)D(\Lambda^{-1}) = \Pi'(\Lambda \hat{p},m) \,. $$

This is the general transformation of Π. Explicit forms of Π will be derived later.

If Π is a first order derivative operator, the only general form it can take is


 * $$\Pi(\hat{p},m) = \pi^\alpha(m)\hat{p}_\alpha + C(m) $$

where the n×n matrices πα(m) and C(m) may all depend on the particle mass m. Since the four momentum is a covariant four vector, the πα form the components of a contravariant four vector so that Π is invariant under Lorentz transformations of momentum. The matrix C(m) must be invariant in all frames, which can only be possible if it is the identity matrix multiplied by a relativistic scalar invariant.

The transformed Π operator in the original momentum operators is


 * $$\Pi'(\Lambda\hat{p},m) = D(\Lambda)\pi^\alpha(m)D(\Lambda^{-1})\hat{p}_\alpha + D(\Lambda)C(m)D(\Lambda^{-1}) $$

hence each πα must transform as


 * $$ {\pi'}^\alpha = \Lambda^\alpha{}_\beta \pi^\beta \,, \quad D(\Lambda) \pi^\alpha(m)D(\Lambda^{-1}) = {\pi'}^\alpha(m) \,. $$

Starting from the rest frame of the particle, the coordinates, momenta, wave function, and RWE in this rest frame can all be transformed appropriately to obtain the corresponding quantities or operators in any other boosted frame. It will be easier to use the momentum representation because in the rest frame, p = (p0, 0), so there are fewer variables to keep track of, and the differential equation will be converted to an algebraic equation.

Take the Fourier transform of the original equation to obtain the momentum space equation


 * $$\Pi(p,m) \phi(p) = 0$$

where the FT of ψ(x) is


 * $$ \phi(p) = \int \frac{N d^4 x}{(2\pi)^3} e^{-i p_\alpha x^\alpha}\psi(x) \quad $$

(including an extra normalization factor N to be adjusted later).

The FT of the Π operator is still Π, but all derivative operators are replaced by momentum components. For the first order case,


 * $$\Pi(p,m) = \pi^\alpha(m)p_\alpha + C(m) $$

In the rest frame of the particle, the momentum space wavefunction φ(prest) or φ(p0, 0) only has energy dependence. Now transform to a boosted frame with momentum p, so the new wave function is φ&prime;(p&prime;) or


 * $$\phi'(p_0, \mathbf{p}) = D(L)\phi(p_0, \boldsymbol{0})$$

where L is a Lorentz transformation for the boost, and D(L) a corresponding representation.

The momentum space Π operator in the rest frame is (given in Tung's book, can't find in Weinberg. Can't understand this formula, how do we arrive at two J and two σ values?? It must have something to do with the direct products in expression for J = A&otimes;I + I&otimes;B which each need two a and b values. Need to explain origin properly... )


 * $$ \Pi(p_\text{rest},m)^{J'\sigma'}{}_{J\sigma} \propto \delta^{J'}{}_{s}\delta^{J}{}_{s}\delta^{\sigma'}{}_{\sigma} \,, $$

which filters only those J values which equal the spin s of the particle (σ = −s, ..., s is the z-component projection spin quantum number for s). In the particle's rest frame, the only degrees of freedom are rotations. The wave function can transform under rotations, with J = A&otimes;I + I&otimes;B the generator.

In terms of the a and b labels this is related to the CG coefficients (will add formula later).

The corresponding operator in the boosted frame is


 * $$ D(L) \Pi(p_\text{rest},m)D(L^{-1}) = \Pi'(L p_\text{rest},m)  $$

in detail


 * $$ D(L)^{k' \ell'}{}_{J'\sigma'} \Pi(p_\text{rest},m)^{J'\sigma'}{}_{J\sigma} D(L^{-1})^{J\sigma}{}_{k\ell} = \Pi'(L p_\text{rest},m)^{k'\ell'}{}_{k\ell}  $$

so that


 * $$ D(L)^{k' \ell'}{}_{s\sigma} D(L^{-1})^{s\sigma}{}_{k\ell} = \Pi'(L p_\text{rest},m)^{k'\ell'}{}_{k\ell}  $$

Massless particles with spin s
=Wave function=

General representations
Let $α = (α_{1}, α_{2}, ..., α_{n})$ be dimensionless discrete-valued observables, and $ω = (ω_{1}, ω_{2}, ..., ω_{m})$ be continuous-valued observables (not necessarily dimensionless). All $α$ are in an $n$-dimensional set $A = A_{1} × A_{2} × ... A_{n}$ where each $A_{i}$ is the set of allowed values for $α_{i}$, likewise all $ω$ are in an $m$-dimensional "volume" $Ω ⊆ ℝ^{m}$ where $Ω = Ω_{1} × Ω_{2} × ... Ω_{m}$ and each $Ω_{i} ⊆ ℝ$ is the set of allowed values for $ω_{i}$, a subset of the real numbers $ℝ$. For generality $n$ and $m$ are not necessarily equal.

Then, $Ψ(α, ω, t)$ is referred to as the "wave function" of the system.

For example, for a single particle in 3d with spin s, neglecting other degrees of freedom, using Cartesian coordinates, we could take $α = (s_{z})$ for the spin quantum number of the particle along the z direction, and $ω = (x, y, z)$ for the particle's position coordinates. Here $A = {−s, −s + 1, ..., s − 1, s}$ is the set of allowed spin quantum numbers and $Ω = ℝ^{3}$ is the set of of all possible particle positions throughout 3d position space. An alternative choice is $α = (s_{y})$ for the spin quantum number along the y direction and $ω = (p_{x}, p_{y}, p_{z})$ for the particle's momentum components. In this case $A$ and $Ω$ are the same.

In the Copenhagen interpretation, the probability density of finding the system in any state is


 * $$\rho=|\Psi(\boldsymbol{\alpha},\boldsymbol{\omega},t)|^2$$

The probability of finding system with $α$ in some or all possible discrete-variable configurations, $D ⊆ A$, and $ω$ in some or all possible continuous-variable configurations, $C ⊆ Ω$, is the sum and integral over the density,


 * $$P=\sum_{\boldsymbol{\alpha}\in D}\int_C \rho \, d^m\boldsymbol{\omega}$$

where $d^{m}ω = dω_{1}dω_{2}...dω_{m}$ is a "differential volume element" in the continuous degrees of freedom. The units of the wavefunction are then such that $ρ d^{m}ω$ is dimensionless, by dimensional analysis $Ψ$ must have the same units as $(ω_{1}ω_{2}...ω_{m})^{−1/2}$. Since the sum of all probabilities must be 1, the normalization condition


 * $$1=\sum_{\boldsymbol{\alpha}\in A}\int_{\Omega} \rho \, d^m\boldsymbol{\omega}$$

must hold at all times during the evolution of the system. The interpretation is the system will be in a particular state, all the $α_{i}$ and $ω_{j}$ will have particular values at the time $t$ the system is measured.

Every value of the wave function is accumulated into a single vector in Dirac notation


 * $$|\Psi\rangle=\sum_{\boldsymbol{\alpha}}\int d^m\boldsymbol{\omega}\Psi(\boldsymbol{\alpha},\boldsymbol{\omega},t)|\boldsymbol{\alpha},\boldsymbol{\omega}\rangle$$

in which $(α, ω)$ index the components of the vector, and $|α, ω\rangle$ are the basis vectors in this representation.

One particle states in 3d position space
The position-space wave function of a single particle without spin in three spatial dimensions is similar to the case of one spatial dimension above:
 * $$\Psi(\mathbf{r},t)$$

where $r$ is the position vector in three-dimensional space, and $t$ is time. As always $Ψ(r, t)$ is a complex-valued function of real variables. As a single vector in Dirac notation


 * $$|\Psi(t)\rangle = \int d^3 \mathbf{r} \Psi(\mathbf{r},t) |\mathbf{r}\rangle $$

All the previous remarks on inner products, momentum space wave functions, Fourier transforms, and so on extend to higher dimensions.

For a particle with spin, ignoring the position degrees of freedom, the wave function is a function of spin only (time is a parameter);


 * $$\xi(s_z,t)$$

where $s_{z}$ is the spin projection quantum number along the $z$ axis. (The $z$ axis is an arbitrary choice; other axes can be used instead if the wave function is transformed appropriately, see below.) The $s_{z}$ parameter, unlike $r$ and $t$, is a discrete variable. For example, for a spin-1/2 particle, $s_{z}$ can only be $+1/2$ or $−1/2$, and not any other value. (In general, for spin $s$, $s_{z}$ can be $s, s − 1, ..., −s + 1, −s$). Inserting each quantum number gives a complex valued function of space and time, there are $2s + 1$ of them. These can be arranged into a column vector


 * $${ \xi = \begin{bmatrix} \xi(s,t) \\ \xi(s-1,t) \\ \vdots \\ \xi(-(s-1),t) \\ \xi(-s,t) \\ \end{bmatrix} = \xi(s,t) \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ \end{bmatrix} + \xi(s-1,t)\begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \\ 0 \\ \end{bmatrix} + \cdots + \xi(-(s-1),t) \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ \end{bmatrix} + \xi(-s,t) \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \\ \end{bmatrix} } $$

In bra ket notation, these easily arrange into the components of a vector


 * $$|\xi (t)\rangle = \sum_{s_z=-s}^s \xi(s_z,t) | s_z \rangle $$

The entire vector $|s_{z}\rangle$ is a solution of the Schrödinger equation (with a suitable Hamiltonian), which unfolds to a coupled system of $|s_{z}\rangle$ ordinary differential equations with solutions $ξ$. This contrasts the solutions to position space wave functions, the position coordinates being continuous degrees of freedom, because then the Schrödinger equation does take the form of a wave equation.

More generally, for a particle in 3d with any spin, the wave function can be written in "position–spin space" as:
 * $$\Psi(\mathbf{r},s_z,t)$$

and these can also be arranged into a column vector


 * $$\Psi(\mathbf{r},t) = \begin{bmatrix} \Psi(\mathbf{r},s,t) \\ \Psi(\mathbf{r},s-1,t) \\ \vdots \\ \Psi(\mathbf{r},-(s-1),t) \\ \Psi(\mathbf{r},-s,t) \\ \end{bmatrix}$$

in which the spin dependence is placed in indexing the entries, and the wave function is a complex vector-valued function of space and time only.

All values of the wave function, not only for discrete but continuous variables also, collect into a single vector


 * $$|\Psi(t)\rangle = \sum_{s_z}\int d^3 \mathbf{r} \Psi(\mathbf{r},s_z,t) |\mathbf{r}, s_z\rangle $$

For a single particle, the tensor product $2s + 1$ of its position state vector $ξ(s, t), ξ(s − 1, t), ..., ξ(−s, t)$ and spin state vector $&otimes;$ gives the composite position-spin state vector


 * $$|\psi(t)\rangle \otimes |\xi(t)\rangle = \sum_{s_z}\int d^3 \mathbf{r} \psi(\mathbf{r},t)\xi(s_z,t) |\mathbf{r}\rangle \otimes |s_z\rangle $$

with the identifications


 * $$|\Psi (t)\rangle = |\psi(t)\rangle \otimes |\xi(t)\rangle $$
 * $$\Psi(\mathbf{r},s_z,t) = \psi(\mathbf{r},t)\xi(s_z,t) $$
 * $$|\mathbf{r},s_z \rangle= |\mathbf{r}\rangle \otimes |s_z\rangle $$

The tensor product factorization is only possible if the orbital and spin angular momenta of the particle are separable in the Hamiltonian operator underlying the system's dynamics (in other words, the Hamiltonian can be split into the sum of orbital and spin terms ). The time dependence can be placed in either factor, and time evolution of each can be studied separately. The factorization is not possible for those interactions where an external field or any space-dependent quantity couples to the spin; examples include a particle in a magnetic field, and spin-orbit coupling.

The preceding discussion is not limited to spin as a discrete variable, the total angular momentum J may also be used. Other discrete degrees of freedom, like isospin, can expressed similarly to the case of spin above.

Thomas rotation
In the composition of two Lorentz boosts (see Wigner rotation), we encounter


 * $$ \begin{align}\mathbf{M} & =\gamma_\mathbf{u}\gamma_\mathbf{v}\frac{\mathbf{vu}^\mathrm{T}}{c^2}+\left(\mathbf{I}+\dfrac{\gamma_\mathbf{v}^2}{\gamma_\mathbf{v}+1}\frac{\mathbf{vv}^\mathrm{T}}{c^2}\right)\left(\mathbf{I}+\dfrac{\gamma_\mathbf{u}^2}{\gamma_\mathbf{u}+1}\frac{\mathbf{uu}^\mathrm{T}}{c^2}\right) \\

& =\mathbf{I}+\gamma_\mathbf{u}\gamma_\mathbf{v}\frac{\mathbf{vu}^\mathrm{T}}{c^2}+ \dfrac{\gamma_\mathbf{u}^2}{\gamma_\mathbf{u}+1}\frac{\mathbf{uu}^\mathrm{T}}{c^2} +\dfrac{\gamma_\mathbf{v}^2}{\gamma_\mathbf{v}+1}\frac{\mathbf{vv}^\mathrm{T}}{c^2}+\dfrac{\gamma_\mathbf{v}^2}{\gamma_\mathbf{v}+1}\dfrac{\gamma_\mathbf{u}^2}{\gamma_\mathbf{u}+1}\frac{\mathbf{vv}^\mathrm{T}}{c^2}\frac{\mathbf{uu}^\mathrm{T}}{c^2} \\

&=\mathbf{I}+\dfrac{\gamma_\mathbf{u}^2}{\gamma_\mathbf{u}+1}\frac{\mathbf{uu}^\mathrm{T}}{c^2} +\dfrac{\gamma_\mathbf{v}^2}{\gamma_\mathbf{v}+1}\frac{\mathbf{vv}^\mathrm{T}}{c^2}+\gamma_\mathbf{u}\gamma_\mathbf{v}\left(1 + \dfrac{\gamma_\mathbf{v}}{\gamma_\mathbf{v}+1}\dfrac{\gamma_\mathbf{u}}{\gamma_\mathbf{u}+1}\frac{\mathbf{v}^\mathrm{T}\mathbf{u}}{c^2}\right) \frac{\mathbf{vu}^\mathrm{T}}{c^2} \end{align}$$

The above formulae constitute the relativistic velocity addition and the Thomas rotation explicitly in the general Lorentz transformations. Throughout, the important formula


 * $$\mathbf{M}=\mathbf{R}+\frac1{\gamma+1}\mathbf{b}\mathbf{a}^\mathrm{T} $$

holds, allowing the rotation matrix (and hence the axis and angle) to be defined completely in terms of the relative velocities $|ψ\rangle$ and $|ξ\rangle$.

In the axis–angle representation, the general 3d rotation matrix is


 * $$\mathbf{R} = \mathbf{I} + \mathbf{E}\sin\epsilon + \mathbf{E}^2(1-\cos\epsilon) \,, $$

where the components of a unit vector $u$ parallel to the axis are arranged into the antisymmetric matrix


 * $$\mathbf{E} = \begin{bmatrix} 0 & -e_z & e_y \\ e_z & 0 & -e_x \\ -e_y & e_x & 0 \end{bmatrix} $$

(which should not be confused for the Cartesian unit vectors $v$). Here the right-handed convention for the spatial coordinates is used (see orientation (vector space)), so that rotations are positive in the anticlockwise sense according to the right-hand rule, and negative in the clockwise sense. This matrix rotates any 3d vector about the axis $D$ through angle $u, v$ anticlockwise (an active transformation), which has the equivalent effect of rotating the coordinate frame clockwise about the same axis through the same angle (a passive transformation).

Starting from $tom[u, v]$, the matrix $gyr[u, v]$ rotates this into $R$ anticlockwise, it follows their cross product (in the right-hand convention)


 * $$\mathbf{a}\times\mathbf{b}= \frac{\gamma_\mathbf{u}\gamma_\mathbf{v}(\gamma^2 -1)(\gamma+\gamma_\mathbf{v}+\gamma_\mathbf{u}+1)}{(\gamma_\mathbf{v}+1)(\gamma_\mathbf{u}+1)(\gamma+1)c^2} \mathbf{u}\times\mathbf{v}$$

defines the axis correctly, therefore the axis is also parallel to $R$, geometrically this is perpendicular to the plane of the boost velocities. Since the magnitude of $e$ is neither interesting nor important, only the direction is, it is customary to normalize the vector into the unit vector above thus


 * $$\mathbf{e} = \frac{\mathbf{u}\times\mathbf{v}}{|\mathbf{u}\times\mathbf{v}|}$$

which still completely defines the direction of the axis without loss of information.

The angle $e_{x}, e_{y}, e_{z}$ between $e$ and $ε$ is not the same as the angle $a$ between $R$ and $b$.

The two cross products are


 * $$\mathbf{a}\times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\epsilon \mathbf{e} $$
 * $$\mathbf{u} \times \mathbf{v} = |\mathbf{u}||\mathbf{v}|\sin\alpha \mathbf{e} $$

hence the angles are related by


 * $$\sin\epsilon = |\mathbf{u}||\mathbf{v}|\sin\alpha \frac{\gamma_\mathbf{u}\gamma_\mathbf{v} (\gamma+\gamma_\mathbf{v}+\gamma_\mathbf{u}+1)}{(\gamma_\mathbf{v}+1)(\gamma_\mathbf{u}+1)(\gamma+1)}$$

so they cannot be equal in general.

The angle of a rotation matrix in the axis–angle representation can be found from the trace of the rotation matrix, the general result for any axis is


 * $$ \mathrm{tr}(\mathbf{R}) = 1+2\cos\epsilon $$

which in turn is equivalent to, by the linearity of the trace


 * $$ \mathrm{tr}(\mathbf{R}) = \mathrm{tr}(\mathbf{M}) - \frac{1}{\gamma+1} \mathrm{tr}(\mathbf{ba}^\mathrm{T}) $$

The trace of $u×v$ appears to be messy, but can be reduced to a compact expression using the facts that


 * $$\mathrm{tr}(\mathbf{I})=3\,,\quad\mathrm{tr}\left(\frac{\mathbf{uu}^\mathrm{T}}{c^2}\right)=\frac{u^2}{c^2}=\frac{\gamma_\mathbf{u}^2-1}{\gamma_\mathbf{u}^2}\,,\quad\mathrm{tr}\left(\frac{\mathbf{vv}^\mathrm{T}}{c^2}\right)=\frac{v^2}{c^2}=\frac{\gamma_\mathbf{v}^2-1}{\gamma_\mathbf{v}^2}$$


 * $$\mathrm{tr}(\mathbf{v}\mathbf{u}^\mathrm{T})=\mathbf{v}\cdot\mathbf{u}=\frac{\gamma-\gamma_\mathbf{u}\gamma_\mathbf{v}}{\gamma_\mathbf{u}\gamma_\mathbf{v}}\,,\quad\mathbf{v}^\mathrm{T}\mathbf{u}=\mathbf{v}\cdot\mathbf{u}$$

which result from


 * $$\gamma_\mathbf{u}=\frac{1}{\sqrt{1-\frac{\mathbf{u}^2}{c^2}}} \quad \gamma_\mathbf{v}=\frac{1}{\sqrt{1-\frac{\mathbf{v}^2}{c^2}}} \quad \gamma = \gamma_\mathbf{u}\gamma_\mathbf{v}\left(1+\frac{\mathbf{u}\cdot\mathbf{v}}{c^2}\right)$$

and for reference


 * $$\mathrm{tr}(\mathbf{M})=\frac{(1+\gamma_\mathbf{u}+\gamma_\mathbf{v}+\gamma)^2}{(\gamma_\mathbf{v}+1)(\gamma_\mathbf{u}+1)}-\gamma$$

The trace of $u×v$ is numerically equal to the dot product of the vectors $ε$ and $a$, and can be immediately connected to the angle $b$ between them and the overall Lorentz factor through their magnitudes, therefore


 * $$\mathrm{tr}(\mathbf{ba}^{\mathrm{T}})=\mathbf{a}\cdot\mathbf{b}=(\gamma^2-1)\cos\epsilon$$

Combining these trace formulae gives an established result


 * $$\cos\epsilon = \frac{(1+\gamma+\gamma_\mathbf{u}+\gamma_\mathbf{v})^2}{(1+\gamma)(1+\gamma_\mathbf{u})(1+\gamma_\mathbf{v})} - 1 $$

The rotation is simply a "static" rotation and there is no relative rotational motion between the frames, there is relative translational motion in the boost. However, if the frames accelerate, then the rotated frame rotates with an angular velocity. This effect is known as the Thomas precession, and arises purely from the kinematics of successive Lorentz boosts.

Interpreting the axes of rotation
In frame Σ&prime;, Σ moves with velocity $α$ and F moves with velocity $u$, neither of which are rotated. This observer can define the angle $v$ between $M$ and $ba^{T}$, so the angle between $a$ and $b$ is $ε$, and the cross product of them in this frame is


 * $$(-\mathbf{u})\times\mathbf{v}=-\mathbf{u}\times\mathbf{v}=-|\mathbf{u}||\mathbf{v}|\sin\epsilon \mathbf{e}$$

where $−u$ is a unit vector parallel to $v$.

In this frame Σ&prime;, it is valid that


 * $$\mathbf{u}\times\mathbf{v} = |\mathbf{u}||\mathbf{v}|\sin\epsilon \mathbf{e}$$

The other observers do not measure the angle $π − ε$ (or $−u$) because they only measure one of $v$ or $u$ (or their negatives), as well as one of the composite velocities $v$ or $ε$ (or their negatives). Also, the relativity of simultaneity means the angles between $e$ and $u×v$ is meaningless in the other frames. As perceived from the other frames Σ and Σ&prime;&prime;, is the rotation axis the same as in Σ&prime;, in other words in the same direction of $−u$? Or are there different directions of rotation axes which somehow accounts for all/part of the Wigner rotation? The answer is the direction of the rotation axis is the same in all frames, but only the relative velocities of the other frames may be used to define the axis.

In frame Σ, Σ&prime; moves with velocity $v$, Σ&prime;&prime; moves with velocity


 * $$\mathbf{u}\oplus\mathbf{v}=\frac{c}{\gamma}\mathbf{a}$$

and is rotated clockwise about an axis parallel to


 * $$\mathbf{u}\times\frac{c}{\gamma}\mathbf{a}=\mathbf{u}\times(\mathbf{u}\oplus\mathbf{v})=\frac{\gamma_\mathbf{v}}{\gamma}\mathbf{u}\times\mathbf{v}$$

In frame Σ&prime;&prime;, Σ&prime; moves with velocity $π − ε$, Σ moves with velocity


 * $$-\mathbf{v}\oplus\mathbf{u}=-\frac{c}{\gamma}\mathbf{b} $$

and is rotated clockwise about an axis parallel to


 * $$-\frac{c}{\gamma}\mathbf{b}\times\mathbf{v}=-(\mathbf{v}\oplus\mathbf{u})\times\mathbf{v} =-\frac{\gamma_\mathbf{u}}{\gamma}\mathbf{u}\times\mathbf{v}$$

but since the axis is reversed for the same angle, this corresponds to a rotation in the opposite sense about $ε$, here anticlockwise.

The rotation axes in each frame are all proportional to $ε$, only the direction of the rotation axis is important, the proportionality factors are not because they have no effect on direction. (They do indicate what happens to the rotation axes in the nonrelativistic limit, when relative velocities are much less than light). In all we can conclude


 * $$\mathbf{e}=\frac{\mathbf{u}\times\mathbf{v}}{|\mathbf{u}\times\mathbf{v}|}$$

is the axis of the Thomas rotation.

Six-angular momentum (orbital)
Add to the tensor section.

The transformation of boost components are


 * $$\begin{align}

M'^{k0} & =\Lambda^k{}_\mu\Lambda^0{}_\nu M^{\mu\nu}\\ & =\Lambda^k{}_0\Lambda^0{}_0M^{00}+\Lambda^k{}_i \Lambda^0{}_0M^{i0}+\Lambda^k{}_0\Lambda^0{}_j M^{0j}+\Lambda^k{}_i\Lambda^0{}_j M^{ij} \\ & = (\Lambda^k{}_i \Lambda^0{}_0-\Lambda^k{}_0\Lambda^0{}_i )M^{i0} +\Lambda^k{}_i \Lambda^0{}_j M^{ij} \\ \end{align}$$

as for the orbital angular momentum


 * $$\begin{align}

M'^{k\ell} & =\Lambda^k{}_\mu\Lambda^\ell{}_\nu M^{\mu\nu} \\ & =\Lambda^k{}_0\Lambda^\ell{}_0M^{00}+\Lambda^k{}_i \Lambda^\ell{}_0M^{i0}+\Lambda^k{}_0\Lambda^\ell{}_j M^{0j}+\Lambda^k{}_i \Lambda^\ell{}_j M^{ij} \\ & = (\Lambda^k{}_i \Lambda^\ell{}_0-\Lambda^k{}_0\Lambda^\ell{}_i )M^{i0} +\Lambda^k{}_i \Lambda^\ell{}_j M^{ij} \end{align}$$

Since


 * $$M^{0i}=-M^{i0}=-cN^i$$
 * $$M^{ij}=x^ip^j-x^jp^i =L^{ij}=\varepsilon^{ijk}L_k$$

and


 * $$\begin{align}

\Lambda^0{}_0 & = \gamma \\ \Lambda^i{}_0 & = - \gamma \beta^i \quad \Lambda^0{}_i = - \gamma \beta_i \\ \Lambda^i{}_j & = \delta^i{}_j + \frac{(\gamma-1)}{\beta^2} \beta^i\beta_j \end{align}$$

we have


 * $$\begin{align}\Lambda^k{}_i\Lambda^\ell{}_0-\Lambda^k{}_0\Lambda^\ell{}_i & =\left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right)(-\gamma\beta^\ell)-(-\gamma\beta^k)\left(\delta^\ell{}_i+\frac{(\gamma-1)}{\beta^2}\beta^\ell\beta_i\right)\\

& =\gamma\left(\beta^k\delta^\ell{}_i-\beta^\ell\delta^k{}_i\right) \end{align}$$


 * $$\begin{align}\Lambda^k{}_i\Lambda^0{}_0-\Lambda^k{}_0\Lambda^0{}_i & =\left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right)\gamma-(-\gamma\beta^k)(-\gamma\beta^i)\\

& =\gamma\left[\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i-\gamma\beta^k\beta^i\right]\\ & =\gamma\left[\delta^k{}_i+\left(\frac{(\gamma-1)}{\beta^2}-\gamma\right)\beta^k\beta_i\right]\\ & =\gamma\left[\delta^k{}_i+\left(\frac{\gamma-\gamma\beta^2-1}{\beta^2}\right)\beta^k\beta_i\right]\\ & =\gamma\left[\delta^k{}_i+\left(\frac{\gamma^{-1}-1}{\beta^2}\right)\beta^k\beta_i\right]\\ & =\gamma\delta^k{}_i-\left(\frac{\gamma-1}{\beta^2}\right)\beta^k\beta_i \end{align}$$


 * $$\Lambda^k{}_i\Lambda^\ell{}_0 = \Lambda^k{}_i\Lambda^0{}_\ell = -\gamma\beta^\ell \left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right) $$


 * $$\begin{align}

\Lambda^k{}_i\Lambda^0{}_0 & =\left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right)\gamma \\ \Lambda^k{}_0\Lambda^0{}_i & = \gamma^2 \beta^k \beta^i \\ \end{align}$$


 * $$\begin{align}

\Lambda^k{}_i\Lambda^\ell{}_j & =\left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right)\left(\delta^\ell{}_j+\frac{(\gamma-1)}{\beta^2}\beta^\ell\beta_j\right)\\ & =\delta^k{}_i\delta^\ell{}_j+\frac{(\gamma-1)}{\beta^2}\delta^k{}_i\beta^\ell\beta_j+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\delta^\ell{}_j+\frac{(\gamma-1)}{\beta^2}\frac{(\gamma-1)}{\beta^2}\beta^\ell\beta_j\beta^k\beta_i \end{align}$$

gives


 * $$\begin{align}

cN'^k & = (\Lambda^k{}_i \Lambda^0{}_0-\Lambda^k{}_0\Lambda^0{}_i )cN^i +\Lambda^k{}_i \Lambda^0{}_j \varepsilon^{ijn} L_n \\ & = \left(\gamma\delta^k{}_i-\left(\frac{\gamma-1}{\beta^2}\right)\beta^k\beta_i\right)cN^i + -\gamma\beta^j \left(\delta^k{}_i+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\right) \varepsilon^{ijn} L_n \\

& = \gamma cN^k -\left(\frac{\gamma-1}{\beta^2}\right)\beta^k (\beta_i cN^i) -\gamma\beta^j \delta^k{}_i \varepsilon^{ijn} L_n -\gamma\frac{(\gamma-1)}{\beta^2}\beta^j \beta^k\beta_i \varepsilon^{ijn} L_n \\

& = \gamma cN^k -\left(\frac{\gamma-1}{\beta^2}\right)\beta^k (\beta_i cN^i) -\gamma \beta^j \varepsilon^{kjn} L_n \\ \end{align}$$

or in vector form, dividing by c


 * $$\mathbf{N}' = \gamma \mathbf{N} -\left(\frac{\gamma-1}{\beta^2}\right)\boldsymbol{\beta} (\boldsymbol{\beta}\cdot \mathbf{N}) -\frac{1}{c}\gamma \boldsymbol{\beta}\times\mathbf{L}$$

or reinstating β = v/c,


 * $$\mathbf{N}' = \gamma \mathbf{N} -\left(\frac{\gamma-1}{v^2}\right)\mathbf{v} (\mathbf{v}\cdot \mathbf{N}) - \gamma \mathbf{v}\times\mathbf{L}$$

and


 * $$\begin{align}

L'^{k\ell} & = (\Lambda^k{}_i \Lambda^\ell{}_0-\Lambda^k{}_0\Lambda^\ell{}_i )cN^i +\Lambda^k{}_i \Lambda^\ell{}_j L^{ij} \\ & = \gamma c \left(\beta^k\delta^\ell{}_i-\beta^\ell\delta^k{}_i\right) N^i + \left(\delta^k{}_i\delta^\ell{}_j+\frac{(\gamma-1)}{\beta^2}\delta^k{}_i\beta^\ell\beta_j+\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i\delta^\ell{}_j+\frac{(\gamma-1)}{\beta^2}\frac{(\gamma-1)}{\beta^2}\beta^\ell\beta_j\beta^k\beta_i\right) L^{ij} \\

& = \gamma c ( \beta^k N^\ell -\beta^\ell N^k ) + L^{k\ell} +\frac{(\gamma-1)}{\beta^2} \beta^\ell \beta_j L^{kj} +\frac{(\gamma-1)}{\beta^2}\beta^k\beta_i L^{i\ell} \\

& = \gamma c ( \beta^k N^\ell -\beta^\ell N^k ) + L^{k\ell} + \frac{(\gamma-1)}{\beta^2} \left(\beta^\ell \beta_i (-L^{ik}) + \beta^k\beta_i L^{i\ell} \right) \end{align}$$

or converting to pseudovector form


 * $$\begin{align}

\varepsilon^{k\ell n}L'_n & = \gamma c ( \beta^k N^\ell -\beta^\ell N^k ) + \varepsilon^{k\ell n}L_n + \frac{(\gamma-1)}{\beta^2} \left(\beta^\ell \beta_i (-\varepsilon^{ikn}L_n) + \beta^k\beta_i \varepsilon^{i\ell n}L_n \right) \\

& = \gamma c ( \beta^k N^\ell -\beta^\ell N^k ) + \varepsilon^{k\ell n}L_n + \frac{(\gamma-1)}{\beta^2} \left(\beta^\ell (\varepsilon^{kin}\beta_i L_n) - \beta^k \varepsilon^{\ell i n}\beta_i L_n \right) \\

\end{align} $$

in vector notation


 * $$\mathbf{L}' = \gamma c \boldsymbol{\beta}\times\mathbf{N} + \mathbf{L} + \frac{(\gamma-1)}{\beta^2} \boldsymbol{\beta}\times(\boldsymbol{\beta}\times\mathbf{L})$$

or reinstating β = v/c,


 * $$\mathbf{L}' = \gamma \mathbf{v}\times\mathbf{N} + \mathbf{L} + \frac{(\gamma-1)}{v^2} \mathbf{v}\times(\mathbf{v}\times\mathbf{L})$$

=Tensors and spinors=

Co-/contra-variance
This table summarizes how the manipulation of covariant and contravariant indices fit in with invariance under a passive transformation between bases, and with the inner product. The barred indices refer to the final coordinate system after the transformation.


 * {| class="wikitable"

! Geometry/Algebra ! Covariance ! Contravariance !A... ! ...has a basis... ! ...to coordinate... ! ...in which... ! The coordinate vector transformation is... ! ...while the basis transformation is... !...which are invariant because... ! The inner product is...
 * +Summary of above
 * vector (or contravariant vector)
 * covector (or 1-form, covariant vector, dual vector)
 * $$\mathbf{e}_\alpha = \partial_\alpha \mathbf{r} = \frac{\partial \mathbf{r}}{\partial x^\alpha}$$, which are tangent
 * $$\mathbf{e}^\alpha = \nabla x^\alpha $$, which are normal
 * curves
 * surfaces
 * one coordinate varies, all others are constant.
 * one coordinate is constant, all others vary.
 * $$a^\bar{\alpha} = a^\beta L^\bar{\alpha}{}_\beta $$
 * $$\Omega_\bar{\alpha} = \Omega_\gamma L^\gamma{}_\bar{\alpha} $$
 * $$\mathbf{e}_\bar{\alpha} = L^\gamma{}_\bar{\alpha} \mathbf{e}_\gamma $$
 * $$\mathbf{e}^\bar{\alpha} = L^\bar{\alpha}{}_\beta \mathbf{e}^\beta $$
 * $$a^\bar{\alpha}\mathbf{e}_\bar{\alpha} = a^\beta L^\bar{\alpha}{}_\beta L^\gamma{}_\bar{\alpha} \mathbf{e}_\gamma = a^\beta \delta_\beta{}^\gamma \mathbf{e}_\gamma = a^\gamma \mathbf{e}_\gamma $$
 * $$\Omega_\bar{\alpha}\mathbf{e}^\bar{\alpha} = \Omega_\gamma L^\gamma{}_\bar{\alpha} L^\bar{\alpha}{}_\beta \mathbf{e}^\beta = \Omega_\gamma \delta^\gamma{}_\beta \mathbf{e}^\beta = \Omega_\beta \mathbf{e}^\beta $$
 * colspan="2"|

$$\begin{align}\langle \mathbf{a},\boldsymbol{\Omega} \rangle & = \langle a^\alpha\mathbf{e}_\alpha, \Omega_\beta \mathbf{e}^\beta \rangle \\ & = a^\alpha \langle \mathbf{e}_\alpha, \mathbf{e}^\beta \rangle \Omega_\beta \\ & = a^\alpha \delta^\beta{}_\alpha \Omega_\beta \\ & = a^\alpha \Omega_\alpha \\ \end{align}$$
 * }
 * }