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

• 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

Start with two Lorentz boostsB(ζ₁) andB(ζ₂) with rapidities

${\displaystyle {\mathit{\zeta }}_1={\mathbf{n}}_1{\tanh }^{-1}{\beta }_1,{\mathit{\zeta }}_2={\mathbf{n}}_2{\tanh }^{-1}{\beta }_2}$

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

${\displaystyle {\mathbf{n}}_1=\frac{{\mathit{\beta }}_1}{{\beta }_1},{\mathbf{n}}_2=\frac{{\mathit{\beta }}_2}{{\beta }_2}}$

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

${\displaystyle {\mathit{\beta }}_1=\frac{{\mathbf{v}}_1}{c},{\mathit{\beta }}_2=\frac{{\mathbf{v}}_2}{c}}$

There are two inequivalent composite Lorentz transformationsB(ζ₁)B(ζ₂) andB(ζ₂)B(ζ₁). There are two composite boosts

${\displaystyle {\mathit{\beta }}_1\oplus {\mathit{\beta }}_2=\frac{1}{1+{\mathit{\beta }}_1\cdot {\mathit{\beta }}_2}\left[{\mathit{\beta }}_1+\frac{{\mathit{\beta }}_2}{{\gamma }_1}+\frac{{\gamma }_1}{1+{\gamma }_1}({\mathit{\beta }}_1\cdot {\mathit{\beta }}_2){\mathit{\beta }}_1\right]}$

${\displaystyle {\mathit{\beta }}_2\oplus {\mathit{\beta }}_1=\frac{1}{1+{\mathit{\beta }}_2\cdot {\mathit{\beta }}_2}\left[{\mathit{\beta }}_2+\frac{{\mathit{\beta }}_1}{{\gamma }_2}+\frac{{\gamma }_2}{1+{\gamma }_2}({\mathit{\beta }}_2\cdot {\mathit{\beta }}_1){\mathit{\beta }}_2\right]}$

where ⊕ is the relativisticvelocity-addition, defined by the first expression above (second one follows immediately by switching betas).

The composite rapidities are

${\displaystyle \mathit{\zeta }=\mathbf{n}{\tanh }^{-1}\beta ,\overline{\mathit{\zeta }}=\overline{\mathbf{n}}{\tanh }^{-1}\overline{\beta }}$

The unit vectors

${\displaystyle \mathbf{n}=\frac{{\mathit{\beta }}_1\oplus {\mathit{\beta }}_2}{|{\mathit{\beta }}_1\oplus {\mathit{\beta }}_2|},\overline{\mathbf{n}}=\frac{{\mathit{\beta }}_2\oplus {\mathit{\beta }}_1}{|{\mathit{\beta }}_2\oplus {\mathit{\beta }}_1|}}$

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

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

In summary

${\displaystyle \mathit{\zeta }=\frac{{\mathit{\beta }}_1\oplus {\mathit{\beta }}_2}{\beta }{\tanh }^{-1}\beta ,\overline{\mathit{\zeta }}=\frac{{\mathit{\beta }}_2\oplus {\mathit{\beta }}_1}{\beta }{\tanh }^{-1}\beta }$

so these vectors have the same magnitude

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

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

In terms of rapidity

${\displaystyle {\mathit{\beta }}_1\oplus {\mathit{\beta }}_2=\frac{1}{1+{\mathbf{n}}_1\cdot {\mathbf{n}}_2\tanh {\zeta }_1\tanh {\zeta }_2}\left[{\mathbf{n}}_1\tanh {\zeta }_1+\frac{{\mathbf{n}}_2\tanh {\zeta }_2}{\cosh {\zeta }_1}+\frac{\cosh {\zeta }_1}{1+\cosh {\zeta }_1}({\mathbf{n}}_1\cdot {\mathbf{n}}_2){\mathbf{n}}_1{\tanh }^2{\zeta }_1\tanh {\zeta }_2\right]}$

${\displaystyle {\mathit{\beta }}_2\oplus {\mathit{\beta }}_1=\frac{1}{1+{\mathbf{n}}_2\cdot {\mathbf{n}}_1\tanh {\zeta }_2\tanh {\zeta }_1}\left[{\mathbf{n}}_2\tanh {\zeta }_2+\frac{{\mathbf{n}}_1\tanh {\zeta }_1}{\cosh {\zeta }_2}+\frac{\cosh {\zeta }_2}{1+\cosh {\zeta }_2}({\mathbf{n}}_2\cdot {\mathbf{n}}_1){\mathbf{n}}_2{\tanh }^2{\zeta }_2\tanh {\zeta }_1\right]}$

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

so in full,

${\displaystyle \mathit{\zeta }=\frac{{\mathit{\beta }}_1\oplus {\mathit{\beta }}_2}{\beta }{\tanh }^{-1}\beta ,\overline{\mathit{\zeta }}=\frac{{\mathit{\beta }}_2\oplus {\mathit{\beta }}_1}{\beta }{\tanh }^{-1}\beta }$

Ifζ₁ andζ₂ are separated by an angle, and have magnitudesζ₁ andζ₂, these are related toζ by (add formula later...)

Line element and energy-momentum relation

In the following, the four spacetime coordinates of the particle arex = (ct,x) (a.k.a. "four position vector"), and itsfour momentum isp = (E/c,p). TheMinkowski metricη throughout is (+, −, −, −). Natural unitsc =ħ = 1 will be used as standard in RQM.

Using the relativistic invariance of the infinitesimalline element

${\displaystyle d{x}_{\mu }d{x}^{\mu }=d{\tau }^2}$

and the definition offour momentum

${\displaystyle p_{\mu }=m\frac{d{x}_{\mu }}{d\tau }}$

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

${\displaystyle p_{\mu }{p}^{\mu }=m^2}$

or explicitly in terms of energy and 3-momentum

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

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

Wigner's classification

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

${\displaystyle p^2=m^2>0,p^0>0}$

${\displaystyle p^2=0,p\ne 0,p^0>0}$

${\displaystyle p^2=p=0}$

Postulates of QM

All RWEs must also be consistent with theSchrödinger 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 thefour gradient with components

${\displaystyle {\hat{p}}_{\mu }=i{\mathrm{\partial }}_{\mu }}$

${\displaystyle \hat{p}=(\hat{E},\hat{\mathbf{p}})=i\left(\frac{\mathrm{\partial }}{\mathrm{\partial }t},-\mathrm{\nabla }\right)}$

Spin and angular momentum

Further, in non-relativistic quantum mechanics a particle with spins is described by a wavefunctionψ with 2s + 1 componentsψ_σ indexed by the spin projection quantum numberσ = −s, −s + 1, ...,s − 1,s, often these are arranged into a column vector

${\displaystyle \psi =\left(\begin{array}{c}{\psi }_s\\ {\psi }_{s-1}\\ ⋮\\ {\psi }_{-s+1}\\ {\psi }_{-s}\end{array}\right)}$

Still in non-relativistic quantum mechanics, thespin operator for a particle of spins is a vectors of matricess_x,s_y,s_z, each are (2s + 1)×(2s + 1) matrices.

There is also therelativistic angular momentum tensorM, and thePauli-lubanski pseudovector is defined fromM andp. 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.

${\displaystyle [p_{\mu },p_{\nu }]=0}$

${\displaystyle \frac{1}{i}[M_{\mu \nu },p_{\rho }]={\eta }_{\mu \rho }{p}_{\nu }-{\eta }_{\nu \rho }{p}_{\mu }}$

${\displaystyle \frac{1}{i}[M_{\mu \nu },M_{\rho \sigma }]={\eta }_{\mu \rho }{M}_{\nu \sigma }-{\eta }_{\mu \sigma }{M}_{\nu \rho }-{\eta }_{\nu \rho }{M}_{\mu \sigma }+{\eta }_{\nu \sigma }{M}_{\mu \rho },}$

For any representation of the Poincare group, the infinitesimal unitary operators corresponding to boostsU(B), rotationsU(R), and translationsU(a) on the Hilbert space for the system are

${\displaystyle U(B)\approx I-i\mathit{\zeta }\cdot \mathbf{K}}$

${\displaystyle U(R)\approx I-i\mathit{\theta }\cdot \mathbf{J}}$

${\displaystyle U(a)\approx I+ia\cdot p}$

or

${\displaystyle U(\mathrm{\Lambda })\approx I-\frac{i}{2}{\omega }_{\alpha \beta }{M}^{\alpha \beta }}$

${\displaystyle U(a)\approx I+i{a}^{\mu }{p}_{\mu }}$

whereω is an antisymmetric parameter matrix for boosts and rotations.

In the finite case, one obtains the matrix exponentials. The generatorsJ,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.

Combining the four momentum operator and energy momentum relation leads to the KG eqn

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

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

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

for suitable quantitiesγ^μ. Following LL vol 4 (at top of my head), expand the brackets

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

and split the productγ^μγ^ν into symmetric and antisymmetric products

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

The summation of the antisymmetric tensorγ^μγ^ν −γ^νγ^μ with the symmetric tensorp_μp_ν is zero (this applies generally for the summation of any antisymmetric and symmetric tensors). Since the KG eqn must be recovered, the quantitiesγ^μ must satisfy the anticommutation relation

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

One of the factors

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

is theDirac equation, which applies to any spin-1/2 particle, massive or massless. Ifψ satisfies this equation, it will also satisfy the KG equation, but the converse is not true because it can be shown theγ^μ, known as thegamma 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 theBargmann-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 theJoos–Weinberg equation.

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.

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 theJ,K andA,B generators.Done
• Link theA,B generators to representations of the Lorentz group (not just the labels (A,B) for classifying representations, but actually connect the generatorsA andB themselves to theD matrices)
• Plainly write down how spinors corresponding to a particle of spinj 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 parametersα =θ +iζ andα* =θ −iζ 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

For every Lorentz transformation Λ on the coordinates in Minkowski spacetimeM, there is a unitary operatorD(Λ) on the Hilbert spaceH of allowed quantum statesψ of the physical system. The representationsD depend on the system.

Any Λ can be decomposed into pure boostsB and pure rotationsR. Boosts have parametersζ and generatorsK. Rotations have parametersθ and generatorsJ. The generatorsJ andK can be expressed as the following 4×4 matrices

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

${\displaystyle \mathrm{\Lambda }(\mathit{\zeta },\mathit{\theta })=e^{-\mathit{\zeta }\cdot \mathbf{K}+\mathit{\theta }\cdot \mathbf{J}}}$

Individual boosts and rotations are

${\displaystyle R(\mathit{\theta })=e^{\mathit{\theta }\cdot \mathbf{J}}}$

${\displaystyle B(\mathit{\zeta })=e^{-\mathit{\zeta }\cdot \mathbf{K}}}$

Explicit formulae for the representations for ann component object are

${\displaystyle D(\mathrm{\Lambda }(\mathit{\zeta },\mathit{\theta }))=}$

${\displaystyle D(R(\mathit{\theta }))=}$

${\displaystyle D(B(\mathit{\zeta }))=}$

Another basis for the Lorentz generators takes complex conjugates of theJ andK generators:A = (J +iK)/2 andB = (J −iK)/2. The corresponding parameters areα =θ +iζ andα* =θ −iζ.

Using the above 4×4 matrices forJ andK,A andB take the form

which satisfy the commutation relations

${\displaystyle [A_x,A_y]=i{A}_z,[B_x,B_y]=i{B}_z}$

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

${\displaystyle [A_i,A_j]=i{\epsilon }_{ijk}{A}_k,[B_i,B_j]=i{\epsilon }_{ijk}{B}_k,[A_i,B_j]=0}$

The components ofA andB each satisfy the commutation relations of the Lie algebra su(2). Taken together,A andB satisfy the commutation relations of the direct sum of their Lie algebras, su(2) ⊕ su(2). The corresponding Lie group is the tensor product SU(2) ⊗ SU(2).

The finite Lorentz transformation is

${\displaystyle \mathrm{\Lambda }(\mathit{\alpha },{\mathit{\alpha }}^{\ast })=e^{\mathit{\alpha }\cdot \mathbf{A}+{\mathit{\alpha }}^{\ast }\cdot \mathbf{B}}}$

The generatorsA andB do not have to be the 4×4 matrices matrices above. The commutation relations they satisfy are exactly those of thespin operator, indicating they can be spin matrices. LetS^(j) be the spin operator corresponding to spinj. ThenA =S^(A), a vector of three (2A + 1)×(2A + 1) matrices, andB =S^(B), a vector of three (2B + 1)×(2B + 1) matrices. The z-component spin projection quantum numbers forA_z area = −A, −A + 1, ...,A − 1,A, likewise forB_z they areb = −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 areJ =A +B andiK =A −B. In the extreme cases,

• A =0 (henceA = 0) whileB arbitrary:B = −iK =J, in other wordsB =J =S^(B) whileK =iS^(B).
• B =0 (henceB = 0) whileA arbitrary:A =iK =J, in other wordsA =J =S^(A) whileK = −iS^(A).

NoticeJ 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 momentumJ is more accurately written using the direct or tensor product ⊗ as follows

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

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

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

${\displaystyle [\mathbf{J}{]}_{a^{\prime }{b}^{\prime }ab}=[\mathbf{A}{]}_{a^{\prime }a}{\delta }_{b^{\prime }b}+{\delta }_{a^{\prime }a}[\mathbf{B}{]}_{b^{\prime }b}}$

where the multiple indices select the components of spinors the operatorJ acts on. Explicitly, ifξ_a is a 2A + 1 component spin wave function andη_b a 2B + 1 component spin wave function,J acts on the their tensor product as follows:

${\displaystyle [\mathbf{J}{]}_{a^{\prime }{b}^{\prime }ab}{\xi }_a{\eta }_b=([\mathbf{A}{]}_{a^{\prime }a}{\xi }_a){\eta }_{b^{\prime }}+{\xi }_{a^{\prime }}([\mathbf{B}{]}_{b^{\prime }b}{\eta }_b)}$

useful link

another

another

The tensor product representation is denoted and defined by [SU Particles and Fields p.231 eq 8.1.8 (in notation used here)]

${\displaystyle D^{(A,B)}(\mathit{\alpha },{\mathit{\alpha }}^{\ast })=D^{(A)}({\mathit{\alpha }}^{\ast })\otimes D^{(B)}(\mathit{\alpha })}$

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

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

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

${\displaystyle D^{(A,0)}({\mathit{\alpha }}^{\ast })\stackrel{?}{=}{e}^{i{\mathit{\alpha }}^{\ast }\cdot \mathbf{A}}\leftrightarrow \mathbf{A}={\mathbf{S}}^{(A)},{\mathit{\alpha }}^{\ast }}$

${\displaystyle D^{(0,B)}(\mathit{\alpha })\stackrel{?}{=}{e}^{i\mathit{\alpha }\cdot \mathbf{B}}\leftrightarrow \mathbf{B}={\mathbf{S}}^{(B)},\mathit{\alpha }}$

${\displaystyle D^{(j)}(\mathit{\alpha })\stackrel{?}{=}{e}^{i\mathit{\alpha }\cdot {\mathbf{S}}^{(j)}}\leftrightarrow {\mathbf{S}}^{(j)},\mathit{\alpha }}$

One can findD^(1/2,0) andD^(0,1/2) (to be derived soon).

General transformation of 2-component spinors

LL vol 4 "Fermions" chapter

A 2 component left handed spinorζ and right handed spinorη generally transform according to

${\displaystyle {\xi }^{\prime }=M\xi ,{\eta }^{\prime }=M^{\ast }\eta }$

in matrix notation

in spinor index notation

${\displaystyle {{\xi }^{\prime }}_{\lambda }=M_{\lambda }{}^{\mu }{\xi }_{\mu },{{\eta }^{\prime }}_{\dot{\lambda }}={M^{\ast }}_{\dot{\lambda }}{}^{\dot{\mu }}{\eta }_{\dot{\mu }}}$

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

D^(1/2,0) andD^(0,1/2)

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

${\displaystyle {\xi }^{\prime }=D^{(1/2,0)}\xi ,{\eta }^{\prime }=D^{(0,1/2)}\eta }$

where

${\displaystyle D^{(1/2,0)}=e^{i{\mathbf{S}}^{(1/2)}\cdot {\mathit{\alpha }}^{\ast }},D^{(0,1/2)}=e^{i{\mathbf{S}}^{(1/2)}\cdot \mathit{\alpha }}}$

and

${\displaystyle {\mathbf{S}}^{(1/2)}=\frac{1}{2}\mathit{\sigma }}$

is the spin-1/2 operator, directly proportional to thePauli matricesσ.

D^(1/2,0) ⊕D^(0,1/2)

Dirac spinors transform as

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

explicitly in matrix form

(D^(1/2,0) ⊕D^(0,1/2))^⊗n

For a tensor product ofn Dirac spinors

in whichD =D^(1/2,0) ⊕D^(0,1/2), the product transforms as then "tensor power" (tensor product of representation with itselfn times)

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

Likewise, a multicomponent spinor transforms in the same way

${\displaystyle {{\psi }^{\prime }}_{{\rho }_1{\rho }_2\cdots {\rho }_n}=D_{{\rho }_1{\sigma }_1}{D}_{{\rho }_2{\sigma }_2}\cdots D_{{\rho }_n{\sigma }_n}{\psi }_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}}$

Bargmann–Wigner spinors transform like this.

D^(j,0) andD^(0,j)

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

${\displaystyle {\xi }^{\prime }=D^{(j,0)}\xi ,{\eta }^{\prime }=D^{(0,j)}\eta }$

Need to find correspondence

${\displaystyle D^{(j,0)},D^{(0,j)}\leftrightarrow {\mathbf{S}}^{(j)},\mathit{\alpha },{\mathit{\alpha }}^{\ast }}$

D^(j,0) ⊕D^(0,j)

The 2(2j + 1)-component spinor

${\displaystyle \psi =\left(\begin{array}{c}\xi \\ \eta \end{array}\right)}$

transforms as

${\displaystyle {\psi }^{\prime }=D^{(j,0)}\oplus D^{(0,j)}\psi }$

where

The Joos-Weinberg wave function transforms like this.

Start by writing the RWE in general form

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

where Π is a linear differential operator, and depends on the massm of the particle. The linearity is required for consistency with the SE. (If the RWE equation was a nonlinear PDE, Π andψ would not be separable from each other, because Π would be a function ofψ). For generality letψ haven components, and Π be ann×n matrix to act on all the components ofψ. The relation betweenn and the spin quantum numbers will be found later.

The four spacetime coordinatesx change under aLorentz transformation Λ, while then components of the wave function collectively transform according to an×n transformation matrixD(Λ),

${\displaystyle x^{\prime }=\mathrm{\Lambda }x,{\psi }^{\prime }(x^{\prime })=D(\mathrm{\Lambda })\psi (x),}$

explicitly for all the components

${\displaystyle {x^{\prime }}^{\alpha }={\mathrm{\Lambda }}^{\alpha }{}_{\beta }{x}^{\beta },{\psi }^{\prime }(x^{\prime }{)}_{\rho }=D(\mathrm{\Lambda }{)}_{\rho \sigma }\psi (x{)}_{\sigma }.}$

Notice the four spacetime coordinatesx have spacetime (or "Lorentz") indicesα andβ which take values 0, 1, 2, 3. The wave functionsψ and transformation matrixD(Λ) simply have matrix indicesρ andσ which take the values 1, 2, ...n. The transformed wave function in terms of the original wave function, in the transformed coordinates throughout, is

${\displaystyle {\psi }^{\prime }(x^{\prime })=D(\mathrm{\Lambda })\psi ({\mathrm{\Lambda }}^{-1}{x}^{\prime }),}$

The transformation matricesD(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:${\displaystyle D({\mathrm{\Lambda }}_1{\mathrm{\Lambda }}_2)=D({\mathrm{\Lambda }}_1)D({\mathrm{\Lambda }}_2)}$

Identity element:${\displaystyle D(I)=I}$

Inverse element:${\displaystyle D(\mathrm{\Lambda }{)}^{-1}=D({\mathrm{\Lambda }}^{-1})}$

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

The transformed wave equation is

${\displaystyle {\mathrm{\Pi }}^{\prime }({\hat{p}}^{\prime },m){\psi }^{\prime }(x^{\prime })=0.}$

Projecting theHermitian conjugate wave functionsψ^† onto their own wave equations (temporarily suppressing arguments for clarity),

${\displaystyle {\psi }^{†}\mathrm{\Pi }\psi =0,}$

${\displaystyle {{\psi }^{\prime }}^{†}{\mathrm{\Pi }}^{\prime }{\psi }^{\prime }={\psi }^{†}{D}^{†}{\mathrm{\Pi }}^{\prime }D\psi =0,}$

IfD(Λ) is unitary (D(Λ)^† =D(Λ)⁻¹), comparing these equations leads to thesimilarity transformation

${\displaystyle \mathrm{\Pi }=D^{-1}{\mathrm{\Pi }}^{\prime }D,}$

in full

${\displaystyle \mathrm{\Pi }(\hat{p},m)=D(\mathrm{\Lambda }{)}^{-1}{\mathrm{\Pi }}^{\prime }({\hat{p}}^{\prime },m)D(\mathrm{\Lambda }),}$

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

${\displaystyle D(\mathrm{\Lambda })\mathrm{\Pi }(\hat{p},m)D({\mathrm{\Lambda }}^{-1})={\mathrm{\Pi }}^{\prime }(\mathrm{\Lambda }\hat{p},m).}$

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

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 theFourier transform of the original equation to obtain the momentum space equation

${\displaystyle \mathrm{\Pi }(p,m)\varphi (p)=0}$

where the FT ofψ(x) is

${\displaystyle \varphi (p)=\int \frac{N{d}^{4}x}{(2\pi {)}^3}{e}^{-i{p}_{\alpha }{x}^{\alpha }}\psi (x)}$

(including an extra normalization factorN 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,

${\displaystyle \mathrm{\Pi }(p,m)={\pi }^{\alpha }(m)p_{\alpha }+C(m)}$

In the rest frame of the particle, the momentum space wavefunctionφ(p_rest) orφ(p₀,0) only has energy dependence. Now transform to a boosted frame with momentump, so the new wave function isφ′(p′) or

${\displaystyle {\varphi }^{\prime }(p_0,\mathbf{p})=D(L)\varphi (p_0,\mathbf{0})}$

whereL is a Lorentz transformation for the boost, andD(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 attwoJ andtwoσ values?? It must have something to do with the direct products in expression forJ =A⊗I +I⊗B which each need twoa andb values. Need to explain origin properly...)

${\displaystyle \mathrm{\Pi }(p_{\text{rest}},m{)}^{J^{\prime }{\sigma }^{\prime }}{}_{J\sigma }\propto {\delta }^{J^{\prime }}{}_s{\delta }^J{}_s{\delta }^{{\sigma }^{\prime }}{}_{\sigma },}$

which filters only thoseJ values which equal the spins of the particle (σ = −s, ...,s is the z-component projection spin quantum number fors). In the particle's rest frame, the only degrees of freedom are rotations. The wave function can transform under rotations, withJ =A⊗I +I⊗B the generator.

In terms of thea andb labels this is related to the CG coefficients (will add formula later).

The corresponding operator in the boosted frame is

${\displaystyle D(L)\mathrm{\Pi }(p_{\text{rest}},m)D(L^{-1})={\mathrm{\Pi }}^{\prime }(L{p}_{\text{rest}},m)}$

in detail

${\displaystyle D(L{)}^{k^{\prime }{\ell }^{\prime }}{}_{J^{\prime }{\sigma }^{\prime }}\mathrm{\Pi }(p_{\text{rest}},m{)}^{J^{\prime }{\sigma }^{\prime }}{}_{J\sigma }D(L^{-1}{)}^{J\sigma }{}_{k\ell }={\mathrm{\Pi }}^{\prime }(L{p}_{\text{rest}},m{)}^{k^{\prime }{\ell }^{\prime }}{}_{k\ell }}$

so that

${\displaystyle D(L{)}^{k^{\prime }{\ell }^{\prime }}{}_{s\sigma }D(L^{-1}{)}^{s\sigma }{}_{k\ell }={\mathrm{\Pi }}^{\prime }(L{p}_{\text{rest}},m{)}^{k^{\prime }{\ell }^{\prime }}{}_{k\ell }}$

Letα = (α₁,α₂, ...,α_n) be dimensionless discrete-valued observables, andω = (ω₁,ω₂, ...,ω_m) be continuous-valued observables (not necessarily dimensionless). Allα are in ann-dimensionalsetA =A₁ ×A₂ × ...A_n where eachA_i is the set of allowed values forα_i, likewise allω are in anm-dimensional "volume"Ω ⊆ ℝ^m whereΩ = Ω₁ × Ω₂ × ... Ω_m and eachΩ_i ⊆ ℝ is the set of allowed values forω_i, asubset of thereal numbersℝ. For generalityn andm 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 spins, neglecting other degrees of freedom, using Cartesian coordinates, we could takeα = (s_z) for the spin quantum number of the particle along the z direction, andω = (x,y,z) for the particle's position coordinates. HereA = {−s, −s + 1, ...,s − 1,s} is the set of allowed spin quantum numbers andΩ = ℝ³ is the set of of all possible particle positions throughout 3d position space. An alternative choice isα = (s_y) for the spin quantum number along the y direction andω = (p_x,p_y,p_z) for the particle's momentum components. In this caseA andΩ are the same.

In theCopenhagen interpretation, theprobability density of finding the system in any state is

${\displaystyle \rho =|\mathrm{\Psi }(\mathit{\alpha },\mathit{\omega },t){|}^2}$

The probability of finding system withα in some or all possible discrete-variable configurations,D ⊆A, andω in some or all possible continuous-variable configurations,C ⊆ Ω, is the sum and integral over the density,^{[nb 1]}

${\displaystyle P=\sum _{\mathit{\alpha }\in D}{\int }_C\rho d^m\mathit{\omega }}$

whered^mω =dω₁dω₂...dω_m is a "differential volume element" in the continuous degrees of freedom. The units of the wavefunction are then such thatρ d^mω is dimensionless, bydimensional analysisΨ must have the same units as(ω₁ω₂...ω_m)^−1/2. Since the sum of all probabilities must be 1, thenormalization condition

${\displaystyle 1=\sum _{\mathit{\alpha }\in A}{\int }_{\mathrm{\Omega }}\rho d^m\mathit{\omega }}$

must hold at all times during the evolution of the system. The interpretation is the system will be in a particular state, all theα_i andω_j will have particular values at the timet the system is measured.

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

${\displaystyle |\mathrm{\Psi }⟩=\sum _{\mathit{\alpha }}\int d^m\mathit{\omega }\mathrm{\Psi }(\mathit{\alpha },\mathit{\omega },t)|\mathit{\alpha },\mathit{\omega }⟩}$

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

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:

${\displaystyle \mathrm{\Psi }(\mathbf{r},t)}$

wherer is theposition vector in three-dimensional space, andt is time. As alwaysΨ(r, t) is a complex-valued function of real variables. As a single vector in Dirac notation

${\displaystyle |\mathrm{\Psi }(t)⟩=\int d^3\mathbf{r}\mathrm{\Psi }(\mathbf{r},t)|\mathbf{r}⟩}$

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

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

${\displaystyle \xi (s_z,t)}$

wheres_z is thespin projection quantum number along thez axis. (Thez axis is an arbitrary choice; other axes can be used instead if the wave function is transformed appropriately, see below.) Thes_z parameter, unliker andt, is adiscrete variable. For example, for aspin-1/2 particle,s_z can only be+1/2 or−1/2, and not any other value. (In general, for spins,s_z can bes,s − 1, ... , −s + 1, −s). Inserting each quantum number gives a complex valued function of space and time, there are2s + 1 of them. These can be arranged into acolumn vector^{[nb 2]}

${\displaystyle \xi =\left[\begin{array}{c}\xi (s,t)\\ \xi (s-1,t)\\ ⋮\\ \xi (-(s-1),t)\\ \xi (-s,t)\end{array}\right]=\xi (s,t)\left[\begin{array}{c}1\\ 0\\ ⋮\\ 0\\ 0\end{array}\right]+\xi (s-1,t)\left[\begin{array}{c}0\\ 1\\ ⋮\\ 0\\ 0\end{array}\right]+\cdots +\xi (-(s-1),t)\left[\begin{array}{c}0\\ 0\\ ⋮\\ 1\\ 0\end{array}\right]+\xi (-s,t)\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ 1\end{array}\right]}$

In bra ket notation, these easily arrange into the components of a vector^{[nb 3]}

${\displaystyle |\xi (t)⟩=\sum _{s_z=-s}^s\xi (s_z,t)|s_z⟩}$

The entire vectorξ is a solution of the Schrödinger equation (with a suitable Hamiltonian), which unfolds to a coupled system of2s + 1 ordinary differential equations with solutionsξ(s,t),ξ(s − 1,t), ...,ξ(−s,t). 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:

${\displaystyle \mathrm{\Psi }(\mathbf{r},s_z,t)}$

and these can also be arranged into a column vector

${\displaystyle \mathrm{\Psi }(\mathbf{r},t)=\left[\begin{array}{c}\mathrm{\Psi }(\mathbf{r},s,t)\\ \mathrm{\Psi }(\mathbf{r},s-1,t)\\ ⋮\\ \mathrm{\Psi }(\mathbf{r},-(s-1),t)\\ \mathrm{\Psi }(\mathbf{r},-s,t)\end{array}\right]}$

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

${\displaystyle |\mathrm{\Psi }(t)⟩=\sum _{s_z}\int d^3\mathbf{r}\mathrm{\Psi }(\mathbf{r},s_z,t)|\mathbf{r},s_z⟩}$

For a single particle, thetensor product⊗ of its position state vector|ψ⟩ and spin state vector|ξ⟩ gives the composite position-spin state vector

${\displaystyle |\psi (t)⟩\otimes |\xi (t)⟩=\sum _{s_z}\int d^3\mathbf{r}\psi (\mathbf{r},t)\xi (s_z,t)|\mathbf{r}⟩\otimes |s_z⟩}$

with the identifications

${\displaystyle |\mathrm{\Psi }(t)⟩=|\psi (t)⟩\otimes |\xi (t)⟩}$

${\displaystyle \mathrm{\Psi }(\mathbf{r},s_z,t)=\psi (\mathbf{r},t)\xi (s_z,t)}$

${\displaystyle |\mathbf{r},s_z⟩=|\mathbf{r}⟩\otimes |s_z⟩}$

The tensor product factorization is only possible if the orbital and spin angular momenta of the particle are separable in theHamiltonian operator underlying the system's dynamics (in other words, the Hamiltonian can be split into the sum of orbital and spin terms^[1]). 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 amagnetic field, andspin-orbit coupling.

The preceding discussion is not limited to spin as a discrete variable, the totalangular momentumJ may also be used.^[2] Other discrete degrees of freedom, likeisospin, can expressed similarly to the case of spin above.