Relativistic invariance Flashcards
Klein-Gordon equation
What is the Klein-Gordon equation and how did it come to be? What problems arise?
It’s the generalization of the Schrödinger equation for relativistic cases. Can be obtained by substituting E² – p² = m² into the Schrödinger equation:
(□ + m²)ψ = 0
Arising problems:
- non-positive definite probability density
- negative energy solutions
Dirac equation
What’s the Dirac Hamiltonian and how does it solve one of the KG problems?
H(D) = -i∑(k)α(k)∂(k)ψ + βmψ,
where α(k) and β are constant 4x4 matrices and ψ is a bispinor.
- This gives a positive definite probability density.
- Energies are still negative.
Dirac equation
What’s the Dirac equation? What’s the solution (interpretation) for the negative energy problem?
By multiplying the “Dirac-Schrödinger” eq. with β and relabeling the matrices to get the Dirac matrices we get:
(iγ^μ∂(μ) – m)ψ = 0
Both for a particle at rest and for nonzero p, the negative energy solutions are still there.
Interpretation: Dirac sea
- electrons at E < 0 levels, creating holes when excited above 0
- holes «—» antiparticles
- ψ is a field from which the particle-antiparticle content emerges naturally
The Dirac matrices behave according to Clifford algebra (anticommutators).
Rel. invariance, fields and groups
Why do we need fields to describe particles?
Because there’s a relation between the spin and the relativistic invariance of the fields. (fields «—» particles —» spin)
- spin 0: scalar field
- spin 1/2: fermion field
- spin 1: (massless) vector field
The vector fields have 3 or 2 physical components based on whether they have mass or not.
Rel. invariance, fields and groups
What’s a particle in terms of group theory? What do the terms mean?
Particle = unitary irrep of the Poincaré group
- Poincaré group: isometry group of spacetime
- Poincaré transformations: Lorentz transformations, translation in spacetime (a particle needs to have this)
- irrep: there are no reps which would only mix among themselves (P(ij) is not diagonal)
- unitarity: P^(-1) = P^+, so the scalar products commute regardless of the frame
Poincaré group
What are the properties of the rotation group?
R(V): 3x3 real matrices
- (R^T)R = 1: orthogonal —» forms O(3)
- detR = +/–1: leaves/changes orientation —» +: forms SO(3)
- transformations: R = 1 + iθ(i)J(i) —» R = exp(iθ(i)J(i))
- generator: components of angular momentum, hermitian
- commutator: [J(i), J(j)] = iε(ijk)J(k) —» Lie algebra
- tells us how to rotate a 3 component vector
R(S): 2x2 commplex matrixes
- other rep that satisfies the Lie algebra: Pauli matrices —» forms SU(2)
- acts on two components spinors
Hermitian generator = unitary transformation
The det = –1 option doesn’t form a subgroup, just a subset.
Poincaré group
What are the field representations of the rotation group?
Scalar field: R0 = 1, just a point, doesn’t rotate
Fermion field: R(S), spinor transformation
Vector field: R(V), vector transformation
Poincaré group
What are the properties of vector Lorentz transformations?
Λ(V): 4x4 real matrices
- (Λ^T)gΛ = g: forms O(1,3) subgroup
- detΛ = +/-1 (proper/improper) —» +: forms SO(1,3)
- (Λ^0(0))^2 ≥ 1: Λ^0(0) ≥ 1 is orthocronous (parity), Λ^0(0) ≤ 1 is non-orthocronous (time reversal)
- transformations: Λ = 1 + iθ(i)J(i) + iβ(i)K(i) —» Λ = exp(iθ(i)J(i) + iβ(i)K(i))
- K is non-hermitian (not a phyisical quantity) —» Λ is not unitary
- commutators: [J(i), J(j)] = iε(ijk)J(k), [K(i), K(j)] = -iε(ijk)K(k), [J(i), K(j)] = iε(ijk)J(k)
The orthocronous makes a subgroup, the non-orthocronous doesn’t.
Poincaré group
What are the properties of spinor Lorentz transformations?
Λ(S): 2x2 complex matrices
- J(i)+/- = (1/2)[J(i) +/– iK(i)] (diff. basis)
- part of SU(2)
- commutators: [J(i)+, J(j)+] = iε(ijk)J(k)+, [J(i)–, J(j)–] = iε(ijk)J(k)–, [J(i)+, J(j)–] = 0
- transformation: adding spins, (j1, j2) are reps of the Lorentz group
- (1/2, 0): ψ(L) —» J(i) = iK(i), J = σ(i)/2, K -iσ(i)/2
- (0, 1/2): ψ(R) —» J(i) = –iK(i), J = σ(i)/2, K iσ(i)/2
- tells us how a bispinor field behaves under a Lorentz boost: Λ(S)(ψ(L), ψ(R))
j1, j2 = 0, 1/2, 1
(1/2, 0) and (0, 1/2) are two inequivalent irreps of the Lorentz group, they correspond to left- and right-handed components
Poincaré group
How does the Lorentz group imply that we need to employ fields?
We want to associate particles with unitary fields but Lorentz trafos (that are finite) are not unitary. Theorem: the Lorentz group has no finite dimensional unitary reps. The reps of translations are infinite dimenional, so we need to look at infinite reps which is basically saying that we need fields.
Poincaré group
What’s the Lie algebra of the Poincaré group?
Generators: V^(μν) for rotations and boosts, P^ρ (momentum) for translations
- these two don’t commute w/ each other
- we are looking for unitary reps of translations which are characterized by momentum states
- Lorentz trafo leaves the sign of p^0 unchanged so we can specify the states: insert táblázat
- Boosts don’t commute with the Hamiltonian (H = P^0): [K, H] ≠ 0 because of time independent quantities.
- The Poincaré group tells us what particles can exist.
Poincaré group
How can we represent the physical momentum states of the Poincaré group?
With the Little group which is a subgroup of SO(1,3) that leaves p^μ invariant (the unitary, infinite dimenional reps of boosts, so the rest of the Lorentz group). Theorem: the Little group is the same for all p^μ ε {p^μ}.
Massive particles:
- representant momentum: p^μ = (m, 0,0,0)
- little group = SO(3)
- reps labelled by: j* = 0, 1/2, 1, …*
- # of d.o.f: 2j + 1 (= 1,2,3 for scalar, spinor, vector quantities)
Massless particle:
- representant momentum: p^μ = (κ, 0,0, κ)
- little group = E(2) (“translations” and rotations on/of a plane)
- reps labelled by: λ helicity, |λ| = 0, 1/2, 1
- # of d.o.f: 2, because we have to consider the change in sign if parity is applied
- fermions: |λ| = +/- 1/2, massless vector bosons: |λ| = +/-1
