The Klein-Gordon Equation, The Maxwell and Proca Equations and The Dirac Equation Flashcards
What is the Klein-Gordon equation?
- a relativistic form of the the Schrodinger equation
- it correctly describes spin 0 particles
The Klein-Gordon Equation
(∂² + m² ) φ = 0
-the Klein-Gordon equation is true for any system
Derivation of the Klein-Gordon Equation
-start with: E² = p_² + m² -or E² = pipi + m² or pμp^μ = m² -let: p^μ = i∂^μ and introduce wavefunction φ, then: (i∂μi∂^μ - m²) φ = 0 (-∂μ∂^μ - m²) φ = 0 (∂² - m²) φ = 0
Solutions to the Klein-Gordon Equation
-plane wave solutions: φ = Ae^(-ik.x) -where p^μ=k^μ, i.e. plane waves where k is the momentum four vector -this means is a solution if: E² = p_² + m² OR E = ±√[p_²+m²] -meaning half of the valid solutions have negative energy
The Klein-Gordon Equation
Conserved Current for φ
j^μ = i(φ∂^μφ - φ∂^μφ)
The Klein-Gordon Equation
Probability Problem
-recall that the Schrodinger equation had probability density ρ=ΨΨ and probability density current
ji=1/2mi [Ψ∂iΨ - Ψ∂iΨ]
-for the Klein-Gordon equation, ρ=j^o=i(φ∂^μφ - φ∂^μφ*)
-but this can be negative!!
The Klein-Gordon Equation
Negative Energy and Negative Probability
-the conserved current for the Klein-Gordon Equation can be expressed:
j^μ = 2k^μ |A|²
-then density:
ρ = jo = 2E|A|²
-so negative energy solutions correspond to negative probability
The Klein-Gordon Equation
Explaining Negative Energy Solutions
- take the complex conjugate of a negative energy solution
- this recovers a positive energy solution with negative momentum
- it is equivalent to the positive energy solution travelling in the opposite direction, an antiparticle solution
- the negative energy particle solution becomes a positive energy antiparticle
The Klein-Gordon Equation
Explaining Negative Probability
- you can multiply j^μ by any constant and it will still satisfy the continuity equation
- multiplying by e.g. charge, then ∂μj^μ=0 now implies conservation of charge
- let Je=Qj^μ, then Jo is the charge density and J1,2,3 are the charge current
- in this case, charge can be thought of as any conserved quantum quantity
The Klein-Gordon Equation
Electromagnetism
-make substitution: p^μ = p^μ - qA^μ -or equivalently: ∂^μ = ∂^μ + iqA -sub into the Klein-Gordon Equation
Crossing Symmetry
Definition
- emission of a particle with energy E, momentum p_ and charge q is equivalent to absorption of an antiparticle with energy -E, momentum -p_ and charge -q
- again charge represents any quantum quantity
- these two processes have the same transition amplitude
Crossing Symmetry
Feynman Diagrams
- equivalence of the two processes in Feynman diagram representation is guaranteed by CPT theorem
- remember that lines on a Feynman diagram indicate flow of a quantum number, not actual trajectories
Transition Amplitude and Transition Rate
-equal transition amplitude does NOT indicate equal transition rate since transition amplitude does not account for phase space
Normalisation
- in non-relativistic quantum mechanics, normalisation is typically normalisation to one particle per box (system)
- this is not possible for relativistic quantum mechanics since different observers can disagree on the size of the box
- e.g. if one observer sees one particle in volume V but a second observer moves relative to V with speed u then the second observer sees volume V/√[1-u²] due to Lorentz contraction in direction parallel to the observers motion
- we normalise to 2E particles per volume V
- this way for a moving observer scaling of E and V cancels out and all observers can agree
Deriving an Equation for Spin 1 Particles
- the Maxwell equation is a good place to start since it is the combination of electromagnetism and quantum mechanics that leads to photons which are spin 1
- start with the Maxwell equation and reinterpret A^μ as a wavefunction
Maxwell Equation
∂²A^μ - ∂^μ ∂.A = j^μ
-where A^μ=(∇.A_) is the 4-potential and j^μ=(ρj_) is the 4-current
Maxwell Equation
Degrees of Freedom of A^μ
-A^μ appears to have 4 degrees of freedom since it is a four vector
-but we also have gauge invariance meaning many different values of A can give the same E and B fields
-this means that some constraints can be placed on A without loss of generality
-apply the Lorenz condition to reduce the Maxwell equation:
∂²A^μ = j^μ
-OR in free space:
∂²A^μ = 0
-so any one of the components of A^μ can be written in terms of the other three
-the number of degrees of freedom is therefore reduced by 1 to 3
Lorenz Condition
∂.A = ∂μA^μ = 0
Plane Wave Solutions to the Maxwell Equation
A^μ = N ε^μ e^[-i(k.x)]
-where ε is a constant polarisation vector
-subbing into to the reduced form of the Maxwell equation
=>
-the plane wave is ony a solution if k=0 i.e. if the particle being described is massless
-subbing into the Lorenz condition gives:
k.ε = 0
-so one component of the polarisation vector is uniquely determined by the others again implying 3 degrees of freedom
Gauge Invariance of the Maxwell Equation and Lorenz Condition
-can make a gauge transformation:
A^μ -> A^μ’ = A^μ + ∂^μχ
-as long as ∂²χ=0
-without altering the Maxwell Equation of the Lorenz Condition
Degrees of Freedom for Polarisation in Plane Wave Solutions to the Maxwell Equation
-can impose the condition: εo = 0 -since there is no time-like polarisation -suppose the photon is travelling in the z-direction, then: k^μ = (ε,0,0,k) -but the Lorenz condition gives: ε.k = 0 => εo E - ε1 (0) - ε2 (0) - ε3 k = 0 ε3 k = 0 -or more generally, for travel in any direction: ε_.k_ = 0 -polarisation is orthogonal to momentum -only two transverse degrees of freedom
Polarisation Basis Vectors
-there are two degrees of freedom for polarisation in plane wave solutions to the Maxwell equation but we can still make different choices of basis vectors
-for case:
k^μ = (ε,0,0,k)
-can choose a linear polarisation e.g.
{(0,1,0,0), (0,0,1,0)}
-OR circular polarisation
{1/√2 (0,1,i,0), 1/√2 (0,1,-i,0)}
-which corresponds to the helicity states of a photon, can also show that this is spin 1
Helicity
- helicity measures spin along the momentum axis
- it is a conserved quantity
- the 2s+1=DoF only applies for particles with mass
What does the Maxwell equation describe?
-massless spin 1 particles - photons
How can mass be introduced to the Maxwell equation?
-the Maxwell equation under the Lorenz condition:
∂²A^μ = 0
-is essentially 4 coupled, massless Klein-Gordon equations
-suppose that we can include mass by adding an m²A^μ term
-this gives the Proca Equation
The Proca Equation
∂²A^ν - ∂ ∂.A^ν + m²A^ν = j^ν -OR in free space: ∂²A^ν - ∂ ∂.A^ν + m²A^ν = 0 -this describes massive spin 1 particles -NOT photons
The Proca Equation Under Gauge Transformation
-consider the gauge transform:
A^μ -> A^μ’ = A^μ + ∂^μχ
-substitute into the RHS of the Proca Equation
=>
m² ∂^μ χ
-this is not equal to zero so there is no gauge invariance for massive spin 1 particles
-only for massless spin 1 particles
The Proca Equation and the Lorenz Condition
-differentiate the Proca Equation with respect to x^μ
=>
-either m=0 or ∂.A=0
-so if the particles has mass, m≠0, then the Lorenz condition must hold for spin 1 particles therefore reducing the degrees of freedom to 3
-for massless particles we choose to apply the Lorenz condition but it is not strictly the only condition that could be placed on A
Klein-Gordon Equation in the Presence of an Electromagnetic Field
-in free space, the Klein-Gordon Equation:
(∂² + m²)φ = 0
-in the presence of an EM field the RHS is equal to a set of source terms for φ, a spin 0 particle
-each of the three source terms describes an allowed interaction that can be represented by a Feynman diagram that results in the production of a φ
Modified Maxwell Equation
- source terms can be added to the Maxwell equation as well as the Klein-Gordon equation
- each of the three source terms describes an allowed interaction that results in the production of a photon
Probability Amplitude From Feynman Diagrams
- there is an expression associated with each vertex and imaginary exchange particle
- when multiplied together they form an algebraic representation of the Feynman diagram from which the probability amplitude can be determined
Motivation for the Dirac Equation
-find a first order Schrodinger-like equation that obeys the relativistic energy-momentum relation:
E² = p_² + m²
The Dirac Equation
(i γ^μ ∂μ - m) Ψ = 0
-where γ^μ are 4x4 matrices such that:
{γ^μ, γ^ν} = 2g^μν
-and Ψ is a Dirac spinor
Deriving the Dirac Equation
Equation
-start with:
i ∂/∂t Ψ = H^ Ψ
-assume H^ is a linear combination of momentum operator and mass:
H^ = α_.p^_ + βm
-divide through by β and define; 1/β = γo and αi/β = γi
=>
γ^μ pμ^ Ψ = mΨ
-so Ψ is an eigenstate of γ^μ pμ^ with eigenvalue m
-substituting in the momentum operator pμ^=i∂μ
=>
(i γ^μ ∂μ - m)Ψ = 0
Deriving the Dirac Equation
Defining the Gammas
-act with the γ^μ pμ^ operator on the first equation
=>
(γ^μ pμ^)(γ^μ pμ^) Ψ = m²Ψ
=>
(γ^νγ^μ+γ^μγ^ν)pμ^pμ^ Ψ = 2m²Ψ
-but from the relativistic energy-momentum relation, we know that:
p^² Ψ = m²Ψ
=>
γ^νγ^μ+γ^μγ^ν = 2g^μν
-OR
{γ^μ,γ^ν} = 2g^μν
-where {} indicates the anticommutatorGamma Anticommutator Relations
{γo,γo} = 2γoγo = 2
{γi,γi} = 2γiγi = -2
{γi,γj} = 0
=because of these relations, the gammas can’t be numbers, we can represent them with matrices, at least 4x4
Dirac Representation of the Gamma Matrices
γo = 2x2 with entries; II, 0, 0, II
-where II is a 2x2 identity matrix and o is a 2x2 zero matrix
γi = 2x2 with entries 0, σi, -σi, 0
-where σi are the Pauli matrices
Plane Wave Solutions to the Dirac Equation
Description
-assume a plane wave solution of form:
Ψ = N u(p) e^{-ip.x]
-where N is for normalisation and u(p) is a constant
-sub into Dirac equation
-have a plane wave solution if:
p²=m²
i.e. if
E = ± √[p²_+m²]
-STILL HAVE NEGATIVE ENERGY SOLUTIONSPlane Wave Solutions to the Dirac Equation
Constraint on u(p)
(γ^μ pμ - m) u(p) = 0 -sub in gamma matrices => equation for the two components of u(p), uA(p) and uB(p) uA(p) = [σ_.p_]/[E-m] uB(p) OR uB(p) = [σ_.p_]/[E+m] uA(p) -so the two components are NOT independent
Plane Wave Solutions to the Dirac Equation Choosing uA(p)
- there are two independent cases,
1) choose uA(p)=(1 0)t
2) choose uA(p)=(0 1)t - then calculate corresponding uB(p)
- together, these correspond to the two spin states of a positive energy particle
Plane Wave Solutions to the Dirac Equation Choosing uB(p)
- as for uA(p), there are two independent case;
1) choose uB(p)=(1 0)t
2) choose uB(p)=(0 1)t - then calculate corresponding uA(p)
- together, these correspond to the two spin states of a negative energy particle
Plane Wave Solutions to the Dirac Equation
Reinterpreting Negative Energy Solutions
- want to reinterpret the two negative energy states as positive energy states of an antiparticle
- use a charge conjugation operation to do this
- let Ψ^c = CΨ* for some matrix C
- in particular for basis spinor u, let v be an antiparticle basis spinor, v = Cu*
- v should be a solution so substituting u(p)=v in the plane wave solution of the Dirac equation
- also starting from the Dirac equation with u, take complex conjugate and substitute u*=C^(-1)v
- this recover the same restraint on v
- so take v1=Cu1* and v2=Cu2* as basis states for antiparticle
