The Klein-Gordon Equation, The Maxwell and Proca Equations and The Dirac Equation Flashcards Preview

PHYS3543 Theoretical Elementary Particle Physics > The Klein-Gordon Equation, The Maxwell and Proca Equations and The Dirac Equation > Flashcards

Flashcards in The Klein-Gordon Equation, The Maxwell and Proca Equations and The Dirac Equation Deck (42)
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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
E² = pipi + m²
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
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=Ψ*Ψ 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

-make substitution:
p^μ = p^μ - qA^μ
-or equivalently:
∂^μ = ∂^μ + iqA
-sub into the Klein-Gordon Equation


Crossing Symmetry

-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



-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 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