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

Question 1

What is the Klein-Gordon equation?

Answer 1

• a relativistic form of the the Schrodinger equation

- it correctly describes spin 0 particles

Question 2

The Klein-Gordon Equation

Answer 2

(∂² + m² ) φ = 0

-the Klein-Gordon equation is true for any system

Question 3

Derivation of the Klein-Gordon Equation

Answer 3

-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

Question 4

Solutions to the Klein-Gordon Equation

Answer 4

-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

Question 5

The Klein-Gordon Equation

Conserved Current for φ

Answer 5

j^μ = i(φ∂^μφ - φ∂^μφ)

Question 6

The Klein-Gordon Equation

Probability Problem

Answer 6

-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!!

Question 7

The Klein-Gordon Equation

Negative Energy and Negative Probability

Answer 7

-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

Question 8

The Klein-Gordon Equation

Explaining Negative Energy Solutions

Answer 8

• 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

Question 9

The Klein-Gordon Equation

Explaining Negative Probability

Answer 9

• 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

Question 10

The Klein-Gordon Equation

Electromagnetism

Answer 10

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

Question 11

Crossing Symmetry

Definition

Answer 11

• 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

Question 12

Crossing Symmetry

Feynman Diagrams

Answer 12

• 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

Question 13

Transition Amplitude and Transition Rate

Answer 13

-equal transition amplitude does NOT indicate equal transition rate since transition amplitude does not account for phase space

Question 14

Normalisation

Answer 14

• 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

Question 15

Deriving an Equation for Spin 1 Particles

Answer 15

• 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

Question 16

Maxwell Equation

Answer 16

∂²A^μ - ∂^μ ∂.A = j^μ

-where A^μ=(∇.A_) is the 4-potential and j^μ=(ρj_) is the 4-current

Question 17

Maxwell Equation

Degrees of Freedom of A^μ

Answer 17

-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

Question 18

Lorenz Condition

Answer 18

∂.A = ∂μA^μ = 0

Question 19

Plane Wave Solutions to the Maxwell Equation

Answer 19

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

Question 20

Gauge Invariance of the Maxwell Equation and Lorenz Condition

Answer 20

-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

Question 21

Degrees of Freedom for Polarisation in Plane Wave Solutions to the Maxwell Equation

Answer 21

-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

Question 22

Polarisation Basis Vectors

Answer 22

-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

Question 23

Helicity

Answer 23

• helicity measures spin along the momentum axis
• it is a conserved quantity
• the 2s+1=DoF only applies for particles with mass

Question 24

What does the Maxwell equation describe?

Answer 24

-massless spin 1 particles - photons

Question 25

How can mass be introduced to the Maxwell equation?

Answer 25

-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

Question 26

The Proca Equation

Answer 26

∂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

Question 27

The Proca Equation Under Gauge Transformation

Answer 27

-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

Question 28

The Proca Equation and the Lorenz Condition

Answer 28

-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

Question 29

Klein-Gordon Equation in the Presence of an Electromagnetic Field

Answer 29

-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

Question 30

Modified Maxwell Equation

Answer 30

• 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

Question 31

Probability Amplitude From Feynman Diagrams

Answer 31

• 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

Question 32

Motivation for the Dirac Equation

Answer 32

-find a first order Schrodinger-like equation that obeys the relativistic energy-momentum relation:
E² = p_² + m²

Question 33

The Dirac Equation

Answer 33

(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

Question 34

Deriving the Dirac Equation

Equation

Answer 34

-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

Question 35

Deriving the Dirac Equation

Defining the Gammas

Answer 35

-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 anticommutator

Question 36

Gamma Anticommutator Relations

Answer 36

{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

Question 37

Dirac Representation of the Gamma Matrices

Answer 37

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

Question 38

Plane Wave Solutions to the Dirac Equation

Description

Answer 38

-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 SOLUTIONS

Question 39

Plane Wave Solutions to the Dirac Equation

Constraint on u(p)

Answer 39

(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

Question 40

Plane Wave Solutions to the Dirac Equation
Choosing uA(p)

Answer 40

• 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

Question 41

Plane Wave Solutions to the Dirac Equation
Choosing uB(p)

Answer 41

• 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

Question 42

Plane Wave Solutions to the Dirac Equation

Reinterpreting Negative Energy Solutions

Answer 42

• 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