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

1
Q

What is the Klein-Gordon equation?

A
  • a relativistic form of the the Schrodinger equation

- it correctly describes spin 0 particles

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2
Q

The Klein-Gordon Equation

A

(∂² + m² ) φ = 0

-the Klein-Gordon equation is true for any system

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3
Q

Derivation of the Klein-Gordon Equation

A
-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
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4
Q

Solutions to the Klein-Gordon Equation

A
-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
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5
Q

The Klein-Gordon Equation

Conserved Current for φ

A

j^μ = i(φ∂^μφ - φ∂^μφ)

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6
Q

The Klein-Gordon Equation

Probability Problem

A

-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!!

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7
Q

The Klein-Gordon Equation

Negative Energy and Negative Probability

A

-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

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8
Q

The Klein-Gordon Equation

Explaining Negative Energy Solutions

A
  • 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
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9
Q

The Klein-Gordon Equation

Explaining Negative Probability

A
  • 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
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10
Q

The Klein-Gordon Equation

Electromagnetism

A
-make substitution:
p^μ = p^μ - qA^μ
-or equivalently:
∂^μ = ∂^μ + iqA
-sub into the Klein-Gordon Equation
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11
Q

Crossing Symmetry

Definition

A
  • 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
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12
Q

Crossing Symmetry

Feynman Diagrams

A
  • 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
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13
Q

Transition Amplitude and Transition Rate

A

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

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14
Q

Normalisation

A
  • 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
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15
Q

Deriving an Equation for Spin 1 Particles

A
  • 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
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16
Q

Maxwell Equation

A

∂²A^μ - ∂^μ ∂.A = j^μ

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

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17
Q

Maxwell Equation

Degrees of Freedom of A^μ

A

-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

18
Q

Lorenz Condition

A

∂.A = ∂μA^μ = 0

19
Q

Plane Wave Solutions to the Maxwell Equation

A

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

20
Q

Gauge Invariance of the Maxwell Equation and Lorenz Condition

A

-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

21
Q

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

A
-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
22
Q

Polarisation Basis Vectors

A

-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

23
Q

Helicity

A
  • helicity measures spin along the momentum axis
  • it is a conserved quantity
  • the 2s+1=DoF only applies for particles with mass
24
Q

What does the Maxwell equation describe?

A

-massless spin 1 particles - photons

25
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
26
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 ```
27
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
28
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
29
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
30
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
31
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
32
Motivation for the Dirac Equation
-find a first order Schrodinger-like equation that obeys the relativistic energy-momentum relation: E² = p_² + m²
33
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
34
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
35
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 anticommutator ```
36
Gamma 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
37
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
38
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 SOLUTIONS ```
39
Plane 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 ```
40
``` 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
41
``` 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
42
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