Flashcards in The Klein-Gordon Equation, The Maxwell and Proca Equations and The Dirac Equation Deck (42)

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1

## What is the Klein-Gordon equation?

###
-a relativistic form of the the Schrodinger equation

-it correctly describes spin 0 particles

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## The Klein-Gordon Equation

###
(∂² + m² ) φ = 0

-the Klein-Gordon equation is true for any system

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## 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

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## 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

5

##
The Klein-Gordon Equation

Conserved Current for φ

### j^μ = i(φ*∂^μφ - φ∂^μφ*)

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##
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!!

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##
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

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##
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

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##
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

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##
The Klein-Gordon Equation

Electromagnetism

###
-make substitution:

p^μ = p^μ - qA^μ

-or equivalently:

∂^μ = ∂^μ + iqA

-sub into the Klein-Gordon Equation

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##
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

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##
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

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## Transition Amplitude and Transition Rate

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

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## 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

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## Deriving an Equation for Spin 1 Particles

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-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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## Maxwell Equation

###
∂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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##
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

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## Lorenz Condition

### ∂.A = ∂μA^μ = 0

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## 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

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## 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

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## 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

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## 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

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## Helicity

###
-helicity measures spin along the momentum axis

-it is a conserved quantity

-the 2s+1=DoF only applies for particles with mass

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## What does the Maxwell equation describe?

### -massless spin 1 particles - photons

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## 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

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## 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

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## 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

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## 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

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## 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

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