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Covariant form of the Dirac equation. A μ = ( A , i φ ) , x μ = ( r , ict) and p μ = -i ħ∂/∂x μ = (-iħ ▼, -( ħ/c) ∂ /∂t) = ( p , iE/c). and γ μ = (-i β α , β ) Definition of the Dirac Matrices. Then H Ψ = i ħ∂ Ψ /∂t Equation (1)
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Covariant form of the Dirac equation Aμ = (A, iφ) , xμ = (r, ict) and pμ = -iħ∂/∂xμ = (-iħ▼, -(ħ/c) ∂ /∂t) = (p, iE/c) and γμ = (-iβα, β) Definition of the Dirac Matrices Then H Ψ = i ħ∂Ψ/∂t Equation (1) where H = βmc2 + α∙ (cp + e A) By substitution: [ (pμ + e Aμ/c)∙γμ - imc] Ψ = 0 let πμ = pμ +(e/c) AμDefinition of “Dirac Momentum” then [πμ γμ - imc] Ψ = 0 Equation (2) This is covariant! It can be reduced to the Klein-Gordon form by multiplying by [πμ γμ + imc]
Reduction to Klein-Gordon form (1) Thus [πμ γμ + imc] [πμ γμ - imc] Ψ = 0 multiplying out => [Σ π2μ + m2c2 + Σ´ πμ γμπν γν]Ψ = 0 But πμ γμπν γν + πν γνπμ γμ = γμγνπμπν+ γνγμπν πμ = γμγν ( πμ πν- πν πμ) Equation (3) But πμπν = (pμ + e Aμ/c) (pν + e Aν/c) = pμ pν +(e2/c2) AμAν +(e/c)[pμ Aν + Aμpν ]
Reduction to Klein-Gordon form (2) Substituting in the last term => (e/c) [ …] = (e/c)[ (-iħ) (∂ /∂μ )(Aν) + Aμ (-iħ) ∂ /∂ ν ] = (-iħe/c) ∙ Aν∙ ∂ /∂μ + (-iħe/c) ∙ Aμ ∙ ∂ /∂ν + (-iħe/c) ∙ ∂ (Aν)/∂μ Only the last term survives the commutator {Eq (3)} And we are left with the electromagnetic field tensor: Fμν = ∂Aν/∂xμ - ∂Aμ/∂xν Thus [Σ(πμ2 + m2c2) + ħe/ic) ΣγμγνFμν]Ψ = 0 Represents a charged particle in an electromagnetic field
Reduction to an exact non-covariant equation • First term • πμ2 = {(pμ + (e/c) Aμ}2 • = pμ2 + (e2/c2)Aμ2 + (e/c) (Aμpμ + pμAμ) • =p2 – (E2/c2) + e2 A2/c2 + (e2/c2)(-φ2) + (2e/c) A∙ p + (2e/c)(iE/c)(i φ) Substituting operators into the first term and dividing by -2m, & including m2 term, & collecting terms, gives => (E-mc2) + eφ + (ħ2/2m)∙▼2 + 1/(2mc2) (E-mc2 + e φ )2 +ieħ/(mc) ∙ A∙▼- e2A2/(2mc2)
(b) 2nd term ΣγμγνFμν∙{ħe/(2ic)} Spacelike terms: γjγj = iσi Space-time terms: γkγ4 = iαk The 4-component Pauli terms are σi = (τi 0 ) and α k = (0τk ) ( 0 τi) (τk 0 ) Note: the electromagnetic tensor terms are: spacelike Hi = Fjk ; and the time-like Fi4 = (1/i) Ei Substituting, and dividing by (-2m) => (-eħ/2mc) σ∙H + ieħ/(2mc) ∙α∙E Noting that the Bohr magneton is μ0 = eħ/(2mc), and W= (E-mc2) the whole equation becomes: [W + eφ + {ħ2/(2m)}▼2 + {1/(2mc2)}(W + eφ)2 +{ieħ/(mc)}∙A∙▼ -e2A2/(2mc2) -μ0∙σ∙H + iμ0∙α∙E]Ψ = 0
Pauli Approximation The Pauli approximation leads to a similar non-covariant equation, but with only reduced 2-component matrices… Assume that W= E – mc2 is << mc2 define E0 = mc2 or | E – E0 | ≤ E0 then <v> << c and <p> <<mc Letting α = ( 0 τ ) and β = ( 1 0 ) and ( UA ) = ( ψ1 ) ( τ 0 ) ( 0 1 ) ( UB ) ( ψ2 ) ( ψ3 ) ( ψ4 ) UA & UB are each 2 component wavefunctions We can write the Dirac equation [α∙(cp + eA) + βE0 - eφ] Ψ = EΨ which becomes the 4-component equation: [ ( 0 τ) ∙ (cp + eA) + ( 1 0 ) E0 -e φ ] ( UA ) = E ( UA) [ ( τ 0 ) ( 0 -1) ] ( UB ) = ( UB) giving 2 coupled Equations for UA & UB τ·(cp +eA) UB + (E0 – eφ) UA =EUA τ·(cp +eA) UA - (E0 + eφ) UB = EUB
In the second equation, we can make another approximation: E0 + E >> eφ Hence, E0 + E + e φ≈ 2E0 = 2mc2 Then UB ≈ {τ /(2mc)} ∙ ( p +eA/c) UA Now substitute in the first equation to give: [ E-E0 + e φ -(1/2m) (p + eA/c)2 ] UA = 0 This is exactly the Schrodinger equation for hydrogen – Hence Ψ1 = Ψ2, (same equation), no cross terms) and UA is the hydrogen wavefunction. Then do an iterative process…. calculate a new UB and substitute again for a new UB equation, solve etc… BUT, there is an easier way… Go back to the exact non-covariant solutions – the only term which couples the first 2 components (UA) with the second 2 components is the last term (ieħ/2mc) ( α• E ) Ψ
Continuing to the Pauli approximation Use the approximation for UB (αU)A = τUB≈ {1/(2mc)} [p + i(p x τ)] UA most easily justified by looking at component by component. Ψ Now the last term can be rewritten without any cross linkage between UA and UB: [ ieħ/2mc α•E ] Ψ => [ {iμ0/(2mc)} E• p + {μ0/(2mc)} τ•( E x p ) ]Ψ which is just an equation for ( ψ1 ) = UA ( ψ2 ) The E • p term has no classical analog, and is just the Darwin term The E x pterm is the Lorentz force on the moving electron
The Dirac energy Expanding this equation in powers of (αZ)2 yields the Pauli energy as the first 2 terms….
On the aZ expansion for the Dirac energy of a one-electron atom L J Curtis, Department of Physics, University of Lund, S-223 62 Lund, Sweden J. Phys. B10, L641 (1977) Abstract. A procedure for directly prescribing a term of arbitrary order in an CLZ-expansion of the Dirac energy of a one-electron atom is presented, and utilised to obtain higher-order corrections to the Dirac fine-structure formula. These can then be combined with terms not included in the Dirac formalism and applied, for example, to semi-empirical charge-screening parametrisations of multi-electron atoms.
Expanding the 4 terms using the binomial coefficients The table gives CPQ for each P and Q; The row Norm is the common denominator for each column. P=0 gives CPQ=1, the rest energy mc2. P=1 is the Balmer energy.
Radiative corrections: the Lamb shift Removes the degeneracy between states of the same J, but different L angular momenta. • Two terms – both calculated to infinite order • the photon self-energy term • The vacuum polarization term
QED - theory - short history See your QM field theory text for more details…. First calculation – 1947 - Bethe - 1 photon emission & absorption within a few percent of experiment. Precision… Feynman, Schwinger, Tomonaga – 1950s – developed calculational techniques for the higher-order diagrams – often by numerical integration Mohr - developed analytical theories to include all order diagrams of the 1-photon exchange and the pair production, plus parts of multiphoton exchange and multiple pair production… Sapirstein – continuing calculations in muonic systems, and higher-order terms – an active program. ΔE(Lamb) ~ Z4α5mc2. F(Zα)/π where F(Zα) = A40 + A41 ln(Zα)-2 + A50 + A60(Zα)2 +…{higher powers of Zα}
QED - Experimental work – short history Late 1930’s – spectroscopy: Several measurements of Balmer-α suggested that the 2s1/2 and 2p1/2 levels had a different energies (10-30% precision) 1947 – microwaves – Lamb & Retherford – measured the 2s-2p difference directly to a few parts in 10,000 – ΔE=1058 MHz 1950-60s–gradual improvement in H(2s-2p), other measurements in higher Z 1-electron ions – e.g. in hydrogen: Lundeen & students (at Harvard & Notre Dame) - in hi-Z ions –use of accelerators – up to chlorine (Z=17) 1970s onwards – lasers – measurements of Lyman α – 1s-2p, later 1s-2s
The Munich group’s precision measurements http://www.mpq.mpg.de/mpq.html The Hydrogen Spectrometer Hydrogen atoms are excited by longitudinal Doppler-free two photon excitation at 243 nm from a frequency doubled ultrastable dye laser at 486 nm. The UV radiation is then resonantly enhanced in a linear cavity inside a vacuum chamber . A small electric field that mixes the metastable 2S state (lifetime 1/7 sec) with the fast decaying 2P state is applied. We set an upper limit on the second order Doppler-shift below 1 kHz. Setup for exciting the 1s-2s transition • Results: • Test of QED • proton radius • Variation of fundamental constants • Phys. Rev. Lett. 92, 230802 (2004)