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) Aμ Definition 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 γμ πν γν ] Ψ = 0

But πμ γμ πν γν + πν γν πμ γμ

= γμ γν πμ πν + γν γμ πν πμ = γμ γν ( πμ πν - πν πμ) Equation (3)

But πμ πν = (pμ + e Aμ/c) (pν + e Aν/c)

= pμ pν +(e2/c2) Aμ Aν +(e/c)[pμ Aν + Aμ pν ]

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

(a) First term

πμ2 = {(pμ + (e/c) Aμ}2

= pμ2 + (e2/c2)Aμ

2 + (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)

The 4-component Pauli terms are σi = (τi 0 ) and α k = (0 τk ) ( 0 τi) (τk 0 )

(b) 2nd term

ΣγμγνFμν∙{ħe/(2ic)}

Spacelike terms: γjγj = iσi

Space-time terms: γkγ4 = iαk

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∙ENoting 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

2 coupledEquations for UA & UB

UA & UB areeach 2 componentwavefunctions

τ·(cp +eA) UB + (E0 – eφ) UA =EUA

τ·(cp +eA) UA - (E0 + eφ) UB = EUB

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 )

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

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

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

Ψ

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.

Continuing to the Pauli approximation

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 p term is the Lorentz force on the moving electron

Now solve for a central potential

Calculating the small terms…

3/4n)

Adding the small terms together

Angular momenta & normalization

Exact solution for the Coulomb potential case

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.

The n=2 levels in hydrogen

Radiative corrections: the Lamb shift

Two terms – both calculated to infinite order

1. the photon self-energy term2. The vacuum polarization

term

Removes the degeneracy between states of the same J, but different L angular momenta.

QED - theory - short historySee 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)/πwhere 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 MHz1950-60s –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:1. Test of QED2. proton radius3. Variation of

fundamental constants

Phys. Rev. Lett. 92, 230802 (2004)