0.1 Klein-Gordon Equation 0.2 Dirac Equation

More documents

Recommendations

Info

$' \^' I$

ii.) ⃗σ · ˆr commutes with ⃗ J and changes parity, so you can prove ⃗σ · ˆrY m ωj = −Y m −ωj iii.) Since ⃗σ · ⃗L = J 2 − L 2 − 1 4 ⃗σ2 , it is easy to get ⃗σ · ⃗L Y m ωj = (ωj + 1 2 ω − 1) Ym ωj Putting the Hamiltonian and the wave function together, we obtain the coupled equations 0.6.4 Solution f ′ − ω(j + 1 2 ) r g ′ + ω(j + 1 2 ) g − r At r = ∞ both f and g satisfy the equation ( f − m + E + α ) g = 0 (14) r ( m − E − α ) f = 0 (15) r f ′′ − (m 2 − E 2 )f = 0 which has decaying solution f = e −√ m 2 −E 2r , suggesting the new dimensionless radial variable ρ = √ m 2 − E 2 r Introducing this variable and new independent variables such that f = F e −ρ and g = Ge −ρ brings the equations to the form dF dρ − F − βF ρ − 1 ν G − αG ρ dG dρ − G + βG ρ − νF + αF ρ = 0 = 0 (16) in which ν stands for √ (m − E)/(m + E) and β stands for ω(j+ 1 2 ). Examination of the equations near ρ = 0 shows F and G must behave as ρ s where s = √ β 2 − α 2 , a power which is less than |β| (≥1) by something of the order of α 2 . Introducing F (ρ) = ∑ a p ρ s+p p=0 10
and a similar series for G with coefficients b p leads to the following recursion relation (s + p + 1)a p+1 − a p − βa p+1 − 1 ν b p − αb p+1 = 0 (s + p + 1)b p+1 − b p + βb p+1 − νa p + αa p+1 = 0 The series must terminate to avoid generating the exponentially increasing solution. If p is the index of the highest power of ρ before termination, then we can deduce that a p = −νb p from the fact that a p+1 and b p+1 are both zero while a p and b p are non-zero. If we use this relation in the recursion relations above with p → p − 1, we find that all the coefficients can be eliminated and a quadratic equation for ν obtained: The appropriate solution is positive: αν 2 + 2(p + s)ν − α = 0. ν = − p + s α √ (p ) 2 + s + + 1 α It is then a matter of algebra to deduce from this that E = m √ 1 + α2 (p+s) 2 To make sense of this and to compare it to the NR result, we can expand it in powers of α = e 2 /4π ≈ 1/137, noting the dependence of s = √ (j+ 1 2) 2 − α 2 on α. The final result is E = m (1 − α2 2n 2 − α4 2n 4 ( n j+ 1 2 − 3 4 ) ) + O(α 6 ) , in which n stands for p + j + 1 2. After the rest energy, the α 2 term is the Rydberg energy calculated in NR QM and the α 4 term is the relativistic correction. In all orders the energy depends only on j and n, so some but not all of the ’accidental’ degeneracy of the H atom remains. DISCUSSION 11
Page 1 and 2: [revised Oct 15, 2010] 0.1 Klein-Go
Page 3 and 4: where ψ 0 is a constant 4-componen
Page 5 and 6: and for boosts Σ 0i = − 1 2 ( 0
Page 7 and 8: In higher order, the equivalence of
Page 9: is conserved. The operator P define
Page 13 and 14: expressed in terms of the fields. A
Page 15: This Lagrangian leads to solutions

0.1 Klein-Gordon Equation 0.2 Dirac Equation

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?