Advanced quantum mechanics by Freeman J Dyson; David Derbes

By Freeman J Dyson; David Derbes

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E. e. e.  1 0 β = γ0 =  0 0  0  0 γ1 =   0 −1  1 0  0 0 0 1 0 0 0 1 0 0 0 0 −1 0  0  0 α2 =  0 i  1 0 0 0 0 −1   0  0 0 −1 0  0 0   0  −1 0 1 0 0  0  0 γ3 =   −1 0 O I I O ρ2 = O iI −iI O  0 0 = 1 0  0 0 0 1 0 0 = i αk = ρ 1 σk σk = −iρ2 γk ρ2 = iρ1 β γk = iρ2 σk αk σ + σ αk = 2δ k ρ1 0 0 0 i  0 0 −i 0 1 0 0 0  0 1  = γ5 0 0  −i 0 0 −i   0 0  0 0 0 i 0 0 ρ1 = −iα1 α2 α3 ρ1 = −iρ2 β γk σ + σ γk = −2δ k ρ2 σk σ = αk α = −γk γ = iσm O −I O −σk γk = 0 0 γ2 =  0 i  0 1 0 0 0 −1   0 0 0  1 0 0 0 σk = ρ 1 αk I O β=  −i 0   0  0 0 i 0 0 0  1 0  0 0 ρ1 = 0 0 −i 0 σk O  −i 0   0  0 ρ21 = ρ22 = I ρ2 = −α1 α2 α3 β βσk − σk β = 0 k, , m = (1, 2, 3) cyclicly permuted 30 Advanced Quantum Mechanics αk ρ1 − ρ 1 αk = γ µ ρ1 + ρ 1 γµ = σ k ρ1 − ρ 1 σk = 0 αk ρ2 + ρ2 αk = γk ρ2 − ρ2 γk = σk ρ2 − ρ2 σk = βρ2 + ρ2 β = 0  αk σ = iαm  k, , m = (1, 2, 3) cyclicly permuted σk γ = iγm  γk α = iβσm Latin indices: 1, 2, 3.

Now in the non-relativistic approximation i ∂ ie + A4 ∂x4 c 2 − ∂ = mc2 + O(1) ∂t m2 c2 2 = = 1 2 c2 1 2 c2 × = −i 1 2 c2 × 2 −i ∂ − eΦ ∂t −i ∂ − eΦ − mc2 ∂t ∂ − eΦ + mc2 ∂t {−2mc2 + O(1)} −i ∂ − eΦ + mc2 ∂t − m 2 c4 (98) 25 The Dirac Theory Hence h2 ∂ −i − eΦ + mc2 ψ − ∂t 2m + 3 ie ∂ + Ak ∂xk c k=1 e [σ · H − iα · E]ψ + O 2mc 1 mc2 2 ψ =0 The non-relativistic approximation means dropping the terms O 1/mc2 . Thus the non-relativistic Schr¨odinger equation is h2 2m 3 ie ∂ + Ak ∂xk c 2 e (σ · H − iα · E) ψ 2mc k=1 (99) The term α · E is really relativistic, and should be dropped or treated more exactly.

Calculate the total rate of emission of pairs, and the angular and momentum distributions. Solution Let ∆E be the excitation energy, ρN and j N the charge and current density operators of the nucleus. Then for the transition we are interested in ρN and j N are functions of position r with the time-variation of the single matrix element given by exp {−i∆E/ }. Also ∇ · jN = − ∂ρH ∆E =i ρN ∂t (119) The electrostatic potential V of the nucleus has the matrix element given by ∇2 V = −4πρN (120) The states being spherically symmetric, ρ N is a function of r only, and so the general solution of Poisson’s equation simplifies to 12 V (r) = − 6π r r 0 r12 ρN (r1 ) dr1 (121) Outside the nucleus V (r) = Ze2 /r is constant in time, and so the matrix element of V (r) for this transition is zero.

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