130_notes.dvi

   A 1 (−E+mc^2 ) +pzc A 2 (−E+mc^2 ) + (px+ipy)c −A 1 pzc−A 2 (px−ipy)c+ (E+mc^2 ) −A 1 (px+ipy)c+A 2 pzc   = 0 A 1 = −pz ...

We willnormalize the statesso that there is one particle per unit volume. ψ†ψ= 1 V N^2 ( 1 + p^2 c^2 (|E|+mc^2 )^2 ) = 1 V N^2 ( ...

u(~pr)†u(r ′) ~p = |E| mc^2 δrr′ Are the free particle states still eigenstates of Σz= ( σz 0 0 σz ) as were the states of a par ...

wave,ψ(3)~p. ψ~p(1) = N     1 0 pzc E+mc^2 (px+ipy)c E+mc^2    e i(~p·~x−Et)/ ̄h jμ(1) = N^2 c([B 2 +B∗ 2 ,−i[B 2 −B 2 ∗ ...

~vgroup= dω dk ˆk=dE dp pˆ=± pc^2 √ p^2 c^2 +m^2 c^4 pˆ Clearly, we want waves that propagate in the right direction. Perhaps th ...

equation that wasfirst order in the time derivative, he hoped to have an equation that behaved like the Schr ̈odinger equation, ...

the non-relativistic limit, thenormalization condition is a bit unnaturalin the two component theory. The normalization correcti ...

The transformed equation will be the same as the Dirac equation ifS−^1 γμSaμν=γν. Multiply by the inverse Lorentz transformation ...

x′ 1 = x 1 coshχ+ix 4 sinhχ x′ = γx+βγi(ict) x′ 4 = x 4 coshχ−ix 1 sinhχ We verify that a boost along theidirection is like a ro ...

cosh^2 χ 2 −sinh^2 χ 2 = 1 4 ((e χ 2 +e − 2 χ )^2 −(e χ 2 −e − 2 χ )^2 ) = 1 4 (eχ+ 2 +e−χ−eχ+ 2−e−χ) = 1 γ 1 coshχ+iγ 4 sinhχ = ...

Forμ= 3 or 4,aμν=δμνand the requirement is fairly obviously satisfied. Checking the requirement forμ= 1, we get. γ 1 cos^2 θ 2 + ...

= √ E+mc^2 2 EV′     1 0 0 βγmc^2 γmc^2 +mc^2    e i(~p·~x−Et)/ ̄h = √ E+mc^2 2 EV′     1 0 0 βγ γ+1    e i(~p·~ ...

ψ~p(1) =    coshχ 2 0 0 sinhχ 2 0 coshχ 2 sinhχ 2 0 0 sinhχ 2 coshχ 2 0 sinhχ 2 0 0 coshχ 2    1 √ V    1 0 0 0   ei ...

∂ ∂x′ 4 = ∂ ∂x 4 ( −γj ∂ ∂xj +γ 4 ∂ ∂x 4 ) SPψ+ mc ̄h SPψ = 0 SP−^1 ( −γj ∂ ∂xj +γ 4 ∂ ∂x 4 ) SPψ+ mc ̄h ψ = 0 Sinceγ 4 commutes ...

S−boost^1 = cosh χ 2 −iγiγ 4 sinh χ 2 S†rot= cos θ 2 +γjγisin θ 2 = cos θ 2 −γiγjsin θ 2 Sboost† = cosh χ 2 −iγ 4 γisinh χ 2 = c ...

For rotations and boosts,γ 5 commutes withSsince it commutes with the pair of gamma matrices. For a parity inversion, it anticom ...

The components of orbital angular momentum do not commute withH. [H,Lz] =icγ 4 [γjpj,xpy−ypx] = ̄hcγ 4 (γ 1 py−γ 2 px) The compo ...

(~γ·~p)(~Σ·J~) = γipiΣjJj= −i 2 piJjγiγmγnǫmnj γiγmγnǫmnj = 2(δijγ 5 γ 4 +ǫijkγk) (~γ·~p)(~Σ·J~) = −ipiJj(δijγ 5 γ 4 +ǫijkγk) =− ...

Removing theψ†from the left andψfrom the right and dotting intoA, we have the interaction Hamiltonian. Hint=ieγ 4 γμAμ Note the ...

(λ^2 −1) = 0 λ=± 1 vx=±c Thus, if wemeasure the velocity component in any direction, we should either get plus or minusc. This s ...