130_notes.dvi

and compare it to the original derivative. v ̇j(t) = i ̄h (− 2 Evj+ 2c^2 pj) = i ̄h (− 2 E(c^2 pj/E+ (vj(0)−c^2 pj/E)e−^2 iEt/ ̄ ...

36.14.3The Expected Velocity and Zitterbewegung The expected value of the velocity in a plane wave state can be simply calculate ...

QM. We have already seen that (even with no applied fields), while thetotal angular momentum operator commutes with the Dirac Ha ...

κ^2 ̄h^2 − ̄h^2 4 = j(j+ 1) ̄h^2 κ^2 = j^2 +j+ 1 4 κ = ±(j+ 1 2 ) Kψ = ±(j+ 1 2 )ψ The eigenvalues ofKare κ=± ( j+ 1 2 ) ̄h. We ...

Note that the eigenvalues for the upper and lower components have the same possible values, but are opposite for energy eigensta ...

= 1 r σixi r 1 2 (σjσnxjpn+ (σjσn+ 2iǫnjkσk)xnpj) = 1 r σixi r ( 1 2 (σjσnxjpn+σjσnxnpj) +iǫnjkσkxnpj) = 1 r σixi r ( 1 2 (σjσnx ...

The phase factor depends on the conventions we choose for the statesYjℓmj. For our conventions, the factor is−1. ~σ·~x r YjℓmAj= ...

  ( ∂F ∂ρ− κF ρ ) ( ∂G ∂ρ+ κG ρ )  =   (√ k 2 k 1 − Zα ρ ) G (√ k 1 k 2 + Zα ρ ) F     ( ∂ ∂ρ− κ ρ ) F− (√ k 2 k 1 − Z ...

look at them=− 1 recursion relationswitham=bm= 0 and solve the equations. (s−κ)a 0 +Zαb 0 = 0 −Zαa 0 + (s+κ)b 0 = 0 ( (s−κ) Zα − ...

At this point we take the difference between the two equations to get one condition. ( (s+nr−κ) √ k 1 k 2 +Zα ) anr+ ( Zα √ k 1 ...

Relativistic corrections become quite important for highZatoms in which the typical velocity of electrons in the most inner shel ...

For~p= 0 and~k= 0, a delta function requires thatp~′′= 0. It turns out thatu(r ′′)† 0 γ^4 γnu (r) 0 = 0, so that the cross secti ...

The matrix element is to be taken with the initialelectron at rest, ̄hk << mc, the final electron (approximately) at rest, ...

|c(2)p~′,r′;k~′ǫˆ′(t)|^2 = ( e^2 2 mV ) 2 1 ω′ω |ǫˆ·ǫˆ′|^2 δrr′ 2 πtδ(E′/ ̄h−E/ ̄h+ω′−ω) Γ = ( e^2 2 mV ) 2 2 π ω′ω ∫ V k′^2 dk′ ...

charge mom. Energy Sz ~j(EM) ~v spinor spin up, positive energy electron −e pˆz + √ p^2 c^2 +m^2 c^4 + ̄h 2 −zˆ +ˆz u(1)(~p) spi ...

Take thecomplex conjugatecarefully remembering thatx 4 andA 4 will change signs. ( ∂ ∂xi + ie ̄hc Ai ) γi∗S∗Cψ+ ( − ∂ ∂x 4 − ie ...

ψ (3) ~p = √ mc^2 |E|V u (3) ~p e i(~p·~x+|E|t)/ ̄h → √ mc^2 |E|V u (2) −~pe i(−~p·~x−|E|t)/ ̄h ψ~p(4)= √ mc^2 |E|V u(4)~p ei(~p ...

t t t 1 3 2 No reference to the “negative energy” sea is needed. No change in the “negative energy” solutions is neededalthough ...

( γμ ∂ ∂xμ + mc ̄h ) ψ= 0 Thisgives us the Dirac equationindicating that this Lagrangian is the right one. The Euler- Lagrange e ...

Writing theHamiltonian in terms of these fields, the formula can be simplified as follows H = ∫ ψ† ( ̄hcγ 4 γk ∂ ∂xk +mc^2 γ 4 ) ...