130_notes.dvi

(Frankie) #1

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 ∗],[B 1 +B∗ 1 ],i[1 +B∗ 1 B 1 +B∗ 2 B 2 ])

jμ(1) = N^2

c
E+mc^2

(

[2pxc],[2pyc],[2pzc],i[E+mc^2 +

p^2 c^2
E+mc^2

]

)

jμ(1) = N^2

2 c
E+mc^2
(pxc,pyc,pzc,iE)

ψ~p(3) = N





−pzc
−E+mc^2
−(px+ipy)c
−E+mc^2
1
0




e

i(~p·~x−Et)/h ̄

jμ(3) = N^2 c([A 2 ∗+A 2 ],−i[−A∗ 2 +A 2 ],[A∗ 1 +A 1 ],i[A∗ 1 A 1 +A∗ 2 A 2 + 1])

jμ(3) = N^2

c
−E+mc^2

(

− 2 pxc,− 2 pyc,− 2 pzc,i[−E+mc^2 +

p^2 c^2
−E+mc^2

]

)

jμ(3) = −N^2

2 c
−E+mc^2

(pxc,pyc,pzc,iE)

SinceEis negative for the “negative energy” solution, theprobability density is positivebut
the probabilityflux is in the opposite direction of the momentum.


36.6.3 Alternate Labeling of the Plane Wave Solutions


Start from the four plane wave solutions: 1 and 2 with positive energy and 3 and 4 with negative.
There are four solutions for each choice of momentum~p.


ψ(1)~p =


|E|+mc^2
2 |E|V





1

0

pzc
E+mc^2
(px+ipy)c
E+mc^2




e

i(~p·~x−Et)/ ̄h ψ(2)
~p =


|E|+mc^2
2 |E|V





0

1

(px−ipy)c
E+mc^2
−pzc
E+mc^2




e

i(~p·~x−Et)/ ̄h

ψ(3)~p =



|E|+mc^2
2 |E|V





−pzc
−E+mc^2
−(px+ipy)c
−E+mc^2
1
0




e

i(~p·~x−Et)/ ̄h ψ(4)
~p =


|E|+mc^2
2 |E|V





−(px−ipy)c
−Ep+zmcc^2
−E+mc^2
0
1




e

i(~p·~x−Et)/ ̄h

Concentrate on the exponential which determines the wave property. For solutions 3 and 4, both
the phase and group velocity are in the opposite direction to the momentum, indicating we have a
problem that was not seen in non-relativistic quantum mechanics.


~vphase=

ω
k

ˆk=E
p

pˆ=±


p^2 c^2 +m^2 c^4
p

ˆp
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