u(~pr)†u(r′)
~p =|E|
mc^2δrr′Are the free particle states still eigenstates of Σz=
(
σz 0
0 σz)
as were the states of a particle atrest? In general, they are not. To have an eigenvalue of +1, a spinor must have zero second and
fourth components and to have an eigenvalue of -1, the first and third components must be zero. So
boosting our Dirac particle to a frame in which it is moving, mixes up the spin states.
There is one case for which these are still spin eigenstates. If the particle’s momentum is in the z
direction, then we have just the spinors we need to be eigenstatesof Σz. That is, if we boost along
the quantization axis, the spin eigenstates are preserved. Theseare calledhelicity eigenstates.
Helicity is the spin component along the direction of the particle. While itis possible to make
definite momentum solutions which are eigenstates of helicity, it is notpossible to make definite
momentum states which are eigenstates of spin along some other direction (except in the trivial case
of~p= 0 as we have shown).
To further understand these solutions, we can compute the conserved probability current. First,
compute it in general for a Dirac spinor
ψ =
A 1
A 2
B 1
B 2
.
jμ = icψγ ̄μψ
jμ = icψ†γ 4 γμψγ 4 =
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 − 1
jμ = ic(A∗ 1 A∗ 2 −B 1 ∗ −B∗ 2 )γμ
A 1
A 2
B 1
B 2
γ 1 =
0 0 0 −i
0 0 −i 0
0 i 0 0
i 0 0 0
γ 2 =
0 0 0 − 1
0 0 1 0
0 1 0 0
−1 0 0 0
γ 3 =
0 0 −i 0
0 0 0 i
i 0 0 0
0 −i 0 0
jμ = c
A∗ 1 B 2 +A 1 B∗ 2 +A∗ 2 B 1 +A 2 B∗ 1
−i[A 1 ∗B 2 −A 1 B 2 ∗−A∗ 2 B 1 +A 2 B 1 ∗]
A∗ 1 B 1 +A 1 B∗ 1 −A∗ 2 B 2 −A 2 B∗ 2
i[A∗ 1 A 1 +A∗ 2 A 2 +B∗ 1 B 1 +B∗ 2 B 2 ]
Now compute it specifically for a positive energy plane wave,ψ(1)~p , and a “negative energy” plane