0.1 Klein-Gordon Equation 0.2 Dirac Equation

which can be identified with the spin. The total angular momentum J i =
L i + S i commutes with H.
As in NR QM, The plane wave solutions above are not eigenstates of
any of the angular momentum operators. (For eigenstates of ⃗ J see the H-
atom below.) There is however an important operator that commutes with
H and can be well-defined, the helicity h = ⃗σ · ⃗p/|⃗p|. It has eigenvalues ±1,
and by choosing χ to be one of its eigenstates, the full ψ will also be an
eigenstate. The helicity is not Lorentz invariant, however, because a boost
in the direction of the momentum can bring the particle to rest and reverse
its momentum without changing its spin.
0.4 Transformation
If we do an LT, the components of ψ get mixed like the components of a
3-vector under rotation. The new wave function satisfies
ψ ′ (x ′ ) = S(Λ)ψ(Λ −1 x ′ )
by analogy with the similar equation for rotations. Note first that S must
be a representation of the Lorentz group up to a phase. Starting with the
Dirac equation in x, change variables to x ′ = Λx and multiply by S:
−iSγ µ S −1 (∂ µ x ′ν )∂ ′ νSψ(Λ −1 x) + mSψ(Λ −1 x) = 0,
where we used the chain rule and inserted S −1 S remembering that S acts in
the same space as the γ’s and may not commute with them. The result is
the Dirac equation in the new frame with the new wave function identified
as above, provided S(Λ) satisfies the condition
S −1 γ µ S = Λ µ ν γν (7)
Since S(Λ) is a representation of the Lorentz Group, we can find it by finding
its infinitesimal elements. Putting S = 1 + ɛ ab Σ ab ,
(1 − ɛ ab Σ ab )γ µ (1 + ɛ ab Σ ab ) = (δ µ ν + ɛ abg µν G ab
µν )γν
By ansatz, the solution is found to be Σ ab = − 1 2γ a γ b . It is interesting to note
the explicit forms of Σ for rotations
( ) ⃗σ
Σ ij = i k
0
2
0 ⃗σ k .
4