The Lorentz transformation operator acting on an undotted, i.e. right-handed, spinor can be expressed as e−12σ⋅ϕ+i12σ⋅θ.
There is a very cool, almost childlike, derivation of this expression in Landau Vol. 4 S. 18, I've never seen anywhere else, deriving e−12σ⋅ϕ first, then ei12σ⋅θ.
When deriving the second term, a cross product arises in the calculation, and I can't make sense of what to do with it. To properly explain the calculation, I have derived e−12σ⋅ϕ first to set the notation, and hopefully pique your interest, and then tried to derive the second term. My question will be: can you finish the calculation, and explain the cross product issue?
Given a position vector r=(x,y,z)=(x1,y1,z1), define σ⋅r=[zx−iyx+iy−z]=xiσi=x[0110]+y[0−ii0]+x[100−1]
so that
xi=12tr[(σ⋅r)σi],
abbreviated as
r=12tr[(σ⋅r)σ].
Adding tI=x0σ0 to this gives T(t,r)=T(t,x1,x2,x3)=T(t,x,y,z)=T=xiσi=[t+zx−iyx+iyt−z]
so that
t=12tr(T) and xi=12tr(Tσi),
i.e.
r=12tr(Tσ).
If we perform an infinitesimal Lorentz boost on (t,r) with infinitesimal velocity δV the Lorentz transformation of (t,r) becomes
t′=t−r⋅δV,r′=r−tδV
and that of T becomes T′=BTB+=(I+λ)T(I+λ+)=T+λT+Tλ+
so that our goal is to find B, i.e. λ, via
t′=t−r⋅δV=t−12tr(Tσ⋅δV)=12tr(T′)=12tr(BTB+)12tr[(I+λ)T(I+λ+))=t+12tr[T(λ+λ+)]
so that λ+λ+=−σ⋅δV implies λ=λ+=−12σ⋅δV.
(Can justify this fully by expanding r′ in the same way and solving both equations for λ,λ+)
giving, for δV=ϕ
B=I+λ=I−12σ⋅ϕ=e−12σ⋅ϕ,
the first part of our Lorentz transformation operator. Calculating the second part is the issue, hence my question. The cross product complicates things.
Under an infinitesimal rotation δθ we see
r′=r−δθ×r=12tr(Tσ)−δθ×12tr(Tσ)=12tr(T′σ)=12tr(BTB+σ)12tr[(I+λ)T(I+λ+)σ)=12tr(Tσ)+12tr[λTσ+Tλ+σ]
and solving for λ in the equality
−δθ×12tr(Tσ)=12tr[λTσ+Tλ+σ]
is unmanageable to me, but the answer is λ=12iσ⋅δθ.
How do you deal with this cross product and finish the calculation, to get the answer?
Edit: I think I found the answer. Using the relation for the cross product given here:
X=σ⋅x,Y=σ⋅y,iσ⋅(x×y)=12(XY−YX)
I should have written
r′=r−δθ×r=12tr(Tσ)−12tr(δθ×r)=12tr(T′σ)=12tr(BTB+σ)12tr[(I+λ)T(I+λ+)σ)=12tr(Tσ)+12tr[λTσ+Tλ+σ]
and so I think that settles it, phew!!!