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  Spinor Lorentz Transform via Vectors - Cross Product Issue

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The Lorentz transformation operator acting on an undotted, i.e. right-handed, spinor can be expressed as e12σϕ+i12σθ.

There is a very cool, almost childlike, derivation of this expression in Landau Vol. 4 S. 18, I've never seen anywhere else, deriving e12σϕ first, then ei12σθ.

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 e12σϕ 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 σr=[zxiyx+iyz]=xiσi=x[0110]+y[0ii0]+x[1001]
so that
xi=12tr[(σr)σi],
abbreviated as
r=12tr[(σr)σ].


Adding tI=x0σ0 to this gives T(t,r)=T(t,x1,x2,x3)=T(t,x,y,z)=T=xiσi=[t+zxiyx+iytz]
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=trδV,r=rtδV
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=trδV=t12tr(TσδV)=12tr(T)=12tr(BTB+)12tr[(I+λ)T(I+λ+))=t+12tr[T(λ+λ+)]
so that λ+λ+=σδV implies λ=λ+=12σδV.
(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=I12σϕ=e12σϕ,
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=rδθ×r=12tr(Tσ)δθ×12tr(Tσ)=12tr(Tσ)=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=12iσδθ.
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=σx,Y=σy,iσ(x×y)=12(XYYX)

I should have written

r=rδθ×r=12tr(Tσ)12tr(δθ×r)=12tr(Tσ)=12tr(BTB+σ)12tr[(I+λ)T(I+λ+)σ)=12tr(Tσ)+12tr[λTσ+Tλ+σ]

and so I think that settles it, phew!!!

asked Nov 20, 2016 in Theoretical Physics by bolbteppa (120 points) [ revision history ]
edited Nov 20, 2016 by bolbteppa

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