Physics Asked on June 19, 2021
I’m working in 3,1 Minkowski spacetime, representing null vectors as a product of two commuting spinors so that eg. $$p_i^{dot{alpha}alpha} = |i]^{dot{alpha}}langle i|^alpha.$$
I know that special conformal transformations act in terms of the spinors as $$K_{idot{alpha}alpha} = frac{partial}{partial|i]^{dot{alpha}}} frac{partial}{partiallangle i|^alpha}.$$
Is it known how to give a finite transformation of $K_i$ acting on the spinors? So of the form $$e^{bcdot K_i}|irangle = f_b(|irangle)$$ for some function $f_b$ and vector $b$?
It looks intuitively to me like it should be straight-forward given that $$K_i |irangle =0,$$ however I imagine there are some difficulties in taking the exponential of a second derivative operator.
I found the answer I was looking for in twistor space, where the conformal group acts linearly. Under a Fourier transform back to momentum space, we can write a special conformal transformation acting on $|jrangle$ as $$|jrangle^alpha mapsto |jrangle^alpha + i, b^{alphadot{alpha}}frac{partial}{partial|j]^{dot{alpha}}}.$$ I'm not really sure how useful this statement is, but I think it makes sense. I would be interested to know if it's correct to write that $$(e^{bcdot K_j}|jrangle)^alpha = |jrangle^alpha + i ,b^{alphadot{alpha}}frac{partial}{partial|j]^{dot{alpha}}},$$ and if that is correct, how to get to the right hand side from the exponential on the left.
Correct answer by Joe on June 19, 2021
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