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 $$K_{i\dot{\alpha}\alpha} = \frac{\partial}{\partial|i]^{\dot{\alpha}}} \frac{\partial}{\partial\langle i|^\alpha}.$$

Is it known how to give a finite transformation of $K_i$ acting on the spinors? So of the form $$e^{b\cdot K_i}|i\rangle = f_b(|i\rangle)$$ for some function $f_b$ and vector $b$?

It looks intuitively to me like it should be straight-forward given that $$K_i |i\rangle =0,$$ however I imagine there are some difficulties in taking the exponential of a second derivative operator.


1 Answer 1


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 $|j\rangle$ as $$|j\rangle^\alpha \mapsto |j\rangle^\alpha + i\, b^{\alpha\dot{\alpha}}\frac{\partial}{\partial|j]^{\dot{\alpha}}}.$$ 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 $$(e^{b\cdot K_j}|j\rangle)^\alpha = |j\rangle^\alpha + i \,b^{\alpha\dot{\alpha}}\frac{\partial}{\partial|j]^{\dot{\alpha}}},$$ and if that is correct, how to get to the right hand side from the exponential on the left.


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