2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.