Sunday, 24 November 2019

quantum field theory - Spin Up with Indefinite Helicity


Imagine we are studying the spin quantization along the same axis as the momentum. What if I have a Dirac spinor with a spin up but no definite helicity ($\psi_L,\psi_R\neq0$):


$$ u(p)= \left(\begin{array}{c} \sqrt{p ·\sigma}\xi_{\uparrow} \\ \sqrt{p ·\bar\sigma}\xi_{\uparrow} \end{array}\right) = \left(\begin{array}{c} \psi_L \\ \psi_R \end{array}\right)$$


How should I understand this? If the spin along the $p$-direction is positive, shouldn’t the particle have definite right-hand helicity and have 0 left-hand component? In other words, shouldn’t we have $ u(p) \sim \left(\begin{array}{c} 0 \\ \rm something \end{array}\right)$ ?




Answer



So the problem is that I did not understand that $\psi_L$and $\psi_R$ are chirality eigenstates not helicity eigenstates.


In the high energy limit where helicity and chirality are basically the same thing, the state does in fact become an eigenstate (I did the math but it’s too long to post here unless someone asks for it.)


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