We know that in Lorentzian signature, fermions are representations of \begin{equation} Spin(3,1)\cong SL(2,\mathbb{C})\cong SU(2)\times SU(2)^* \end{equation} where crucially left/right handed fermions are exchanged under complex conjugation. This is what one means by the $SU(2)^*$ above. From an operational perspective, we write undotted and dotted indices, $\psi^\alpha$ and $\bar{\chi}_{\dot{\alpha}}$ to keep track of the left/right handed Weyl spinors. The statement that left/right handed spinors are related by conjugation is then captured by $(\psi_\alpha)^\dagger=\bar{\psi}_{\dot{\alpha}}$. An important consequence here is that one can’t write a $U(1)$ invariant mass term such as $\psi \bar{\psi}$.

Now if we move to Euclidean signature, fermions transform under \begin{equation} Spin(4)\cong SU(2)\times SU(2) \end{equation} and this time we have two independent copies of $SU(2)$. This means that left/right handed spinors now do not transform into each other under conjugation. This leads me to the following two questions:

  1. Why can I not write a $U(1)$ invariant mass term $\psi \psi^\dagger$ in Euclidean signature? This to me seems fine as it is just the contraction of a fundamental/anti-fundamental $SU(2)$ index?

  2. How does this conjugation difference between the two signatures get resolved under Wick rotation? Are there any issues with degrees of freedom on the two sides?



Your Answer

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

Browse other questions tagged or ask your own question.