4
$\begingroup$

I'm reading this paper by Alfredo Iorio and I have a doubt concerning the anticommutation relations he uses for the Dirac field.

Around eq. (2.25), he wants to find the unitary operator $U$ that implements quantum mechanically a Weyl transformation: $$\psi'(x)=U\psi(x)U^{-1}=e^{d_\psi\sigma(x)}\psi(x),~~\text{with }d_\psi=\frac{1-n}{2}.$$

He uses the anticommutation relations $$\{\psi(x),\psi^\dagger(y)\}=\delta^{(n)}(x-y)\tag{A}$$ $$\{\psi(x),\psi(y)\}=0\tag{B}$$ to find that the unitary operator is $$U=\exp\left\{-d_\psi\int d^ny~\sigma(y)\psi^\dagger(y)\psi(y)\right\}.$$

However, are those commutation relations correct? I thought that the correct ones are with equal time, i.e. $$\{\psi_\alpha(t,\mathbf{x}),\psi^\dagger_\beta(t,\mathbf{y})\}=\delta_{\alpha\beta}\delta^{(n-1)}(\mathbf{x}-\mathbf{y}).$$

Besides, I don't see that $U$ is unitary, that would require that $$\left(\psi^\dagger(y)\psi(y)\right)^\dagger=-\psi^\dagger(y)\psi(y)$$

$\endgroup$

1 Answer 1

3
$\begingroup$

You are perfectly correct, on both issues. That part of the paper, in fact, is only illustrative of what, in general terms, one should expect when implementing Weyl symmetry in a quantum context. Indeed, even though the anticommutation relations and the actual form of U are not correct, the discussion that follows is correct. Just consider U there as an unknown operator, whose effect is to produce the Weyl transformations on fields and (vacuum) states. After that paper, with collaborators, we actually spent some time on trying identifying the correct operator and quantum procedure, to the extent that a MSc thesis of a student of mine has been fully dedicated to that, see

https://dspace.cuni.cz/handle/20.500.11956/179240

There you will see some of the questions answered, but not to the point of having in our hands a satisfactory U that could be proposed as the solution to the problem. I might have published an Erratum to the paper, but opted for a new clarifying paper on that specific topic, that never came. I apologize for that. Your input may actually trigger a fresh re-start to eventually solve the problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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