The Dirac field is quantized as: $$\psi(x^\mu)=\int{d^3 p\over(2\pi)^3\sqrt{2\omega_p}}[a_s(p)u_s(p)e^{-ipx}+b_s^{\dagger}(p)v_s(p)e^{ipx}]$$

In the title:$$[\psi,\hat{J_z}]=J_z\psi+i(x{\partial \psi\over\partial y}-y{\partial\psi\over\partial x})$$

This equation is a result of: $$U(\Lambda)\psi(x)U^{\dagger}(\Lambda)=\Lambda^{-1}_{spinor}\psi(\Lambda x)$$

$\hat{J_z}$is an operator in the Hilbert space, while $J_z$ is the z-axis rotation generator for Dirac spinor.

we have: $\ \ \ \ [\hat{J_z},a_s(p)]=-s\ a_s(p)\ \ \ \ $and$\ \ \ \ [\hat{J_z},b_s^{\dagger}(p)]=s\ b_s^{\dagger}(p)$

and accordingly: $\ \ \ \ J_zu_s(p)=s\ u_s(p)\ \ \ \ $and$\ \ \ \ J_zv_s(p)=-s\ v_s(p)$

which gives:$\ \ \ \ [\psi,\hat{J_z}]=J_z\psi\ \ \ \ $and$\ \ \ \ i(x{\partial \psi\over\partial y}-y{\partial\psi\over\partial x})=0\ \ \ \ $, the latter is obviously incorrect.

There is no way for the "angular momentum" term $\ \ i(x{\partial \psi\over\partial y}-y{\partial\psi\over\partial x})\ \ $to pop out. What is going wrong here?


1 Answer 1


Well if it walks like a duck, if it quaks like a duck, it is a duck. Okay, let's nuance it a bit. $J_z$ is the spin operator, not the orbit angular momentum. It is defined like:

$$ J_z = \int d^3x\, \psi^\dagger(x)\frac{\Sigma^z}{2} \psi(x)$$

where $\Sigma$ is

$$ \mathbf{\Sigma} = \left(\begin{array}{cc} \mathbf{\sigma} & 0 \\ 0 & \mathbf{\sigma} \end{array}\right)$$

where the small $\sigma$'s are the Pauli-matrices. In order to make it work the $\psi$'s have to be considered as operators acting on the Fock space, they have to be developed in annihilation and creation operators.

The proof that the $J_z$ defined like this fulfills the stated relations in the post are popular exercises in many textbooks, but the post Again on Spin Operator in Dirac Field Theory (Peskin & Schroeder) might help.

  • $\begingroup$ Thanks! But this is not what I meant by the question. I understand that $\hat{J_z}$ on the LHS of the equation is the operator you described, the $J_z$ on the RHS is a purely classical quantity. And by angular momentum I mean the term $i(x{\partial \psi\over\partial y}-y{\partial\psi\over\partial x})$ ,this term is certainly not zero. $\endgroup$
    – Bababeluma
    Oct 11, 2023 at 9:41
  • 2
    $\begingroup$ The term $i(x \partial_y - y \partial_x)$ arises from the orbital angular momentum piece. What @FredericThomas is saying is that your ${\hat J}_z$ ONLY captures the spin angular momentum part so there is no reason it should reproduce the extra term $i(x \partial_y - y \partial_x)$. $\endgroup$
    – Prahar
    Oct 11, 2023 at 10:41
  • $\begingroup$ @Prahar, thanks! I understand now! $\endgroup$
    – Bababeluma
    Oct 11, 2023 at 11:15
  • $\begingroup$ @Bababeluma: Usually in this type of computations the orbital angular momentum is not considered. I think, the trick is to consider the system at rest at zero coordinates. $\endgroup$ Oct 11, 2023 at 11:25
  • $\begingroup$ @FredericThomas Thanks! This is actually a series of problems originating from the post: physics.stackexchange.com/q/783341 which you also gave an answer. At last, I found it might originates from the fact that Schwartz got the spinor solution wrong in his textbook as I pointed out in this post:physics.stackexchange.com/q/783948 Can you confirm this typo in Schwartz? $\endgroup$
    – Bababeluma
    Oct 11, 2023 at 11:38

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