I have been reading this pdf (it is released on https://www.physicsforums.com/threads/angular-momentum-in-qft.448715/)about the total angular operator $J_{Z}$ of Dirac field \begin{equation} \psi(x)=\int\frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}\sqrt{\frac{m}{E_{p}}}\sum_{s}b(\vec{p},s)u(\vec{p},s)e^{-ipx}+d^{\dagger}(\vec{p},s)v(\vec{p},s)e^{ipx} \end{equation} At the end of the passage, the author gets the result

\begin{equation} J_{Z}=-i\sum_{s=\pm1}\int d^{3}p\left[b^{\dagger}(\vec{p},s)P_{xy}b(\vec{p},s)+d(\vec{p},s)P_{xy}d^{\dagger}(\vec{p},s)\right]+\frac{1}{2}\sum_{s=\pm1}s\int d^{3}p\left[b^{\dagger}(\vec{p},s)b(\vec{p},s)-d(\vec{p},s)d^{\dagger}(\vec{p},s)\right] \end{equation} where $b$ and $d$ satifies $b|0>=0$ and $d|0>=0$ and $P^{xy}=p^{x}\frac{\partial}{\partial p^{y}}-p^{y}\frac{\partial}{\partial p^{x}}$.

It is easy to see from the expression for $J_{Z}$ that the $b$'s term vanishes when applied to vacuum since $b|0>=0$. However, from this expression, I can not see why $J_{Z}$|0>=0 because $d^{\dagger}|0> \neq 0$. It seems that we need to calculate the anti-commutation relation $\left\{d(\vec{p},s),d^{\dagger}(\vec{p},s)\right\}=\delta^{3}(0)$, but if the anti-commutator is not zero, then I think $J_{Z}$ will not be zero either.

Can anyone explain what to do next to see the zero result?

  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Jun 22, 2022 at 16:58
  • $\begingroup$ I haven't thought hard about this, but have you thought about the Dirac Sea kind of stuff? (I have many holes in my knowledge, but is it possible you have it the other way around, and d^dag |0> = 0?) (where |0> is the free vacuum filled with antiparticles) $\endgroup$ Jun 22, 2022 at 21:11
  • $\begingroup$ @JoshuaLin If Dirac sea is considered, we have to handle negative energy, I think it is a much harder task. $\endgroup$
    – Joe Di.
    Jun 23, 2022 at 2:31
  • 2
    $\begingroup$ quantum operator products have to be normal ordered. $\endgroup$
    – Prahar
    Jun 23, 2022 at 5:51

1 Answer 1


The only step you missed is $:J_{z}:$ the normal ordering.

For the second term, it should be $$\int d^{3}p\left[b^{\dagger}(\vec{p},s)b(\vec{p},s)+d^{\dagger}(\vec{p},s)d(\vec{p},s)\right]$$

instead because quantum observables should be normal ordered. The plus sign is due to the fact that Dirac fields are fermions who anti-commmute.

Similarly, the first term should also be normal ordered:

$$\int d^{3}p\left[b^{\dagger}(\vec{p},s)P_{xy}b(\vec{p},s)-(P_{xy}d^{\dagger}(\vec{p},s))d(\vec{p},s)\right].$$

Then you find that the angular momentum of the vacuum state is indeed zero because $b|\mathrm{VAC}\rangle=0$ and $d|\mathrm{VAC}\rangle=0$.

  • $\begingroup$ Thank you for replying. But I am a little confused, since we already normal order the Hamiltonian, can we still make another normal order about angular momentum? Is all normal ordering independent of each other? Besides, it seems that we can't freely normal order an operator otherwise all operators with $b$ vanishes for vacuum? $\endgroup$
    – Joe Di.
    Jun 24, 2022 at 15:50
  • $\begingroup$ @JoeDi. You have to normal order every physical observables otherwise there's always an ambiguity. For example, if you have two classical bosons $f(x)g(x)$. Classically, it doesn't matter if you write it as $fg$ or $gf$, but quantum mechanically there's a huge difference. Normal ordering is nothing but a choice of conventions. If you already chose a specifit convention for the Hamiltonian, you'd have to keep it for other charges so that the theory is consistent. $\endgroup$
    – Valac
    Jun 24, 2022 at 17:02
  • $\begingroup$ @JoeDi. The expression of the Hamiltonian doesn't determine the entire theory. For example, in QED you have a set of independent conserved charges, such as Spin $\mathbf{S}$, Hamiltonian $H$, Momentum $\mathbf{P}$, and Electric Charge $Q$. These operators are Hermitian, and are totally independent. Each one of them should be normal ordered in canonical quantization. $\endgroup$
    – Valac
    Jun 24, 2022 at 17:07
  • 1
    $\begingroup$ @JoeDi. It is an axiom (or rather, the definition of the vacuum state) that Hermitian operators vanish when acting on vacuum states in QFT. The vacuum has the maximal number of symmetries. i.e. a vacuum state should be invariant under a Lorentz transformation or any internal (such as global U(1) and global SU(N) in gauge theories) symmetries. It follows that a vacuum state must be annihilated by the corresponding generators of these symmetries, aka the associated Hermitian operators. $\endgroup$
    – Valac
    Jun 24, 2022 at 17:10
  • $\begingroup$ sorry for reply late.Your comments is really helpful!Thank you! $\endgroup$
    – Joe Di.
    Jul 4, 2022 at 5:42

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