I have a little confusion about Dirac field operator. Field operator can be written as $$\psi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(a^s_pu^s(p)e^{-ip\cdot x}+b^{s\dagger}_pv^s(p)e^{ip\cdot x}).$$ So I thought that Hermitian conjugate would be $$\psi^{\dagger}(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(u^{s\dagger}(p)a^{s\dagger}_pe^{ip\cdot x}+v^{s\dagger}(p)b^{s}_pe^{-ip\cdot x}),$$ but the book says $$\psi^{\dagger}(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(a^{s\dagger}_pu^{s\dagger}(p)e^{ip\cdot x}+b^{s}_pv^{s\dagger}(p)e^{-ip\cdot x}).$$ How can I resolve this?

  • 3
    $\begingroup$ They are the same expression. What is the problem? $\endgroup$
    – mike stone
    Commented Jul 8, 2021 at 15:28

1 Answer 1


The order of $a_p^{s\dagger}$ and $u^{s\dagger}$ doesn't matter because $a_p^{s\dagger}$ doesn't act on $u^{s\dagger}$. $u^{s\dagger}$ is a finite dimensional vector, whereas $a_p^{s\dagger}$ is an operator that acts on the infinite dimensional space of field states.

  • $\begingroup$ Thanks for the answer. Then the commutator will be $$[\psi,\psi^{\dagger}]=\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}(a^r_pu^r(p)a^{s\dagger}_pu^{s\dagger}(p)-a^{s\dagger}_pu^{s\dagger}(p)a^r_pu^r(p)+\cdot\cdot\cdot)$$ How does it become $$\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}([a^r_p,a^{s\dagger}_p]u^r(p)u^{s\dagger}(p)+\cdot\cdot\cdot)?$$ $\endgroup$
    – Asung
    Commented Jul 8, 2021 at 15:58
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    $\begingroup$ @Asung The commutator acts on the creation and annihilation operators. Try to write down the full expression of the field operators inside the commutator. $\endgroup$ Commented Jul 8, 2021 at 17:11
  • $\begingroup$ @Davide Morgante Can you explain more specifically? $\endgroup$
    – Asung
    Commented Jul 9, 2021 at 1:57
  • $\begingroup$ @Asung Since this is a different question from the one asked in the beginning, you should open another thread. As per the rules, I cannot answer in the comments. $\endgroup$ Commented Jul 9, 2021 at 7:26

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