The annihilation operator in the Dirac field could be written as

$$ a_p^s = \frac{e^{iE_pt}}{\sqrt{2E_p}}u^s(p)^\dagger\int d^3xe^{-ipx}\psi(t,x) $$

Where \begin{equation*} \begin{split} \psi(t, x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_{s = 1,2}\left[a_p^su^s(p)e^{-ipt}+ b_{-p}^{s\dagger}v^s(-p)e^{ipt}\right]e^{ipx} \end{split} \end{equation*}

My question is how do we find $a_p^{s\dagger}$? For operators $A, B, C$, we have $(ABC)^{\dagger} = C^{\dagger}B^{\dagger}A^{\dagger}$. Can we split $a^\dagger$ into 4 parts and use this identity to get

$$ a_p^{s\dagger} = \int d^3x\psi^{\dagger}(t,x)e^{+ipx}u^s(p)\frac{e^{-iE_pt}}{\sqrt{2E_p}} $$

  • $\begingroup$ This is essentially right, but following your argument, I don't know if you have the right intuition or if you simply followed blindly the formula $(ABC)^\dagger = C^\dagger B^\dagger A^\dagger$... $\endgroup$ Jan 7, 2023 at 17:31
  • $\begingroup$ @Jeanbaptiste Roux Thanks for the comment! I don't think I'm quite comfortable with finding this hermitian conjugate, as I'm unsure if I can just follow that formula, or if I can also follow some rules like flipping the signs of the exponential and use $(a^\dagger)^\dagger = a$. In the second case, I'm worried about if this will give the wrong result because some terms in $a^\dagger$ and $\psi$ might not commute. $\endgroup$
    – IGY
    Jan 7, 2023 at 17:44

1 Answer 1


Your final formula is right, but I'm confused why you used $(ABC)^\dagger = C^\dagger B^\dagger A^\dagger$ since there is just a product of two objects, namely $u^s(p)^\dagger \psi(x)$. Moreover, it is important to justify the formula here, because the dagger on $u^s(p)$ is just the $\mathbb{C}^4$ dagger while on $\psi(x)$ it is both the $\mathbb{C}^4$ dagger and the Hilbert space dagger combined.

In other words, I believe your problem is that an object like $\psi(x)$ is both a $\mathbb{C}^4$ column vector and an operator in a Hilbert space. A simple way to proceed is to observe that in $u^s(p)^\dagger \psi(x)$ the dagger is the $\mathbb{C}^4$ one. Now $u^s(p)^\dagger$ is a row vector whose entries are numbers, while $\psi(x)$ is a column vector whose elements are operators. The matrix product $u^s(p)^\dagger \psi(x)$ is therefore a single operator.

Let us denote $u^s(p) = (u^s_\alpha(p))$ where $\alpha$ is the $\mathbb{C}^4$ index and also $\psi(x) = (\psi_\alpha(x))$. Then we have $$u^s(p)^\dagger \psi(x)=\sum_\alpha u^s_\alpha(p)^\ast \psi_\alpha(x).$$

Now it is easy to take the adjoint of this operator

$$[u^s(p)^\dagger \psi(x)]^\dagger=\sum_\alpha u_\alpha^s(p) \psi_\alpha^\dagger(x)=\psi^\dagger(x)u^s(p).$$

Everything else on the $a_p^s$ formula are numbers, so that the adjoint becomes just the complex conjugate. Indeed rewrite $a_p^s$ by bringing $u^s(p)^\dagger$ into the integral

$$a_p^s = \frac{e^{iE_pt}}{\sqrt{2E_p}}\int d^3xe^{-ipx}u^s(p)^\dagger\psi(t,x). $$

Now take the adjoint. It is

$$(a_p^s)^\dagger = \frac{e^{-iE_pt}}{\sqrt{2E_p}}\int d^3xe^{ipx}[u^s(p)^\dagger\psi(t,x)]^\dagger=\frac{e^{-iE_pt}}{\sqrt{2E_p}}\int d^3xe^{ipx}\psi^\dagger(x)u^s(p), $$

as you proposed.

  • $\begingroup$ Thanks so much for the answer! $\psi$ is a column vector because it's a linear combination of column vectors $u$ and $v$, is that correct? Should we treat the creation and annihilation operators as numbers? $\endgroup$
    – IGY
    Jan 7, 2023 at 22:30
  • 1
    $\begingroup$ Yes, that's right, that's why $\psi$ is a column vector. We should treat the creation and annihilation operators as operators, not numbers. In particular the dagger acts on them like the Hilbert space dagger. $\endgroup$
    – Gold
    Jan 7, 2023 at 22:37
  • $\begingroup$ Thanks!! What's the difference between a Hilbert space dagger and a $C-4$ dagger? $\endgroup$
    – IGY
    Jan 7, 2023 at 22:39
  • 1
    $\begingroup$ Well, the $\mathbb{C}^4$ dagger acts on column vectors by transposing them to row vectors and taking the complex conjugate of the entries. If the entries are operators, it takes the Hilbert space dagger of the entries. The Hilbert space dagger is defined as follows: for every ket $|\psi\rangle$ there exists an associated bra $\langle \psi|$. If $A$ is an operator, the bra corresponding to $A|\psi\rangle$ is $\langle \psi|A^\dagger$. It is, in fact, the same definition in finite dimensions, except that for operators acting on a Hilbert space. $\endgroup$
    – Gold
    Jan 7, 2023 at 22:42

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