In A. Zee's QFT in a Nutshell, he defines the field for the Klein-Gordon equation as

$$ \tag{1}\varphi(\vec x,t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}[a(\vec k)e^{-i(\omega_kt-\vec k\cdot\vec x)} + a^\dagger(\vec k)e^{i(\omega_kt-\vec k\cdot\vec x)}] $$

When calculating $\pi=\partial_0\varphi^\dagger$, I came to

$$ \tag{2}\varphi^\dagger(\vec x,t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}[a^\dagger(\vec k)e^{i(\omega_kt-\vec k\cdot\vec x)} + a(\vec k)e^{-i(\omega_kt-\vec k\cdot\vec x)}] $$

But this would imply that $\varphi^\dagger=\varphi$. Is that correct?

(Intuitively it would make sense, because in QM we also consider self-adjoint operators.)

If it's correct, then why do we explicitly write $\pi=\partial_0\varphi^\dagger$ instead of just $\pi=\partial_0\varphi$? Why bother distinguishing $\varphi$ from $\varphi^\dagger$ at all?

In case it is not correct, then the first two equations of this answer are most likely wrong.

  • 2
    $\begingroup$ For neutral fields that is correct. If you want to keep the formalism as general as possible and include charged particles you really want to allow for non-selfadjoint fields. $\endgroup$
    – Phoenix87
    Commented Nov 11, 2015 at 11:07
  • $\begingroup$ Oh OK, makes sense. (if you want some rep, write that as an answer). $\endgroup$
    – Bass
    Commented Nov 11, 2015 at 11:09

1 Answer 1


For a real scalar field I think what you have written is correct..But if you want to describe a complex scalar field then we need to distinguish between $\phi$ and $\phi^{\dagger}$...


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.