# Is the scalar field operator self-adjoint?

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.

• 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. – Phoenix87 Nov 11 '15 at 11:07
• Oh OK, makes sense. (if you want some rep, write that as an answer). – Bass Nov 11 '15 at 11:09

## 1 Answer

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 $\phi^{\dagger}$...