# Quantization of Klein-Gordon Field

I have a question about the quantization procedure of the Klein-Gordon field as presented in Peskin&Schroeder.

The field is expressed as a Fourier decomposition $$\phi(x,t) = \int \frac{d^3p}{(2\pi)^3} e^{ipx} \phi(p,t),$$ with $\phi^*(p,t) = \phi(-p,t)$ so that $\phi(x)$ is real.

To continue one introduces ladder operators: $$\phi(p) = \frac{1}{\sqrt{2\omega_p}} (a_p + a^\dagger_{-p}).$$

But now $\phi^*(p) = \frac{1}{\sqrt{2\omega_p}} (a_p + a^\dagger_{-p})^* = \frac{1}{\sqrt{2\omega_p}} (a^*_p + a^T_{-p}) = \phi^T(-p) \neq \phi(-p) = \frac{1}{\sqrt{2\omega_p}} (a_{-p} + a^\dagger_{p}).$

So why is the difference between $\phi^T(-p)$ and $\phi(-p)$ not important (is there even a difference?)?

For a scalar field the operators $^*$ and $\dagger$ are the same, as for a scalar $\phi^T=\phi$.