# QFT field expansion: Positive/negative energies and exponentials

I'm trying to understand the relationship between the exponentials in the field expansion and the energy of the of the particle being positive or negative.

• Is it necessary or is it a convention that the exponential $e^{ip^{\mu}x_{\mu}}$ ($e^{-ip^{\mu}x_{\mu}}$) must be by the side of the creation (annihilation) operator $a_{\vec{p}}^{\dagger}$ ($a_{\vec{p}}$)?

• How can I see that $a_{\vec{p}}^{\dagger}$ creates a positive energy particle in Klein-Gordon? And in Dirac's equation?

In Dirac's case, when you plug $u\left(p\right)e^{-ip^{\mu}x_{\mu}}$ in $h_{D}=-i\partial_{0}$, where $h_{D}=-i\vec{\alpha}\cdot\vec{\nabla}+m\beta$ is Dirac's Hamiltonian in QM, you find that it is an eigenfunction with eigenvalue $+E_{\vec{p}}$. From this I would think that this exponential tells me that, if the creation operator is beside it, it creates a positive-energy particle (this is taken from Peskin&Schroeder Section 3.5)(the positive-frequency exponential has a negative eigenvalue). However, in pg. 55 of the same section, P&S say "if we want $\bar{\psi}_{a}\left|0\right>$ to be made up of only positive-energy components, then $a_{\vec{p}}^{\dagger}u\left(p\right)e^{ip^{\mu}x_{\mu}}$ must create the particle state. This contradicts what he said before (or, at least, it seems to me it does)!

In Klein-Gordon's case, any exponential has eigenvalue $E_{\vec{p}}^{2}$ because it is a second-order differential equation, so how do I associate exponential with positive or negative energies?

Hope my question is understandable.

Thanks