Why aren't four-vectors used in the definition of a Klein-Gordon quantum field? I am a beginner who is learning QFT. When I was going through the quantisation of Klein-Gordon real-field. I got confused about something:
The solution to Klein-Gordon equations are of the form $ \psi(x^\mu) \sim e^{ik_\mu x^\mu} $. Now the solutions in Peskin and Schroeder have no time dependence. It gives a Fourier transform of this kind:
$$ \psi(\vec x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{i\vec k. \vec x} + a^\dagger(\vec p) e^{-i\vec k. \vec x} \Bigr] $$
My problem here is, why is the solution is a superposition of $ e^{i\vec k. \vec x} $ and not $ e^{ik_\mu x^\mu} $ ?
[EDIT]
For example in these notes in equations 90 and 113 the solutions are superposition of  $ e^{ik_\mu x^\mu} $ and I don't where these two things disagree.
 A: As I mentioned in the comments, P&S are working in the Schrodinger picture which means that the operator fields are time-independent. Of course, in the Heisenberg picture, the solution of the Klein-Gordon equation is dependent on time (and then it will have four-vectors). In order to see this, let us write down the Klein-Gordon equation:
\begin{equation}
\left(\partial^2 + m^2 \right) \phi(x)=0
\end{equation}
where $g=\mathrm{diag}(+1,-1,-1,-1)$. Then the solutions in the Heisenberg picture can be written as:
\begin{equation}
\phi(x) = e^{\pm i p_\mu x^\mu}
\end{equation}
which can be easily verified:
\begin{equation}
\begin{aligned}
\partial^2 \phi & = \partial_\mu \partial^\mu \left(e^{\pm i p_\nu x^\nu}\right) \\&
= \partial_\mu \left(\pm i p^\mu\right) \left(e^{\pm i p_\nu x^\nu}\right) \\&
= \left(\pm i p^\mu\right) \left(\pm i p_\mu\right) e^{\pm i p_\nu x^\nu} \\&
= - p_\mu p^\mu e^{\pm i p_\nu x^\nu} \\&
= -(E^2 - \mathbf{p}^2) e^{\pm i p_\nu x^\nu} \\&
= -m^2 e^{\pm i p_\nu x^\nu} \\&
= -m^2 \phi
\end{aligned}
\end{equation}
and so:
\begin{equation}
\left(\partial^2 +m^2\right)\phi = \left(-m^2 +m^2\right)\phi = 0
\end{equation}
It is normal to write the solution in terms of positive frequency solutions and negative frequency solutions:
\begin{equation}
\phi(x)=\phi_+(x) + \phi_-(x) = a e^{- i p_\nu x^\nu} + b e^{+ i p_\nu x^\nu} \tag{1}
\end{equation}
Of course, we also need to sum over all energy-momentum values $p_\mu$ (because equation $(1)$ is a solution for any value of $p_\mu$). Hence, the general solution is:
\begin{equation}
\phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + b(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right]
\end{equation}
where $N$ is a normalization constant.
In order to see how to switch between the Dirac and Schrodinger picture, I refer you to section $2.4$ of P&S.
Edit I couldn't help my self and will quickly add this:
P&S are discussing the real Klein-Gordon field, which means:
$$ \phi = \phi^* $$
and so:
\begin{equation}
\int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i p^\mu x_\mu} + b(\mathbf{p}) e^{ip^\mu x_\mu}\right] = \int \frac{\mathrm{d}^3 \mathbf{p}}{N^*} \; \left[ a^*(\mathbf{p}) e^{ ip^\mu x_\mu} + b^*(\mathbf{p}) e^{-i p^\mu x_\mu}\right]
\end{equation}
which implies:
\begin{equation}
\begin{array}{cc}
a(\mathbf{p}) = b^*(\mathbf{p}) \; ,&  b(\mathbf{p}) = a^*(\mathbf{p}) 
\end{array}
\end{equation}
and $N$ must be a real. So the real field can be written as:
\begin{equation}
\phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + a^*(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right] 
\end{equation}
A: After the comments and answer from Hunter, I think that the difference between these two things lie in the fact that, in Schrodinger picture the $ \psi (\vec x) $ satisfies the equation $$ \bigg (\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2 \bigg)\psi (x) = 0 $$ where $ \omega_p = \sqrt{|\vec p|^2 + m^2} $.
$$ \hat\psi(\vec x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{i\vec k. \vec x} + a^\dagger(\vec p) e^{-i\vec k. \vec x} \Bigr] $$
In the case of Heisenberg picture however, the Schrodinger ladder operators transform like (refer P&S section 2.4)
$$ a_h(\vec p) = e^{iHt}a(\vec p)e^{-iHt} = a(\vec p)e^{-iE_pt}$$
$$ a^\dagger_h(\vec p) = e^{iHt}a^\dagger(\vec p)e^{-iHt} = a(\vec p)e^{iE_pt}$$
Now putting this into the expression for obtaining Heisenberg representation of the field,
$$ \hat\psi_h(\vec x ,t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[a(\vec p) e^{ip_\mu x^\mu} + a^\dagger(\vec p) e^{-ip_\mu x^\mu} \Bigr]  $$ where $ p^0 = E(\vec p) $
Now this operator satisfies the equation 
$$ i\frac{\partial}{\partial t} \hat\psi_h = [\hat\psi_h,\hat H] $$
