In most references I've seen (see, for example, Peskin and Schroeder problem 2.2, or section 2.5 here), one constructs the field operator $\hat{\phi}$ for the complex Klein-Gordon field as follows:
First, you take the Lagrangian density for the classical Klein-Gordon field
$$ \mathcal{L}=\partial_\mu \phi^\dagger\partial^\mu\phi-m^2\phi^\dagger\phi \tag{1} $$ and find the momentum conjugate to the field $\phi$ via
$$ \pi=\frac{\partial\mathcal L}{\partial \dot\phi}=\dot\phi^\dagger.\tag{2} $$ Then, one imposes the usual canonical commutation relations on $\hat\phi$ and $\hat\pi$:
$$ [\hat\phi(x),\hat\pi(y)]=i\delta^3(x-y).\tag{3} $$ So, one needs to find operators $\hat{\phi}$ and $\hat\pi$ such that they obey the above commutation relations, and such that $\hat\pi=\dot\phi^\dagger$. The textbooks then go on to show that defining
$$ \hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}[a_p^\dagger e^{-i p_\mu x^\mu}+b_pe^{i p_\mu x^\mu}]\tag{4} $$ $$ \hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}[a_p e^{i p_\mu x^\mu}-b_p^\dagger e^{-i p_\mu x^\mu}]\tag{5} $$ where $a$ and $b$ are bosonic annihilation operators, satisfies these properties.
My question is: Why do we need two different particle operators to define $\hat\phi$ and $\hat\pi$? It seems to me that one could simply define
$$ \hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}a_p e^{-i p_\mu x^\mu}\tag{6} $$ $$ \hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}a_p^\dagger e^{i p_\mu x^\mu}\tag{7} $$ with $\hat{a}_p$ a single bosonic annihilation operator. Then clearly $\hat{\pi}=\dot{\hat{\phi}}^\dagger$, and also
$$ \begin{array}{rcl} [\hat\phi(x),\hat\pi(y)]&=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}[a_p,a_q^\dagger]\\ &=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}(2\pi)^3\delta^3(p-q)\\ &=&i\int\frac{d^3p}{(2\pi)^3}\frac{1}{2}e^{ip_\mu (y^\mu-x^\mu)}\\ &=&\frac{i}{2}\delta^3(y-x)\\ \end{array}\tag{8} $$
which is, up to some details about normalizing the $\hat{a}_p$, correct. We would then have a Klein-Gordon field with just one kind of excitation, the $\hat{a}_p$ excitation. Why do all textbooks claim we need two separate bosonic excitations, $\hat{a}_p$ and $\hat{b}_p$?