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 $\hat{b}_p$?
Best Answer
There is a conceptually simple (but fiddly) way of relating this expansion to the usual Fourier expansion. TL;DR: Requiring $\phi$ to satisfy the Klein-Gordon equation divides the nonzero Fourier components into two classes, corresponding to particles and antiparticles.
For a general complex scalar field defined on spacetime, $$ \phi(x) = \int\frac{d^4 p}{(2\pi)^4} \hat{\phi}(p) e^{-ip \cdot x}. \label{fourier}\tag{1} $$ This is the ordinary four-dimensional Fourier expansion. Now, let us impose the Klein-Gordon equation $$ 0 = (\partial^2 + m^2) \phi = \int \frac{d^4 p}{(2\pi)^4} (-p^2 + m^2) \hat{\phi}(p) e^{-ip \cdot x}. $$ A general field satisfying this must consist of only modes where $p^2 = m^2$, i.e. on-shell modes. Thus, $$ \hat{\phi}(p) = 2\pi \delta(p^2 - m^2) f(p) $$ for some function $f$ (the factor $2\pi$ is convenient for comparing with the standard expansion). Writing $p = (p^0, \mathbf{p})$, $$ \delta(p^2 - m^2) = \delta\left((p^0)^2 - (\mathbf{p}^2 + m^2)\right). $$ The argument of the $\delta$-function has two zeros $p^0 = \pm \sqrt{\mathbf{p}^2 + m^2}$ (for fixed $\mathbf{p}$), so we use the general rule $$ \delta(f(x)) = \sum_{f(x_i) = 0} \frac{1}{|f'(x_i)|} \delta(x - x_i) $$ to find $$ \delta(p^2 - m^2) = \frac{1}{|2 p^0|} \left[ \delta\left(p^0 - \sqrt{\mathbf{p}^2 + m^2}\right) + \delta\left(p^0 + \sqrt{\mathbf{p}^2 + m^2}\right) \right]. $$ Putting this back into (\ref{fourier}) and performing the $p^0$ integral (abbreviating $\sqrt{\mathbf{p}^2 + m^2} = E_\mathbf{p}$): $$ \begin{align} \phi(x) &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{-i(-E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{i(E_\mathbf{p}, -\mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, -\mathbf{p}) e^{i(E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ \end{align} $$ where we have swapped $\mathbf{p} \mapsto -\mathbf{p}$ in the second term. We identify the usual (Heisenberg picture) expansion $$ \phi(x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left[ a_\mathbf{p} e^{-ip \cdot x} + b^\dagger_\mathbf{p} e^{ip \cdot x} \right] $$ with $$ \begin{align} a_\mathbf{p} &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(E_\mathbf{p}, \mathbf{p}) \\ b_\mathbf{p}^\dagger &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(-E_\mathbf{p}, -\mathbf{p}) \end{align} $$ ($f(E_\mathbf{p}, \mathbf{p})$ and $f(-E_\mathbf{p}, -\mathbf{p})$ are the relativistically normalized annihilation and creation operators). If $\phi$ is real, we know from Fourier analysis that $\hat{\phi}(p) = \hat{\phi}(-p)^\dagger$, which immediately translates to $a_\mathbf{p} = b_\mathbf{p}$.
Up until now, the field is entirely classical ($a_\mathbf{p}$ and $b_\mathbf{p}$ are simply complex numbers, and $\dagger$ is complex conjugation). Thus even the classical field has two types of excitations: Positive-frequency solutions (with coefficients $a_\mathbf{p}$) and negative-frequency solutions (with coefficients $b_\mathbf{p}^\dagger$). After quantizing, they correspond to particles and antiparticles.