Quantization of complex scalar field I'm learning Peskin's qft now and I'm a little confused about problem 2.2 . 
Suppose I write the field $\phi(x)$ as:
$\phi(x) =\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{p}}} (a_{p}e^{-ipx}+b_{p}e^{ipx})$
I know that $b_p$ should be written as $b_p^\dagger$ because it annihilate antiparticle, otherwise $b_p$ creates particle with negative energy.
However, when I calculate the Hamiltonian, all I got is:
$\int \frac{d^3p}{(2\pi)^3}E_{p}(a^\dagger_{p}a_{p}+b^\dagger_{p}b_{p})$
In my result, the $b$ particles create positive energy as $a$ particle did. I'm not sure if I did something wrong in calculation or there are some other explanation in the result.
 A: To begin with, Peskin gives the following action:
$$
\mathcal{S} = \int d^4 \left[ \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi \right].
$$
Let's begin by considering the classical field theory. The equation of motion is
$$
\left( \partial^2 + m^2 \right) \phi = 0
$$
(and its complex conjugate). A general solution to this equation is given by
$$
\phi(\mathbf{x},t) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left( a_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x} + i E_{\mathbf{p}} t} + b_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x} - i E_{\mathbf{p}} t} \right),
$$
where $E_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}$. At this point, the quantities $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$ are just some complex functions of momentum. We can also write down the classical Hamiltonian of this system, where the conjugate momentum is
$$
\pi(\mathbf{x},t) = \partial_t \phi^{\ast}(x,t) = - i \int \frac{d^3 p}{(2 \pi)^3} \sqrt{\frac{E_\mathbf{p}}{2}} \left( a_{\mathbf{p}}^{\ast} e^{-i \mathbf{p} \cdot \mathbf{x} - i E_{\mathbf{p}} t} - b_{\mathbf{p}}^{\ast} e^{-i \mathbf{p} \cdot \mathbf{x} + i E_{\mathbf{p}} t} \right)
$$
(and its complex conjugate).
Now we go to the quantum theory, where the fields $\phi$ and $\pi$ (and therefore the coefficients $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$) become operators. The starting point of quantization is the equal-time canonical commutation relations:
$$
[\phi(\mathbf{x},t),\pi(\mathbf{y},t)] = [\phi^{\ast}(\mathbf{x},t),\pi^{\ast}(\mathbf{y},t)] = i \delta^3(\mathbf{x} - \mathbf{y})
$$
We cannot assume at this point that $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$ are bosonic annihilation operators, we need to actually calculate their commutation relations from the above equations. Inverting the above formulae, we have
$$
a_{\mathbf{p}} = e^{- i E_{\mathbf{p}} t} \int d^3 x \, e^{- i \mathbf{p} \cdot \mathbf{x}} \left( \sqrt{\frac{E_{\mathbf{p}}}{2}} \, \phi(\mathbf{x},t) - i \frac{1}{\sqrt{2E_{\mathbf{p}}}} \, \pi^{\ast}(\mathbf{x},t) \right),
$$
$$
b_{\mathbf{p}} = e^{- i E_{\mathbf{p}} t} \int d^3 x \, e^{- i \mathbf{p} \cdot \mathbf{x}} \left( \sqrt{\frac{E_{\mathbf{p}}}{2}} \, \phi(\mathbf{x},t) + i \frac{1}{\sqrt{2E_{\mathbf{p}}}} \, \pi^{\ast}(\mathbf{x},t) \right).
$$
Now we can compute the commutators directly. They are
$$
[a_{\mathbf{p}},a^{\dagger}_{\mathbf{p}'}] = - (2 \pi)^3 \delta^3(\mathbf{p} + \mathbf{p}'), \qquad [b_{\mathbf{p}},b^{\dagger}_{\mathbf{p}'}] = (2 \pi)^3 \delta^3(\mathbf{p} + \mathbf{p}').
$$
As you can see, one of these commutators is the wrong sign compared to the usual bosonic commutation relations, and as a result the construction of the usual bosonic Fock space proceeds differently. (In particular, $[a^{\dagger} a, a^{\dagger}] \propto - a^{\dagger} $ so $a^{\dagger}$ is actually a lowering operator.) To get the usual commutation relations, we should define
$$
\tilde{a}_{\mathbf{p}} = a^{\dagger}_{\mathbf{p}}, \quad \tilde{a}^{\dagger}_{\mathbf{p}} = a_{-\mathbf{p}}, \quad \tilde{b}_{\mathbf{p}} = b_{\mathbf{p}}, \quad \tilde{b}^{\dagger}_{\mathbf{p}} = b^{\dagger}_{-\mathbf{p}}.
$$
This gives us the usual result. In textbooks they often write down the initial expansion in such a way so that the answer comes out correctly, but in practice one needs to go through the above procedure (until you've done it enough times that it has become automatic).
A: I am afraid what is wrong is in your initial writing.  
A complex scalar field should be written as
$\phi(x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} (a_p e^{-ipx} + b_p^\dagger e^{ipx})$
And, by complex conjugation,
$\phi^*(x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} (a_p^\dagger e^{ipx} + b_p e^{-ipx})$
Clearly $a_p^\dagger \neq b_p^\dagger$ as these operators create particles of opposite charge. However in both cases $\omega_p = \sqrt{\vec p^2 +m^2} \gt 0$.  
To get the normal ordering of the $b's$ operators, at the end of the demonstration you have to apply the equal-time commutation relations
$[b_p, b_{p'}^\dagger] = (2 \pi)^3 \delta^3 (\vec p - \vec p')$ 
Note: Your notation of the $b$ operator as creator is confusing. The notation used in literature for a complex scalar field and reported here is a generalization of the notation applied to a real scalar field.
