Q:

Is the ground state energy of a quantum field actually zero?

I start by outlining the little I know about the basics of quantum field theory.
The simplest relativistic field theory is described by the Klein-Gordon equation of motion for a scalar field $\phi(\vec{x},t)$:
$$\frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0.$$
We can decouple the degrees of freedom from each other by taking the Fourier transform:
$$\phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t).$$
Substituting back into the Klein-Gordon equation we find that $\phi(\vec{p},t)$ satisfies the simple harmonic equation of motion
$$\frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi.$$
Therefore, for each value of $\vec{p}$, $\phi(\vec{p},t)$ solves the equation of a harmonic oscillator vibrating at frequency
$$\omega_\vec{p}=+\sqrt{\vec{p}^2+m^2}.$$
Thus the general solution to the Klein-Gordon equation is a linear superposition of simple harmonic oscillators with frequency $\omega_\vec{p}$. When these harmonic oscillators are quantized we find that each has a set of discrete positive energy levels given by
$$E^p_n=\hbar\omega_\vec{p}(n+\frac{1}{2})$$
for $n=0,1,2\ldots$ where $n$ is interpreted as the number of particles with momentum $\vec{p}$.
My question is what about the harmonic oscillator solutions that vibrate at negative frequency
$$\bar{\omega}_\vec{p}=-\sqrt{\vec{p}^2+m^2}?$$
When these harmonic oscillators are quantized we get a set of discrete negative energy levels given by
$$\bar{E}^p_n=\hbar\bar{\omega}_\vec{p}(n+\frac{1}{2})$$
for $n=0,1,2\ldots$ where $n$ can now be interpreted as the number of antiparticles with momentum $\vec{p}$.
If this is correct then the total energy of the ground state, per momentum $\vec{p}$, is given by 
\begin{eqnarray}
T^p_0 &=& E^p_0+\bar{E}^p_0\\
&=& \frac{\hbar\sqrt{\vec{p}^2+m^2}}{2} + \frac{-\hbar\sqrt{\vec{p}^2+m^2}}{2}\\
&=& 0.
\end{eqnarray}
Thus the total ground state energy, $T_0$, is zero; there is no zero-point energy.
Does this interpretation of the negative frequency solutions make sense?

A:

No, this doesn't make any sense. There are no negative momentum oscillators here. In momentum space, the Hamiltonian of a free real scalar field $\phi$ is 
$$ H = \int \left(\frac{1}{2} \lvert \Pi(\vec p) \rvert^2 + \frac{\omega_p^2}{2} \lvert \phi(\vec p) \rvert^2 \right)\frac{\mathrm{d}^3 p}{(2\pi)^3},$$
where $\omega_p = \sqrt{\vec p^2 + m^2}$. There is no sign ambiguity: $\omega_p$ is always positive, and the free scalar field can be seen as a collection of such oscillators with positive frequency, one for each momentum $\vec p$.
The "negative frequency solutions" you have likely heard about are something different: In the mode expansion for the field in position space, we have
$$ \phi( x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}} \left( a(\vec p) \exp(\mathrm{i}px) + a^\dagger(\vec p)\exp(-\mathrm{i}px)\right)$$
and the pre-quantum field theoretic interpretation of $\phi(x)$ as a wavefunction would now identify $a(\vec p)\exp(\mathrm{i}px)$ as a "negative frequency solution" since a Hamiltonian eigenstate evolves as $\exp(-\mathrm{i}\omega_pt)$ but this contains the term $\exp(\mathrm{i}\omega_p t)$. Since quantum field theory does not identify $\phi(x)$ as a solution to the Schrödinger equation, there is no problem with this term here.

