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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?

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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.

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There are no negative energy levels. The energy levels belonging to negative frequencies are positive as well. The Noether energy is proportional to $\omega^2$ divided by the square of a norm proportional to $\sqrt{\omega} $, so it is positive definite. The angular frequency can be positive or negativebut its sign determines the sign of the charge.

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    $\begingroup$ Why is the energy proportional to $\omega^2$? where did you get this from? And how does that imply that it is positive-definite? $A=-\omega^2$ is also proportional to $\omega^2$, but it is negative-definite. $\endgroup$ – AccidentalFourierTransform Aug 21 '18 at 19:24
  • $\begingroup$ @accidentalfouriertransform You are entirely correct. - E is negative definite. The point is that changing the sign of $\omega$ does not change the sign of the energy, only that of the charge. $\endgroup$ – my2cts Aug 22 '18 at 13:28

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