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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 ##\large \phi(\vec{x},t)##:

$$\large \frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0.$$

We can decouple the degrees of freedom from each other by taking the Fourier transform:

$$\large \phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t).$$

Substituting back into the Klein-Gordon equation we find that ##\large \phi(\vec{p},t)## satisfies the simple harmonic equation of motion

$$\large \frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi.$$

Therefore, for each value of ##\large \vec{p}##, ##\large \phi(\vec{p},t)## solves the equation of a harmonic oscillator vibrating at frequency

$$\large \omega_\vec{p}=+\sqrt{\vec{p}^2+m^2}.$$

Thus the general solution to the Klein-Gordon equation is a linear superposition of simple harmonic oscillators with frequency ##\large \omega_\vec{p}##. When these harmonic oscillators are quantized we find that each has a set of discrete positive energy levels given by

$$\large E^p_n=\hbar\omega_\vec{p}(n+\frac{1}{2})$$

for ##\large n=0,1,2\ldots## where ##\large n## is interpreted as the number of particles with momentum ##\large \vec{p}##.

My question is what about the harmonic oscillator solutions that vibrate at negative frequency

$$\large \bar{\omega}_\vec{p}=-\sqrt{\vec{p}^2+m^2}?$$

When these harmonic oscillators are quantized we get a set of discrete negative energy levels given by

$$\large \bar{E}^p_n=\hbar\bar{\omega}_\vec{p}(n+\frac{1}{2})$$

for ##\large n=0,1,2\ldots## where ##\large n## can now be interpreted as the number of antiparticles with momentum ##\large \vec{p}##.

If this is correct then the total energy of the ground state, per momentum ##\large \vec{p}##, is given by

\begin{eqnarray*}

\large T^p_0 &=& \large E^p_0+\bar{E}^p_0\\

&=& \large \frac{\hbar\sqrt{\vec{p}^2+m^2}}{2} + \frac{-\hbar\sqrt{\vec{p}^2+m^2}}{2}\\

&=& \large 0.

\end{eqnarray*}

Thus the total ground state energy, ##\large T_0##, is zero; there is no zero-point energy.

Does this interpretation of the negative frequency solutions make sense?

The simplest relativistic field theory is described by the Klein-Gordon equation of motion for a scalar field ##\large \phi(\vec{x},t)##:

$$\large \frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0.$$

We can decouple the degrees of freedom from each other by taking the Fourier transform:

$$\large \phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t).$$

Substituting back into the Klein-Gordon equation we find that ##\large \phi(\vec{p},t)## satisfies the simple harmonic equation of motion

$$\large \frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi.$$

Therefore, for each value of ##\large \vec{p}##, ##\large \phi(\vec{p},t)## solves the equation of a harmonic oscillator vibrating at frequency

$$\large \omega_\vec{p}=+\sqrt{\vec{p}^2+m^2}.$$

Thus the general solution to the Klein-Gordon equation is a linear superposition of simple harmonic oscillators with frequency ##\large \omega_\vec{p}##. When these harmonic oscillators are quantized we find that each has a set of discrete positive energy levels given by

$$\large E^p_n=\hbar\omega_\vec{p}(n+\frac{1}{2})$$

for ##\large n=0,1,2\ldots## where ##\large n## is interpreted as the number of particles with momentum ##\large \vec{p}##.

My question is what about the harmonic oscillator solutions that vibrate at negative frequency

$$\large \bar{\omega}_\vec{p}=-\sqrt{\vec{p}^2+m^2}?$$

When these harmonic oscillators are quantized we get a set of discrete negative energy levels given by

$$\large \bar{E}^p_n=\hbar\bar{\omega}_\vec{p}(n+\frac{1}{2})$$

for ##\large n=0,1,2\ldots## where ##\large n## can now be interpreted as the number of antiparticles with momentum ##\large \vec{p}##.

If this is correct then the total energy of the ground state, per momentum ##\large \vec{p}##, is given by

\begin{eqnarray*}

\large T^p_0 &=& \large E^p_0+\bar{E}^p_0\\

&=& \large \frac{\hbar\sqrt{\vec{p}^2+m^2}}{2} + \frac{-\hbar\sqrt{\vec{p}^2+m^2}}{2}\\

&=& \large 0.

\end{eqnarray*}

Thus the total ground state energy, ##\large T_0##, is zero; there is no zero-point energy.

Does this interpretation of the negative frequency solutions make sense?

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