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jcap
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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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