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?


2 Answers 2


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.


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.

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Aug 21, 2018 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
    Commented Aug 22, 2018 at 13:28

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