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According to Dark Energy and the Accelerating Universe quantum field theory says that the energy density of the vacuum, $\rho_{vac}$, should be given by $$\rho_{vac}=\frac{1}{2}\sum_{\rm fields}g_i\int_0^{\infty}\sqrt{k^2+m^2}\frac{d^3k}{(2\pi)^3}\approx\sum_{\rm fields}\frac{g_i k^4_{max}}{16\pi^2}$$ where $g_i$ is positive/negative for bosons/fermions and $k_{max}$ is some momentum cutoff.

My question is why do we only take the positive square root terms?

According to the Feynman-Stueckelberg interpretation a positive energy antiparticle going forward in time is equivalent to a negative energy particle going backwards in time. Maybe we cannot rule out negative energy virtual particles moving backwards in time?

Therefore, in order to include antiparticles in the above sum, maybe we should include the negative square root terms? If we do then we find that the energy density $\rho_{vac}=0$.

Addendum ok I accept that I was wrong - antiparticles have positive energy. But my original question still stands. Why do we ignore the negative energy modes in the above vacuum energy density calculation?

John Eastmond
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    Both particles and anti-particles have energy $E=+\sqrt{\boldsymbol p^2+m^2}$. Anti-particles do not have negative energy. – AccidentalFourierTransform Jan 21 '18 at 20:06
  • Ok but why are the negative energy terms ignored in the vacuum energy density calculation? – John Eastmond Jan 22 '18 at 05:26
  • Why do we ignore the negative energy modes in the above vacuum energy density calculation? – John Eastmond Jan 22 '18 at 05:43
  • @JohnEastmond there's no such thing as "negative energy mode". There's negative frequency modes. Energy is a well-defined operator on the Fock space of the QFT, and its spectrum is bounded from below, meaning that every single elementary particle contributes with positive energy. – Prof. Legolasov Jan 22 '18 at 12:10
  • @JohnEastmond note that there's plenty of reasons to believe that the calculation above doesn't make much sense (not the last one is: the resulting prediction is off by some factor of $10^{120}$ from experimentally measured value, lol). But your explanation doesn't seem like one of them, sorry. – Prof. Legolasov Jan 22 '18 at 12:19

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