Hamiltonian for a Klein-Gordon field can be written as - $$H= \int \frac{d^3p}{(2\pi)^3 \omega_{\vec{p}}}[a^{\dagger}_\vec{p}a_\vec{p}+\frac{1}{2}(2\pi)^3\delta^{(3)}(0).\tag{1}]$$ In one of my lecture notes on QFT, It is written that - In absence of gravity, we can neglect the second term in above equation which will lead us to- $$H= \int \frac{d^3p}{(2\pi)^3 \omega_{\vec{p}}}a^{\dagger}_\vec{p}a_\vec{p}.\tag{2}$$ It is obvious that second term in the first equation will make vacuum energy infinite. But Neglecting this term is the best we can do? Something is infinite and just for our convenience, we are putting it to zero. How is this logically and mathematically justifiable? Secondly, What is the role of gravity in neglecting this term? Why can we not neglect this term if gravity is present?

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    $\begingroup$ The way physicists usually justify it is to say that only diferences of energy matter, and so the neglected term would be canceled anyway, so we can drop it. The issue with gravity is that when GR comes into play, energy itself acts as a source of gravity by virtue of Einstein's equation. Then not only differences of energy matter and this argument would not apply. $\endgroup$
    – Gold
    Oct 21, 2017 at 19:34
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    $\begingroup$ Generally, the process of quantization (i.e. passing from classical theory to the quantum theory of the same phenomenon) is ambiguous. The two forms (eqs (1) and (2) in your question) are related by reordering of $a$ and $a^{\dagger}$. Classical physics in insensitive to operator reordering, so both terms are equally good for describing the classical behavior. In QFT, however, the second form leads to a well-defined theory, while the first form doesn't. It is only logical to assume that in the quantum theory, the second form is the correct expression. $\endgroup$ Oct 22, 2017 at 1:14
  • $\begingroup$ possible duplicate: physics.stackexchange.com/q/296803 $\endgroup$ Oct 22, 2017 at 11:53
  • $\begingroup$ @Prof.Legolasov Can you expand a bit on this? Is it a settled issue that the second term should be dropped? What is the basis for this? In principle, using the ordering ambiguity one can rearrange any expression and then promote it to any given operator. Is there any systematics about this? $\endgroup$
    – Dr.Yoma
    Feb 25, 2023 at 20:18
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    $\begingroup$ @Dr.Yoma it's a long story, won't possibly fit in the character limit on comments here. I would ask this as a separate question if i were you. In this particular case yes, this is settled. The ambiguities are usually resolved in QFT where most orderings give inconsistent (read, plagued with infinity) definitions, like the first one in the OP's question $\endgroup$ Feb 25, 2023 at 22:05

2 Answers 2


The vacuum energy is the simplest and earlist example of a phenomenon that plagues many quantum field theories: Divergences. In fact, it exhibits two kinds of divergences. We start from the expression for the Hamiltonian: $$ H = \int \left(\omega_p a^\dagger(p)a(p) + \frac12\omega_p (2\pi)^3 \delta(0)\right)\frac{\mathrm{d}^3 p}{(2\pi)^3}$$ and introduce the shorthand $E_0 = \frac12 \int \omega_p \delta(0)\mathrm{d}^3 p$ for the second term.

  1. Infrared (IR) divergence: $E_0$ is divergent because we are computing the energy in an infinite volume! A better quantity to look at is the energy density $$ \epsilon_0 = \frac12 \int \omega_p \frac{\mathrm{d}^3p}{(2\pi)^3}$$.

  2. Ultraviolet (UV) divergence: Unfortunately, $\epsilon_0$ is still divergent because $$ \epsilon_0 = \frac{1}{(2\pi)^2}\int_0^\infty \sqrt{\lvert p\rvert^2 + m^2} \lvert p \rvert^2\mathrm{d}\lvert p\rvert$$ does not converge. However, it is clearly finite for any upper boundary $\Lambda < \infty$ of the integral. This is the first hint that QFT should generally be thought of as an effective theory approximating some other, different underlying fundamental theory. Therefore, QFTs typically come with some sort of "cutoff" $\Lambda$ on the allowed momenta/energies.

The vacuum energy also allows us a glimpse at another feature of QFTs: Getting rid of such divergences via renormalization. Without changing the dynamics, we may add a term $\int\mathrm{d}^3 V_0$ to the classical Hamiltonian, i.e. add a constant energy density. In the quantum theory with a cutoff $\Lambda$, the total vacuum energy density is now $$ \epsilon_0(\Lambda) + V_0,$$ but since $V_0$ was arbitrary, we are allowed to set $V_0 = -\epsilon_0(\Lambda)$, making the total vacuum energy density zero *even if we lift $\Lambda\to\infty$ again. This is the reason we are allowed to neglect the vacuum energy in theories without gravity - it can be renormalized away very easily.

However, in a theory with gravity, $V_0$ participates in the dynamics - it is essentially the cosmological constant! Therefore we are not allowed to choose its value as we see fit, and we cannot get rid of the vacuum energy through such a choice.


I usually don't attempt answering questions on the topics that I am not quite familiar with but this would be an exception.

As far as I have read, the reason is that, in the experiments, we can only measure the exchanges of energy, i.e. the difference of energies. This means you can make your datum anywhere you want. In this case, we choose the exact infinity that is produced by the second term as our datum and thus, in this redefined sense, only the first term is our energy.

Gravity makes it all complicated because gravity sees everything and everyone sees gravity. That is to say that in gravity, you cannot set your datum of energy where you wish. There is a meaning to the absolute zero energy because that would mean no gravitational effects. But if there is some energy then it means there would be some gravitational effect. Thus, if you consider gravity, then the second term will create measurable effects via producing gravity and thus, you can not just ignore it. But I would like to add my two cents by saying that since we got the actual equation via methods not concerning gravity, it might be the case that it already shifts the datum by some amount so that it doesn't match the expression for energy that one would have gotten if she had used the datum as zero in the sense of gravitational effects. Thus, it might be the case that the whole of the second term is actually produced due to this inherent shift of datum that is made by the non-gravitational method that we used to derive the expression in the first place. But as I said, these are my two cents - I do not claim to have read them anywhere. The rest of the answer is based on David Tong's lecture notes on QFT.

  • $\begingroup$ So you mean infinities will not appear if we take gravity into account and we will have a measurable vacuum energy? $\endgroup$
    – physics101
    Oct 21, 2017 at 19:47
  • $\begingroup$ @physics101 No, no. I do not claim that. I just speculate that it might be the case. The answer to your question lies in my statements before I added my "two cents". $\endgroup$
    – user87745
    Oct 21, 2017 at 19:48

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