I said: As far as I understand it quantum field theory says that the vacuum has an infinite energy density.

r/AskPhysics RedditorAbstractAlgebruh said: But wouldn't that be due to the way we do the math rather than a physical consequence? We could either use normal ordering to get rid of the infinite terms or plug the classical field expansion into the hamiltonian, and then impose commutation relations which gives a finite hamiltonian. Rather than imposing first, then plugging the field expansion into the hamiltonian which gives an infinite contribution.

Is the infinite vacuum energy density physical or is it a mathematical artifact?


4 Answers 4


Quantum field theory does not say that "the vacuum has infinite energy density" because part of our modern conception of quantum field theory is the idea of renormalization.

Renormalization is a necessary step in extracting actual, physical predictions out of the formalism of quantum field theory, and the naive divergence ("infiniteness") of the vacuum energy density is before renormalization. I explain the rough idea in this answer of mine, but the upshot is that, in non-gravitational theories, the energy density is unobservable and so can be renormalized to any value we want, in particular zero, while in a gravitational theory you need to pick the value we actually, experimentally, observe.

Therefore, quantum field theory actually makes no prediction at all for the vacuum energy density, just like it does not predict the values of the fundamental coupling constants (like the fine structure constant) - these are values that must be determined by experiment and used as inputs to the theory, not as predictions.

  • $\begingroup$ I do remember, during my renormalization courses, a funny computation where we changed the expected value of the void by changing the renormalization prescription, i.e. the order of operators in a product (time-ordered product, for instance). It's a bit far in my memory, though. $\endgroup$
    – Miyase
    Commented May 17, 2022 at 15:44
  • 1
    $\begingroup$ We can redefine the Hamiltonian $H$ so the ground state has zero expectation value $\langle H \rangle$, but we can't get rid of non-zero field amplitude variances $\langle P_{\mathbf k,\lambda}^+ P_{\mathbf k,\lambda} \rangle$ . The infinite energy is spurious, but the zero point fluctuations aren't. $\endgroup$ Commented May 18, 2022 at 1:36
  • $\begingroup$ @JánLalinský You're right of course, but I don't quite understand if (and if so how) this is intended as criticism of my answer. $\endgroup$
    – ACuriousMind
    Commented May 18, 2022 at 7:43
  • 1
    $\begingroup$ @ACuriousMind not really meant as a big criticism of your answer, more as an addition that may help someone interested in these issues. Because sometimes there is a misleading sentiment that the Hamiltonian redefinition solves all "UV catastrophy"-like problems in QFT. $\endgroup$ Commented May 18, 2022 at 18:05
  • $\begingroup$ In other words, through renormalisation the theoretical value of infinity is subtracted from "the theoretical value of infinity + an experimental value". $\endgroup$ Commented Jul 30, 2022 at 17:45

In reality the vacuum energy is zero. It must be so to be in line with the measured flat geometry of space-time of the universe. Dark Energy can be explained by the cosmological constant, which is separate from the energy content of space-time.

So how then do we deal with the mathematical prediction of an infinite value? As mentioned in other answers, one such "solution" is renormalization. I put that in quotes because that process doesn't actually resolve infinites, it just moves them elsewhere. While I know most physicists wouldn't claim to be mathematicians, I can tell you that from a mathematical perspective this isn't really a valid solution. It's only hotly defended because the end result "works". But it's pretty much the equivalent of hand waving and ignoring the issue... which again is defended because ignoring it gives predictions that work.

What's amazing to me, though, is how easy it is to remove the infinity via a simple change in the program of quantization... indicating that it's existence is just a mistake of program / mathematics and not something "real".

As an example, let's look at the simplest of cases: the harmonic oscillator. The typical prescription is to write our energy equation, then convert everything to operators:

$$ E = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 x^2 \\ E \rightarrow \hat{H} \\ p \rightarrow \hat{p} \\ x \rightarrow \hat{x} $$

We then create creation and destruction operators using the position and momentum operators per the usual story, all culminating in the usual problematic equation:

$$ \hat{H} = \hbar \omega (a^{\dagger} a + \frac{1}{2}) $$

But now let's do something different. Let's change variables BEFORE the replacement with operators:

$$ q \equiv x + \frac{i}{m \omega}p \\ E = \frac{1}{2} m \omega^2 q^* q $$

And now we replace our new variables with operators:

$$ E \rightarrow \hat{H} \\ q \rightarrow \hat{q} = \hat{x} + \frac{i}{m \omega} \hat{p} \\ q^* \rightarrow \hat{q}^{\dagger} = \hat{x} - \frac{i}{m \omega} \hat{p} $$

If you do this, and then create creation and destruction operators again, you instead get:

$$ \hat{H} = \hbar \omega a^{\dagger} a $$

It's worth asking here, what have we really done? Isn't it just a mathematical trick to change variable definitions before second quantization? Does this mean quantum theory "prefers" complex variables for quantization over real ones? BTW, if you apply this same trick to photons and the electromagnetic field, the infinities disappear there as well. Perhaps this mechanism is just a "better" way to renormalize the theory... where we change quantized variables as opposed to moving infinities into bare masses and charges.

From my point of view, since it's so easy to remove these infinites via a simple variable change, they aren't real / a real prediction of the theory and can, for all intents and purposes, be ignored... The only time this really matters is when trying to introduce gravity into the theory anyway.


Indeed the present theory predicts a very high or diverging vacuum energy of the universe. This is known the cosmological constant problem or vacuum catastrophe. It is a consequence of the zero point energy of quantum electrodynamics and quantum field theory. Some physicists believe that renormalisation solves the problem solutions but after checking the existing literature there is no sign of consensus on the solution. It is an unsolved problem in physics.

An interesting new development is described here.

  • $\begingroup$ While I'm not disputing your answer, you should specify what you're calling "the present theory". The original post says quantum field theory, and the problem you mention doesn't exist there (see @AcuriousMind's answer). $\endgroup$
    – Miyase
    Commented May 17, 2022 at 18:42
  • $\begingroup$ @Miyase I edited my answer in response to your comment. There is no consensus that this problem is solved. $\endgroup$
    – my2cts
    Commented May 17, 2022 at 19:43
  • $\begingroup$ I think there is consensus that renormalization works in the sense suggested in my answer. The problem this answer refers to (but doesn't spell out) is that if one views quantum field theory as an effective theory with an intrinsic cutoff $\Lambda$, so that the $\epsilon_0(\Lambda)$ (notation from the answer linked in my answer) becomes a prediction of the theory instead of a diverging quantity, then dimensional analysis "suggests" very large values for this $\epsilon_0(\Lambda)$. It is disingenuous to suggest that this is some sort of disagreement. $\endgroup$
    – ACuriousMind
    Commented May 17, 2022 at 21:03
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    $\begingroup$ "I think there is consensus' This opinion requires references. $\endgroup$
    – my2cts
    Commented May 25, 2022 at 5:53

The energy of the vacuum is also known as "dark energy", or the cosmological constant. The value of this in Planck units is about $10^{-120}$, so not only is it non-infinite it is nearly zero.

The only effect this energy has is through gravitation. Since energy (or mass) gravitates.

The energy of the vacuum corresponds to pairs of virtual particles being created and destroyed near instantaneously. In non-gravitational QFT these can simply be ignored. But they can't be ignored when we bring in gravity.

Since the force mediating gravity can be thought of as a graviton in some respects. This means gravitons pass between those pairs. So in fact dark energy can be thought of as quantum corrections to the graviton field. (It can't be got rid of).

Are these quantum corrections infinite? Well maybe, but only in the sense that a lot of non-renormalised things are infinite such as the mass of the electron.

The added problem with gravity is that renormalisation relates quantities to the scale at which you are measuring them. But in gravity that scale is also related to the metric tensor which is a dynamic quantity and can even change from place to place. Therefore renormalisation in the usual sense becomes undefined.

In conclusion the question is unanswerable to a greater degree of satisfaction without first knowing a complete theory of quantum gravity.


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