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Ján Lalinský
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This is not a problem with cosmological constant in models of universe. The "cosmological constant problem" is a bad name for a particular (bad) interpretation of a certain calculation procedure in QFT. You can find the usual account of this calculation e.g. in this answer by Nihar Karve: https://physics.stackexchange.com/a/626958/31895

When we canonically quantize e.g. free EM field, we get a field Hamiltonian that is a sum over Fourier expansion modes (where mode $\mathbf k\lambda$ is a multiindex indexing discrete basis for wave vector and polarization), in which there is a contribution $\frac{1}{2}\hbar \omega$ per mode:

$$ H_{cq} = \sum_{\mathbf k,\lambda} \bigg( \hbar\omega_{\mathbf k} a_{\mathbf k,\lambda}^+ a_{\mathbf k,\lambda} + \frac{1}{2}\hbar \omega_{\mathbf k}\bigg). $$

This zero point term implies all eigenvalues of $H_{cq}$ are infinite, and expectation value of $H_{cq}$ is infinite for any state.

Having all eigenvalues infinite breaks calculations. The correct conclusion from this is that we have a defective (wrong) Hamiltonian. This is not a big deal, nobody expects canonical quantization to always give the correct Hamiltonian. We have a system with infinite number of degrees of freedom, not a finite set of harmonic oscillators, so maybe the Hamiltonian is something non-trivially different than what we have for a finite set of simple harmonic oscillators. So we should at least look for a better Hamiltonian (if not for a completely different formulation of the theory).

However, there are several different ways to improve on the above Hamiltonian, here are two of them:

  1. high frequency cutoff in the above sum applied to zero point term, so the Hamiltonian has finite eigenvalues;

  2. so-called normally ordered Hamiltonian, which has no zero-point term in any mode, so the Hamiltonian has finite eigenvalues, and zeroth eigenvalue is zero, so ground state has zero eigenvalue;

  3. ...

The first way introduces a modification of the Hamiltonian which "cuts off" (suppresses to zero) contribution of terms with frequency that is higher than some arbitrary cutoff frequency $\omega_c$; however, many calculations' results consistent with experiments are affected by choice of value of $\omega_c$ in a negligible way, as long as we don't put it too low. This way of using the cutoff is, albeit ugly, arguably fine in practical sense, since the results are correct and do not depend on the choice strongly.

However, when this cutoff frequency is selected in a particular way, connected to gravity constant $G$ (which is just additional assumption, foreign to QFT), we get a definite but very high value for the zeroth eigenvalue, so-called ground state energy or zero point energy. Notice that in this terminology, the Hamiltonian eigenvalues are identified with possible values of energy of the field. This is strictly speaking not necessary, because Hamiltonian need not equal energy, as there is infinity of valid Hamiltonians, but energy is defined uniquely (e.g. by one particular Hamiltonian from the set). But in quantum theory it is customary to use only that Hamiltonian, from infinity of options, which we use to define energy; and we can always choose this Hamiltonian arbitrarily from the set of valid Hamiltonians, because absolute value of energy has no effect in quantum theory.

Then, some people think:

  • maybe the GR stress-energy tensor in vacuum has contribution due to expectation values of stress-energy tensor in QFT, based on the chosen Hamiltonian and momenta (linking the concept of energy in QFT to concept of 00 component of stress-energy tensor in GR);

  • maybe we can calculate value of this contribution using the QFT formalism and get a finite value;

  • then we use the fact the result of above is non-zero as a basis for a new idea: that the non-zero contribution of the cosmological term in Einstein's equations is actually due to this non-zero Hamiltonian eigenvalue of the ground state.

The cosmological term in Einstein's equations is actually not a part of the stress-energy tensor in GR proper, but formally it can be put into it to get the "effective stress tensor", so with that, the above thought train makes sense.

But of course, it is well-known that this assumption-loaded procedure (the calculation and its interpretation in relation to GR) produces energy density value that is many orders greater than the observed value of the effective energy density in GR (if we adopt the cosmological constant term into the stress-energy tensor). This very bad disagreement should give us pause and make us realize the procedure was obviously quite dubious right from the start, irrespective of the immense disagreement on numerical value. There is no reliable theory involving gravity constant $G$ implying there should be such $G$-dependent cutoff in Fourier expansions. A theory of gravity on discrete space would do it, but we have no indication for such discrete space, and no viable theory for it. Also, there is no necessity to fix the problem with the infinite Hamiltonian using a cutoff or discrete space, there are other ways (below).

Some people talk about the procedure as if it was a QFT prediction of the value of cosmological constant, but it is not. Ground state eigenvalue of Hamiltonian in QFT is not easily argued to be connected to the GR stress-energy tensor. Energy in GR is objective, measurable thing, with effect on geometry; but in quantum theory, energy value is arbitrary, it depends on the choice of Hamiltonian that we use to define energy. So we can't extract GR energy from the quantum theory and the chosen Hamiltonian there.

If we believe there is a link between QFT field energy and GR stress-energy tensor, from observations, we can extract constraints on the proper definition of energy in QFT, based on the requirement that this energy should be consistent with GR and observations of its gravity effects. So really what is a correct statement here, assuming the belief, is that the observational constraints on the value of the cosmological constant imply constraints on the choice of energy-defining Hamiltonian in QFT.

Sabine Hossenfelder put it well: let me ask a rhetorical question: which theory was falsified and rejected as a result of observations showing this calculated value of cosmological constant is completely wrong? None, because it is obvious the above procedure is full of dubious assumptions. What has been actually falsified is the procedure in all its different variations, because they all produce wrong result. Since this procedure, with its dubious assumptions, is not an inescapable implication of QFT, nothing about QFT itself has been falsified.

There is the second way to fix the original problem - just use the Hamiltonian without the zero point energy terms:

$$ H_{no} = \sum_{\mathbf k,\lambda} \hbar\omega_{\mathbf k} a_{\mathbf k,\lambda}^+ a_{\mathbf k,\lambda}~~~. $$ Sometimes this is called the normal ordering Hamiltonian, and it is the one used in practical calculations of light-matter interactions. The lowest eigenvalue of this is zero - the most plausible value ever for the lowest possible energy of EM field in a theory where the concept of energy is based on the Poynting expressions, so this Hamiltonian is also the best, in the class of theories based on Poynting expressions, and consistent with GR constraints, to define field energy. Expectation values for any non-pathological quantum state are finite, no infinite expectation values appear. There is no arbitrary cutoff, no heuristics with gravity constant. Just a Hamiltonian-based theory of EM field, which correctly reproduces many experiments involving light-matter interaction. Similarly, one should strive for similar state of things for other quantum fields, for finite Hamiltonian eigenvalues, preferably with zero ground state eigenvalue.

BUT, since in this formulation of QFT, the field ground state energy is zero, there is no good reason for why it should be connected to the concept of non-zero cosmological term in GR in any direct way, except if we want them to be so connected due to some other motivations. But non-zero cosmological constant term is really a concept from GR and GR-based cosmological models and consistent with super-macroscopic observations; there is no real logical necessity or real need in physics to have this macroscopic concept linked to field ground state Hamiltonian eigenvalue in QFT. I should say many people don't agree on this last point, they think these things should be linked, but none of them know how to formulate a functional QFT of gravity where the above procedure would produce a reasonable value of cosmological constant. All known attempts produce wrong results without doctoring the parameters, so it is very likely the whole enterprise is ill-conceived, and there is no link to be found between the cosmological constant in GR and ground state energy in QFT.

Ján Lalinský
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