Can Vacuum Energy Alter Our Understanding of Gravity?

[swampwiz]

I was reading this, but am confused:
https://nautil.us/the-remarkable-emptiness-of-existence-256323/

... in Einstein’s general theory of relativity, vacuum energy has the curious ability to generate a repulsive gravitational force.

 

[PeterDonis]

I was reading this, but am confused:
https://nautil.us/the-remarkable-emptiness-of-existence-256323/

As usual, we have a pop science source that doesn't do a very good job of explaining the actual physics.

It is true that vacuum energy, as it appears in the Einstein Field Equation of General Relativity, is equivalent to the stress-energy tensor of a perfect fluid with a pressure equal to minus its energy density, i.e., $p = - \rho$. It is also true that this kind of stress-energy tensor results in geodesics, i.e., worldlines of freely falling objects, diverging, instead of converging as one would expect due to the gravity of an ordinary perfect fluid. In cosmology this appears as the accelerating expansion of the universe.

However, calling this a "repulsive gravitational field" is misleading, because in order to see the divergence of geodesics, you have to be inside the fluid. Since vacuum energy is everywhere, we are "inside the fluid" as far as vacuum energy is concerned; but when we talk about the "gravitational field" of an ordinary massive object like a planet, we are usually talking about the effects on geodesics outside the mass, not inside it.

 

[otennert]

The topic of vacuum energy in general has caused confusion in so many discussions, also in this forum. And the article referenced surely does not help convey clarity to the public.

Quote: "But our quantum understanding of vacuum energy says it should be infinite, or at least incredibly large. Definitely not small. This discrepancy between the theoretical energy of the vacuum and the observed value is one of the greatest mysteries in modern physics."

That's of course not right. The unrenormalized "vacuum energy" seemingly calculated within the context of the Casimir effect, for example, can be taken to any arbitrary value, yes even infinite. But by definition, the renormalized value is zero, by simple subtraction. To my knowledge, there is no finite value that any QFT calculation gives for the vacuum energy as of today.

One has to admit, however, that even the great Steven Weinberg himself once mentioned in his paper
The cosmological constant problem”. In: Rev. Mod. Phys. 61 (1989), pp. 1–23: “Perhaps surprisingly, it was a long time before particle physicists began seriously to worry about this problem, despite the demonstration in the Casimir effect of the reality of zero-point energies.”

Well, what to say...

 

[PeterDonis]

by definition, the renormalized value is zero, by simple subtraction

No, it isn't; if it were, the Casimir effect would not exist. The existence of the Casimir effect means that there is a difference in the vacuum energy between the plates, vs. not between the plates. And that means the renormalization sum cannot be zero in both places.

 

[vanhees71]

The Casmir effect is not about vacuum energies but about (quantum) charge fluctations in the plates. It's more similar to the (quantum description) of van der Waals forces between neutral molecules rather than some "vacuum energies" (whatever that should mean, because the vacuum energy is just the ground-state energy of a special-relativistic QFT, and in SR the absolute value of energies doesn't have a physical meaning but only energy differences, i.e., you can "renormalize" the ground-state-energy value by adding an arbitrary value without changing the physics, and it's of course most convenient to give the vacuum/ground state the energy value of 0).

The standard introductory textbook calculations of the Casmir force between two electrically neutral conducting plates is in fact a limiting case making $e \rightarrow \infty$. For details, see

R. L. Jaffe, The Casimir effect and the quantum vacuum,
Phys. Rev. D 72, 021301 (2005),
https://doi.org/10.1103/PhysRevD.72.021301
https://arxiv.org/abs/hep-th/0503158

 

[PeterDonis]

n SR the absolute value of energies doesn't have a physical meaning

But in GR, it does; and this thread is in the context of GR since it's asking about "repulsive gravitational force" (e.g., the accelerating expansion of the universe).

 

[otennert]

But in GR, it does; and this thread is in the context of GR since it's asking about "repulsive gravitational force" (e.g., the accelerating expansion of the universe).

Yes that's right. In GR, an absolute energy level does matter, but the unrenormalized vacuum energy surely is not it. Otherwise the prediction for the cosmological constant would be off by about 121 orders of magnitude, depending on where you place the cut-off. But I also agree to not deviate too much from the original question in the post, and the topic of dark energy and/or vacuum energy has been discussed often times in several other threads.

I only find it amusing that when even Weinberg himself once fell for that, let's say misleading, interpretation, how then can we blame a popular science journal for adopting that thought?

 

[PeterDonis]

In GR, an absolute energy level does matter, but the unrenormalized vacuum energy surely is not it.

Yes, that is a well known issue and has been ever since the accelerated expansion of the universe was discovered in the late 1990s.

I only find it amusing that when even Weinberg himself once fell for that, let's say misleading, interpretation

I'm not sure it's misleading. It's a description of a genuine disconnect between what quantum field theory appears to be telling us and what GR appears to be telling us. Referring to the Casimir effect is just a way of emphasizing that what quantum field theory appears to be telling us works in other experimental contexts besides trying to estimate the global cosmological constant of our universe.

 

[otennert]

Yes, that is a well known issue and has been ever since the accelerated expansion of the universe was discovered in the late 1990s.I'm not sure it's misleading. It's a description of a genuine disconnect between what quantum field theory appears to be telling us and what GR appears to be telling us. Referring to the Casimir effect is just a way of emphasizing that what quantum field theory appears to be telling us works in other experimental contexts besides trying to estimate the global cosmological constant of our universe.

But that is exactly the point! QFT does not say anything about a finite value for the vacuum energy, at least not to my knowledge. In all QFT calculations by simple subtractions the zero-point energy is renormalized to exactly zero, for the lack of any other sensible value. Even when setting up the Hamiltonian of a QFT in terms of creation/annihilation operators, the infinite sum of terms of 1/2 is just disposed of by normal ordering, and fundamentally the infinity occurring in the Casimir force calculation is exactly of the same type.

Actually this process is the simplest of all renormalization procedures, and if a finite value of that unrenormalized but infinite zero-point energy appears at all in the calculation, it is because of the regularization scheme chosen, hence the so-often-cited disagreement with observation by an order of magnitude of 121.

However, the most frequently applied regularization scheme in textbooks (and the one used by Casimir himself in his 1948 paper) via the Euler--Maclaurin formula does not use any specific cut-off at all, but leaves it completely undefined. Hence nowhere in the entire calculation is there an expression for the vacuum energy at all. Casimir himself speaks of a "zero-point pressure" between the macroscopic plates, and this, on the other hand, is a completely valid concept, at least quantitatively.

The Casimir effect, on the other hand, can be pictured in many ways, like e.g. as a van der Waals force between 2 gigantic molecules, which I think is the most convincing one. Due to the effectively macroscopic setup of the system, there is moreover no fine structure constant $\alpha$ appearing in the formula for the attractive force per unit area. Casimir and Polder in a (I think) preceding paper also calculated the effective force between a charged particle (with intrinsic polarizability) and a conducting plate. Now $\alpha$ implicitly appears linearly in the result for the force. And eventually, when no macroscopic plate is present, the force between 2 polarizable particles implicitly contains $\alpha$ quadratically, as already calculated by Fritz London in 1930, resp. 1937.

In my eyes, this shows that connecting the Casimir force with the vacuum energy (and *not* with any energy gradient) is misleading. And I might be mistaken, but I think this connection had only been made from around the early 1980s on within the astrophysics community, and still lingers on, albeit much less present, leading to so much confusion.

 

[PeterDonis]

In all QFT calculations by simple subtractions the zero-point energy is renormalized to exactly zero

That's because it is assumed that the background spacetime is flat, and therefore the zero point energy must be zero by definition. But, again, in this thread we are working in the context of GR, where spacetime is not assumed to be flat and we know by observation that our universe has a small positive cosmological constant, which leads to spacetime curvature even in the absence of any other matter or energy. Those things invalidate the assumptions that are used to obtain the renormalized value of zero.

 

[vanhees71]

But in GR, it does; and this thread is in the context of GR since it's asking about "repulsive gravitational force" (e.g., the accelerating expansion of the universe).

Yes, and in this respect it's indeed the most unsolved puzzle ever. From the Standard Model of Particle Physics extrapolated to the GUT or Planck scales one predicts a factor of $10^{60}$ or $10^{122}$ higher cosmological constant than observed, and this puzzle is not really solved yet. In this sense it's an open question.

Interestingly Pauli very early on argued against a physical meaning of the "zero-point energy" of harmonic oscillators by exactly making the point that if it were effective as a source of gravity we'd live in a closed universe with a radius smaller than the distance between Earth and the Moon ;-).

 

[strangerep]

[...] it's indeed the most unsolved puzzle ever. [...]

The only puzzle I see in these types of thread is that of human nature...

For the harmonic oscillator, the classical Hamiltonian is of the form:
$$H_{\text{Cl}} ~=~ (p^2 + x^2)/2 ~,$$[This post has been revised from here onwards, correcting a sign error.]

The naively quantized version:$$\hat H_{\text{Cl}} ~=~ (\hat p^2 + \hat x^2)/2$$ has eigenvalues $\hbar(n + \frac12)$, with $n = 0,1,2\dots$, hence vacuum energy $\hbar/2$.

The correct quantum Hamiltonian ("correct" because it gives zero vacuum energy) is of the form $$\hat H_{\text{Q}} ~=~ \frac12 (\hat x - i \hat p)(\hat x+i \hat p)$$with$$(\hat x+i \hat p) |0\rangle ~=~ 0 ~.$$The fact that $\hat H_{\text{Q}}$ can be written as
$$\hat H_{\text{Q}} ~=~ \Big(\hat p^2 ~+~ \hat x^2 ~-~ i [\hat p, \hat x]\Big)/2 ~=~ (\hat p^2 + \hat x^2)/2 ~-~ \hbar/2 ~,$$i.e., $$\hat H_{\text{Cl}} ~=~ \hat H_{\text{Q}} ~+~ \hbar/2 ~,$$simply means that $\hat H_{\text{Q}}$ was indeed the physically correct Hamiltonian all along, not $\hat H_{\text{Cl}}$.

This $\hat H_{\text{Q}}$ has zero vacuum energy, hence doesn't pose a problem in GR, and it matches the classical Hamiltonian in the limit as $\hbar\to 0\,$; -- which is all we can reasonably expect of a quantum-classical correspondence.

I remember a (transcript of) a talk by Dirac in his later life where he said something like: "The Heisenberg equation of motion is something to which we should hold fast. If it gives the wrong results it means we are using the wrong Hamiltonian."

That's what's happening here with all this gigantic vacuum energy nonsense. We were simply trying to use wrong Hamiltonian $\hat H_{\text{Cl}}\,$. All renormalizations are essentially like that, i.e., introducing corrections for an initially-incorrect choice of Hamiltonian, trying to gradually approximate a version that's closer to the physically correct one.

(And that's why the nonsense of gigantic quantum vacuum energy has nothing to do with the cosmological constant.)

 

[vanhees71]

It's the other way round. The original Hamiltonian $\hat{H}_Q$ has the eigenvalues $E_n=\hbar \omega (n+1/2)$ with $n \in \mathbb{N}_0=\{0,1,\ldots \}$, i.e., the lowest energy is $\hbar \omega/2$, and if you have a system with an infinite amount of free harmonic oscillators like a quantum field the zero-point energy diverges.

Within Newtonian mechanics or special relativity this poses no problem since one can always subtract a constant energy value and use the annihilation and creation operators and normal order the Hamiltonian wrt. these to get a 0 ground-state (in the QFT vacuum) energy. Additive constants of the total energy are unobservable, because only energy differences are.

For the single harmonic oscillator both Hamiltonians are thus physically equivalent, and for the (free) quantum fields you simply have renormalized the total energy to a finite value, and the resulting representation of the Poincare group works with the normal ordered definition of the generators when written in terms of the quantum fields.

In GR, however, the absolute value of the energy-momentum tensor (field) of the matter fields enters as the source term of the Einstein-Hilbert field equation and thus additive constants to the total energy of the matter manifest themselves as contributions to the "cosmological constant", which is why it's also called "dark energy", i.e., you can move the cosmological-constant term from the left-hand side to the right-hand side of the Einstein-Hilbert equation and reinterpret it as a contribution to the energy density of matter. When now using the renormalization group to evaluate the vacuum energy of the Standard Model from the low-energy scales to high-energy scales like the Planck scale you get huge constributions, and that's a problem, because obviously the "dark energy content" of the universe is obviously much smaller. For details, see the classic paper by Weinberg:

S. Weinberg, The cosmological constant problem, Rev. Mod.
Phys. 61, 1 (1989),
https://link.aps.org/abstract/RMP/V61/P1

 

[otennert]

Now I have read Weinberg's paper several times, and I still cannot really spin my mind around it.

First of all, do we agree the whole approach so far is semiclassical at best, i.e. by considering quantum fields defined over curved space, right? For the lack of any quantum gravity theory so far, this is the best we can do.

Second: the main issue I have in understanding is eq. (3.5), which shows an unrenormalized value for $\langle \rho\rangle$, hence the cut-off appears explicitly. Now, the whole concept of renormalization is that given a real measurable value of some quantity (say, energy or mass) at some renormalization point, which may be e.g. at momentum $p=p_0$, two terms are supposed to combine to a physical quantity: the one term representing a quantity which is non-measurable in principle (like a bare mass), and the other being a divergent term, as in:
$$m(p=p_0) = m_0 (\text{physically ill-defined}) + f(p) (\text{divergent}) =: m_\textrm{ph}.$$

Can anyone explain to me why renormalization is not necessary in interpreting eq. (3.5)? I do agree that in GR an absolute energy level matters, but renormalization is not just shifting energy levels at will. As just said, renormalization is about combining a physically ill-defined and a divergent term, thus implying a complete re-interpretation of the quantities entering e.g. the (bare) Lagrangian, and eventually leading to the RG equations exactly due to the possibility to shift the renormalization point. I am currently giving Birrell/Davies a quick read, and there is some chapter on why renormalization is not needed, because the regularization scheme chosen leads to convergent terms only, let's see if I can quickly grasp the logic. However, if someone could provide me with some quick & dirty hint, it may spare me reading a lot of pages.

In any case I stick to the opinion that the Casimir effect is not suited to exemplify all this. And Weinberg again uses vacuum energy and energy fluctuations synonymously, and within the context of the Casimir effect. Another issue I am having with this paper.

 

[strangerep]

It's the other way round. [...]

I had a sign error. Please re-read my (now-edited) post #12.

 

[vanhees71]

Yes, but you cannot say that the one or the other Hamiltonian is the "correct" one describing a harmonic oscillator, because both are physically equivalent. The only difference is an additive constant to the total energy operator (=Hamiltonian in this case) and of course its eigenvalues, but this difference is not in any way observable within non-relativistic or special-relativistic physics.

 

[otennert]

I had a sign error. Please re-read my (now-edited) post #12.

As vanhees71 has mentioned, both Hamiltonians are equivalent in QM. The difference only matters in QFT, where an infinite number of quantum harmonic oscillators contribute to the (e.g.) electromagnetic field, and the constant factor then turns into an infinity constant, which is eliminated by defining the vacuum energy density to be zero, by the most simple version of renormalization thinkable, namely normal ordering. In the old days of QFT, people still had no systematic notion of renormalization and described it quite condescendingly as "substraction physics" or the "Klein--Jordan trick".

 

[strangerep]

Yes, but you cannot say that the one or the other Hamiltonian is the "correct" one describing a harmonic oscillator, because both are physically equivalent. The only difference is an additive constant to the total energy operator (=Hamiltonian in this case) and of course its eigenvalues, but this difference is not in any way observable within non-relativistic or special-relativistic physics.

... but that nonzero additive constant does cause a problem in GR (which is the context of this thread). Hence those Hamiltonians are not physically equivalent in the context of GR.

[+ Similar reply to @otennert ...]

 

[vanhees71]

Yes, that's what I already said above in #13.

 

[vanhees71]

Now I have read Weinberg's paper several times, and I still cannot really spin my mind around it.

First of all, do we agree the whole approach so far is semiclassical at best, i.e. by considering quantum fields defined over curved space, right? For the lack of any quantum gravity theory so far, this is the best we can do.

Second: the main issue I have in understanding is eq. (3.5), which shows an unrenormalized value for $\langle \rho\rangle$, hence the cut-off appears explicitly. Now, the whole concept of renormalization is that given a real measurable value of some quantity (say, energy or mass) at some renormalization point, which may be e.g. at momentum $p=p_0$, two terms are supposed to combine to a physical quantity: the one term representing a quantity which is non-measurable in principle (like a bare mass), and the other being a divergent term, as in:
$$m(p=p_0) = m_0 (\text{physically ill-defined}) + f(p) (\text{divergent}) =: m_\textrm{ph}.$$

Can anyone explain to me why renormalization is not necessary in interpreting eq. (3.5)? I do agree that in GR an absolute energy level matters, but renormalization is not just shifting energy levels at will. As just said, renormalization is about combining a physically ill-defined and a divergent term, thus implying a complete re-interpretation of the quantities entering e.g. the (bare) Lagrangian, and eventually leading to the RG equations exactly due to the possibility to shift the renormalization point. I am currently giving Birrell/Davies a quick read, and there is some chapter on why renormalization is not needed, because the regularization scheme chosen leads to convergent terms only, let's see if I can quickly grasp the logic. However, if someone could provide me with some quick & dirty hint, it may spare me reading a lot of pages.

In any case I stick to the opinion that the Casimir effect is not suited to exemplify all this. And Weinberg again uses vacuum energy and energy fluctuations synonymously, and within the context of the Casimir effect. Another issue I am having with this paper.

The divergences of higher-order perturbative contributions to the S-matrix elements in relativistic QFT come from a too sloppy use of the field operators, which are operator-valued distributions and thus cannot be so easily multiplied at the same space-time argument. This is however just sloppyness of the physicists and can be cured by using "smeared operators" as in the Epstein-Glaser approach. Then everything is finite right from the beginning, but that's not the point concerning the dark-energy/cosmological-constant problem.

Even when everything is finite, which you can as well achieve with the more handwavy approach using some other regularization than "smearing" like dim. reg. or just at any stage calculate everything with physical quantities and add appropriate counterterms to the self-energy, proper-vertex functions, and particularly also our case of 1PI vacuum diagrams as in the (generalized) BPHZ approach. The problem is that you have to choose a renormalization scale and a renormalization prescription and at each (loop) order perturbation theory fit the finite parameters (wave-function normalization factors, masses, coupling constants) to some measured cross sections. The quantities like physical masses and couplings are then defined by the renormalized, connected n-point Green's functions. E.g., the masses of particles or masses and widths of resonances are given by the corresponding poles of the one-particle Green's functions. In this way, to the order the perturbation theory is evaluated, the S-matrix elements (cross sections) are invariant under the choice of the energy-momentum scale and the renormalization scheme, but changing the renormalization scheme and/or the energy-momentum scale changes the renormalized parameters including the "zero-point energy", and this "running of parameters", determined by the renormalization-group equations, is the problem. Adapting everything at the low-energy scales needed to describe the outcome of our experiments with accelerators, implies a huge value of the "zero-point energy" when using the RG equations to calculate this parameter at higher energy-momentum scales like the GUT scale or even the Planck-mass scale. Here a tremendous finetuning is needed, and that's considered unnatural.

 

[otennert]

The divergences of higher-order perturbative contributions to the S-matrix elements in relativistic QFT come from a too sloppy use of the field operators, which are operator-valued distributions and thus cannot be so easily multiplied at the same space-time argument. This is however just sloppyness of the physicists and can be cured by using "smeared operators" as in the Epstein-Glaser approach. Then everything is finite right from the beginning, but that's not the point concerning the dark-energy/cosmological-constant problem.

Absolutely: it is much too often not mentioned that field operators are operator-valued singular distributions, hence the infinities occurring in QFT are the result of exactly this "sloppy" use and must be dealt with by the renormalization programme. However, as dirty as it might look, it works, and at least eventually remedies the mathematically ill-defined calculational approach in the end, with spectacular results.

By the way, I am not overly knowledgeable on the Epstein--Glaser approach, but I have intended to give the books by Günter Scharf a read some time.

And also yes: I don't think it has anything to do with the cosmological constant problem.

Even when everything is finite, which you can as well achieve with the more handwavy approach using some other regularization than "smearing" like dim. reg. or just at any stage calculate everything with physical quantities and add appropriate counterterms to the self-energy, proper-vertex functions, and particularly also our case of 1PI vacuum diagrams as in the (generalized) BPHZ approach. The problem is that you have to choose a renormalization scale and a renormalization prescription and at each (loop) order perturbation theory fit the finite parameters (wave-function normalization factors, masses, coupling constants) to some measured cross sections. The quantities like physical masses and couplings are then defined by the renormalized, connected n-point Green's functions. E.g., the masses of particles or masses and widths of resonances are given by the corresponding poles of the one-particle Green's functions. In this way, to the order the perturbation theory is evaluated, the S-matrix elements (cross sections) are invariant under the choice of the energy-momentum scale and the renormalization scheme,

I do agree so far. But all of the above applies to QFT in general, in simple Minkowski space. One would think that it would also apply to QFT defined in curved spacetimes, but here's where additional issues come in.

but changing the renormalization scheme and/or the energy-momentum scale changes the renormalized parameters including the "zero-point energy", and this "running of parameters", determined by the renormalization-group equations, is the problem. Adapting everything at the low-energy scales needed to describe the outcome of our experiments with accelerators, implies a huge value of the "zero-point energy" when using the RG equations to calculate this parameter at higher energy-momentum scales like the GUT scale or even the Planck-mass scale. Here a tremendous finetuning is needed, and that's considered unnatural.

This is where I am not really able to follow. If am I am getting you wrong, my apologies, but are you hinting at the diffculty of extrapolating a QFT to regimes way beyond their scope of applicability? Again, this is an issue not of QFTs in curved spacetimes, but of QFTs in general, and is well-accepted by the philosophy of effective theory. Let alone the fact that no viable candidate of a GUT exists nowadays that is physically sufficiently predictive, or even mathematically well-defined.

Also, one general statement often mentioned is that in a general curved spacetime, normal ordering does not work anymore, as re-defining energy levels contradicts the fact that absolute energy levels count in GR. I agree to the latter, but not to the former per se. Because hang on, here's my first objection: we've just said that all along we are using a mathematically ill-defined procedure to handle infinities, and the logic is they are unphysical artifacts of the sloppy math applied, but now all of a sudden these mathematically ill-defined entities are to become physically meaningful?

What I do have understood though, and this seems quite natural using some basic notion of QFT, is the problem of being able to define a vacuum state in general. Lacking Poincare invariance in a general spacetime, it is simply not possible to have a notion of a "ground state" as being the one having lowest energy. The existence of a vacuum state, however, is necessary for defining a Fock space as we know it, and the vacuum is then unique. Stationary spacetimes, permitting a timelike Killing field, belong to the class that allow for this. Hence as I see it, standard QFT approach is in general not applicable anyway.

Now, as mentioned in a previous post, Birrell/Davies (https://www.cambridge.org/core/books/quantum-fields-in-curved-space/95376B0CAD78EE767FCD6205F8327F4C) seems to be quite a good starting point for understanding the issues. However, repeating and understanding the calculations therein requires much more time and effort than I have been able to put in so far.

They devote a whole chapter §6 on the renormalization of the stress-energy tensor $\langle T^μ{}_ν\rangle$. As said, without having reproduced any calculation in detail, there are indeed renormalization schemes, but they are more complicated than in flat spacetime. But still there are some highly symmetrical spacetimes that allow the calculation of $\langle T^μ{}_ν\rangle_\textrm{ren}$. A massive, free scalar field on a de Sitter space, for example, leads to a renormalized stress-energy tensor as given in eq. (6.183), which also correctly implies the trace anomaly present in that case. The same scalar field defined on an Einstein static spacetime delivers an energy density of (eq. (6.190))
$$ρ=\frac{1}{480π^2 a^4},$$
where $a$ is the radius of the space 3-sphere. This would then lead to ($κ$ is the Einstein gravitational constant)
$$Λ=\frac{1}{a^2}=\frac{κρ}{2},$$
and hence
$$Λ=\frac{κ}{960π^2 a^4}.$$

Now, as I must admit, I need more time to understand renormalization techniques in curved spacetimes. However, there seem to be very few explicit examples and even less mathematically strict results. My main critique is that all over the literature, seemingly simple results are shown, and eq. (3.5) in above-cited Weinberg's review is no exception. There is no way this equation is physically meaningful in any sense if the baseline for doing the respective calculations is as given in Birrell/Davies. For general spacetimes, a Fourier mode decomposition of a field is not even possible. How then is a quantity like $\rho$ even calculated in general?

Quantum effects in general and QFT on a curved spacetime in specific still seems to be a poorly understood concept, and I am wondering if we are actually in a position already to assess whether there is a cosmological constant problem or not.

 

[strangerep]

Yes, that's what I already said above in #13.

...but you didn't progress to what I was advocating: i.e., that (using my notation), $$\hat H_{\text{Q}} ~:=~ \frac12 (\hat x - i \hat p)(\hat x+i \hat p)$$is the physically correct Hamiltonian.

 

[strangerep]

My main critique is that all over the literature, seemingly simple results are shown, and eq. (3.5) in above-cited Weinberg's review is no exception. There is no way this equation is physically meaningful in any sense [...]

Of course not. That's Weinberg's point (IIUC). He's just showing what happens in a naive calculation of vacuum energy using naively quantized harmonic oscillators.

 

[vanhees71]

...but you didn't progress to what I was advocating: i.e., that (using my notation), $$\hat H_{\text{Q}} ~:=~ \frac12 (\hat x - i \hat p)(\hat x+i \hat p)$$is the physically correct Hamiltonian.

Well, it's the correct Hamiltonian modulo an additive constant. We simply have to measure the "cosmological constant" and renormalize the value of the "zero-point energy" to be in accordance with this meaasured value. All the parameters in the QFTs have to be fitted to experiment. In this sense they are "manifestations of our ignorance" as my HEP professor used to say in his QFT lectures ;-).

 

[otennert]

We simply have to measure the "cosmological constant" and renormalize the value of the "zero-point energy" to be in accordance with this meaasured value. All the parameters in the QFTs have to be fitted to experiment.

Indeed! Well, at least that's what the usual procedure would require...