I Is the Cosmological Constant Problem a Misunderstanding of Zero-Point Energy?

[AndreasC]

There is no right or wrong quantization scheme. As I repeatedly tried to explain in special-relativistic QFT the absolute value of the total energy is not observable, and you have to renormalize the total energy anyway. The standard renormalization is such to make the energy of the ground state (vacuum) 0.

In general-relativistic QFT, i.e., QFT in a given "classical background spacetime" you have the same problem, and if you then calculate the energy-momentum tensor of the "matter fields" you have to renormalize it too, and to adjust it to what's expected to be observed at the Planck scale needs finetuning, which is considered "unnatural" by most physicists. That's why one looks for some (symmetry) principle which explains the value of the cosmological constant, but there's no such model, let alone a solution for the notorious problem of quantizing also the gravitational interaction in a consistent way. Until there's no such theory, it's quite speculative to guess, how this problem might be solved.

Ah I see. So, if I've got this right, the argument is that the quantum theory doesn't have a unique "recipe" for calculating the stress energy tensor. About the symmetry principle, what exactly do you mean?

 

[AndreasC]

For the harmonic oscillator, which was the specific case we were considering, it is $x^2 / 2$, where $x$ is the displacement from equilibrium. That's the physical definition: the potential energy is zero at the equilibrium point. To quantize that you just make $x$ the position operator, as with any other potential in QM.

More generally, the "correct" potential energy is the one whose zero point corresponds to the state that is physically picked out as "equilibrium" or something similar. For example, for an isolated gravitating body the potential energy is zero at infinity. That's how you pick out the "correct" potential from all the other mathematically possible options.

There is no reason in the classical case why the potential has to be 0 in equilibrium, it is 100% equivalent, and often used in physical problems. There is no reason in the classical theory to prefer one over the other. And even when you do assume that in the classical case, as I showed quantizing it does not give a unique valid quantum potential energy because different quantization schemes give different results. You assert that the "correct" quantum potential energy has to be 0 at something like the "equilibrium point", but I could assert that the "correct" quantum potential energy is the one such that the ground state has no energy, which also sounds very nice and valid and definitely more convenient. But we can keep having this argument forever because neither of us is making this claim based on a good theoretical reason or physical observation at this point. If anything I think the case where the ground state energy is zero is definitely the "nicest" and most convenient.

 

[vanhees71]

I think the discussion of the non-relativistic harmonic oscillator is off-topic with respect to the cosmological-constant problem, but it's a good example to show, that an arbitrary additive constant added to the potential or the total energy/Hamiltonian is physically irrelevant:

Within non-relativistic physics there is no right or wrong choice for any additive constant to the total energy of the system, because such an additive constant doesn't change the physics. It's a kind of very simple "gauge invariance", i.e., you can use any potential which gives the force $F=-m \omega^2 x$, which is $V(x)=m \omega^2 x^2/2+V_0$ with $V_0=\text{const}$.

Quantizing the system you make $x$ and $p$ operators fulfilling the Heisenberg algebra $[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1}$ and write down the Hamiltonian (where there are no operator-order problems at all):
$$\hat{H}=\frac{1}{2} \hat{p}^2 + \frac{m \omega^2}{2} \hat{x}^2 +V_0 \hat{1}.$$
Then with the usual annihilation operator
$$\hat{a}=\sqrt{\frac{m \omega}{2 \hbar}} \hat{x}+\frac{\mathrm{i}}{\sqrt{2 m \hbar \omega}} \hat{p},$$
you get
$$\hat{H}=\frac{\hbar \omega}{2} (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \hat{1}) +V_0 \hat{1},$$
and the usual argument with that there's a ground state $|\Omega \rangle$ defined by $\hat{a} |\Omega \rangle=0$ and then excited states which are given by $|n \rangle=\hat{a}^{\dagger n}/\sqrt{n!} |\Omega \rangle$ with the eigenvalues $\hat{N} |n \rangle=n |n \rangle$ with $n \in \{0,1,\ldots \}$ the ground state is the eigenstate with $n=0$, i.e., we can also write $|\Omega \rangle=|n=0 \rangle$.

Now, all the Hamiltonian does is to provide the time evolution of the state vectors in the Schrödinger picture,
$$\hat{U}(t)=\exp\left ( -\frac{\mathrm{i} t \hat{H}}{\hbar} \right).$$
The arbitrarily chosen parameter $V_0$ thus just provides a phase factor (as also does the "zero-point energy" $\hbar \omega/2$), i.e., it is unobservable, because pure states are defined by Hilbert-space vectors only up to a non-zero factor (or equivalently for normalized states up to a phase factor).

So you can just choose $V_0=-\hbar \omega/2$, so that
$$\hat{H}=\hbar \omega \hat{N},$$
which makes the ground-state energy 0. That doesn't mean that this were the only "correct choice" of $V_0$, it's just a particularly convenient choice.

 

[AndreasC]

but it's a good example to show, that an arbitrary additive constant added to the potential or the total energy/Hamiltonian is physically irrelevant:

Right, that was my point, I was replying to @Vanadium 50 , they brought it up since they believed we should look at this first and decide whether or not it is physical before looking at the whole universe.

By the way, you varied the energy by adding a constant to the classical Hamiltonian, but what I find kind of interesting is that even if you do decide a certain energy for the classical Hamiltonian, when you quantize it due to the ambiguity in quantization you can still get any zero point energy you want, so there is no way out of it even if you decide it for the classical case.

Anyways, I guess the whole problem is more like a fine tuning problem rather than a real disparity between theories and their predictions, would you agree with this?

 

[vanhees71]

Sigh... First of all: The additive constant is physically irrelevant only within Newtonian physics and special relativity, i.e., when neglecting gravitation. Second, the zero-point energy, i.e., the energy value of the ground state (state of lowest energy) can be arbitrarily chosen (in both classical and quantum mechanics). This choice doesn't change the physics of the system, which is uniquely described by any choice.

Within GR the choice of the "zero-point energy value" is physically relevant, because it appears on the right-hand side of Einstein's field equation, i.e., any contribution to the energy density adds to the gravitational field and thus is physically observable.

We only have an incomplete description of quantum (field) theory when taking into account gravitation, i.e., all we can do so far is to keep the gravitational field classical, defining a "background spacetime". For cosmology that's the Friedmann-Lemaitre-Robertson-Walker spacetime, and then you can describe the matter and radiation by a quantum field theory within this background spacetime. The total energy of these matter and radiation fields is indetermined and has to be "renormalized" such that the corresponding cosmological constant fits the observations of cosmology, and this needs an "artificial finetuning", which is considered unsatisfactory by many physicists.

 

[Vanadium 50]

Is this thread about answering a question, or about advocating a particulalr point of view?

 

[PeterDonis]

There is no reason in the classical case why the potential has to be 0 in equilibrium

Yes, there is: the potential energy comes from whatever energy source produces the restoring force that acts to return the oscillator to equilibrium. If the oscillator is at equilibrium, the restoring force is zero, so the energy stored in whatever energy source produces the restoring force should be zero as well. For example, in the case of a spring, the potential energy is the energy stored in the spring: if the spring is at equilibrium, the stored energy is zero.

 

[PeterDonis]

different quantization schemes give different results

But only one quantization scheme gives a potential energy that matches the classical potential energy $x^2 / 2$. That's what picks out that particular quantization scheme as being preferred. As I've already said.

At this point I think we are just restating our positions. You are of course free to choose what you believe; but you have not given me any reason to doubt the physical arguments I have made for the position I have taken. So at this point I think we will just have to disagree.

 

[AndreasC]

First of all: The additive constant is physically irrelevant only within Newtonian physics and special relativity, i.e., when neglecting gravitation

I know... We've been talking about that...

I think we are saying kind of the same thing basically. If I understand correctly you are saying the problem is more a fine tuning problem than an inconsistency between predictions. I will have to look into the estimation that was carried out and turned out wrong to see what the assumptions were.

 

[AndreasC]

But only one quantization scheme gives a potential energy that matches the classical potential energy x2/2.

Maybe, but it does not give a ground state energy that matches the classical case. Also adding a constant to classical potentials is very common, it's what you are doing every time you are setting a specific potential to 0 when solving circuits or chosing a ground potential for problems with a uniform gravitational field etc. Energy difference matters, not the absolute potential.

If the oscillator is at equilibrium, the restoring force is zero, so the energy stored in whatever energy source produces the restoring force should be zero as well.

Yes, this doesn't change no matter what constant you add to the potential energy.

As you said, it matters when you start introducing general relativity into the mix. My understanding so far is that the quantum theory thus makes no unique prediction about what should be treated as the "right" level that can be confidently combined with general relativity. I think what @vanhees71 is saying is that this is the problem, that it is indetermined and would need fine tuning, not that it is determined some way and is inconsistent. Of course they can correct me if I'm understanding this wrong.

 

[AndreasC]

Is this thread about answering a question, or about advocating a particulalr point of view?

These things are not incompatible. I have a point of view so far that at this point is very fluid and I am not very confident in it and I am investigating what others have to say about it which goes through asking questions where I would have to advocate for my point of view to see what the arguments for their point of view are, and how strong they are.

 

[strangerep]

If the oscillator is at equilibrium, the restoring force is zero, so the energy stored in whatever energy source produces the restoring force should be zero as well.

This does not follow.

$F \propto -dV/dx$, so if $F=0$ one may only conclude that $V$ is constant wrt $x$.

 

[vanhees71]

Yes, there is: the potential energy comes from whatever energy source produces the restoring force that acts to return the oscillator to equilibrium. If the oscillator is at equilibrium, the restoring force is zero, so the energy stored in whatever energy source produces the restoring force should be zero as well. For example, in the case of a spring, the potential energy is the energy stored in the spring: if the spring is at equilibrium, the stored energy is zero.

No, any potential leading to the same restoring force is as good as any other. That's why potentials are only determined up to an additive constant, and this constant is not observable. Concerning total energy only energy differences are observable.

The total energy of a spring is
$$E=\frac{m}{2} \dot{x}^2 + \frac{D}{2} x^2 + E_0,$$
where $E_0$ is arbitrary.

Only when considering gravitation within GR the choice of the value of the ground-state energy plays a role as a source of the gravitational field.

 

[Haelfix]

Im picking up some confusions, so I thought I would briefly sketch a heurestic of the CC argument.

If you wish to quantize Einstein gravity, you need to come to grips with expectation values of the Einstein stress energy tensor. B/c even if gravity doesn't obey quantum mechanics, matter does!
So we need to understand formal objects like <Tuv>.
Well first we impose symmetry principles and local lorentz invariance enforces the following in vacuum.

$$ R_{uv} -\frac{1}{2}g_{uv}R-\lambda g_{uv}=<T_{uv}>=-\frac{1}{M_{pl}^{2}}<\rho >g_{uv}$$
Thats absolutely critical, b/c it tells you that cosmological constant terms appearing on the left hand side is indistinguishable from energy densities computed from quantum fields in their vacuum state.

Moreover, we sort of know how to do this.. These expression arise from the sum of feynman diagrams, where we have vacuum bubble diagrams (with external graviton loops). This will lead to an integral over the momentums (up to some hard cutoff $\Lambda$) for a field of mass M. This is just like summing up harmonic oscillators of energy E (see the Weinberg review for more details)

Which leads to the following badly divergent expression:

$$ <\rho>_{vac} = \frac{1}{2}\int_{0}^{\Lambda}\frac{4\pi k^{2}dk}{2\pi ^{3}}\sqrt{k^{2}+M^{2}} = \frac{\Lambda^{4}}{16\pi ^{2}}+\frac{M^{2}\Lambda^{2}}{32\pi ^{2}}+\frac{M^{4}ln(\frac{\Lambda}{M})}{16\pi ^{2}}+...$$
You will recognize the famous quartic divergence. Putting in a value for the cutoff $\Lambda=Mpl$, gives you the ~120 order of magnitude estimate.

Now there are a few immediate problems you can point to in the above logic.

The first is that these are quantum field theories, and in quantum field theories we were always instructed to view quadratic or quartic divergences of the cutoff with skepticism. That we need to put in appropriate regularization and to introduce counterterms and so forth, and maybe we don't know how to do that for quantum gravity, but if we did it properly things would vanish.
Ok. But you will notice the third term in the above is a logarithmic divergence. Those usually are physical and involves the mass of a particle. We don't know about Planck scale physics, but we do know about standard model fields, like the top quark. Plugging in that value, gives us an estimate that is still some ~52 orders of magnitude to big.

Ok. You might still object and say we don't really understand how to quantize things. Maybe, maybe.. The top quark field gets some strange contribution to its vacuum expectation that somehow makes it zero.
Good.. But that same voodoo you make for the top quark field, somehow needs to be communicated to the equivalent computation for another standard model field. Like the muon.
Which leads to the question.. Why would muons and top quarks conspire together in exactly the same way in whatever new theory you cook up, as they don't manifestly seem related in this context.

Next point. Maybe, you don't believe in these diagrams contributing in the first place. eg Loops of things with gravitons.. Artifacts of perturbation theory you might say.
Well, that's sort of like looking at the expansion of e^x = 1 + .5x^2 + ... and saying something like 'x^2 is unphysical'. Yes, strictly speaking true, but that doesn't mean you can ignore it either when doing the computation.

In particular, if you look at the most famous and earliest example of these virtual QFT effects... the Lamb shift of the hydrogen atom. It arises from the same sort of sums over vacuum diagrams as the above.
And this has a physical effect, it contributes to a shift in energy levels. So, by the equivalence principle it must gravitate, and that means you can couple the theory in the way we just did. So there is in fact some sort of reality to this afterall (and indeed nontrivial computations in this formalism have been done and verified in famous neutron interferometry experiments in the presence of gravitational fields)

Finally, to the extreme die hard skeptic of the above, there is an ultimate coup de grace! And that is the fact that there are perfectly classical contributions to the cosmological constant arising from the confinement scale of QCD, or the electroweak phase transition in the early universe. These things also contribute (by shifting the overall height of the potential energy), and produce exact numbers many times too big.

So to summarize. Strictly speaking, we don't have an exact computation (nor will we ever, as that requires knowledge of all matter particles all the way to the Planck scale). Instead we have a sequence of estimates.
The trouble is that getting around the problem these estimates create involves either some sort of ridiculous conspiracy, or modifying some cherished principle that in many ways is a much bigger deal. (Like a Lorentz invariance violation or the failure of effective field theory or the existence of a new symmetry principle). Clearly something must go wrong in the argument above, but the more you stare at it, the harder and more nontrivial things become..

 

[PeterDonis]

it does not give a ground state energy that matches the classical case.

And the reason for that is that, in the classical case, the ground state is the oscillator sitting at rest at equilibrium; but in the quantum case, that is impossible, because it would be a state that was an eigenstate of both position (at the equilibrium point) and momentum (zero momentum). And there cannot be any such state because the position and momentum operators do not commute. So the quantum ground state has to have a higher energy than the classical ground state, in order to have a finite uncertainty in at least one of position and momentum. (And as we know, the actual ground state is a Gaussian with a finite uncertainty of both position and momentum.)

As you said, it matters when you start introducing general relativity into the mix.

Yes, and as I have already pointed out, this thread is about the cosmological constant problem, which means we are talking about GR. So the fact that in Newtonian physics and SR absolute values of energy don't matter is irrelevant, which is why I keep objecting when that point is brought up in this thread.

My understanding so far is that the quantum theory thus makes no unique prediction about what should be treated as the "right" level that can be confidently combined with general relativity.

I agree, since we don't currently have a good theory of quantum gravity, and I think that is what we would need to resolve the question. I think at our current state of knowledge there are three possibilities that we can't rule out for where the observed small positive cosmological constant in our universe comes from:

(1) It is due to quantum zero point energy, but we don't currently understand how to calculate that correctly.

(2) It is due to some higher order correction or perturbation to quantum zero point energy, which is forced by some symmetry to vanish at lowest order, but can still have higher order small nonzero corrections (perhaps due to the symmetry being slightly broken).

(3) It is due to something else (perhaps some brute property of spacetime itself that has no deeper explanation, or perhaps some other emergent phenomenon from quantum gravity), and there is some symmetry or correction that forces quantum zero point energy to vanish to all orders in the presence of gravity.

My personal opinion is that (1) will turn out to be right (because of a generalization of the physical argument I gave above for why the ground state energy of a harmonic oscillator in the quantum case must be higher than in the classical case), but of course that's not the same as saying I have an actual solution to the problem.

 

[PeterDonis]

Only when considering gravitation within GR the choice of the value of the ground-state energy plays a role as a source of the gravitational field.

Yes, and as I have already pointed out, this thread is about the cosmological constant problem, which means we are talking about GR, which means energy gravitates and we can't just arbitrarily adjust energy values. For the case of a harmonic oscillator, the potential energy at the equilibrium point has to be zero--or more precisely, there must be no additional contribution to the potential energy due to whatever energy storage mechanism is driving the oscillator. For example, in the case of a spring, the spring of course has mass (as does the weight on the end of it), and those masses gravitate, but they gravitate the same no matter what state the oscillator is in. And at the equilibrium point of the oscillator, the only additional energy that gravitates, over and above those masses, is whatever kinetic energy the oscillator has--there is no additional potential energy stored in the spring that gravitates. That is the point I was making, and it stands regardless of how energies can be adjusted in the absence of gravity, since in this thread such cases are irrelevant.

 

[vanhees71]

Yes, and as I have already pointed out, this thread is about the cosmological constant problem, which means we are talking about GR, which means energy gravitates and we can't just arbitrarily adjust energy values. For the case of a harmonic oscillator, the potential energy at the equilibrium point has to be zero--or more precisely, there must be no additional contribution to the potential energy due to whatever energy storage mechanism is driving the oscillator. For example, in the case of a spring, the spring of course has mass (as does the weight on the end of it), and those masses gravitate, but they gravitate the same no matter what state the oscillator is in. And at the equilibrium point of the oscillator, the only additional energy that gravitates, over and above those masses, is whatever kinetic energy the oscillator has--there is no additional potential energy stored in the spring that gravitates. That is the point I was making, and it stands regardless of how energies can be adjusted in the absence of gravity, since in this thread such cases are irrelevant.

Why must the ground state energy of the HO be zero? Nothing tells us that it must be, and that's the problem in connection with the cosmological-constant problem. The cosmological constant simply is one of the many constants in our "standard model" we have to adjust to observations, i.e., there's no 1st principle telling us its value. In addition you have a tremendous fine-tuning problem, as explained in #74.

As you very well know, it's not mass that is the source of gravitation but all forms of energy, momentum, and stress. That means that a spring in an excited state leads to a different gravitational field than when it's in its ground state.

 

[PeterDonis]

Why must the ground state energy of the HO be zero?

That's not what I said. I said that the potential energy due to whatever energy source drives the oscillator is zero at the equilibrium point. That's a different statement.

Classically, the ground state of the HO is the oscillator sitting at rest at the equilibrium point. So in the presence of gravity, this would be zero energy over and above the masses of the oscillator components (which are the same regardless of the state of the oscillator).

Quantum mechanically, however, as I pointed out in post #75 in response to the OP, there is no such state as "sitting at rest at the equilibrium point", since any such state would be a simultaneous eigenstate of position and momentum and no such state exists. So the quantum HO ground state must have a higher energy than the classical HO ground state.

a spring in an excited state leads to a different gravitational field than when it's in its ground state.

Yes, and the difference is due to the kinetic plus potential energy of the oscillator, which is the energy we have been considering.

 

[vanhees71]

That's not what I said. I said that the potential energy due to whatever energy source drives the oscillator is zero at the equilibrium point. That's a different statement.

But that's also not true! The absolute level of the potential is not physically determined. You can add any constant without changing any physics. Particularly the equilibrium point won't change when you add an arbitrary constant to the potential.

Classically, the ground state of the HO is the oscillator sitting at rest at the equilibrium point. So in the presence of gravity, this would be zero energy over and above the masses of the oscillator components (which are the same regardless of the state of the oscillator).

The equivalence class of potentials is
$$V(x)=\frac{m \omega^2}{2} x^2+V_0,$$
which $V_0$ arbitrary. The total energy is
$$E=\frac{1}{2m} \vec{p}^2 + V(x),$$
the ground-state energy is at the stationary point and at $E_0=V_0$.

For the same reason you can also argue about the choice of the zero-level of kinetic energy and add the rest energy, $mc^2$ to make it compatible with the preferred choice of special relativistic kinetic energy, because then you get an energy-momentum four-vector.

Quantum mechanically, however, as I pointed out in post #75 in response to the OP, there is no such state as "sitting at rest at the equilibrium point", since any such state would be a simultaneous eigenstate of position and momentum and no such state exists. So the quantum HO ground state must have a higher energy than the classical HO ground state.

That's of course true, but here you compare the ground-state energy between the classical and the quantum model, and of course you are right that because of the uncertainty relation between position and momentum the ground-state energy in the quantum case is larger than in the classical case when using the same convention concerning the arbitrary energy level.

It is indeed part of the problem when it comes to the cosmological-constant problem that there is no physical theoretical argument for what's the right choice of the arbitrary absolute level of the total energy, and it's only the total energy (density) which enters the Einstein equation. Indeed the universality of the coupling to all forms of energy, momentum, and stress to the gravitational field is due to general covariance of the theory, which is a local gauge symmetry (what's gauged is the Poincare symmetry of SR). At our present status of knowledge, indeed this absolute value (aka "cosmological constant") is a parameter that has to be determined by experiment (as have the other about 25 constants of the HEP standard model and who knows how many more, if we find more particles and have to extend it).

Yes, and the difference is due to the kinetic plus potential energy of the oscillator, which is the energy we have been considering.

Exactly!

 

[PeterDonis]

The absolute level of the potential is not physically determined. You can add any constant without changing any physics.

Please, please, please, please do not keep repeating this when I have already, repeatedly, explained why it is irrelevant to this thread. In the presence of gravity, this statement is false, because energy gravitates, as you have already agreed. In the context of this thread, that is the only case that matters, and that is the only case I am talking about with any of my posts in this thread. It is very frustrating to have to keep repeating this.

there is no physical theoretical argument for what's the right choice of the arbitrary absolute level of the total energy

For the simple case of the harmonic oscillator, there is. I have already given it in this thread, repeatedly--and then I have had to keep reminding people that my argument is made for the case that is relevant to this thread, where gravity matters and energy gravitates, so our physical arguments need to be made based on what gravitating energy we expect to be present.

Similarly, in the more complicated case of trying to calculate the vacuum energy density of a quantum field, it makes no sense to me to just adjust it by arbitrary constants in the presence of gravity, without having some physical argument for where such adjustments to the gravitating energy density come from. It certainly makes no sense to say that "you can add any constant to the energy without changing any physics" in the presence of gravity, which is why I keep objecting when such claims are made in this thread.

 

[AndreasC]

And the reason for that is that, in the classical case, the ground state is the oscillator sitting at rest at equilibrium; but in the quantum case, that is impossible, because it would be a state that was an eigenstate of both position (at the equilibrium point) and momentum (zero momentum). And there cannot be any such state because the position and momentum operators do not commute. So the quantum ground state has to have a higher energy than the classical ground state, in order to have a finite uncertainty in at least one of position and momentum.

Not true, my Hamiltonian has exactly the same physics as your Hamiltonian within the quantum theory, as you can readily check.

(because of a generalization of the physical argument I gave above for why the ground state energy of a harmonic oscillator in the quantum case must be higher than in the classical case)

Even if we accept this as a physical argument, "higher" just means "higher". How much higher? If you just want it to be higher, why do you not like the anti-Wick scheme, which gives a term of +1 rather than a term of +1/2?

Of course gravity matters, but what I'm saying is that we don't have as far as I can tell and going by what you presented a theoretical tool WITHIN quantum theory as it stands to decide between one or the other energy levels, at least when it comes to the QHO. That's all.

 

[PeterDonis]

my Hamiltonian has exactly the same physics as your Hamiltonian

Not in the presence of gravity. Which, as I have repeatedly stated, is the context that matters for this thread.

Of course gravity matters

Which contradicts your claim that the two Hamiltonians have exactly the same physics. They can't, if gravity matters.

I have already made this point repeatedly and you have repeatedly failed to respond to it. if you're not even going to listen to what other people are saying, there's no point in continuing this thread.

 

[PeterDonis]

How much higher?

The physical argument is that we should expect the Hamiltonian to be $p^2 / 2 + x^2 / 2$, with no extra constants, because that is the Hamiltonian that is consistent with my physical argument about the potential energy. That is my answer to this question (which also answers the related questions you are asking).

 

[AndreasC]

Not in the presence of gravity.

Yes, but you have not given an argument that includes gravity. And as far as I can tell nobody has measured the gravitational field of the ground state of a quantum harmonic oscillator.

Which contradicts your claim that the two Hamiltonians have exactly the same physics. They can't, if gravity matters.

But we are talking about the physics predicted by the quantum theory, regardless of general relativity because nobody here has made an argument including general relativity. Your argument is based on plain old quantum theory but it can't hold, because it's not true you can differentiate them based on that. But you are saying that it is because the wave function has to be a certain way, and I'm telling you it is that way regardless of the additive constant being 1/2 or 1 or 2 or -3 or whatever.

I have already made this point repeatedly and you have repeatedly failed to respond to it

I have, you are assuming people are not realizing absolute energy matters when GR is taken into account, but that is not the case. You are trying to make a physical argument WITHOUT reference to GR about what it should be, except if you are not referencing GR you can't make a physical argument in this case because the physics are the same. If you DO reference GR then yes, the physics is different, but you would have to actually reference GR in your argument in some way. Maybe there is a way to make a theoretical argument to uniquely determine that additive constant without looking at GR, but you have not made such an argument. I can make the exact same argument as you are making for any other energy level and tell you you are wrong because in GR additive constants in energy matter, and my additive constant is right therefore yours is wrong. Yes, we all know here it matters in GR so there is no need to keep reiterating it.

Now if you believe your argument for why it should be +1/2 is good enough, alright. But its premises are pretty flimsy, for instance it's just not true that it has to be that in order for the wave function to be a certain desirable way, it doesn't matter what it is for the wave function, it doesn't give physically meaningful results (again, I stress, without reference to GR, to which, I repeat, you did not make any reference in your argument for that level).

 

[AndreasC]

In particular, if you look at the most famous and earliest example of these virtual QFT effects... the Lamb shift of the hydrogen atom. It arises from the same sort of sums over vacuum diagrams as the above.
And this has a physical effect, it contributes to a shift in energy levels. So, by the equivalence principle it must gravitate, and that means you can couple the theory in the way we just did. So there is in fact some sort of reality to this afterall (and indeed nontrivial computations in this formalism have been done and verified in famous neutron interferometry experiments in the presence of gravitational fields)

Now this is a much more interesting example than the QHO, and I will certainly look a bit more into that.

 

[PeterDonis]

you have not given an argument that includes gravity

Sure I have. I have argued that energy gravitates, and therefore the only energies we should include are the gravitating energies we expect to be present on physical grounds.

as far as I can tell nobody has measured the gravitational field of the ground state of a quantum harmonic oscillator.

That's because "a quantum harmonic oscillator" in the idealized sense we are using the term here doesn't exist, and the real systems we have that most closely approximate such a thing, like springs with weights on the ends of them, have gravitational fields too small to measure. (Although some torsion balance techniques seem to be getting close to the point where such a thing might be possible.)

nobody here has made an argument including general relativity.

Sure I have. The argument that energy gravitates is based on GR; that's the theoretical framework we have that makes that prediction.

you are saying that it is because the wave function has to be a certain way

No, I'm not. I'm saying that the Hamiltonian has to be a certain way.

As far as wave functions go, we already know the wave functions for the energy eigenstates of the quantum harmonic oscillator. It is true that those wave functions are the same if you put an arbitrary additive constant into the Hamiltonian (at least, with the quantum theory we have now--possibly a quantum gravity theory might change that, but I'm not assuming that as a basis for my arguments), but that's irrelevant because I am not arguing that the wave functions only take the form we know they take if the Hamiltonian is a certain way. I am arguing that the Hamiltonian has to be a certain way based on physical grounds that, as far as they have to do with anything in the quantum framework, have to do with observables, not wave functions. Energy is an observable, and changing the operator for that observable--the Hamiltonian--changes the observable even if the wave functions stay exactly the same.

you are assuming people are not realizing absolute energy matters when GR is taken into account

I'm not assuming it; I'm seeing it explicitly in the posts you and @vanhees71 keep making, which keep saying that the physics is exactly the same when you add an arbitrary constant to the energy. You both know quite well that that is wrong in the presence of gravity, but you keep saying it anyway. So I keep objecting. Stop saying it and I'll stop objecting.

The rest of your posts just repeats things I've already responded to above.

 

[AndreasC]

Sure I have. I have argued that energy gravitates, and therefore the only energies we should include are the gravitating energies we expect to be present on physical grounds.

Yes, but the way you determine those "physical grounds" makes no reference to GR, so you can't be using that as an argument to "prove" that yours are valid.

No, I'm not. I'm saying that the Hamiltonian has to be a certain way.

You said that the energy has to be "higher than the classical case" (which doesn't uniquely determine it anyways) because you can't have an eigenstate of both position and momentum and the wavefunction is a gaussian. Except none of that uniquely determines the Hamiltonian either, because they are properties of the wave function which pretty much ignores the additive constants, as you said yourself. So these arguments weren't valid.

I'm not assuming it; I'm seeing it explicitly in the posts you and @vanhees71 keep making, which keep saying that the physics is exactly the same when you add an arbitrary constant to the energy.

Yes, within the framework of quantum mechanics, which is what we have been studying in the case of the harmonic oscillator, since nobody has tried to explicitly involve GR in an argument about what the energy level should be set as, the physics are the same, as far as anyone in this thread has been able to argue. Therefore trying to make a physical argument like the ones you have been trying to make will not work, because your arguments basically have two parts. The 1st part is that in GR the absolute energy matters. With this nobody has disagreed, but you have not offered a way to uniquely determine that energy via GR or anything like that. Then there is a 2nd part, which is a physical argument that makes absolutely no reference to GR. These arguments can not work because as far as QM is concerned, the physics doesn't change at all, so you can't construct an argument like that purely from QM. So you are lacking the theoretical tools to make such a determination.

Is there one true energy level that we should be accepting? Very likely, if we make a number of assumptions that may not necessarily be justified (such that saying a complete unification of GR and quantum theory would involve such concepts). This is not the main point of contention, and nobody ever said it doesn't matter in GR, so let's please move on from that. The main point of contention is that you have not offered a valid way to uniquely determine what that is. You did technically make reference to GR, yes, but that was only to say absolute energy matters - which nobody ever questioned - and not to offer a way to actually uniquely determine the energy in the case of the harmonic oscillator and explain why it has to be +1/2 and not something else. When people are telling you the physics doesn't change, they are telling you that within the framework of quantum mechanics the physics of the QHO do not change so you will have to try harder than arguments which do not directly involve GR and make appeal to the physics determined by QM and QM alone to advocate for your chosen energy level. As I said, I can say exactly the same things as you said to advocate for any energy level, and then say that it has to be this way because in GR energy matters, so it can't be anything else. It is justified to take for granted that GR uniquely determines what that level should be, but no way to actually determine that has been presented in this thread as of yet.

I hope this has cleared up my position and I believe probably also the position of @vanhees71 and others who have brought up the same objection to you.

 

[AndreasC]

Im picking up some confusions, so I thought I would briefly sketch a heurestic of the CC argument.

If you wish to quantize Einstein gravity, you need to come to grips with expectation values of the Einstein stress energy tensor. B/c even if gravity doesn't obey quantum mechanics, matter does!
So we need to understand formal objects like <Tuv>.
Well first we impose symmetry principles and local lorentz invariance enforces the following in vacuum.

$$ R_{uv} -\frac{1}{2}g_{uv}R-\lambda g_{uv}=<T_{uv}>=-\frac{1}{M_{pl}^{2}}<\rho >g_{uv}$$
Thats absolutely critical, b/c it tells you that cosmological constant terms appearing on the left hand side is indistinguishable from energy densities computed from quantum fields in their vacuum state.

Moreover, we sort of know how to do this.. These expression arise from the sum of feynman diagrams, where we have vacuum bubble diagrams (with external graviton loops). This will lead to an integral over the momentums (up to some hard cutoff $\Lambda$) for a field of mass M. This is just like summing up harmonic oscillators of energy E (see the Weinberg review for more details)

Which leads to the following badly divergent expression:

$$ <\rho>_{vac} = \frac{1}{2}\int_{0}^{\Lambda}\frac{4\pi k^{2}dk}{2\pi ^{3}}\sqrt{k^{2}+M^{2}} = \frac{\Lambda^{4}}{16\pi ^{2}}+\frac{M^{2}\Lambda^{2}}{32\pi ^{2}}+\frac{M^{4}ln(\frac{\Lambda}{M})}{16\pi ^{2}}+...$$
You will recognize the famous quartic divergence. Putting in a value for the cutoff $\Lambda=Mpl$, gives you the ~120 order of magnitude estimate.

Now there are a few immediate problems you can point to in the above logic.

The first is that these are quantum field theories, and in quantum field theories we were always instructed to view quadratic or quartic divergences of the cutoff with skepticism. That we need to put in appropriate regularization and to introduce counterterms and so forth, and maybe we don't know how to do that for quantum gravity, but if we did it properly things would vanish.
Ok. But you will notice the third term in the above is a logarithmic divergence. Those usually are physical and involves the mass of a particle. We don't know about Planck scale physics, but we do know about standard model fields, like the top quark. Plugging in that value, gives us an estimate that is still some ~52 orders of magnitude to big.

Ok. You might still object and say we don't really understand how to quantize things. Maybe, maybe.. The top quark field gets some strange contribution to its vacuum expectation that somehow makes it zero.
Good.. But that same voodoo you make for the top quark field, somehow needs to be communicated to the equivalent computation for another standard model field. Like the muon.
Which leads to the question.. Why would muons and top quarks conspire together in exactly the same way in whatever new theory you cook up, as they don't manifestly seem related in this context.

Next point. Maybe, you don't believe in these diagrams contributing in the first place. eg Loops of things with gravitons.. Artifacts of perturbation theory you might say.
Well, that's sort of like looking at the expansion of e^x = 1 + .5x^2 + ... and saying something like 'x^2 is unphysical'. Yes, strictly speaking true, but that doesn't mean you can ignore it either when doing the computation.

In particular, if you look at the most famous and earliest example of these virtual QFT effects... the Lamb shift of the hydrogen atom. It arises from the same sort of sums over vacuum diagrams as the above.
And this has a physical effect, it contributes to a shift in energy levels. So, by the equivalence principle it must gravitate, and that means you can couple the theory in the way we just did. So there is in fact some sort of reality to this afterall (and indeed nontrivial computations in this formalism have been done and verified in famous neutron interferometry experiments in the presence of gravitational fields)

Finally, to the extreme die hard skeptic of the above, there is an ultimate coup de grace! And that is the fact that there are perfectly classical contributions to the cosmological constant arising from the confinement scale of QCD, or the electroweak phase transition in the early universe. These things also contribute (by shifting the overall height of the potential energy), and produce exact numbers many times too big.

So to summarize. Strictly speaking, we don't have an exact computation (nor will we ever, as that requires knowledge of all matter particles all the way to the Planck scale). Instead we have a sequence of estimates.
The trouble is that getting around the problem these estimates create involves either some sort of ridiculous conspiracy, or modifying some cherished principle that in many ways is a much bigger deal. (Like a Lorentz invariance violation or the failure of effective field theory or the existence of a new symmetry principle). Clearly something must go wrong in the argument above, but the more you stare at it, the harder and more nontrivial things become..

I think this post is probably a bit more relevant to all we've said but I have to look a bit more into what exactly happens in the Lamb shift.

 

[PeterDonis]

the way you determine those "physical grounds" makes no reference to GR

As I've already said, "energy gravitates" comes from GR. If that's not enough for you, we'll just have to disagree.

You said that the energy has to be "higher than the classical case" (which doesn't uniquely determine it anyways) because you can't have an eigenstate of both position and momentum

Yes. And that's all that's necessary for that claim. But that's not the only claim I was making.

and the wavefunction is a gaussian.

No. I mentioned that the wave function is a Gaussian simply to emphasize that that is consistent with my position. It is not the basis for my position.

none of that uniquely determines the Hamiltonian either

Again, I am not arguing that the Hamiltonian is a certain way on the basis of it being required by the wave function. That would be the tail wagging the dog. The wave function is determined by the Hamiltonian, not the other way around; and the fact that multiple possible Hamiltonians can determine the same wave function does not mean there cannot be other arguments for preferring one Hamiltonian over the other.

Yes, within the framework of quantum mechanics,

No, within the framework of quantum mechanics in the absence of gravity. The fact that we do not currently have a quantum theory of gravity does not mean you can just ignore gravity in contexts where it is relevant, such as this thread. You need to do the best you can at taking into account both gravity (the fact that energy gravitates, per GR) and quantum mechanics, given what we currently know. Adding arbitrary constants to the Hamiltonian does not do that. It just ignores gravity, and you can't do that.

It seems to me that you basically want to ignore gravity when it suits you, while claiming that you agree that gravity matters. But you can't have it both ways. If gravity matters, which it does, then you can't just say "well, ordinary QM says we can add arbitrary constants to the Hamiltonian, so there". That simply is not a valid argument.

The main point of contention is that you have not offered a valid way to uniquely determine what that is.

I'm sorry, but this is simply false. You can be obstinate and say you continue to disagree with the way I have offered. But you can't say I haven't offered one. I don't see the point of continuing to repeat my arguments. If you disagree, well, then we disagree. But that's not the same as me not providing an argument at all.

Again, the rest of your post just repeats points that I've already responded to. I doubt I can add anything more to what I've already said, and I'm not going to respond to further posts from you that just repeat the same things. If you have any new issues to raise, I'll respond to those.

 

[AndreasC]

No, within the framework of quantum mechanics in the absence of gravity

There is no proper framework of quantum mechanics that includes GR. Quantum mechanics is a theory that does not include gravity, except you can try to combine their predictions together. But at any rate it is pointless because the part of your physical argument that is supposed to show how you uniquely determine the Hamiltonian does not involve any gravitational effects.

If there are arguments other than the wave function and the consequences that support your viewpoint over others then please talk about them and not things related to the wave function. I don't understand why you need to bring up the gaussian or the position-momentum uncertainty or whatever, all these things are consistent with any choice of additive constant. And no, no one "ignores" gravity when it "suits" them. Yes, in ordinary QM you can add whatever constant you want. Ordinary QM does not involve gravity. If you want to make an argument for why one energy level is correct and not the other, you have to actually use GR somehow, but you haven't done that. You tried to make a purely ordinary QM argument for which level is better, and it predictably didn't work.

Adding arbitrary constants to the Hamiltonian does not do that. It just ignores gravity, and you can't do that.

You also effectively added an arbitrary constant to your Hamiltonian because you have not shown that your choice is better than any other. That's what I've been saying all this time. Nothing that has been presented has conclusively argued any specific option is right or wrong. If someone could come here and say "hey, this is right because it is the only option that is not in conflict with x other part of the theory and agrees with observations", then it would be settled, but that has clearly not happened yet. That's all and I think it's the last thing I'll say about it.