I Inflation and the false vacuum

[friend]

Your answer, PeterDonis, makes no mention of the vacuum energy that caused inflation being equivalent to any kind of zero point energy. I suppose that is because zero point energy is only concerned with harmonic oscillators, whereas the inflationary vacuum is caused by some non-zero constant field (or changing slowly). Is there no zero point energy (as such) involved with initial inflation?

It seems we do have a zero point energy now associated with the SM fields, and the naive calculation is like 120 orders of magnitude greater than what is observed. Although, I've heard a few people say that there is some cancellation going in with the effects of nearby oscillating fields so that the net effect could be zero, or near zero. Is it even true that the zero point energy of all the SM fields equivalent to the cosmological constant or dark energy. If so, how are we ever going to know whether this zero point energy or dark energy can decrease to a true vacuum energy when we presently have such a terrible disagreement with measurement? And what would that do to the SM fields if that zero point energy did decrease?

But if the present dark energy value is the only scalar field whose energy could fall and cause another phase transition, then I suppose that means we don't have to worry about another round of inflationary expansion. Expansion could only slow down from its present value since the vacuum energy would be less than today.

Just curious: do you know of any physical eschatological theories that could cause the universe to experience another round of rapid expansion?

 

[PeterDonis]

Your answer, PeterDonis, makes no mention of the vacuum energy that caused inflation being equivalent to any kind of zero point energy.

That's because what caused inflation is not vacuum energy, it's a nonzero vacuum expectation value for the inflaton field. They are not the same thing. Vacuum energy is the expectation value of the energy operator--the Hamiltonian--when the field is in a vacuum state. The vacuum expectation value of the inflaton field is the expectation value of the field operator when the field is in a vacuum state. The field operator is not the same as the Hamiltonian operator.

I suppose that is because zero point energy is only concerned with harmonic oscillators

No, it isn't, it's there for any quantum field. "Zero point energy" is just another way of saying that the expectation value of the Hamiltonian operator when the field is in a vacuum state is not zero; or, to put it another way, the Hamiltonian operator has an extra term in it that is independent of the state of the field, it's just a nonzero constant, which is referred to as "zero point energy".

It seems we do have a zero point energy now associated with the SM fields

Yes, as I said above, it's there for any quantum field. But all of the SM fields, except for the Higgs, have nonzero spin, so they cannot have a nonzero vacuum expectation value, even though they have nonzero zero point energy. A nonzero vacuum expectation value is required for a field to be able to cause inflation (but even then it's not sufficient; the Higgs has a nonzero VEV but does not cause inflation).

Is it even true that the zero point energy of all the SM fields equivalent to the cosmological constant or dark energy.

That's one hypothesis, but we don't really know.

do you know of any physical eschatological theories that could cause the universe to experience another round of rapid expansion?

I'm not aware of any, no.

 

[friend]

I feel your explanation in post #48 may not be consistent with your last post. In post #48 you seem to be saying that there are two values of the VEV of the field (not the energy of the field). But as I recall, The inflaton field is a plot of field strength (on the x-axis) vs. energy (on the y-axis). The inflaton energy is not zero when the inflaton field is zero. Instead it has a somewhat constant energy value and there is a slow slope as the field increases. Then it rather sharply decreases at some level of the field (this is where inflation stops and energy is transferred to the SM fields). Then with slightly more field strength, the energy reaches a local minimum and begins to increase again. (The Mexican hat potential, as I recall). So it would then not be the field strength that causes or ends inflation, it's the energy of the field that causes all of this. In fact the field strength itself increases a bit at the end of inflation where the energy falls dramatically IIRC. So perhaps we need to take a look again at whether it is the "vacuum energy"=zero-point-energy that is causing inflation, etc.

 

[PeterDonis]

In post #48 you seem to be saying that there are two values of the VEV of the field (not the energy of the field).

Yes. More precisely, there are two states of the inflaton field, one with a nonzero VEV and another with a zero VEV; the "false vacuum" and "true vacuum" states, respectively.

as I recall, The inflaton field is a plot of field strength (on the x-axis) vs. energy (on the y-axis).

No, it's a plot of the field's value ("strength" if you like) on the x-axis vs. the potential energy on the y-axis. The potential energy is a portion of the Lagrangian (or Hamiltonian), but it is not all of it; if nothing else, it leaves out the kinetic energy associated with the field. So the y-axis on those plots is not the same as the "energy", which is the expectation value of the complete Hamiltonian operator.

The reason the potential energy is important is that the field has a tendency to "move downhill" towards a region of lower potential energy. See further comments below.

The inflaton energy is not zero when the inflaton field is zero. Instead it has a somewhat constant energy value and there is a slow slope as the field increases. Then it rather sharply decreases at some level of the field (this is where inflation stops and energy is transferred to the SM fields). Then with slightly more field strength, the energy reaches a local minimum and begins to increase again. (The Mexican hat potential, as I recall). So it would then not be the field strength that causes or ends inflation, it's the energy of the field that causes all of this. In fact the field strength itself increases a bit at the end of inflation where the energy falls dramatically IIRC.

You're misdescribing the model somewhat, and you're also mixing up two different models of inflation, "old" and "new"--and you're also mixing in models of spontaneous symmetry breaking that have nothing to do with inflation (the "Mexican hat" potential). Here is a better description:

In the "old" inflation model (Guth's original formulation), the potential energy as a function of the state of the field had two minima. One, called the "false vacuum", corresponded to a state of the field with a nonzero vacuum expectation value. This was only a local minimum of the potential energy, i.e., it had a lower potential energy than other "nearby" field states, but a significantly higher potential energy than a more "distant" field state, called the "true vacuum" state. The true vacuum field state had a zero vacuum expectation value of the field.

In this model, the field was hypothesized to start out in the "false vacuum" state (more precisely, to be driven there by the natural dynamics of the field "moving downhill" towards states of lower potential energy, but then getting "stuck" in the local minimum, like a small valley in a mountain range). In this state, the nonzero VEV of the field drove exponential expansion. You could say, I suppose, that the reason the nonzero VEV drove exponential expansion is that it was equivalent to there being a large positive cosmological constant, which can be thought of as an "energy of empty space" that makes empty space exponentially expand. But this only happens because the field itself (i.e,. the field operator, not the energy operator) has a nonzero VEV; at least, that's my understanding of the underlying math. Also, the energy involved here is not well described as "zero point energy" in any case; see below.

In this model, the "false vacuum" state is metastable: classically, the field will stay there forever because it's a local minimum, but when we add quantum fluctuations, there is a nonzero amplitude for the field to quantum tunnel to the "true vacuum" state. When that happens, it causes two things: first, the field's VEV changes from nonzero to zero, which stops inflation; second, the expectation value of the energy operator for the field decreases drastically, because that value is much lower for the "true vacuum" state than for the "false vacuum" state. (Note, though, that it is not zero for the "true vacuum" state; see comments at the end on "zero point energy".) That energy has to go somewhere, and where it goes is into the ordinary SM fields, "reheating" them to a very high temperature and creating the hot, dense, rapidly expanding "Big Bang" state. So at the end of all this, we have the inflaton field in the "true vacuum" state with zero VEV, where it will then remain forever, and the SM fields at very high temperature in the "Big Bang" state.

This model is simple, but it turned out to have a number of issues, and to try and address them, Linde and others came up with a somewhat different model called the "new inflation" model. In this model, the "false vacuum" state is not a local minimum of the potential energy; it is "at the top of a hill", but the potential energy as a function of the field state has a very, very small slope in that region. So the field "moves downhill" very, very slowly when it starts from the "false vacuum" state. While it is "moving downhill" very, very slowly, the field's VEV is almost constant at some nonzero value, and drives inflation as discussed above. (This is called the "slow roll" model of inflation.)

However, in this model, as the field "moves downhill" away from the original "false vacuum" state, the "hill" gradually gets steeper, and so the field "moves downhill" faster. This process ends up at the "bottom of the hill", which is the "true vacuum" state, with zero VEV, and once there, the field stays there forever, just as in the "old inflation" model. While the field is "rolling downhill" faster, inflation is stopping and energy is being transferred from the inflaton field to the SM fields, but this is somewhat more gradual than in the old inflation model where the transition was due to quantum tunneling. In the new inflation model no tunneling is necessary; the ordinary classical dynamics of the inflaton field will take it from the "false vacuum" to the "true vacuum" state.

Note that, as I mentioned above, neither of the potential energy functions in these models (old or new) is of the "Mexican hat" type. That type of potential is associated with a different process, the spontaneous symmetry breaking process that, for example, broke electroweak symmetry and allowed the Higgs field to give other SM fields a nonzero mass. In this kind of process, the field state with a zero VEV (the one at the top of the "Mexican hat") has a higher potential energy than a family of field states with a nonzero VEV (the whole circle of states at the bottom of the trough of the Mexican hat). (Note that in the case of the inflaton field above, the zero VEV "true vacuum" state had a lower potential energy than the nonzero VEV "false vacuum" state.) So the natural dynamics of the field will carry it from the zero VEV state to one of the nonzero VEV states--but it will have to pick one nonzero VEV state out of a whole family of possible ones. Picking one state out of the family breaks the underlying symmetry of the field--SU(2) x U(1) electroweak symmetry, in the case of the Higgs field.

perhaps we need to take a look again at whether it is the "vacuum energy"=zero-point-energy that is causing inflation, etc.

As I noted above, the energy in the inflaton field in the "false vacuum" state is not well described as "zero point energy" in any case. Why not? There are at least two reasons. First, the "false vacuum" state is not a state of lowest energy globally; it is only a state of lowest energy locally (i.e., with respect to "nearby" field states). So it's not a "zero point" state, because that implies a state with globally lowest energy.

Second, as I've noted several times now, all quantum fields have "zero point energy", and this energy is independent of the state of the field; it's an extra term in the Hamiltonian that's just a constant, with no dependence on the field state. So the inflaton field in the "true vacuum" state also has this energy--yet it doesn't cause inflation in that state. So whatever it is that is causing inflation when the field is in the "false vacuum" state, it has to be something else, something that isn't there in the "true vacuum" state--something other than "zero point energy". The obvious difference is the nonzero VEV of the field itself in the "false vacuum" state, as compared with its zero VEV in the "true vacuum" state; as I said above, one could also, I suppose, associate this with the extra energy stored in the field, but this energy would also not be properly described as "zero point energy", as above. (Also, all of the SM fields are present while inflation is happening--they are all in their own vacuum states, all with zero VEV, and all having "zero point energy" associated with them as well--but none of them are causing inflation.)

 

[friend]

(i.e,. the field operator, not the energy operator) has a nonzero VEV;

Thank you for the effort you put into your response. I'm starting to hit the "like" button on your posts, whether I understand them or not.

In both scenarios that you describe, there is a move towards lower "potential" energy on the graph (whether it is a mexican hat or not). This in itself means the field "strength" is not zero by the time the potential energy reaches its true vacuum level. So I don't know what you mean VEV of the field operator goes to zero in the true vacuum state. Where is the VEV of the field operator on the graph of potential energy vs. field strength? Your use of the term VEV seems to imply that we are dealing with quantum theoretic calculation involving quantum fluctuations of the inflaton field. This only begs the question as to how these "fluctuation" are manifest. Are they virtual particles of some sort? Is there a zero-pont-energy associated with them; that would be the average value represented by the line on the graph, right?
Also, the videos I watched where Leonard Susskind lectures on this tells me that there is an upturn with increased field after the local minimum of energy is reached in the potential energy vs inflaton field strength. He explains that there may be some oscillations at the bottom of the hill and that this may be what dark matter is. IIRC.

 

[nikkkom]

whatever it is that is causing the current accelerating expansion of the universe--is itself a nonzero vacuum expectation value of some scalar field, i.e., a false vacuum state.

I think that "falseness" is not related to "nonzero". True vacuum is not necessarily vacuum where all fields are zero (all-zero is not even necessarily a vacuum). True vacuum state is the state with not only local minimum of energy, but with global minimum of energy - there are no "lower vacuums than this one".

 

[nikkkom]

... address them, Linde and others came up with a somewhat different model called the "new inflation" model. In this model, the "false vacuum" state is not a local minimum of the potential energy; it is "at the top of a hill"

IIUC such a state is not a vacuum. Vacuum is a state where potential energy has a local minimum.

 

[PeterDonis]

I think that "falseness" is not related to "nonzero". True vacuum is not necessarily vacuum where all fields are zero (all-zero is not even necessarily a vacuum).

Yes, it's easy to get muddled in the terminology in this area--see below.

IIUC such a state is not a vacuum.

It isn't in the sense you give, yes; but the term "false vacuum" is still used to describe it, probably for historical reasons, because of the way inflation theory developed. Unfortunately this happens often in science: confusing terminology gets established for historical reasons and then can't be dislodged.

 

[PeterDonis]

In both scenarios that you describe, there is a move towards lower "potential" energy on the graph (whether it is a mexican hat or not).

Yes.

This in itself means the field "strength" is not zero by the time the potential energy reaches its true vacuum level.

Not necessarily; it depends on the specifics of how the potential energy varies with the field state (I prefer to use the word "state" rather than "strength" here, because, as you appear to agree since you put "strength" in quotes, the numerical value of the field variable doesn't necessarily have any physical meaning in terms of "field strength"). Also, the "field variable" in these graphs is probably better interpreted as the VEV of the field, not as the "raw value" of the field (see below). In the "Mexican hat" case, involved in spontaneous symmetry breaking, the field state with zero VEV has a higher potential energy than field states with nonzero VEV; but I don't know that this is true in inflationary models in cosmology.

Where is the VEV of the field operator on the graph of potential energy vs. field strength?

The "field strength" in these graphs, as I mentioned above, is really the VEV--at least, I think that's the best interpretation of the graphs. AFAIK the graphs themselves are heuristics and are not intended to be exact representations of the underlying models. And the general role that the "field strength" on the graph--i.e., the horizontal axis variable--appears to play is as modeling the VEV of the field. That's certainly the case in the "Mexican hat" graph that is used as a heuristic model of electroweak symmetry breaking.

Your use of the term VEV seems to imply that we are dealing with quantum theoretic calculation involving quantum fluctuations of the inflaton field.

That's not really a good description of what a VEV is. In quantum field theory, the "field" itself is an operator. The VEV is simply the expectation value of that operator when the operator is applied to a vacuum state. In this more precise terminology, the "state" is not a state of the "field", since the field is an operator; rather, the "state" is some vector in a Hilbert space over which the field operator is defined. Taking an expectation value doesn't involve any "fluctuations"; the state the operator is applied to doesn't change, it is constant.

This only begs the question as to how these "fluctuation" are manifest. Are they virtual particles of some sort? Is there a zero-pont-energy associated with them; that would be the average value represented by the line on the graph, right?

The virtual particle picture is not a good one to use when trying to understand inflation models. Virtual particles arise in perturbation theory, and inflation models are not based on perturbation theory.

Also, even when you are using perturbation theory and thinking in terms of virtual particles, the intuitive picture of zero point energy arising from "fluctuations due to virtual particles" is of limited usefulness.

the videos I watched where Leonard Susskind lectures on this tells me that there is an upturn with increased field after the local minimum of energy is reached in the potential energy vs inflaton field strength. He explains that there may be some oscillations at the bottom of the hill and that this may be what dark matter is.

This is a speculative hypothesis; it's not the same as the basic model of inflationary cosmology. In the basic model, once the inflaton field reaches the true vacuum state, it stays there forever, and any quantum fluctuations in it are assumed to be negligible. That might not be true in reality, of course, but it's the simplest model.

 

[friend]

That's not really a good description of what a VEV is. In quantum field theory, the "field" itself is an operator. The VEV is simply the expectation value of that operator when the operator is applied to a vacuum state. In this more precise terminology, the "state" is not a state of the "field", since the field is an operator; rather, the "state" is some vector in a Hilbert space over which the field operator is defined. Taking an expectation value doesn't involve any "fluctuations"; the state the operator is applied to doesn't change, it is constant.

OK. With the field being an operator that when acting on the vacuum state gives 0 (after inflation), that seems to be an extra step that needs more explanation. I've not seen that calculation before. I've only seen the potential energy vs field curve. So when you say VEV of the field operator being zero at the end of inflation, you seem to be indicating that this is different than the field "strength" (x-axis) that we see on the curve. Obviously that curve is monotonic; lower inflaton energies can only be reached at larger field (larger values on the x-axis). I'm missing the math to get from the field (x-axis) to the VEV that you get once you apply the field (operator) on the vacuum state. How does one get a field operator from the scalar on the x-axis? Where does one get the vacuum state on which the field operator acts upon? Thanks.

 

[PeterDonis]

With the field being an operator that when acting on the vacuum state gives 0 (after inflation)

More precisely, the expectation value of the field operator when acting on the true vacuum state is zero (at least, as I understand it in current inflation theory). Operators themselves don't take states to numbers; they take states to states.

when you say VEV of the field operator being zero at the end of inflation, you seem to be indicating that this is different than the field "strength" (x-axis) that we see on the curve.

Heuristically, I think the x-axis of the graph is trying to represent something like the VEV of the field operator, or at least the change in it as the state goes from the false vacuum to the true vacuum. But as I've said before, the graph is just heuristic. I would be very wary of putting too much emphasis on the details of the graph, such as what numbers appear on the x axis. See further comments below.

How does one get a field operator from the scalar on the x-axis?

One doesn't. The x-axis on the graph is a number, and a number isn't an operator. That is one reason why I say you should be wary of putting too much emphasis on the details of the graph. The graph is not showing you the actual math of the underlying model; it's just showing you a heuristic illustration of certain qualitative features of the model.

Where does one get the vacuum state on which the field operator acts upon?

By looking at the actual math of the underlying model and seeing what Hilbert space it is using, and which state vector in that Hilbert space corresponds to the "true vacuum" state in the model.

 

[nikkkom]

So when you say VEV of the field operator being zero at the end of inflation, you seem to be indicating that this is different than the field "strength" (x-axis) that we see on the curve. Obviously that curve is monotonic; lower inflaton energies can only be reached at larger field (larger values on the x-axis). I'm missing the math to get from the field (x-axis) to the VEV that you get once you apply the field (operator) on the vacuum state.

In QFT, fields are operators, in a simpler theory of Quantum Mechanics, they are functions - this is probably much easier to visualize. In this picture, VEV is nothing special - it is really just the value of the field in the vacuum state. In vanilla Standard Model, VEVs of all fields are zero, except Higgs field VEV (and therefore I'm confused why Peter seems to insist that nonzero VEV causes inflation, and non-inflationary state must have zero VEV. Well, our current vacuum has non-zero VEV yet it does not inflate...)

All particles are just small ripples atop these VEVs.

 

[PeterDonis]

I'm confused why Peter seems to insist that nonzero VEV causes inflation, and non-inflationary state must have zero VEV

I'm only saying that with respect to one field, the inflaton field. And I'm not actually sure that's what the underlying math for that field says; that's my understanding but I could be wrong. I certainly agree that a nonzero VEV does not cause inflation for all fields; I believe I mentioned the Higgs as a counterexample earlier in this thread.

 

[haushofer]

The structure constants of the Lie algebra of any Lie group, as I said before, cannot change; they are inherent properties of the group.

At the risk of confusing, a small sidenote: structure 'constants' need not to be constant, but can be field-dependent. These algebras are called soft algebras, and are very common in supergravity. E.g., in N=1 in four dimensions the local SUSY-algebra is a soft algebra, i.e. one uses field-dependent parameters to close the algebra on the metric and gravitino.

This does not influence your argument, but just as a sidenote ;)

 

[PeterDonis]

structure 'constants' need not to be constant, but can be field-dependent. These algebras are called soft algebras

These algebras are not Lie algebras, correct?

 

[friend]

At the risk of confusing, a small sidenote: structure 'constants' need not to be constant, but can be field-dependent. These algebras are called soft algebras, and are very common in supergravity. E.g., in N=1 in four dimensions the local SUSY-algebra is a soft algebra, i.e. one uses field-dependent parameters to close the algebra on the metric and gravitino.

This does not influence your argument, but just as a sidenote ;)

Thank you, haushofer, for your insight. This is certainly relevant to me. For I was thinking that the higher vacuum energy during inflation may have been caused by a higher value of uncertainty during that time. I keep hearing that the vacuum energy is caused by the zero-point-energy of all the quantum fluctuations, and this depends on ΔEΔt≥ħ. So if ħ had a higher value during inflation, then the energy of the vacuum would be greater than now. And since this is the same ħ as used in the lie algebras of the commutation relations, your comment that the structure constants need not be constant is interesting.

 

[PeterDonis]

since this is the same ħ as used in the lie algebras of the commutation relations, your comment that the structure constants need not be constant is interesting

You are confusing the commutation relations of quantum operators, such as $x$ and $p$, with the commutation relations of the Lie algebra of a symmetry group. They are not the same, and the latter do not include any factors of $\hbar$.

Also, as my question in response to haushofer makes clear, I do not think the "soft algebras" he is referring to are Lie algebras. He is talking about other mathematical structures used in supergravity; he is not talking about the mathematical structures used in standard inflationary cosmology. If you want to talk about supergravity, please start a separate thread.

 

[PeterDonis]

I was thinking that the higher vacuum energy during inflation may have been caused by a higher value of uncertainty during that time.

What does "higher value of uncertainty" mean? "Uncertainty" is not a quantum operator or a field variable.

We are getting to the point where using ordinary language to discuss this subject is causing more problems than it solves; ordinary language is too imprecise. We really should be looking at the underlying math.

 

[PeterDonis]

We really should be looking at the underlying math.

In the spirit of trying to practice what I preach, I am going to give a very simple mathematical model that illustrates the issues I have been talking about. This is not the same model that is used in actual inflationary cosmology; it is much more generic, but it should be enough for a start.

A generic expression for the Hamiltonian operator $\hat{H}$ of a quantum system is:

$$
\hat{H} = H\left( \varphi \right) + K
$$

where $H \left( \varphi \right)$ is an operator whose expectation value depends on the state $\varphi$ of the system, and $K$ is a constant that is independent of the state. (Strictly speaking, the $K$ term should be written $K \hat{I}$, where $\hat{I}$ is the identity operator.) The expectation value $\left< \hat{H} \right>$ of $\hat{H}$ is then given by the sum of the expectation value of $H \left( \varphi \right)$ and the constant $K$ (since the expectation value of $\hat{I}$ is just $1$).

If $\varphi$ is a vacuum (or more precisely "true vacuum", as we will see below) state, then the expectation value of $H \left( \varphi \right)$ is zero. (This is the usual definition of a "vacuum" state in ordinary quantum mechanics.) So the expectation value of $\hat{H}$ as a whole in this state is just $K$. $K$ is therefore referred to as the "zero point energy"; it is the expectation value of the energy when the system is in a vacuum state.

Now let's give $H \left( \varphi \right)$ a little more structure. Suppose we have

$$
H \left( \varphi \right) = \left( \partial \varphi \right)^2 + V \left( \varphi \right)
$$

where $\left( \partial \varphi \right)^2$ is the kinetic energy associated with the state and $V \left( \varphi \right)$ is the potential energy. In this formulation, the kinetic energy is still an operator, but the potential energy $V$ is just a number--more precisely, it's a function that takes a state as input and outputs a number (a nonnegative number, to be precise). (We are still being heuristic here; there are plenty of technicalities that we're not getting into, for example in the physical interpretation of "kinetic energy" and "potential energy". But this will serve to illustrate the basic idea.) So the expectation value of $V$ is just the number $V$ itself.

Now we have a different possible definition for what a "vacuum" state is. We can say that a "vacuum" state is a state in which the expectation value of the kinetic energy $\left( \partial \varphi \right)^2$ is zero, regardless of the value of the potential energy $V$. A vacuum state in which $V = 0$ is then called a "true vacuum" state, and a vacuum state in which $V > 0$ is called a "false vacuum" state. In any vacuum state, the expectation value of $\hat{H}$ is evidently $V + K$.

This is the kind of model that is used in inflationary cosmology. The inflaton field starts out in a vacuum state (i.e., a state in which the kinetic energy has zero expectation value) for which $V$ has some large positive value, so it is a "false vacuum" state. It ends up in a vacuum state in which $V = 0$, i.e., a "true vacuum" state. The difference in $V$ between the two states is the energy that gets transferred to the SM fields, reheating them to a very high temperature and creating the hot, dense, rapidly expanding "Big Bang". But $V$ is not "zero point energy"; it's potential energy. The "zero point energy", as above, is $K$, and doesn't change at all through any of this.

Also, $V$ and $K$ in the above are treated purely classically (with the caveat that we have to use the "new inflation" model for the transition from "false vacuum" to "true vacuum" to be driven by the classical dynamics of the system). No "quantum fluctuations" are involved. "Quantum fluctuations" only come into play when we have to take the expectation value of an operator applied to a state that is not an eigenstate of that operator. But in the model above, $V$ and $K$ are not operators, they're just numbers; the only actual operator is the kinetic energy, and in any vacuum state, the kinetic energy has expectation value zero and we don't need to worry about whether the state is an eigenstate of the operator or not. (The usual assumption is that it is.)

 

[friend]

Thank you again, PeterDonis, for the effort. All good stuff to think about.

But I thought it was simpler than that. There is a higher energy false vacuum state (during inflation) and a lower (possibly zero) energy true vacuum state after inflation. I was assuming that for ANY "vacuum" state the lowest energy is governed by ΔEΔt≥ħ/2. If that is true, then the only thing I can think of to account for the difference in energy between the false and true vacuum energies is that ħ must have changed. That seems kind of obvious, what's wrong with that thinking? (Anyone with an answer is welcome to reply)

This paper describes a Generalized Uncertainty Principle, which might be interpreted as a changing ħ. They apply to Dark Energy but not to inflation:
http://arxiv.org/abs/1310.8065

 

[PeterDonis]

I was assuming that for ANY "vacuum" state the lowest energy is governed by ΔEΔt≥ħ/2.

No, this has nothing to do with zero point energy. Trying to think about ZPE in terms of the uncertainty principle doesn't work when you look at the underlying math, however much it gets described that way in pop science sources.

This paper describes a Generalized Uncertainty Principle, which might be interpreted as a changing ħ.

What in the paper leads you to this interpretation? I don't see anything at all that would justify it. The paper is using "natural" units, in which $\hbar$ does not even appear in the equations; the closest thing to it would be the Planck mass $M_p$, and everything I can see in the paper indicates that $M_p$ is assumed to be constant, which would correspond to $\hbar$ being constant.

Note also that equation (2) in the paper, which is the standard QFT expression for the vacuum energy of a scalar field, does not have anything in it corresponding to an uncertainty principle.

 

[friend]

No, this has nothing to do with zero point energy. Trying to think about ZPE in terms of the uncertainty principle doesn't work when you look at the underlying math, however much it gets described that way in pop science sources.

So are you saying that the standard deviation of a constant inflation energy is zero, right? There are no fluctuations so there is no ΔE, right?

What in the paper leads you to this interpretation? I don't see anything at all that would justify it. The paper is using "natural" units, in which $\hbar$ does not even appear in the equations; the closest thing to it would be the Planck mass $M_p$, and everything I can see in the paper indicates that $M_p$ is assumed to be constant, which would correspond to $\hbar$ being constant.

It does not directly say that ħ changes, although it puts an equation where ħ is normally placed. This strongly suggests that what we call ħ is changing.

Section 1, second sentence reads in part "through deforming the usual Heisenberg
uncertainty principle." I took the word deforming to mean to change.

Also, equation 1 has terms such as <p> and <p²>. I assumed those expectation values would change with given circumstances.

Note that in equation 3 they use a similar equation where ħ normally is in the commutation relation, again indicating what we normally would think is $\hbar$ is being replaced with an equation.

Besides, if the uncertainty principle did not change values for various situations, then what is the point of the paper?

 

[PeterDonis]

So are you saying that the standard deviation of a constant inflation energy is zero, right? There are no fluctuations so there is no ΔE, right?

No; I'm saying something much more general. $\Delta E$ in the uncertainty relation you gave refers to the standard deviation of measurements of $E$ at different times when a quantum system is in a given state. It does not refer to the standard deviation of "fluctuations" in the underlying quantum state; there are no such fluctuations. That's true of quantum states in general, not just states of the inflaton field.

it puts an equation where ħ is normally placed. This strongly suggests that what we call ħ is changing.

What equation?

"through deforming the usual Heisenberg
uncertainty principle." I took the word deforming to mean to change.

It means changing the equation that we call the "uncertainty principle". It does not mean changing $\hbar$.

equation 1 has terms such as <p> and <p2>. I assumed those expectation values would change with given circumstances.

What does that have to do with $\hbar$ changing? Expectation values can change without $\hbar$ changing, if we are taking expectation values of different operators or operators acting on different states.

in equation 3 they use a similar equation where ħ normally is in the commutation relation, again indicating what we normally would think is $\hbar$ is being replaced with an equation.

No, one equation is being replaced with another. The $i$ outside the parentheses on the RHS of equation (3) would be $i \hbar$ in conventional units. The $1$ inside the parentheses is all that would be there in the standard commutation relation. Adding the extra terms inside the parentheses changes the commutation relation, but it doesn't change $\hbar$.

if the uncertainty principle did not change values for various situations, then what is the point of the paper?

On their hypothesis, the uncertainty principle does change values. It just doesn't do so because of any change in $\hbar$. It does so because extra terms involving $M_p$ are added to the commutation relations.

 

[friend]

No; I'm saying something much more general. $\Delta E$ in the uncertainty relation you gave refers to the standard deviation of measurements of $E$ at different times when a quantum system is in a given state. It does not refer to the standard deviation of "fluctuations" in the underlying quantum state; there are no such fluctuations. That's true of quantum states in general, not just states of the inflaton field.

As I recall, there is some description where the whole universe itself came into being because of a quantum mechanical fluctuation. They say that the enormous energy of the big bang could come into existence for only a very brief moment of time, etc. So I wonder, how long did inflation take place? How much did the inflation field energy change from false vacuum to true vacuum energy? Is this comparable to the ħ we normally use?

 

[PeterDonis]

As I recall, there is some description where the whole universe itself came into being because of a quantum mechanical fluctuation.

This is another pop science description. Please, please, take some time to look at the actual theories. A significant portion of this thread has been you saying things based on misconceptions due to pop science, and me having to correct them.

how long did inflation take place?

The usual number quoted for the time of the end of inflation is something like $10^{-35}$ second, but that's not really correct because it's using the time coordinate from a notional FRW spacetime model in which inflation does not take place and there is an "initial singularity". But nobody actually uses that model.

In terms of actual inflation models, this question doesn't really have a well-defined meaning. In at least one family of such models, the "eternal inflation" models, the obvious answer is "for an infinite amount of time", but that isn't really correct; a better answer would be that there is no such thing as "time" in the usual sense in a region which is in a "false vacuum" state; "time" only has meaning in the bubbles of "true vacuum".

How much did the inflation field energy change from false vacuum to true vacuum energy?

I'd have to look up the figures for energy density just after "reheating", but it's a large number.

Is this comparable to the ħ we normally use?

$\hbar$ doesn't have units of energy or energy density. If you mean, what do we get if we multiply the energy density of the inflaton field by the time inflation lasted (and then multiply by some volume, such as the volume of the observable universe, to get something with the right units), I would expect that the answer is a number much, much larger than $\hbar$. But I'm not sure that calculation has any physical meaning. For one thing, any volume you could pick is arbitrary. For another, the time, as I said above, isn't really well-defined.

 

[friend]

This is another pop science description. Please, please, take some time to look at the actual theories. A significant portion of this thread has been you saying things based on misconceptions due to pop science, and me having to correct them.

You do realize, don't you, that these ideas are coming from experts in the field being interviewed in documentaries or in debates. Why would they mention something this specific if there were no truth to it? Although, to be fair, IIRC, they qualify their statements as "possibilities".

I guess the question is whether this change from false vacuum energy to true vacuum energy in such a brief amount of time qualifies as a "fluctuation" for which it is legitimate to ask what its standard deviation would be. Wouldn't that be the full range of the change of energy? And wouldn't the standard deviation of the time be the time it took for this change to occur? Or would the actual fluctuation be larger because the energy may have stayed level for a while before changing very much faster?

 

[PeterDonis]

You do realize, don't you, that these ideas are coming from experts in the field being interviewed in documentaries or in debates.

Sorry, pop science is still pop science. Scientists will say things in interviews, documentaries, and debates that they would never say in a peer-reviewed paper--because they know that in a peer-reviewed paper they would get called on it if they were vague or imprecise, or used ordinary language that was likely to cause misunderstandings, or used analogies that were of limited usefulness, or failed to distinguish properly between known facts, reasonable probabilities, plausible hypotheses, and way-out speculations. This is why we have rules about sources here at PF.

Why would they mention something this specific if there were no truth to it?

It's not a matter of "truth". It's that the actual physics and math that they are referring to when they say, in some pop science venue, that "the universe came into being because of a quantum mechanical fluctuation" is nothing like what those words suggest to the average lay person. Again, if they tried this in a peer-reviewed paper, they wouldn't get away with it.

I guess the question is whether this change from false vacuum energy to true vacuum energy in such a brief amount of time qualifies as a "fluctuation" for which it is legitimate to ask what its standard deviation would be.

Once again, the concept of "standard deviation" applies to a set of measurements of a quantum system in a particular state--more precisely, to a set of measurements on an ensemble of quantum systems all prepared in the same state. It doesn't even apply in the context we're discussing, because no measurements are involved, and we don't have an ensemble of universes anyway; we only have one. We just have a single quantum system undergoing a deterministic change of state from "false vacuum" to "true vacuum". So the answer to your question is no.

 

[friend]

Once again, the concept of "standard deviation" applies to a set of measurements of a quantum system in a particular state--more precisely, to a set of measurements on an ensemble of quantum systems all prepared in the same state. It doesn't even apply in the context we're discussing, because no measurements are involved, and we don't have an ensemble of universes anyway; we only have one. We just have a single quantum system undergoing a deterministic change of state from "false vacuum" to "true vacuum". So the answer to your question is no.

You seem to be implying that the whole notion of false vacuum is not subject to theoretical development because it is not measurable. I think that's being a bit too dismissive. I think it's better to say we don't have enough information to develop theory for it yet.

 

[PeterDonis]

You seem to be implying that the whole notion of false vacuum is not subject to theoretical development because it is not measurable.

I'm saying no such thing. I'm only saying that the state transition at the end of inflation, from false vacuum to true vacuum, does not involve any measurement; it's just a quantum state transition. But there are certainly observable consequences of that transition, at least according to the various inflation models. So there are ways to test the models; they just don't involve direct measurements of the state transition (which would obviously be problematic).

I think it's better to say we don't have enough information to develop theory for it yet.

We are developing theories for it--that's what the various inflation models are. And, as above, those models do make predictions about observable quantities, so they can be tested.

 

[friend]

Once again, the concept of "standard deviation" applies to a set of measurements of a quantum system in a particular state--more precisely, to a set of measurements on an ensemble of quantum systems all prepared in the same state. It doesn't even apply in the context we're discussing, because no measurements are involved, and we don't have an ensemble of universes anyway; we only have one. We just have a single quantum system undergoing a deterministic change of state from "false vacuum" to "true vacuum". So the answer to your question is no.

What is an inflaton? How does that relate to the Heisenberg Uncertainty Principle?

 

[PeterDonis]

What is an inflaton?

"Inflaton" is the term for the scalar quantum field that drives inflation while it is in the "false vacuum" state.

How does that relate to the Heisenberg Uncertainty Principle?

It doesn't.

 

[Paulibus]

Fields are mathematical constructs, useful in physics, that assign scalar, vector, tensor quantities to each and every point in spacetime. The convention has developed in quantum physics of ending the names of the quanta of fields with "-on"; for example photon, electron and so also inflaton. Not to be confused with names like mine... Paul Jackson!

 

[friend]

"Inflaton" is the term for the scalar quantum field that drives inflation while it is in the "false vacuum" state.

It doesn't.

Your replies make it sound as if the inflaton field has NOTHING to do with quantum theory. Where is the quantum nature of the inflaton field taken into account as suggested by the quanta called "inflaton"?

As I recall, the CMB is a result of quantum fluctuations in the inflaton field. And these fluctuations eventually resulted in galaxies. Were these same quantum fluctuations that formed the CMB also present in the inflaton field during inflation? Can we get some idea how large these fluctuation were at various stages of inflation from information in the CMB? Thanks.

 

[PeterDonis]

Your replies make it sound as if the inflaton field has NOTHING to do with quantum theory.

Um, you did see that I said it was a quantum field, didn't you?

You seem to think that "quantum theory" is identical with "uncertainty principle". It isn't. There is a lot in quantum theory, and particularly in quantum field theory, that has nothing whatever to do with the uncertainty principle.

Where is the quantum nature of the inflaton field taken into account as suggested by the quanta called "inflaton"?

"Inflaton" doesn't have to refer to a "quantum" (i.e., particle). Not all quantum field states have useful particle interpretations. The states of the inflaton field that are used in inflationary cosmology models don't.

As I recall, the CMB is a result of quantum fluctuations in the inflaton field.

No. The CMB itself is a result of a large number of photons being emitted when the matter of the universe "recombined" into atoms (where before it had been a plasma, with electrons and ions). The small fluctuations in the CMB (more precisely, some of them, not all) are thought to be due to tiny fluctuations in whatever existed before inflation, that led to tiny fluctuations in the inflaton field at the start of inflation, that were magnified by inflation to a point where they could leave an "imprint" in the distribution of matter and radiation during "reheating", at the end of inflation. That imprint in turn left an imprint in the CMB.

The fluctuations before and during inflation are referred to as "quantum fluctuations" because they are fluctuations in the state of a quantum field from one point in spacetime to another. But that terminology is somewhat ambiguous. The SM fields that received a large energy density at "reheating" are also quantum fields, and the fluctuations in those fields after reheating, which in turn led to fluctuations in the CMB, are also fluctuations in the state of quantum fields from one point in spacetime to another. But they aren't usually referred to as "quantum fluctuations". So you have to be careful when interpreting ordinary language descriptions of what is going on. The only way to be sure is to look at the underlying math.

Can we get some idea how large these fluctuation were at various stages of inflation from what they were in the CMB?

Inflation is supposed to have magnified them by many, many orders of magnitude. The fluctuations we currently observe in the CMB are on the order of 1 part in $10^{5}$ or less, so the fluctuations that existed before inflation would have been many, many orders of magnitude less than that. I don't have a handy quick source for an estimate of how many orders of magnitude, though.

 

[friend]

Is it possible to expand the inflaton field in terms of plane waves? What would be the amplitude of these waves? Would these plane wave count as fluctuations? What would then be ΔE and Δt?

 

[PeterDonis]

Is it possible to expand the inflaton field in terms of plane waves?

How much do you know about perturbation theory and its limitations? That is what "expanding in terms of plane waves" refers to.

 

[friend]

How much do you know about perturbation theory and its limitations? That is what "expanding in terms of plane waves" refers to.

I've read about it years ago, but I haven't had a need to use it. So my understanding would be superficial in that regard. If you would like to write a few words in summary and perhaps flag some issues with respect to my concerns here, that would very helpful. Thank you.

 

[PeterDonis]

my understanding would be superficial in that regard.

Then I strongly suggest taking some time to learn more about it.

If you would like to write a few words in summary and perhaps flag some issues with respect to my concerns here, that would very helpful.

I pointed out before that you appear to think that "quantum" equals "uncertainty principle". It doesn't. There are many quantum phenomena which have nothing to do with the uncertainty principle. The behavior of the inflaton field in inflationary cosmology is one of them.

You appear to have a similar misconception with regard to "expanding in plane waves", namely that any quantum state can be so expanded. That's not correct. There are many quantum systems whose states can't be expanded that way, and the inflaton field as it is used in inflationary cosmology is one of them.

Also, even for cases where the relevant quantum states can be expanded in plane waves--for example, in particle scattering experiments--the amplitudes of the waves are quantum probability amplitudes, i.e., complex numbers whose absolute squares give the probabilities of various measurement results; they aren't amplitudes of anything "real" that is fluctuating.

This is a subject in which pop science sources, even when written by experts in the fields, can be very misleading to non-scientists. IMO it's worth taking the time to actually work your way through a good textbook on quantum mechanics and quantum field theory. Some of the frequent posters in the Quantum Physics forum can probably give good recommendations.