Inflation and the false vacuum

[PeterDonis]

What book would you recommend for details?

If you ask a dozen people who are familiar with QFT that question, you will probably get at least thirteen answers. I find Anthony Zee's Quantum Field Theory In A Nutshell to be a fairly good source, but it presumes that you have a good understanding of ordinary quantum mechanics. Weinberg's classic Quantum Theory Of Fields, in three volumes, seems to me to be kind of the equivalent of Misner, Thorne, & Wheeler in GR: a comprehensive reference, but not necessarily the best introduction to the subject (although at the beginning of his first volume, Weinberg gives an excellent survey of the history of the development of QFT).

 

[friend]

Have we been able, then, to explain inflation in terms of micro-physics or quantum physics? I thought that was the whole point of the inflaton as a quantum field. It seems obvious that global effects must also be explainable in microscopic terms because macro-properties are ONLY explainable as an accumulative effect of micro physics. How, then, does the inflaton cause inflation at the micro scale? Is it possible, for example, that virtual pairs splitting apart and coming together requires more space than if that did not happen?

 

[PeterDonis]

Have we been able, then, to explain inflation in terms of micro-physics or quantum physics?

The current explanation (which, btw, is, AFAIK, still speculative, since we don't really have any way of testing it against evidence) involves a quantum scalar field (the inflaton field), but quantum fluctuations in the field do not play a part in the explanation of why inflation happens; only the field's vacuum expectation value (i.e., its average value in its vacuum state) does. So the explanation is really a classical one; it only makes use of the stress-energy tensor of a classical scalar field, which has an equation of state that causes accelerating expansion of the universe.

(Note that, at least as I understand the model, quantum fluctuations in the field do play a part in explaining how inflation ends; quantum fluctuations are what trigger the phase transition that transfers the energy in the inflaton field to ordinary matter and radiation. But this is different from explaining how the inflaton field causes accelerated expansion before the phase transition occurs.)

Is it possible, for example, that virtual pairs splitting apart and coming together requires more space than if that did not happen?

No; at least, virtual particles/quantum fluctuations play no part in the current explanation of how the inflaton field causes inflation. See above.

 

[friend]

So the explanation is really a classical one; it only makes use of the stress-energy tensor of a classical scalar field, which has an equation of state that causes accelerating expansion of the universe.

Thanks for the reply. Your use of the term equation of state tells me that you are explaining things in terms of global variables, and not in terms of what's happening at the micro level to create or stretch out space so that the overall effect is inflation. Perhaps this is where quantum gravity comes in. Although I'm not sure it's right to talk of gravity during inflation.

 

[PeterDonis]

Your use of the term equation of state tells me that you are explaining things in terms of global variables, and not in terms of what's happening at the micro level to create or stretch out space so that the overall effect is inflation.

The equation of state relates pressure and energy density. Those are not global variables; they are local variables. They lead to inflation via the Einstein Field Equation, which is also local.

Perhaps this is where quantum gravity comes in.

If by the "micro level" you mean something underlying spacetime itself, then yes, this would involve some kind of quantum gravity theory.

 

[friend]

The equation of state relates pressure and energy density. Those are not global variables; they are local variables. They lead to inflation via the Einstein Field Equation, which is also local.

? Pressure is a "local" variable? I thought that was a description of an accumulative effect, an average over many particles.

 

[PeterDonis]

Pressure is a "local" variable? I thought that was a description of an accumulative effect, an average over many particles.

"Local" just means "has a value at each point of spacetime". I think the word you are looking for is "microscopic" as opposed to "macroscopic". Pressure is a macroscopic variable, as is energy density. In the ordinary case of fluid matter, statistical mechanics shows how pressure arises from the microscopic motions of particles (and in at least some idealized cases, the equation of state relating pressure and energy density can also be derived in this way). But not all pressure arises in this way; only what is called "kinetic pressure" does. The pressure due to dark energy is not kinetic pressure; neither is the pressure that appears in the equation of state for a scalar field. It is possible that those kinds of pressure are indeed explainable as arising from statistical mechanics over some set of microscopic degrees of freedom; but if so, nobody currently knows what those microscopic degrees of freedom are.

 

[nikkkom]

How, then, does the inflaton cause inflation at the micro scale? Is it possible, for example, that virtual pairs splitting apart and coming together requires more space than if that did not happen?

My understanding is that in any vacuum, virtual pairs production/annihilation can cause vacuum to have nonzero "zero-point energy", and this translates to nonzero lambda. (My understanding is that currently SM has a problem here: its predicted zero-point energy is way too high, and physicists expect that future developments in (B)SM will fix this: new/fixed theory will predict a tiny nonzero zero-point energy for today's vacuum.)

Inflaton field per se is not required to cause inflation. Inflation happens if vacuum is in a state with large zero-point energy. *Any* mechanism which achieves such result will explain inflation. Not only a mechanism via a new field.

I already posted a link to one of proposed theories, and it happens to _not_ use any inflaton fields:

http://arxiv.org/abs/1307.1848

 

[friend]

My understanding is that in any vacuum, virtual pairs production/annihilation can cause vacuum to have nonzero "zero-point energy", and this translates to nonzero lambda. (My understanding is that currently SM has a problem here: its predicted zero-point energy is way too high, and physicists expect that future developments in (B)SM will fix this: new/fixed theory will predict a tiny nonzero zero-point energy for today's vacuum.)

PeterDosis' point seems to be that these virtual particles are fluctuations necessarily about an average. In the case of the SM (not the higgs or inflaton), this average is zero. In the case of the inflaton field the average is not zero. However, if virtual particles are not responsible for accelerated expansion, then I do not see how a constant average value can be responsible for any kind of dynamics. How can a constant anything be involved with the moving parts of any dynamic process. If nothing is moving or changing (like with virtual particles) then how can things change? What dynamics can there be with a constant? That seems like a contradiction of terms.

In order to answer this question I think somehow quantum processes needs be linked to spacetime itself (and not just have quantum processes occurring in spacetime).

 

[Jimster41]

This has been a great thread to follow and think about, w/respect to something I'm reading, and I've learned from it. It's my default to ask questions in context, rather than start another thread that buries active ones, but if it feels off topic feel free to move itand let me know.

Some questions:

1. Are vacuum scalar fields "observable", or are particles (excited states of those fields) the only observables? I'm confused about whether or not the fundamental fabric of reality is quanta, or something else. If they are considered real what is it scalar fields are thought to be made of?
2. What is the scalar field responsible for the a(t) in the FRLW solution(s) to GR? (this conversation has helped my disambiguate it from the cosmological constant). Since it is a solution to GR field eq. I gather relationships between matter and energy density are considered the "causes" of expansion? And that the "Inflation field" discussed here is really sort of the "first cause" setting the boundary conditions of the FRLW in a way that requires a(t)?

 

[PeterDonis]

My understanding is that in any vacuum, virtual pairs production/annihilation can cause vacuum to have nonzero "zero-point energy", and this translates to nonzero lambda.

First of all, "virtual pairs" is a heuristic description only. A better term would be "quantum fluctuations", and an even better term would be "an extra term in the Hamiltonian that is nonzero in the vacuum state". In other words, the nonzero energy of the vacuum state, in this "naive" theoretical view, is due to the fact that the energy operator, the Hamiltonian, has a term that is independent of the value of the field.

However, as has already been noted in this thread, the "naive" theoretical view gives the answer "infinity" to the question of how much energy is in the vacuum. Even if we adjust the calculation to have a cutoff at some finite frequency, we still get an answer that is more than 120 orders of magnitude larger than the actual observed value of $\Lambda$. So the best answer of all, as the Baez article linked to says, is that we do not know why $\Lambda$ has the value it actually has.

these virtual particles are fluctuations necessarily about an average. In the case of the SM (not the higgs or inflaton), this average is zero. In the case of the inflaton field the average is not zero.

Yes, although, as noted above, "virtual particles" is not the term I would use to describe what is going on. Also, it's worth noting, once again, that the average value of the field in the vacuum state is different from the expectation value of the energy (i.e., of the Hamiltonian) in the vacuum state; the latter can be nonzero even if the former is zero.

if virtual particles are not responsible for accelerated expansion, then I do not see how a constant average value can be responsible for any kind of dynamics.

Why not? I don't understand this at all. The constant average value of any field doesn't just sit there. It has physical effects; for example, the constant nonzero vacuum expectation value of the Higgs field gives mass to the other SM particles.

 

[friend]

I'm wondering if I can re-focus this thread. It appears we may not have really addressed the relationship between inflation and the false vacuum. So let me ask some basic questions. What is the vacuum they are referring to when they say false vacuum? Is this the vacuum energy? Is this the zero point energy of all the SM quantum fields? Does the tunneling to a lower value cause inflation? Or does the lowering of this vacuum energy cause a bubble nucleation of present physics within the larger universe of eternal inflation? And of particular interest, if this vacuum energy is a lowering of the zero point energy of all SM quantum fields, then what quantum properties need to change to cause a lowering of the zero point energy? I thought that was fixed by the Heisenberg Uncertainty principle. Thanks.

 

[PeterDonis]

What is the vacuum they are referring to when they say false vacuum?

In the simplest model of inflation, the inflaton field has two possible states with two different vacuum expectation values: one is nonzero and the other is zero. The "false vacuum" is the state in which the VEV is nonzero; the nonzero VEV causes exponential expansion, i.e., inflation. The "true vacuum" is the state in which the VEV is zero; the end of inflation is a state transition of the inflaton field from the false vacuum state to the true vacuum state (and an associated transfer of energy from the inflaton field to the Standard Model fields).

The inflation models actually under consideration have various complications added, but their basic structure is still the same as the above.

Does the tunneling to a lower value cause inflation?

No; the transition of the inflation field from the false vacuum to the true vacuum (which can be thought of as "tunneling" although that is not the only possible model), as I said above, is the end of inflation. The nonzero VEV of the inflaton field in the false vacuum state is what causes inflation.

Or does the lowering of this vacuum energy cause a bubble nucleation of present physics within the larger universe of eternal inflation?

This is one model being considered, yes.

if this vacuum energy is a lowering of the zero point energy of all SM quantum fields

It isn't. The zero point energy of all the SM fields is constant; it doesn't change.

 

[friend]

Thanks for your answer, PeterDonis.

Is this inflationary lowering of the vacuum of a different quality than that posed to possibly happen in the future? There they also talk of a bubble nucleation with a fall from the presently supposed false vacuum to a lower vacuum energy. I don't know what field energies could fall even lower than today?

 

[PeterDonis]

Is this inflationary lowering of the vacuum of a different quality than that posed to possibly happen in the future?

It's the same general mechanism.

I don't know what field energies could fall even lower than today?

The hypothesis, as I understand it, is that what we currently observe as dark energy--i.e., 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. If that is true, then it should be possible in principle for that field to transition to a true vacuum state with a zero VEV, and we would expect that process to, at the very least, transfer the energy to some other field, and possibly also change the observed laws of physics. This is all speculative, and I'm not aware of any data, other than the existence of dark energy itself, that supports such a hypothesis. It's basically the idea that, if it happened once, perhaps it can happen again.

(Actually, the Higgs field would be another possibility, since it is a scalar field. This is at least open to some investigation, since we can measure the mass of the Higgs boson and try to figure out theoretically what range of masses would correspond to a false vacuum state. AFAIK nothing along these lines so far has indicated that the observed Higgs mass is in an unstable range.)

 

[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