I Inflation and the false vacuum

[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.