What's the energy-spread of the quantum Universe state?

[Robert Shaw]

If the universe was in an energy eigenstate then d<A>/dt = 0 for any dynamic variable A. Stuff moves which implies that the Universe isn't in an eigenstate. What factors drive the energy spread?

 

[kimbyd]

If the universe was in an energy eigenstate then d<A>/dt = 0 for any dynamic variable A. Stuff moves which implies that the Universe isn't in an eigenstate. What factors drive the energy spread?

I don't think this is an informative way to look at it. Rather, I think it makes more sense to consider the fact that the entropy of the universe is increasing over time (indicating a low-entropy state in the past). When entropy is maximized, the universe will no longer by dynamic. So the question becomes: how did that low-entropy state in the early universe occur? Unfortunately, we don't yet know the answer.

 

[Robert Shaw]

I don't think this is an informative way to look at it. Rather, I think it makes more sense to consider the fact that the entropy of the universe is increasing over time (indicating a low-entropy state in the past). When entropy is maximized, the universe will no longer by dynamic. So the question becomes: how did that low-entropy state in the early universe occur? Unfortunately, we don't yet know the answer.

Please would you suggest what law/equation guides the evolution of the quantum state of the Universe?

You've commented on entropy which is one property of the state but not the only property.

...what about the state itself, how does it develop?

 

[Robert Shaw]

If the universe was in an energy eigenstate then d<A>/dt = 0 for any dynamic variable A. Stuff moves which implies that the Universe isn't in an eigenstate. What factors drive the energy spread?

For toy universe models such as the simple harmonic oscillator, non-stationary state solutions exist which are superpositions of energy states.

What about bigger systems, the Universe for instance?

 

[PeroK]

If the universe was in an energy eigenstate then d<A>/dt = 0 for any dynamic variable A. Stuff moves which implies that the Universe isn't in an eigenstate. What factors drive the energy spread?

For toy universe models such as the neo-classical harmonic oscillator, non-stationary state solutions exist which are superpositions of energy states.

What about bigger systems, the Universe for instance?

When you talk about $\frac{d \langle A \rangle}{dt}$ you are dealing with the expected value of a dynamic variable $A$ for a large ensemble of identically prepared systems. This is not the value of a dynamic variable for a single system measured repeatedly over time.

For example, if you prepare a large ensemble of particles in a given state and measure the energy of the each particle at some time $t_0$, for example, then you will get an average value of the energy at time $t_0$. Let's call this $\langle E(t_0) \rangle$.

Then, you repeat the preparation process for another ensemble of particles and take measurements at some later time $t_1$, giving an average energy measurement $\langle E(t_1) \rangle$.

And so on.

By conservation of energy, you have $\langle E(t_0) \rangle = \langle E(t_1) \rangle = \langle E(t_2) \rangle \dots$ From which you may infer:

$\frac{d\langle E(t) \rangle}{dt} = 0$

If, additionally, every measurement of energy at time $t_0$ returns the same value $E$, then you know you have prepared the particle in an energy eigenstate corresponding to energy $E$.

In either case, energy eigenstate or superposition of energy eigenstates, the particle moves!

More importantly, given you cannot experimentally prepare an ensemble of universes, the statistical laws of QM do not simply and directly apply even to hypothetical measurements of the "energy of the universe".

In conclusion, I'd say your question is based on a fundamental misconception of QM.

 

[Robert Shaw]

Does QM apply to things that were not prepared in laboratories?

If so what does QM say about states not prepared in laboratories?

Does the preparation have to be carried out by a trained physicist?

 

[Robert Shaw]

Energy eigenstates "the averages and probabilities of all dynamical variables are independent of time" (Ballentine p72). Any QM textbook will say the same.

So |state>= a|E1> + b|E2> +... if any motion is to occur. It can be isolated so that <state|H|state>=constant yet the state is not an eigenstate.

For example a harmonic oscillator hamiltonian has non stationary solutions that superpose energy eigenstates but the total energy <H> is constant because KE and PE balance out.

On a larger scale, if stuff in the universe is non stationary then the overall state cannot be an eigenstate of total energy, it must be a superposition.

 

[PeroK]

Does QM apply to things that were not prepared in laboratories?

If so what does QM say about states not prepared in laboratories?

Does the preparation have to be carried out by a trained physicist?

QM applies to everything, essentially. The question is how you use QM to study and explain things. You can explain a hydrogen atom, whether or not it was created in a laborary or not.

 

[Robert Shaw]

QM applies to everything, essentially. The question is how you use QM to study and explain things. You can explain a hydrogen atom, whether or not it was created in a laborary or not.

In that case QM can be used to describe the universe. If <universe| H |universe> = constant but <A>≠constant then it follows |universe> is not an eigenstate of H but must be a superposition.

 

[PeroK]

Energy eigenstates "the averages and probabilities of all dynamical variables are independent of time" (Ballentine p72). Any QM textbook will say the same.

"Average and probabilities" imply statistics of a large ensemble of identically prepared systems.

The harmonic oscillator is a good example. In the ground state, or any energy eigenstate, the expected value of momentum $\langle p \rangle (t) = 0$. But, this does not mean that every measurement of momentum returns $p = 0$. Instead, you have:

$\langle p^2 \rangle (t) = \frac{m\hbar \omega}{2}$

Which gives an expected value for the kinetic energy of the particle:

$\langle T \rangle (t) = \frac{\hbar \omega}{4} \ \$ (Ground state)

The particle is moving, in the sense that it has a non-zero expected value of kinetic energy in this state.

The point is that the expected value of KE is constant over time. With a superposition of energy eigenstates, the expected value of KE changes over time.

Note that QM says very little about any single measurement of KE for a single harmonic oscillator. To make sense of these equations involved "expected values", you need to consider an ensemble of identical harmonic oscillators.

 

[PeroK]

In that case QM can be used to describe the universe. If <universe| H |universe> = constant but <A>≠constant then it follows |universe> is not an eigenstate of H but must be a superposition.

Yes, but how? You cannot generate an expected value of $A$ from one system. Even hypothetically, you only have one measurement. And, how do you measure the whole universe if you are part of it? What is this measurement of $A$ in the first place?

Note: these are very different questions from the core of QM. Be aware. You cannot learn QM, although many people seem to try, top down. Once you leave the microscopic arena, it becomes increasingly difficult to say what you are doing or measuring in QM terms.

My sincere advice is to learn QM as it applies to microscopic phenomena. At least at first.

 

[PeroK]

Yes, but how?

Let me give you an example. Suppose you hit a tennis ball. You can measure the velocity, say. That is not an expected value. That is a single velocity, $v$.

But, if you have a machine that fires out tennis balls, you can measure the velocity of each and talk about the expected value of the velocity $\langle v \rangle$.

In QM, when you use the quantity $\langle A \rangle$ you must, by definition, be talking about the second scenario. You must be talking about the expected value of a large number of identically prepared systems. In this case, an ensemble of tennis balls emerging from a machine.

In the case of the universe, therefore, you can talk about a quantity $A$. But, there is literally no way to talk about a quantity $\langle A \rangle$. For that, you would need a "universe machine" spitting out identically prepared universes. And that is at least one problem with trying to apply QM to the whole universe.

 

[Robert Shaw]

Let me give you an example. Suppose you hit a tennis ball. You can measure the velocity, say. That is not an expected value. That is a single velocity, $v$.

But, if you have a machine that fires out tennis balls, you can measure the velocity of each and talk about the expected value of the velocity $\langle v \rangle$.

In QM, when you use the quantity $\langle A \rangle$ you must, by definition, be talking about the second scenario. You must be talking about the expected value of a large number of identically prepared systems. In this case, an ensemble of tennis balls emerging from a machine.

In the case of the universe, therefore, you can talk about a quantity $A$. But, there is literally no way to talk about a quantity $\langle A \rangle$. For that, you would need a "universe machine" spitting out identically prepared universes. And that is at least one problem with trying to apply QM to the whole universe.

There is an extensive literature on Quantum Cosmology in which some authors say the state of the universe is a pure state (vector) and others say it is non-pure (density matrix). The universe has a state and the state is represented by a state function.

There is little support among Quantum foundations people for the ensemble interpretation (Albert, Deutsch, Kent, Saunders, Tegmark, Wallace for example). Even Ballentine the most ardent advocate of ensembles has doubts (see p34 of his book) especially regarding the"preparation" assumption.

Unless you think QM only applies to small things prepared in laboratories then the question about the state of the universe is valid. Clearly the universe does have a state of some kind. Solar systems have states. Planets too. It would be absurd to suggest that there is a scale above which QM is inaccurate...in what way would the inaccuracy be manifested?

For open systems it's not surprising the energy is not sharp. But for a closed system such as the universe it's an interesting question.

 

[Khashishi]

On a larger scale, if stuff in the universe is non stationary then the overall state cannot be an eigenstate of total energy, it must be a superposition.

Energy eigenstates are solutions to the time independent Schrodinger equation. Since the universe is growing, it isn't represented by a time independent Hamiltonian. There is not necessarily a conserved energy function.

Anything macroscopic with a finite temperature is not going to be in a pure state. If we want to treat the universe as a thermodynamic system (and we should), then we need to use a mixed state.

It would be absurd to suggest that there is a scale above which QM is inaccurate

We need a quantum theory of gravity before we can say that.

 

[PeroK]

There is an extensive literature on Quantum Cosmology in which some authors say the state of the universe is a pure state (vector) and others say it is non-pure (density matrix). The universe has a state and the state is represented by a state function.

Well, perhaps. But, given that we don't know how big the universe is, or whether it's infinite, that's a bold assertion. If the universe is infinite, then its total mass and total energy may be infinite. So, what might a measurement of the total energy of an infinite universe actually mean? It's not something you could ever measure and, even if hypothetically you could do it, you may not get a finite answer.

The expected values of all dynamic variables may, therefore, be infinite. And, mathematically, within the context of the linear algebra underpinnning QM, that would mean the universe as a whole is not a well-defined system.

What might be the measurable dynamic quantities of an infinite universe, in your opinion?

 

[PeroK]

Solar systems have states. Planets too. It would be absurd to suggest that there is a scale above which QM is inaccurate...in what way would the inaccuracy be manifested?

I think my point is this:

If you are learning QM to get an accurate model of the hydrogen atom, then fine.

If you are learning QM to get an accurate model of the solar system then you are wasting your time.

 

[Robert Shaw]

If the universe is isolated it can grow spatially but keep energy constant

 

[Robert Shaw]

Energy eigenstates are solutions to the time independent Schrodinger equation. Since the universe is growing, it isn't represented by a time independent Hamiltonian. There is not necessarily a conserved energy function.

Anything macroscopic with a finite temperature is not going to be in a pure state. If we want to treat the universe as a thermodynamic system (and we should), then we need to use a mixed state.We need a quantum theory of gravity before we can say that.

Most quantum cosmologists do not concur with your assertion that the universe as a whole is in a mixed (density matrix) state. The consensus is that it's in a pure state overall but the substates are mixed

 

[Khashishi]

Most quantum cosmologists do not concur with your assertion that the universe as a whole is in a mixed (density matrix) state. The consensus is that it's in a pure state overall but the substates are mixed

There's two kinds of mixed state. If two objects are entangled, a subsystem containing just one object is in a mixed state. You must be referring to this in your post. Perhaps everything in the universe is entangled, so any subsystem is necessarily mixed.

The other kind of mixed state simply represents our ignorance about the system. We don't know exactly which pure state the universe is in, so it's in a mixed state. Maybe in principle, we could follow every particle and trace out the trajectory of the whole universe. But since we can't actually do that, we rely on thermodynamics, which deals with average behavior of systems with incomplete information.

 

[kimbyd]

Please would you suggest what law/equation guides the evolution of the quantum state of the Universe?

That's just the quantum field equations. In the classical approximation, Schrodinger's equation.

You've commented on entropy which is one property of the state but not the only property.

...what about the state itself, how does it develop?

It's the property that is the most relevant to your statement. What you're asking, ultimately, is how the universe is in a state that is not equilibrium. And that's a question related to entropy.

Redefining your question in terms of entropy is a bit broader, but it serves the same point because the equilibrium state for our observable universe is empty space (so far as we can tell), which is an eigenstate of energy.

 

[PeterDonis]

Most quantum cosmologists do not concur with your assertion that the universe as a whole is in a mixed (density matrix) state. The consensus is that it's in a pure state overall but the substates are mixed

The "consensus" is not worth much if there is no experimental support. And as @PeroK has pointed out, we can't run repeated experiments on the universe to collect statistics and determine expectation values for observables that can be compared to theoretical predictions. So all the "consensus" of quantum cosmologists amounts to at this point is personal opinions. At some point we might be able to test some of the theoretical claims being made, but right now we can't.

 

[Robert Shaw]

The "consensus" is not worth much if there is no experimental support. And as @PeroK has pointed out, we can't run repeated experiments on the universe to collect statistics and determine expectation values for observables that can be compared to theoretical predictions. So all the "consensus" of quantum cosmologists amounts to at this point is personal opinions. At some point we might be able to test some of the theoretical claims being made, but right now we can't.

Rather than be defeatist why not start with a toy model and then scale it up?

Begin with an isolated toy universe which is a box of photons. It might be a coherent state (as studied by Glauber et al) and in that case the variance and average total energy are known precisely. Alternatively it might be a Fock state in which case the energy is sharp. This toy universe is amenable to your preparation and ensemble conditions.

Now let's scale the model up, all the time keeping it a closed system.

As we get bigger, how does the sharpness of energy change? If the stuff in our toy universe approximates to the stuff in the real universe, what if anything can we say about the total energy distribution.

 

[Robert Shaw]

Rather than be defeatist why not start with a toy model and then scale it up?

Begin with an isolated toy universe which is a box of photons. It might be a coherent state (as studied by Glauber et al) and in that case the variance and average total energy are known precisely. Alternatively it might be a Fock state in which case the energy is sharp. This toy universe is amenable to your preparation and ensemble conditions.

Now let's scale the model up, all the time keeping it a closed system.

As we get bigger, how does the sharpness of energy change? If the stuff in our toy universe approximates to the stuff in the real universe, what if anything can we say about the total energy distribution.

At its simplest the toy universe could be an isolated harmonic oscillator.

State prep A gives an energy eigenstate.

State prep B gives a coherent state which has a spectrum of total energies; measurement of its energy would result in different values. This is not because the prep method is in some way crude but a quantum phenomenon associated with superposition.

As we scale our procedure up, to real matter, say a cup of tea, what can we say about the spread of total energy of the whole state (I'm not asking about the spread of energy of the components).

 

[Robert Shaw]

At its simplest the toy universe could be an isolated harmonic oscillator.

State prep A gives an energy eigenstate.

State prep B gives a coherent state which has a spectrum of total energies; measurement of its energy would result in different values. This is not because the prep method is in some way crude but a quantum phenomenon associated with superposition.

As we scale our procedure up, to real matter, say a cup of tea, what can we say about the spread of total energy of the whole state (I'm not asking about the spread of energy of the components).

Remember the toy universe is perfectly isolated so there is no outside world with which it can decohere

 

[Robert Shaw]

Remember the toy universe is perfectly isolated so there is no outside world with which it can decohere

Remember, coherent states of the oscillator are pure states.

 

[bhobba]

Does QM apply to things that were not prepared in laboratories?

Of course it does - its any preparation procedure at all.

The real issue is applying it to the universe - what prepares the universe.

Deep question - but best to start another thread to discuss it.

In that case QM can be used to describe the universe.

That I think is debatable - but as said above best to start a new thread.

Thanks
Bill

 

[Robert Shaw]

That's just the quantum field equations. In the classical approximation, Schrodinger's equation.It's the property that is the most relevant to your statement. What you're asking, ultimately, is how the universe is in a state that is not equilibrium. And that's a question related to entropy.

Redefining your question in terms of entropy is a bit broader, but it serves the same point because the equilibrium state for our observable universe is empty space (so far as we can tell), which is an eigenstate of energy.

I can see no reason why the universe should be in a eigenstate of energy.

And the spread of energy in a quantum state has nothing to do with entropy.

 

[Robert Shaw]

Of course it does - its any preparation procedure at all.

The real issue is applying it to the universe - what prepares the universe.

Deep question - but best to start another thread to discuss it.
That I think is debatable - but as said above best to start a new thread.

Thanks
Bill

I can make no sense of your claim that any state "preparation procedure" is relevant.

What do you have in mind as laboratory independent state preparation procedures for:

- the Sun
- mount Everest
- your nose
- etc?

Do they obey the laws of quantum mechanics because you have a secret recipe for preparing their states?

If you have a secret recipe for preparing your own nose, then I'll give up this post.

 

[kimbyd]

I can see no reason why the universe should be in a eigenstate of energy.

And the spread of energy in a quantum state has nothing to do with entropy.

I am very confused as to what you are asking, then.

Your original question seemed to be asking why the universe isn't in an eigenstate, expressed as a statement that the universe changes over time and thus must not be in an eigenstate. I pointed out that the change of the universe over time is better-represented by examining entropy rather than energy.

But now you seem to be suggesting that the universe not being in an eigenstate is the natural state that doesn't need any explanation. So what are you asking?

 

[bhobba]

I can make no sense of your claim that any state "preparation procedure" is relevant.

Depending on interpretation state preparation procedure and state are generally considered synonymous.

What prepares the state of the universe?

A state preparation procedure is something like passing an electron through a slit. We know behind the slit it is in a state of definite position - that is a preparation procedure - you need to interact a quantum system with something to prepare it - what do you interact the universe with?

Here is a state preparation that occurs in nature. It is well known that a few stray photons from the cosmic microwave background radiation is enough to give a dust particle a define position.

Yes even your nose is subject to a preparation procedure - it is constantly interacting with its environment which, without going into technical details, is why your nose acts classically ie is a classical object and not a quantum one - of course like everything is made of quantum stuff, but due to that interaction can be considered classical - like just about everything around us.

But please, as previously requested, if you want to discuss that start a new thread.

Thanks
Bill

 

[Robert Shaw]

I am very confused as to what you are asking, then.

Your original question seemed to be asking why the universe isn't in an eigenstate, expressed as a statement that the universe changes over time and thus must not be in an eigenstate. I pointed out that the change of the universe over time is better-represented by examining entropy rather than energy.

But now you seem to be suggesting that the universe not being in an eigenstate is the natural state that doesn't need any explanation. So what are you asking?

For any state to manifest motion, there must be a mixture of energy eigenstates (energy eigenstates are by definition stationary).

From this it follows that the state of the Universe is a superposition of energy states. I'd like to restrict the discussion by assuming the total state begins in a pure state.

So the interesting question is whether the width of the energy variance has any physical significance.

 

[Robert Shaw]

Depending on interpretation state preparation procedure and state are generally considered synonymous.

What prepares the state of the universe?

A state preparation procedure is something like passing an electron through a slit. We know behind the slit it is in a state of definite position - that is a preparation procedure - you need to interact a quantum system with something to prepare it - what do you interact the universe with?

Here is a state preparation that occurs in nature. It is well known that a few stray photons from the cosmic microwave background radiation is enough to give a dust particle a define position.

Yes even your nose is subject to a preparation procedure - it is constantly interacting with its environment which, without going into technical details, is why your nose acts classically ie is a classical object and not a quantum one - of course like everything is made of quantum stuff, but due to that interaction can be considered classical - like just about everything around us.

But please, as previously requested, if you want to discuss that start a new thread.

Thanks
Bill

Happy to start a new thread on state preparation procedures

 

[PeroK]

For any state to manifest motion, there must be a mixture of energy eigenstates (energy eigenstates are by definition stationary).
.

In your desire to apply the laws of QM to macroscopic objects, you are effectively ignoring the basis of those QM laws.

For example, you are part of the universe. When you observe something move, you are not making a measurement of the universe that corresponds to the QM definition of a measurement.

You also make the mistake of assuming QM terminology maps directly to classic terminology, e.g "stationary".

In your terms, I could say. My car is currently sitting outside"stationary". It is therefore in a stationary state, i.e. an energy eigenstate.

But, when I'm driving my car it is "moving", hence in a superposition of energy eigenstates.

Which is nonsense and has nothing to do with stationary states in QM.

Most of your posts involve this imprecise confusion of quantum and classical concepts and terminology.

 

[Robert Shaw]

For any state to manifest motion, there must be a mixture of energy eigenstates (energy eigenstates are by definition stationary).

From this it follows that the state of the Universe is a superposition of energy states. I'd like to restrict the discussion by assuming the total state begins in a pure state.

So the interesting question is whether the width of the energy variance has any physical significance.

To be clear I am making the simplifying assumptions:

1) Universe is a closed quantum system
2) there is motion in the universe

From these it follows that the state of the Universe cannot be an energy eigenstate.

My question is whether the spread of the mixture of energy levels has physical significance.

 

[PeroK]

To be clear I am making the simplifying assumptions:

1) Universe is a closed quantum system
2) there is motion in the universe

From these it follows that the state of the Universe cannot be an energy eigenstate.

My question is whether the spread of the mixture of energy levels has physical significance.

If you take the hydrogen atom in an energy eigenstate, that does not mean that the electron"is not moving". It means the atom has a defined energy: every measurement of energy returns the same value. The electron, when measured, has a non-zero kinetic energy.

If you try to extend the concept of motion - in the model of the solar system, for example - to an atom, the numbers and equations simply do not make sense. You could argue that at the quantum level the concept of motion no longer makes sense.

Motion at the macroscopic level is fundamentally a concept that arises when the quantum mechanical behaviour of a large number of particles is statistically amalgamated.

If you are studying QM and you are thinking in terms of classical motion, you are off on the wrong foot altogether.

 

[Robert Shaw]

If you take the hydrogen atom in an energy eigenstate, that does not mean that the electron"is not moving". It means the atom has a defined energy: every measurement of energy returns the same value. The electron, when measured, has a non-zero kinetic energy.

If you try to extend the concept of motion - in the model of the solar system, for example - to an atom, the numbers and equations simply do not make sense. You could argue that at the quantum level the concept of motion no longer makes sense.

Motion at the macroscopic level is fundamentally a concept that arises when the quantum mechanical behaviour of a large number of particles is statistically amalgamated.

If you are studying QM and you are thinking in terms of classical motion, you are off on the wrong foot altogether.

By motion I mean that for some observable operator A then d<A>/dt is non zero.

For an energy eigenstate, all observables have d<A>/dt=0. That is what I mean by no motion.

Clearly the state of the Universe has d<A>/dt not =0. Hence it is not in an energy eigenstate.

 

[Robert Shaw]

If you take the hydrogen atom in an energy eigenstate, that does not mean that the electron"is not moving". It means the atom has a defined energy: every measurement of energy returns the same value. The electron, when measured, has a non-zero kinetic energy.

If you try to extend the concept of motion - in the model of the solar system, for example - to an atom, the numbers and equations simply do not make sense. You could argue that at the quantum level the concept of motion no longer makes sense.

Motion at the macroscopic level is fundamentally a concept that arises when the quantum mechanical behaviour of a large number of particles is statistically amalgamated.

If you are studying QM and you are thinking in terms of classical motion, you are off on the wrong foot altogether.

In your example of the hydrogen energy eigenstate, for all observables A,B,C etc

d<A>/dt=0, etc.

Nothing is in motion

 

[bhobba]

1) Universe is a closed quantum system
2) there is motion in the universe

I can accept 1 - but it may or may not be true - eg see the latest theories of continuous inflation.

I think what you mean by 2 needs clarification. Assuming you can give a meaning to the quantum state of the universe (many into quantum cosmology adhere to many worlds for this reason - you can do it in that interpretation) I am not sure what you mean by motion. If you mean Schrodinger's equation applies - sure - but beyond that I am a bit perplexed.

Thanks
Bill

 

[PeroK]

By motion I mean that for some observable operator A then d<A>/dt is non zero.

For an energy eigenstate, all observables have d<A>/dt=0. That is what I mean by no motion.

Clearly the state of the Universe has d<A>/dt not =0. Hence it is not in an energy eigenstate.

This is where we came in! You're still using the expected value - a statistical value - and confusing it with a single measurement.

There is nothing clear about your claims - they are simply a mixture of classical and quantum thinking and a fundamental confusion over the meaning of $\langle A \rangle$.

The only thing that is clear is that there is no convincing you.

 

[PeroK]

In your example of the hydrogen energy eigenstate, for all observables A,B,C etc

d<A>/dt=0, etc.

Nothing is in motion

And, yet, if you measure the electron it has a non-zero kinetic energy. How can something that is not moving have non-zero Kinetic Energy and non-zero Angular Momentum?

The fact that those values don't change over time has more of an analogy with the electron is not accelerating!

 

[bhobba]

This is where we came in! You're still using the expected value - a statistical value - and confusing it with a single measurement.

I don't even get that - yes given a state and an observable you can predict an average - but without actual measurement, observation etc, its hard to understand its significance.

Hopefully he is referring to Schrodinger's equation which if you have an interpretation like Many Worlds is perfectly OK to discuss, and generally that's what those into quantum cosmology seem to favor. But we have more recent theories such as continuous inflation where perhaps you don't need MW, but this is not my area of expertise and I would like someone more familiar with it than me to comment.

Thanks
Bill

 

[PeroK]

I don't even get that - yes given a state and an observable you can predict an average - but without actual measurement, observation etc, its hard to understand its significance.

Hopefully he is referring to Schrodinger's equation which if you have an interpretation like Many Worlds is perfectly OK to discuss, and generally that's what those into quantum cosmology seem to favor. But we have more recent theories such as continuous inflation where perhaps you don't need MW, but this is not my area of expertise and I would like someone more familiar with it than me to comment.

Thanks
Bill

MWI or not, if you measure the kinetic energy of the moon at different times, then that gives you $\frac{dA}{dt}$. It does not give you $\frac{d\langle A \rangle}{dt}$.

The OP simply adds the expectation notation to mistakenly align classical thinking and calculations with QM.

Take some classical observable, add some angled brackets and you have QM. That's about the extent of it.

 

[bhobba]

MWI or not, if you measure the kinetic energy of the moon at different times, then that gives you $\frac{dA}{dt}$. It does not give you $\frac{d\langle A \rangle}{dt}$.

Would you consider the moon a wavepacket of very small variance?

In practice of course you are correct - our current technology does not allow us to detect quantum effects with an object the size of the moon - no issue with that at all - but as a matter of principle I am not quite so sure.

Thanks
Bill

 

[PeroK]

Would you consider the moon a wavepacket of very small variance?

In practice of course you are correct - our current technology does not allow us to detect quantum effects with an object the size of the moon - no issue with that at all - but as a matter of principle I am not quite so sure.

Thanks
Bill

So, what's your definition of $\langle A \rangle$?

Are you also defining this as a single measurement?

Or the expected value of a set of measurements taken on a large number of identically prepared systems. Which is my definition.

 

[bhobba]

So, what's your definition of $\langle A \rangle$? Are you also defining this as a single measurement? Or the expected value of a set of measurements taken on a large number of identically prepared systems. Which is my definition.

<A> is the average of a number of measurements so large for all practical purposes it can be considered infinite.

However for the moon our measurement technology is not good enough to have any difference between measurements.

But I now see your point - you were not referring to A as an observable - which is what I thought you meant - you are referring to the actual measurement itself in which case your point is trivially true.

Thanks
Bill

 

[Robert Shaw]

And, yet, if you measure the electron it has a non-zero kinetic energy. How can something that is not moving have non-zero Kinetic Energy and non-zero Angular Momentum?

The fact that those values don't change over time has more of an analogy with the electron is not accelerating!

Returning to my original question:

Please would you clarify do you think a closed Universe is in an energy eigenstate?

To reduce the problem to something simple and analytically solvable, let's treat the Universe as a two qubit closed system where there is an interaction. For any partial measurement to be non-static (e.g. the energy of particle 1) the total state cannot be in a single energy eigenstate.

Single observations are out of scope as QM can say nothing whatever about them.

 

[Robert Shaw]

Would you consider the moon a wavepacket of very small variance?

In practice of course you are correct - our current technology does not allow us to detect quantum effects with an object the size of the moon - no issue with that at all - but as a matter of principle I am not quite so sure.

Thanks
Bill

My question is about how small is the variance.

While you may perhaps be right that it very small, I'm interested to understand the scale of "very small" and what factors contribute to its size.

 

[PeterDonis]

let's treat the Universe as a two qubit closed system where there is an interaction. For any partial measurement to be non-static (e.g. the energy of particle 1) the total state cannot be in a single energy eigenstate.

This is false. Write the total Hamiltonian schematically as $H = H_1 + H_2 + H_i$, where $H_1$ is the free particle Hamiltonian of particle 1, $H_2$ is the free particle Hamiltonian of particle 2, and $H_i$ is the interaction Hamiltonian. If the system is closed, it will be in an eigenstate of $H$ (but see below for a caveat to this), but it will not, in general, be in an eigenstate of $H_1$ or $H_2$. That means that measurements of the energies of the individual particles can give changing values, because those are measurements of $H_1$ or $H_2$. (Heuristically, this is because the interaction creates a potential energy between the two particles, whose value is not contained in measurements of the energies of the individual particles.) But a measurement of $H$ will always give the same value.

Of course in this simple closed system, we can imagine measuring $H$, because we know the real universe consists of a lot more than a closed two-qubit system. But if that closed two-qubit system were the entire universe, there would be no way to measure $H$, because that would require the system to interact with something outside it. So there would be no way to test the claim that the system as a whole was, or was not, in an eigenstate of $H$. However, if you try, you will see that it is impossible to write down a consistent state for the closed two-qubit system that is not an eigenstate of $H$. Heuristically, this is because the system is closed and can't interact with anything else, so there is nowhere for it to gain or lose energy to. Any real system that is not in an energy eigenstate is not closed; it is interacting with something else (for example, an electron in an atom interacts with the electromagnetic field, which is what allows it to change energy levels).

 

[kimbyd]

For any state to manifest motion, there must be a mixture of energy eigenstates (energy eigenstates are by definition stationary).

From this it follows that the state of the Universe is a superposition of energy states. I'd like to restrict the discussion by assuming the total state begins in a pure state.

So the interesting question is whether the width of the energy variance has any physical significance.

This is an answer that might be solvable for the very early universe, but it's also trivial and not very interesting (at least for simpler models):
The very early universe was in thermal equilibrium, which would indicate that the spread of energy eigenstates would have been drawn from the spread determined by the sum of energy of individual particles which follow the thermal distribution.

After stars and galaxies started to form, however, the question becomes much harder to answer. With matter no longer in thermal equilibrium, there aren't any simple calculations that can be done, and quantum gravity might have some influence on the result anyway.

As for the spread of energy states having any physical significance, I don't see how.

 

[Robert Shaw]

This is false. Write the total Hamiltonian schematically as $H = H_1 + H_2 + H_i$, where $H_1$ is the free particle Hamiltonian of particle 1, $H_2$ is the free particle Hamiltonian of particle 2, and $H_i$ is the interaction Hamiltonian. If the system is closed, it will be in an eigenstate of $H$ (but see below for a caveat to this), but it will not, in general, be in an eigenstate of $H_1$ or $H_2$. That means that measurements of the energies of the individual particles can give changing values, because those are measurements of $H_1$ or $H_2$. (Heuristically, this is because the interaction creates a potential energy between the two particles, whose value is not contained in measurements of the energies of the individual particles.) But a measurement of $H$ will always give the same value.

Of course in this simple closed system, we can imagine measuring $H$, because we know the real universe consists of a lot more than a closed two-qubit system. But if that closed two-qubit system were the entire universe, there would be no way to measure $H$, because that would require the system to interact with something outside it. So there would be no way to test the claim that the system as a whole was, or was not, in an eigenstate of $H$. However, if you try, you will see that it is impossible to write down a consistent state for the closed two-qubit system that is not an eigenstate of $H$. Heuristically, this is because the system is closed and can't interact with anything else, so there is nowhere for it to gain or lose energy to. Any real system that is not in an energy eigenstate is not closed; it is interacting with something else (for example, an electron in an atom interacts with the electromagnetic field, which is what allows it to change energy levels).

Most states can all have interactions internally. All that's required is a Hamiltonian with off diagonal terms (and Hermitian). A simple example is a precessing qubit state. It isn't in a single energy state which is why it precesses.

The the two qubit case is interesting because solutions can be found where the energy oscillates between the two internal components with no energy flowing elsewhere.

An isolated harmonic oscillator has solutions that superpositions of energy levels (most textbooks cover this case).

An isolated atom can be in a superposition of energy states. It just sits in this superposed state, neither gaining nor losing energy.

For an isolated atom with spin-orbit coupling there are precessing solutions that are not energy eigenstates (this case can be found in many textbooks).

More complex versions crop up in quantum optics in connection with Glauber states and they certainly can be closed states.

So it is not correct to say "Any real system that is not in an energy eigenstate is not closed"