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

[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"

 

[PeterDonis]

A simple example is a precessing qubit state. It isn't in a single energy state which is why it precesses.

It precesses because it's in an external field. In other words, the full state is not just the state of the qubit, but the state of the qubit and the state of the field. And the full Hamiltonian is not just the Hamiltonian of the qubit, but the Hamiltonian of the qubit plus the Hamiltonian of the field plus the interaction between them. The qubit's state is not an eigenstate of the qubit-only Hamiltonian, but the full state of the system (qubit and field) is an eigenstate of the full Hamiltonian. And it's the latter that is analogous to the state of the universe as a whole, since the universe as a whole has nothing else outside it to interact with.

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.

Do you have a reference?

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

A harmonic oscillator does. An isolated harmonic oscillator doesn't, at least not in any meaningful sense; you can mathematically write down a superposition of its energy eigenstates, but energy will not be conserved in such a system, so it's not physically possible.

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

No, it doesn't. A superposition of energy eigenstates will change with time, and so will the probabilities of measuring it to be in different energy eigenstates if its energy is measured. Also see above.

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).

They are not eigenstates of the Hamiltonian for just the electron. They are eigenstates of the full Hamiltonian of the entire atom, including the interaction between the electron and the nucleus. See above. (Technically, the full Hamiltonian would include the electromagnetic field, since that's what actually mediates the interaction between the electron and the nucleus; but for most purposes the approximation in which the Coulomb potential of the nucleus stands in for the EM field works fine.)

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

I'm not sure what you mean. A quantum optics system is never "closed" in the sense of being isolated; there are all kinds of devices in the system (beam splitters, etc.) whose states are not captured in the quantum state of the qubits.

 

[kimbyd]

A harmonic oscillator does. An isolated harmonic oscillator doesn't, at least not in any meaningful sense; you can mathematically write down a superposition of its energy eigenstates, but energy will not be conserved in such a system, so it's not physically possible.

I don't know if that's a valid conclusion.

There is no definite energy for a harmonic energy in a superposition of states. The expectation value of the energy is well-defined and changes over time, but the expectation value of energy isn't a conserved quantity. The change over time of the expectation value of energy will obey the time-energy uncertainty relation.

 

[PeterDonis]

There is no definite energy for a harmonic energy in a superposition of states.

Yes, I agree. Mathematically, that's what the expression for the state and operating on it with the Hamiltonian says. My point is not that you can't write down the math. My point is that the math doesn't describe anything that is physically possible; a fully isolated harmonic oscillator can't exchange energy with anything, so energy would not be conserved if it were not in an energy eigenstate. I don't think you can get around energy conservation just by saying that the energy of a non-eigenstate isn't well-defined, if that state represents an entire isolated system. In cases where an actual physical system is in a superposition of energy eigenstates, it's because it isn't isolated; there is an interaction with something else.

 

[Robert Shaw]

Yes, I agree. Mathematically, that's what the expression for the state and operating on it with the Hamiltonian says. My point is not that you can't write down the math. My point is that the math doesn't describe anything that is physically possible; a fully isolated harmonic oscillator can't exchange energy with anything, so energy would not be conserved if it were not in an energy eigenstate. I don't think you can get around energy conservation just by saying that the energy of a non-eigenstate isn't well-defined, if that state represents an entire isolated system. In cases where an actual physical system is in a superposition of energy eigenstates, it's because it isn't isolated; there is an interaction with something else.

The state only has to obey:

ih d|state>/dt = H |state> where H is the Hamiltonian for a closed state.

This does not imply that the only solution is an energy eigenstate. Superpositions are also solutions.

For example a wave packet is a multi-energy superposition of plane-wave energy eigenstates. It obeys the above equation.

If you believe otherwise, please would you show how the equation above leads to your conclusion. Begin by showing for wave packets.

 

[PeterDonis]

Superpositions are also solutions.

Mathematically, yes, as I have already said in response to @kimbyd. Math is not the same as physics. Physically, I do not think these solutions are reasonable for closed systems, for reasons I have already explained.

 

[kimbyd]

Mathematically, yes, as I have already said in response to @kimbyd. Math is not the same as physics. Physically, I do not think these solutions are reasonable for closed systems, for reasons I have already explained.

I think you're mixing up classical and quantum concepts.

Ignoring the complexities introduced by gravity for a moment, a classical, closed system will have $dE/dt = 0$. But quantum-mechanically, the value $E$ does not exist. You have energy eigenvalues, and an expectation value. The expectation value is not a physical quantity, and cannot be conserved. I'm pretty sure it's possible to prove that the expectation value of energy must be periodic, but there is no "E" to be conserved in the first place.

 

[PeterDonis]

there is no "E" to be conserved in the first place.

There is if the entire closed system is in an eigenstate of the full Hamiltonian; then $E$ is just the eigenvalue. And my physical (not mathematical) argument is that only such states are reasonable for modeling entire closed systems like the universe. Yes, mathematically you can write down a state for the full closed system that is not an eigenstate of the full Hamiltonian; but does that make sense physically? I am arguing that it doesn't.

 

[kimbyd]

There is if the entire closed system is in an eigenstate of the full Hamiltonian; then $E$ is just the eigenvalue. And my physical (not mathematical) argument is that only such states are reasonable for modeling entire closed systems like the universe. Yes, mathematically you can write down a state for the full closed system that is not an eigenstate of the full Hamiltonian; but does that make sense physically? I am arguing that it doesn't.

I don't think you have a solid argument here.

Fundamentally, the universe is mathematical. Physical intuition can only get you so far. Much about the universe is extremely unintuitive. If you want to prove your position, you're going to have to show the math that demonstrates a cyclical closed system is unphysical.

That math would probably have be rooted in Noether's theorem, but I'm pretty darned sure that a quantum system in a mixed energy state fully complies with Noether's theorem as long as you do the calculations properly (i.e., don't attempt to impose a classical notion of energy onto the system).

 

[PeterDonis]

Fundamentally, the universe is mathematical.

At this point we're out of physics and into speculation which is off topic for this discussion.

I'm pretty darned sure that a quantum system in a mixed energy state fully complies with Noether's theorem

Noether's Theorem isn't about the state, it's about the Lagrangian. If the Lagrangian has a continuous symmetry, there will be an associated conserved current. It's a mathematical theorem; a system can't not "comply" with it (except in the trivial sense of a system whose Lagrangian has no continuous symmetries).

None of this has anything to do with the argument I was making.

 

[kimbyd]

Noether's Theorem isn't about the state, it's about the Lagrangian. If the Lagrangian has a continuous symmetry, there will be an associated conserved current. It's a mathematical theorem; a system can't not "comply" with it (except in the trivial sense of a system whose Lagrangian has no continuous symmetries).

None of this has anything to do with the argument I was making.

It does, because it means an isolated quantum system in a mixed state conserves energy.