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

[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.

 

[PeterDonis]

ated quantum system in a mixed state

How can an isolated quantum system be in a mixed state? Again, I understand that you can write down such a state mathematically. But how does it make sense physically?

 

[Robert Shaw]

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.

Ok let's take the simplest quantum Hamiltonian, zero potential energy everywhere.

Then form a wavepacket.

Wavepackets can exist in isolated closed systems.

Yet they are superpositions of energy levels

 

[Robert Shaw]

Ok let's take the simplest quantum Hamiltonian, zero potential energy everywhere.

Then form a wavepacket.

Wavepackets can exist in isolated closed systems.

Yet they are superpositions of energy levels

Reality is too complex for physics to model.

Football, a game of football, obeys the laws of physics. Physicists cannot model it however, not quantum nor classical.

Knowledge is the problem. We don't know enough about almost everything to write equations.

Look around you. Mostly complex molecules. Air is mostly molecules.

That's for engineers who do a great job using approximations of physics.

 

[Robert Shaw]

Reality is too complex for physics to model.

Football, a game of football, obeys the laws of physics. Physicists cannot model it however, not quantum nor classical.

Knowledge is the problem. We don't know enough about almost everything to write equations.

Look around you. Mostly complex molecules. Air is mostly molecules.

That's for engineers who do a great job using approximations of physics.

Closed systems are idealisations.

Reality seldom approximates to a closed system.

On rare occasions real systems can be created that approximate to an idealised closed system.

Our knowledge of such systems is from the mathematics of the idealised closed system, for which we can write down equations.

Quantum physics textbooks are mostly concerned with closed systems. They tend to focus on eigenstates because they are mathematically simple and good for teaching purposes. They say a little about superpositions - the wavepacket, the Gaussian superposition of oscillator states, etc. but these are more difficult mathematically so get limited coverage.

 

[weirdoguy]

Why don't you write it all in ONE post? There is edit button.

 

[PeterDonis]

@Robert Shaw you are simply repeating your position without responding to what anyone else has said. That is not going to lead to a productive discussion.

Thread closed.