Can a quantum system with internal memory be in an energy eigenstate?

  • #1
mephistomunchen
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If a system is in an eigenstate of the hamiltonian operator, the state of the system varies with time only with a "j exp(w t)" phase factor. So, the system is in a "stationary state": no variation with time of observable properties.

But the system could in theory (for what I understand) be complex enough to contain subsystems that interact with each-other and have a memory of their relative "state". So, from the point of view of a "subsystem observer", the other subsystems are changing with time: so they are not in a stationary state. So, we have a system in a stationary state that is composed by subsystems that see each-other in states that are not stationary.

Is this description correct, or is there an error in what I said?

Are there in the literature some examples of interacting subsystems of a closed system that contain memory and are described in quantum mechanics?
 

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  • #2
vanhees71
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The larger topic is "open quantum systems", i.e., you have some large closed system, usually consisting of a many-body system, often assumed to be in thermal equilibrium, and some micro- or mesoscopic system you are interested in. The total system of course fulfills the unitary time evolution of quantum theory, but usually you can't solve the equations of motion, because they are too complicated and also you are not interested in the microscopic details of the total system but only on the behavior of some "relevant observables" of the subsystem of interest. Formally you get a description of this subsystem by calculating the statistical operator of the total system and then trace out the uninteresting part (in my example above the "heat bath"). Of course, you still can't solve this problem by first calculating the time evolution of the full statistical operarator but you can derive equations of motion for the statistical operator of the subsystem or expectation values and/or correlation functions for the "relevant opertors" of interest.

There are many approaches to approximations of this kind. A famous one is the influence-functional technique due to Vernon and Feynman:

R. P. Feynman and F. L. Vernon, Jr., The Theory of a
general quantum system interacting with a linear dissipative
system, Annals Phys. 24, 118 (1963),
https://doi.org/10.1016/0003-4916(63)90068-X

A similar approach is by Caldeira and Leggett

A. O. Caldeira and A. J. Leggett, Path integral approach to
quantum Brownian motion, Physica A 121, 587 (1983),
https://doi.org/10.1016/0378-4371(83)90013-4

There are of course many more approaches. It's hard to say, which one is of interest for you, because it also depends a bit on the concrete application you are after, which approach is the most natural to use.
 
  • #3
mephistomunchen
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I was more interested in understanding what an observer that is part of a system in a stationary state "sees" when he observes other subsystems of the same stationary state.

I had in mind for example a quantum computer used to model a system composed by two subsystems:

- subsystem A: contains an n-bit registry whose value is updated every T nanoseconds with predefined ciclic arbitrary sequence of values: let's say the sequence of values 8 - 20 - 4 - 31 - 2 - 19 - 8 - 20 - 4 - 31 ... and so on

- subsystem B (the observer) contains a memory of 10 n-bit registries that are used to store the temporal sequences of numbers observed on the subsystem A.

Let's say that we simulate in the quantum computer a weak interaction between the subsystems A and B that is "used" by the subsystem B to "observe" the values present at each time T on the registry of the subsystem A and store them in it's registries (so, B will have a 10 steps memory).

Now, from the point of view of B, the system A is evolving in a deterministic way and there is a definite direction of time, determined by his memory (he remembers 10 instants of time in the past).
But from the point of view of the experimenter that sees the whole quantum system, the system can be prepared in a definite energy eigenstate where there is no time evolution at all (except from the unitary phase factor).

If this is possible (and there is no reason why what I described cannot be done), I wander if it would be possible for the observer B to detect in some way that the "universe" that he sees is in reality in a stationary state. For example, he could "discover" that some of the parameters of his universe seem to be "fine tuned", due to the fact that they are the result of the superposition of an infinite set of possible states of his universe.

Not sure if what I explained here make sense... :-)
 
  • #5
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I had in mind for example a quantum computer used to model a system composed by two subsystems

This system as you describe it can't be closed, because the updating of system A every T nanoseconds has to be done by something outside systems A and B. You haven't included that something in your analysis.

a weak interaction between the subsystems A and B that is "used" by the subsystem B to "observe" the values present at each time T on the registry of the subsystem A and store them in it's registries

Any such interaction will disturb the state of system A so it will no longer be evolving in the way you described before.
 
  • #6
mephistomunchen
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This system as you describe it can't be closed, because the updating of system A every T nanoseconds has to be done by something outside systems A and B. You haven't included that something in your analysis.

The system A is simply going through a cyclic set of states, such as a mechanical device with an internal clock. The "updating logic" is built into the structure of the system. As it does not loose energy due to inelastic processes, I don't see why it can't be a closed system.

Any such interaction will disturb the state of system A so it will no longer be evolving in the way you described before.

This is true in principle, but if the system A is stable and has energy levels much bigger than the energy of interaction between A and B, I think the interaction would cause only small differences in the behaviour of A (for example jumping two numbers at the same time, or not jumping at all from time to time), but the overall behaviour of the system will not be much different.

If this is not the case, do you have a theoretical argument why a system in a stationary state should not contain an observer that see the system itself as evolving in time?
 
  • #7
PeroK
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I was more interested in understanding what an observer that is part of a system in a stationary state "sees" when he observes other subsystems of the same stationary state.

Now, from the point of view of B, the system A is evolving in a deterministic way and there is a definite direction of time, determined by his memory (he remembers 10 instants of time in the past).
But from the point of view of the experimenter that sees the whole quantum system, the system can be prepared in a definite energy eigenstate where there is no time evolution at all (except from the unitary phase factor).

The system A is simply going through a cyclic set of states, such as a mechanical device with an internal clock. The "updating logic" is built into the structure of the system. As it does not loose energy due to inelastic processes, I don't see why it can't be a closed system.

If this is not the case, do you have a theoretical argument why a system in a stationary state should not contain an observer that see the system itself as evolving in time?
None of this seems very quantum mechanical. It might be an idea to try to describe these systems in QM terms. For example:

If system A is a closed system, how can you say what state it's in unless you measured it?

How do you prepare a mechanical device with an internal clock in a stationary state?

How do you prepare an "observer" in a stationary state?

What you have seems to be essentially a classical description of a system, with a hypothesis that "this classical system is in a QM energy eigenstate". Which leads to a confusion of concepts that is difficult to analyse effectively.
 
  • #8
mephistomunchen
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None of this seems very quantum mechanical. It might be an idea to try to describe these systems in QM terms. For example:

If system A is a closed system, how can you say what state it's in unless you measured it?

How do you prepare a mechanical device with an internal clock in a stationary state?

How do you prepare an "observer" in a stationary state?

What you have seems to be essentially a classical description of a system, with a hypothesis that "this classical system is in a QM energy eigenstate". Which leads to a confusion of concepts that is difficult to analyse effectively.

Well, actually I don't know if it's possible to prepare a quantum system in such way, and how to do it. Probably this should have been the question to ask :oops:

However, if the system was a semiconductor crystal (doped with a p impurity) and the "observer" the electron in the conduction band, then it would decay to the ground state (emitting photons) after a while if the temperature is very low. If the electron had more internal states all with the same energy, maybe they could be used to "record" its position as it moves through the crystal. However, electrons have only two internal states and no "camera" to see where they are going...
 
  • #9
PeroK
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Well, actually I don't know if it's possible to prepare a quantum system in such way, and how to do it. Probably this should have been the question to ask :oops:

However, if the system was a semiconductor crystal (doped with a p impurity) and the "observer" the electron in the conduction band, then it would decay to the ground state (emitting photons) after a while if the temperature is very low. If the electron had more internal states all with the same energy, maybe they could be used to "record" its position as it moves through the crystal. However, electrons have only two internal states and no "camera" to see where they are going...
Part of the issue is that an electron cannot be an observer in any meaningful way. Any meaningful information about a system must come from interaction with a (macroscopic) measuring device. This is part of the measurement problem.

For example, we can talk about the different energy levels of the hydrogen atom, but we cannot directly observe the energy state of an atom. Instead, we measure the wavelength of a photon absorbed by or emitted by the atom and infer that it has transitioned from one energy level to the other. There is no way to double-check that by taking a closer look at the atom before and after the photon emission or absorption: the emitted or absorbed photon is all we get. That is the measurement (and it's not even of the atom itself directly).
 
  • #10
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The system A is simply going through a cyclic set of states, such as a mechanical device with an internal clock.

Which, if you try to model it quantum mechanically, you will find is a stationary system--a system that is in an eigenstate of its Hamiltonian. (Basically, it's a harmonic oscillator in an energy eigenstate.) But you wanted your system A to be a non-stationary system.

As it does not loose energy due to inelastic processes, I don't see why it can't be a closed system.

A closed stationary system. (Think about what "does not lose energy" means in quantum mechanics.) But you wanted your system A to be a non-stationary system.

if the system A is stable and has energy levels much bigger than the energy of interaction between A and B

Then B won't be able to record the successive states of A, because A will never change state--it will always be in the same eigenstate of its Hamiltonian. See above.

In order for B to record anything about A, the interaction between A and B has to be able to change A's state, which means kicking A out of the eigenstate of its Hamiltonian in which it is executing the cyclic process you describe.

do you have a theoretical argument why a system in a stationary state should not contain an observer that see the system itself as evolving in time?

No. I am simply pointing out that your scenario, as you describe it, cannot be used to address that question because it is inconsistent. You want subsystem A to execute the same cyclic process forever, which makes it stationary, but you also want it to interact with subsystem B. You can't have both.
 
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  • #11
mephistomunchen
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In order for B to record anything about A, the interaction between A and B has to be able to change A's state, which means kicking A out of the eigenstate of its Hamiltonian in which it is executing the cyclic process you describe.

But look: take two hydrogen atoms in the fundamental state and make them interact weekly by putting them at a distance such that the two electrons start to feel each-other's electric field. Now the two electrons' wave functions will synchronize with each-other (such as two weakly interacting classic pendulums) becoming entangled. Then the interaction between A and B has changed A's state, and from the point of view of atom B, atom A is no more in a stationary state. But from the point of view of the external observer, the full system is now in the stationary state of the system made of two atoms. Isn't it an analogous situation of what I described?

The assumption that one of the two electrons (let's say B) could in theory be substituted by a "observer" subsystem with an internal state capable of recording the evolution of the whole two-body system, instead, (as pointed out by PeroK) is probably wrong.

But that would mean that for some reason I can't simulate the whole two-subsystem with an analogous two-subsystem quantum computer that behaves at the same way, and I don't quite understand why that shouldn't be possible. I know that I can't simply erase the observer's memory in a closed system, but in that case both the system and the observer would follow a cyclic evolution in time - and the two systems would end up being synchronized with each-other, such as in the two-hydrogen-atoms system.
 
  • #12
Vanadium 50
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Words, words, words. This is unlikely to help. You need to write down the math.

Let's start with your "system with internal memory". For concreteness, let's assume in regularly moves from State A to State B to State C and back again to A, neither gaining nor losing energy.That's what you want, right? Please write down a Hamiltonian and a wavefunction that does this.

I don't believe such a state will be stationary. The ABC pattern has the same energy as the CBA pattern. The ground (and stationary) state will surely be (√3|A> + √3|B> + √3|C>)/3.
 
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  • #13
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take two hydrogen atoms in the fundamental state

I assume you mean "ground state", i.e., each atom's electron is in the 1s orbital.

make them interact weekly by putting them at a distance such that the two electrons start to feel each-other's electric field

Won't make a difference; each atom is in the ground state, which means it can't give up any energy since there's no lower energy state to go to. So these atoms cannot interact. At least one would have to be in an excited state for there to be any possibility of interaction. Or the atoms could be in some superposition of energy eigenstates. But not both in the ground state.

Now the two electrons' wave functions will synchronize with each-other (such as two weakly interacting classic pendulums) becoming entangled.

Not if the atoms remain separate atoms. If the two atoms bond to form an H2 molecule, then that does involve entangling the electrons' states, but only if the two-atom system emits radiation, giving up enough energy to put it into the H2 molecule bound state. None of this helps at all if you are trying to get one electron to "observe" the other.

the interaction between A and B has changed A's state, and from the point of view of atom B, atom A is no more in a stationary state.

Wrong. Both atoms are now in the two-atom H2 molecule stationary state, which is also a stationary state of the two-atom system as a whole. The process of changing state involved emitting radiation, as I said above, but the transition was from one stationary state to another for each atom.
 
  • #14
atyy
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If a system is in an eigenstate of the hamiltonian operator, the state of the system varies with time only with a "j exp(w t)" phase factor. So, the system is in a "stationary state": no variation with time of observable properties.

But the system could in theory (for what I understand) be complex enough to contain subsystems that interact with each-other and have a memory of their relative "state". So, from the point of view of a "subsystem observer", the other subsystems are changing with time: so they are not in a stationary state. So, we have a system in a stationary state that is composed by subsystems that see each-other in states that are not stationary.

Is this description correct, or is there an error in what I said?

There is an error in what you said. In an energy eigenstate, only the global phase changes, as you said. The relative phases between the different components of the quantum state do not vary with time. In quantum mechanics a change in global phase has no physical meaning. Changes in relative phases have physical meaning. Since only the global phase changes, the relationship of the various subsystems relative to each other do not change with time.
 

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