3 ms·

> To go smoothly from 0 to 1, the system transitions through a series of superpositions of both states, i.e. it starts out entirely in state 0, and then transit

by kgwgk 7y ago

> To go smoothly from 0 to 1, the system transitions through a series of superpositions of both states, i.e. it starts out entirely in state 0, and then transitions smoothly to being mostly in state 0 and a little bit in state 1, to being half in each state, to being mostly in 1 and a little bit in 0, to being entirely in 1. Not really. An atom can be described by its quantum state only if it's isolated and in that case its energy is constant. If the states s_1 and s_2 have different energies E_1 and E_2 then the quantum state may be a superposition of the states s_1 and s_2 with an expected value for the energy between E_1 and E_2 but no "smooth transition" is possible. For the transition to happen the atom has to be coupled with an external field. One way to think of it is as the evolution of the system composed of the atom plus a photon, where the transition between the states "excited atom and no emitted photon" and "unexcited atom and emited photon" happens at constant energy. Edit: See https://en.m.wikipedia.org/wiki/Spontaneous_emission#Theory https://en.m.wikipedia.org/wiki/Spontaneous_emission#Theory

duality 7y ago
"An atom can be described by its quantum state only if it's isolated and in that case its energy is constant." How do you figure? As a contradiction, take your atom+electromagnetic field system, describe the transition from excited atom to unexcited atom + photon state, and project out the E&M field. Voila, now you have a quantum description of an atom transitioning between different energy states. Its dynamics may look funny, i.e. they may appear nonlocal, they may not conserve energy, etc. but that's different from saying "there is not a quantum description of these dynamics" which is what you're claiming.
- kgwgk 7y ago
  https://en.m.wikipedia.org/wiki/Quantum_state https://en.m.wikipedia.org/wiki/Quantum_state > project out the E&M field. Voila, now you have a quantum description There is no wavefunction describing the state of the subsystem because the system is not separable. > Its dynamics may look funny, i.e. they may appear nonlocal, they may not conserve energy, etc. but that's different from saying "there is not a quantum description of these dynamics" which is what you're claiming. What is the “quantum description of these dynamics”? Quantum mechanics is usually based on something like the following postulate: “The state of a physical system is described by a well-behaved function of the coordinates and time, Ψ(q, t). The function contains all the information that can be known about the system.”
  - duality 7y ago
    Not sure exactly what you're linking the Wikipedia article on quantum states, but if you like it as a reference, check out https://en.wikipedia.org/wiki/Quantum_state#Mixed_states https://en.wikipedia.org/wiki/Quantum_state#Mixed_states. The operation of "projecting out" the E&M field in my previous comment would be realized on the density matrix as a partial trace over the electromagnetic field. You can also take the partial trace of this field of the Hamiltonian operator to see the effective dynamics of the atom when "ignoring" the E&M field. This is a fully quantum description of the state, so I stand by the statement that your claim "An atom can be described by its quantum state only if it's isolated and in that case its energy is constant." is incorrect.
    - kgwgk 7y ago
      Do you agree that “The state of a physical system is described by a well-behaved function of the coordinates and time, Ψ(q, t). The function contains all the information that can be known about the system.”? When the atom is coupled with the electromagnetic field and the state of the system is not separable there is no complete description of the atom given by a wavefunction defining its quantum state. You can have an incomplete description by tracing out the rest of the system, I agree. Let’s say then that "An atom can be completely described by its quantum state only if it's isolated and in that case its energy is constant." Edit: in any case, my point was (and I think that we will agree) that it is misleading to say “Consider a system that transitions from energy state 0 to an adjacent energy state 1. [...] To go smoothly from 0 to 1, the system transitions through a series of superpositions of both states”. The atom goes from the state 0 to the state 1 but during the transition it’s not described by a superposition of those states (that would be a pure state). If anything, it is described by an (improper) mixture of those states, obtained by tracing out the rest of the system.
      - lisper 7y ago
        Yes, but note that I deliberately used the word "system" rather than "atom". The system is the combination of the atom plus whatever it's absorbing energy from or emitting energy to. And that (entangled) system is in a superposition.
lisper 7y ago
That's picking a pretty fine nit. Yes, obviously if an atom undergoes an energy transition then that energy has to come from (or go to) somewhere. That doesn't change the fundamental point, which is that the process doesn't happen instantaneously or discontinuously, and the continuity is provided by transition through a series of superposition states. Yes, to be strictly correct you'd have to say that the superposition is a joint state of the atom and something else and not the atom in isolation. But this post was targeted to a lay audience, and that sort of detail would be more of a distraction than a clarification IMHO.
ConcernedCoder 7y ago
Not if we're really living in a simulation: e.g. if you initiate the state change of an atom ( the example was the addition of one quanta of energy ), then eventually the state will change from quanta 0 to quanta 1, with possibly many superpositions of both states during the described finite lifetime of the transition. This, to me, sounds so much like eventual consistency ( for observers ), that it's a bit scary.
- ttctciyf 7y ago
  The whole business of superposition and measurement so reeks of lazy initialization that you might want to worry about what performance constraints the simulation is trying to beat too.
wwarner 7y ago
What puzzles me is where the photon goes when it bumps up the energy of the electron, and what does it look like when the electron shell drops back down. I realize too much "realism" in QM is just a waste of effort, but I keep seeing the field sort of pinching off a photon.
- qtplatypus 7y ago
  The photon goes into the energy of the electron. Because mass and energy are the same thing the mass of the photon is converted into the energy of the electrons new state.
effie 7y ago
You are right for spontaneous emission, there we need to connect the system to something else, or to formulate a detailed-enough relativistic model with retarded EM fields that is able to lose energy (ordinary Hamiltonian of charged particle variables only cannot describe loss of energy to radiation). But for spontaneous absorption/emission, you are not right. Describing atoms by psi function is extremely successful, it is what Schroedinger did in his papers, what got him the right results for positions and intensities of spectral lines and what put his equation to the center of interest. It is true this successful description has some problems, such as the question of consistency of different world cuts (into system and environment), but the ordinary way to do the split for atom-radiation interaction (time-independent atom Hamiltonian and time-dependent external EM field) describes stimulated emission and absorption very well. Stimulated absorption/emission does require presence of external field, but this field can be taken into account in the usual psi function formulation - this is called semi-classical theory of radiation, it is what Schroedinger proposed in 1920's for simple atoms and is still heavily used for description of interaction of radiation with atoms and molecules.
- kgwgk 7y ago
  You’re right, I was thinking of a time-independent Hamiltonian. I may be mistaken but I thought that there are no stationary states otherwise (so it wouldn’t make sense either to say that the state is a superposition of the states 0 and 1).
- lisper 7y ago
  > You are right for spontaneous emission > But for spontaneous absorption/emission, you are not right Was one of those instances of "spontaneous" supposed to be "stimulated"?
  - effie 7y ago
    Yes, the second one should have been "stimulated". Thanks for pointing that out.
gaze 7y ago
This is incorrect. See the stochastic schroedinger equation. A quantum system will stay pure under measurement as long as the measurement process is completely efficient.
- kgwgk 7y ago
  I’m not familiar with this extension of QM. What is the “system” here? Would the atom stay in a pure state or would a larger system stay in a pure state?
  - gaze 7y ago
    The whole system does. The trick is you never trace anything out. You needn't. A spin interacting with a flying field is itself a pure system. It's just when you trace out the flying field that the system may become impure. If the spin interacts with the field, and the field is measured with measurements local to the field subsystem, the entire system will stay pure, and you'll see the qubit's wavefunction "collapse."
    - kgwgk 7y ago
      Thanks. In that case I don’t think there is a contradiction with what I said, if there is no wave function corresponding to the atom (alone). I understood the original comment as saying that the quantum state of the atom during the transition between energy levels was a superposition of the corresponding pure states (i.e. a pure state) and I was objecting to that.
      - gaze 7y ago
        Yeah I mean that’s still fine. The entanglement between system and field never persists since the field is taken to be measured immediately. The systems are only non-separable as long as they’re entangled. See the notes by Steck on the derivation of the SME and SSE. I agree that if the field is taken to be a degree of freedom entangled with the system under measurement, things are funnier. This experiment isn’t really like spontaneous or stimulated emission... where yeah, DURING the process, the photon pooped out is entangled with your subsystem. In the case of the present experiment there’s a much more subtle thing going on with a three level system where the bright-zero manifold is used as a witness to the dark-zero manifold, where the presence or lack thereof of fast jumps in the bright-zero manifold betray information about the dark-zero manifold, necessarily.

by kgwgk 7y ago

duality 7y ago

"An atom can be described by its quantum state only if it's isolated and in that case its energy is constant." How do you figure? As a contradiction, take your atom+electromagnetic field system, describe the transition from excited atom to unexcited atom + photon state, and project out the E&M field. Voila, now you have a quantum description of an atom transitioning between different energy states. Its dynamics may look funny, i.e. they may appear nonlocal, they may not conserve energy, etc. but that's different from saying "there is not a quantum description of these dynamics" which is what you're claiming.

kgwgk 7y ago
https://en.m.wikipedia.org/wiki/Quantum_state https://en.m.wikipedia.org/wiki/Quantum_state > project out the E&M field. Voila, now you have a quantum description There is no wavefunction describing the state of the subsystem because the system is not separable. > Its dynamics may look funny, i.e. they may appear nonlocal, they may not conserve energy, etc. but that's different from saying "there is not a quantum description of these dynamics" which is what you're claiming. What is the “quantum description of these dynamics”? Quantum mechanics is usually based on something like the following postulate: “The state of a physical system is described by a well-behaved function of the coordinates and time, Ψ(q, t). The function contains all the information that can be known about the system.”
- duality 7y ago
  Not sure exactly what you're linking the Wikipedia article on quantum states, but if you like it as a reference, check out https://en.wikipedia.org/wiki/Quantum_state#Mixed_states https://en.wikipedia.org/wiki/Quantum_state#Mixed_states. The operation of "projecting out" the E&M field in my previous comment would be realized on the density matrix as a partial trace over the electromagnetic field. You can also take the partial trace of this field of the Hamiltonian operator to see the effective dynamics of the atom when "ignoring" the E&M field. This is a fully quantum description of the state, so I stand by the statement that your claim "An atom can be described by its quantum state only if it's isolated and in that case its energy is constant." is incorrect.
  - kgwgk 7y ago
    Do you agree that “The state of a physical system is described by a well-behaved function of the coordinates and time, Ψ(q, t). The function contains all the information that can be known about the system.”? When the atom is coupled with the electromagnetic field and the state of the system is not separable there is no complete description of the atom given by a wavefunction defining its quantum state. You can have an incomplete description by tracing out the rest of the system, I agree. Let’s say then that "An atom can be completely described by its quantum state only if it's isolated and in that case its energy is constant." Edit: in any case, my point was (and I think that we will agree) that it is misleading to say “Consider a system that transitions from energy state 0 to an adjacent energy state 1. [...] To go smoothly from 0 to 1, the system transitions through a series of superpositions of both states”. The atom goes from the state 0 to the state 1 but during the transition it’s not described by a superposition of those states (that would be a pure state). If anything, it is described by an (improper) mixture of those states, obtained by tracing out the rest of the system.
    - lisper 7y ago
      Yes, but note that I deliberately used the word "system" rather than "atom". The system is the combination of the atom plus whatever it's absorbing energy from or emitting energy to. And that (entangled) system is in a superposition.

lisper 7y ago

That's picking a pretty fine nit. Yes, obviously if an atom undergoes an energy transition then that energy has to come from (or go to) somewhere. That doesn't change the fundamental point, which is that the process doesn't happen instantaneously or discontinuously, and the continuity is provided by transition through a series of superposition states. Yes, to be strictly correct you'd have to say that the superposition is a joint state of the atom and something else and not the atom in isolation. But this post was targeted to a lay audience, and that sort of detail would be more of a distraction than a clarification IMHO.

ConcernedCoder 7y ago

Not if we're really living in a simulation: e.g. if you initiate the state change of an atom ( the example was the addition of one quanta of energy ), then eventually the state will change from quanta 0 to quanta 1, with possibly many superpositions of both states during the described finite lifetime of the transition. This, to me, sounds so much like eventual consistency ( for observers ), that it's a bit scary.

ttctciyf 7y ago
The whole business of superposition and measurement so reeks of lazy initialization that you might want to worry about what performance constraints the simulation is trying to beat too.

wwarner 7y ago

What puzzles me is where the photon goes when it bumps up the energy of the electron, and what does it look like when the electron shell drops back down. I realize too much "realism" in QM is just a waste of effort, but I keep seeing the field sort of pinching off a photon.

qtplatypus 7y ago
The photon goes into the energy of the electron. Because mass and energy are the same thing the mass of the photon is converted into the energy of the electrons new state.

effie 7y ago

You are right for spontaneous emission, there we need to connect the system to something else, or to formulate a detailed-enough relativistic model with retarded EM fields that is able to lose energy (ordinary Hamiltonian of charged particle variables only cannot describe loss of energy to radiation). But for spontaneous absorption/emission, you are not right. Describing atoms by psi function is extremely successful, it is what Schroedinger did in his papers, what got him the right results for positions and intensities of spectral lines and what put his equation to the center of interest. It is true this successful description has some problems, such as the question of consistency of different world cuts (into system and environment), but the ordinary way to do the split for atom-radiation interaction (time-independent atom Hamiltonian and time-dependent external EM field) describes stimulated emission and absorption very well. Stimulated absorption/emission does require presence of external field, but this field can be taken into account in the usual psi function formulation - this is called semi-classical theory of radiation, it is what Schroedinger proposed in 1920's for simple atoms and is still heavily used for description of interaction of radiation with atoms and molecules.

kgwgk 7y ago
You’re right, I was thinking of a time-independent Hamiltonian. I may be mistaken but I thought that there are no stationary states otherwise (so it wouldn’t make sense either to say that the state is a superposition of the states 0 and 1).
lisper 7y ago
> You are right for spontaneous emission > But for spontaneous absorption/emission, you are not right Was one of those instances of "spontaneous" supposed to be "stimulated"?
- effie 7y ago
  Yes, the second one should have been "stimulated". Thanks for pointing that out.

gaze 7y ago

This is incorrect. See the stochastic schroedinger equation. A quantum system will stay pure under measurement as long as the measurement process is completely efficient.

kgwgk 7y ago
I’m not familiar with this extension of QM. What is the “system” here? Would the atom stay in a pure state or would a larger system stay in a pure state?
- gaze 7y ago
  The whole system does. The trick is you never trace anything out. You needn't. A spin interacting with a flying field is itself a pure system. It's just when you trace out the flying field that the system may become impure. If the spin interacts with the field, and the field is measured with measurements local to the field subsystem, the entire system will stay pure, and you'll see the qubit's wavefunction "collapse."
  - kgwgk 7y ago
    Thanks. In that case I don’t think there is a contradiction with what I said, if there is no wave function corresponding to the atom (alone). I understood the original comment as saying that the quantum state of the atom during the transition between energy levels was a superposition of the corresponding pure states (i.e. a pure state) and I was objecting to that.
    - gaze 7y ago
      Yeah I mean that’s still fine. The entanglement between system and field never persists since the field is taken to be measured immediately. The systems are only non-separable as long as they’re entangled. See the notes by Steck on the derivation of the SME and SSE. I agree that if the field is taken to be a degree of freedom entangled with the system under measurement, things are funnier. This experiment isn’t really like spontaneous or stimulated emission... where yeah, DURING the process, the photon pooped out is entangled with your subsystem. In the case of the present experiment there’s a much more subtle thing going on with a three level system where the bright-zero manifold is used as a witness to the dark-zero manifold, where the presence or lack thereof of fast jumps in the bright-zero manifold betray information about the dark-zero manifold, necessarily.