Entanglement after measurement

[nomadreid]

TL;DR Summary

(a) If A and B are entangled for t<t0, and measured at t0, do they return to entanglement at t>t0? (b) trivially at t=t0?
(c) For A,B being two electrons in the same energy level in an atom, is the smallest E to measure spin the E to go to another energy level?

Three related questions:
(a) In a pair of entangled particles, after one is measured/observed/determined/collapsed, my understanding is that the measurement breaks the entanglement so that after the measurement, unless something happens to re-entangle them, they are no longer entangled. Correct?

(b) If you have entanglement at t<t0 and not at t>t0, what does one say for t=t0? Or is the question meaningful?

(c) If we have two electrons in the same energy level without being observed, their spin states are automatically entangled, right? (If wrong, ignore what follows.) So it should be impossible to determine the spin states while leaving them in the same energy level, because that would give the contradiction of them being both determined and indeterminate at the same time. Therefore one has to have the electron go to another level, at least temporarily (and if it drops back after the measurement, this would be a case of what I meant in (a) by "unless something happens to re-entangle them") in order to measure the spin, i.e., the minimum energy needed to measure would be the difference between two energy levels. Is this reasoning correct?

It is likely that even my fundamental understanding of entanglement is seriously flawed, and I will be happy to be corrected.

 

[DennisN]

(a) In a pair of entangled particles, after one is measured/observed/determined/collapsed, my understanding is that the measurement breaks the entanglement so that after the measurement, unless something happens to re-entangle them, they are no longer entangled. Correct?

Yes, correct.

 

[nomadreid]

Good; one down, two to go...

 

[DennisN]

(b) If you have entanglement at t<t0 and not at t>t0, what does one say for t=t0? Or is the question meaningful?

I would say it is, or could be, a meaningful question*, since as far as I know decoherence is still being studied. How disentanglement, and thus, observation/measurement is understood, is through the process of decoherence; an observed system, of let's say two particles, looses its initial coherence and the particles instead get entangled with the environment/measuring apparatus.

This happens extremely fast. I don't remember the timescales at the moment, though.

* Edit: What I mean is that it is my understanding that decoherence is a very fast, yet continuous rather than discrete process.

 

[PeroK]

(c) If we have two electrons in the same energy level without being observed, their spin states are automatically entangled, right?

The two electrons form a two-particle system. They do not have individual spin states (I'm sure we've been through this before). You might describe a two-particle system as a combination of one-particle states. But, it's wrong to see one particle as having one definite state and the other particle having the other state.

Here's a way of thinking about that.

Imagine you have a piece of paper that lists all 1-particle states. You have one particle. The state of your particle must be on that list.

Now, imagine you have another piece of paper that lists all two-particle states. You have two particles. The state of your system must be on that list.

In this second case, you are looking at the list and you ask "what state is one of my particles in?" The answer is that the list describes two particle states, so the question is meaningless. There are no one-particle states on that list; only two particle states. You ask "doesn't each of my particles have a definite one-particle state?" The answer from QM is no. They do not have a well-defined state as individual particles. But, between them, they have a two-particle state: something on the list.

Note: the list of two particle states might look like a list of combinations of two one-particle states, but that's a false analysis if you are looking (in general) for well-defined one-particle states in a two-particle system.

If you want to get one of your particles into a one-particle state, then you must extract it from the two particle system. You can do that by measuring it. Then, the state you find for your one-particle system is, indeed, somewhere on the first list of one-particle states.

Now, to the second point. Some of the states on the list of two-particle states are so-called "entangled" states. The allowed states for a two-electron system with the same energy-level are, indeed, entangled. The reason is the Pauli exclusion principle: they must have opposite spins when measured.

1) You have an entangled state.
2) You cannot talk about one particle being in this state and the other being in that state. It's not right to talk about "entangled one-particle states". You have an entangled two-particle state.
3) If you separate the electrons without disturbing the spin state, then the spins when measured must be opposite.
4) To say the spins are opposite before you measure them is not strictly correct. Then you are making the mistake of reading from the first list of single-particle states, when you should be reading from the second list of two-particle states.

 

[Demystifier]

(a) In a pair of entangled particles, after one is measured/observed/determined/collapsed, my understanding is that the measurement breaks the entanglement so that after the measurement, unless something happens to re-entangle them, they are no longer entangled. Correct?

Correct.

(b) If you have entanglement at t<t0 and not at t>t0, what does one say for t=t0? Or is the question meaningful?

It depends on the interpretation of the so called "wave function collapse". A modern view is that the "collapse" is somehow related to decoherence (see the post by @DennisN above), in which case the question is meanigless because we really have a continuous transition that lasts a finite time.

(c) If we have two electrons in the same energy level without being observed, their spin states are automatically entangled, right? (If wrong, ignore what follows.) So it should be impossible to determine the spin states while leaving them in the same energy level, because that would give the contradiction of them being both determined and indeterminate at the same time. Therefore one has to have the electron go to another level, at least temporarily (and if it drops back after the measurement, this would be a case of what I meant in (a) by "unless something happens to re-entangle them") in order to measure the spin, i.e., the minimum energy needed to measure would be the difference between two energy levels. Is this reasoning correct?

Correct.

 

[Demystifier]

I would say it is, or could be, a meaningful question*, since as far as I know decoherence is still being studied. How disentanglement, and thus, observation/measurement is understood, is through the process of decoherence; an observed system, of let's say two particles, looses its initial coherence and the particles instead get entangled with the environment/measuring apparatus.

This happens extremely fast. I don't remember the timescales at the moment, though.

* Edit: What I mean is that it is my understanding that decoherence is a very fast, yet continuous rather than discrete process.

More on decoherence (including the timescales) one can find in the 3rd attachment in https://www.physicsforums.com/threads/reading-materials-on-quantum-foundations.963543/#post-6270768

 

[nomadreid]

Magnificent; thanks, DennisN, PeroK and Demystifier! All very helpful explanations and also from Demystifier an interesting-looking link. (which led to other links...)

 

[atyy]

Just to clarify that in the orthodox interpretation, decoherence (the continuous process) does not do away with the need for measurement (discontinuous). If there is no measurement, there is no outcome. However, a lot can be put into pre-measurement decoherence.

 

[nomadreid]

Thanks, atyy, but I don't quite understand the phrase "pre-measurement decoherence". I thought that decoherence was an integral part of the process of measurement, so that this would translate into "pre-measurement measurement". Sorry for the confusion; could you elaborate?

 

[Nugatory]

I thought that decoherence was an integral part of the process of measurement

Decoherence is caused by interaction with the environment so it’s happening more or less continuously, unless we’ve managed to isolate our quantum system from the environment (put it in vacuum, chill it to cryogenic temperatures, ...).

Measurement doesn’t cause decoherence. Instead decoherence causes us to get a measurement result; it’s why the macroscopic needle on our macroscopic measuring instrument points to one number on the macroscopic dial or another, but is never in a coherent superposition of pointing to different numbers.

 

[vanhees71]

Of course, measurement causes decoherence, because it couples the measured (quantum) objects to the macroscopic measurement device. A measurement is just the interaction of the measured object with the device leading to an entangled state between the object and the measurment device. Projecting to the state of the corresponding "pointer variable" describes the measurement. There's no need for an interpretation of a "collapse/state reduction" as a physical process. It's just interaction between the measured object and the measurement device and Bayesian update of the description based on the reading of the pointer state by the observer. A naive collapse interpretation leads to contradictions with relativistic causality and is (fortunately for the best theory ever, which is the Standard Model of elementary-particle physics) completely unnecessary to be assumed.

 

[Fleurrose]

Just to clarify that in the orthodox interpretation, decoherence (the continuous process) does not do away with the need for measurement (discontinuous). If there is no measurement, there is no outcome. However, a lot can be put into pre-measurement decoherence.

I don't understand the phrase "pre-measurement decoherence".

 

[nomadreid]

The point that decoherence does not completely solve the measurement problem was well made in the link that Demysifier gave in post #7, and that decoherence is not the discrete phenomenon that measurement is was pointed out in vanhee's explanation in post#12. (However, I do not know whether one can yet make a definitive case that decoherence is continuous over time; is there any proof that it is not something weaker than full continuity, such as piecewise semicontinuous?) So, in the sense it appears that "pre-measurement decoherence" makes sense, but I will be glad to see any more clarifications to Fleurrose's doubts in post #13.

 

[Nugatory]

However, I do not know whether one can yet make a definitive case that decoherence is continuous over time

It’s just the unitary evolution described by Schrodinger’s equation; any theory in which it’s not a continuous process isn’t QM.

 

[nomadreid]

Ah, that makes sense. Thanks, Nugatory.

 

[Fleurrose]

It’s just the unitary evolution described by Schrodinger’s equation; any theory in which it’s not a continuous process isn’t QM.
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Ah, that makes sense. Thank you so much for the explain , Nugatory.

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