No-communication theorem question

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Observations on entangled particles do not allow for instantaneous communication, as interacting with one particle only results in a random state change of the other. When Bob applies a magnetic field to his particle, it alters its state, but Alice's particle will also collapse to a corresponding state only upon observation, maintaining the randomness of outcomes. The discussion highlights that while external influences can change one particle's state without causing decoherence, entanglement can still be preserved in other degrees of freedom until a measurement is made. Ultimately, once a degree of freedom collapses, the entanglement for that aspect ceases to exist. The complexities of quantum mechanics ensure that no deterministic communication can occur through entangled particles.
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Hey all,

Performing observations on 1 of 2 entangled particles does not allow instantaneous transmission of information between them. But since the entangled particles compromise a single quantum system, why can't we interact with one particle to influence the other?

Example: Bob and Alice each have one of two spin entangled particles, which are in isolating boxes, and are still coherent. They are very far away. Before "observing" his particle, Bob applies a magnetic field across his box, in the up direction. Since he is changing the quantum state of his particle, Alice's particle must change as well, since they are a single system, right?

I'm sure there has been a thread about this already, I just couldn't find it :confused:
 
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Smacal1072 said:
Example: Bob and Alice each have one of two spin entangled particles. They are very far away. Before "observing" his particle, Bob applies a magnetic field to his particle. Since he is changing the quantum state of his particle, Alice's particle must change as well, since they are a single system, right?

If you change the state by doing something that "reveals" an observable's value, then the entangled partner particle will change accordingly. But you will simply have it jump into a random eigenstate. That is why no communication is possible. (If you act on the particle in such a way as to not cause a wave function collapse, then the other particle does not experience that. Example is changing its direction of travel.)
 
DrChinese said:
If you change the state by doing something that "reveals" an observable's value, then the entangled partner particle will change accordingly. But you will simply have it jump into a random eigenstate. That is why no communication is possible. (If you act on the particle in such a way as to not cause a wave function collapse, then the other particle does not experience that. Example is changing its direction of travel.)

Thks DrChinese - So suppose after applying this magnetic field, Bob opens his box and observes his particle. Since the magnetic field is up, the particle will surely collapse to the "up" eigenstate? And if so, (assuming that the entangled particles were created from a no initial net spin system), Alice's must collapse to the "Down" eigenstate to ensure conservation of spin?
 
Smacal1072 said:
Thks DrChinese - So suppose after applying this magnetic field, Bob opens his box and observes his particle. Since the magnetic field is up, the particle will surely collapse to the "up" eigenstate? And if so, (assuming that the entangled particles were created from a no initial net spin system), Alice's must collapse to the "Down" eigenstate to ensure conservation of spin?

The field is oriented in a direction so that the result can be UP or DOWN for Alice. And you would then see DOWN or UP for Bob accordingly. You can orient in any direction to get the UP/DOWN random results.
 
DrChinese said:
The field is oriented in a direction so that the result can be UP or DOWN for Alice. And you would then see DOWN or UP for Bob accordingly. You can orient in any direction to get the UP/DOWN random results.

I'm not sure I understand - the magnetic field is aligned on the same axis that bob and alice are making their observations. You would most likely never see a "down" observation for Bob, or a corresponding "up" observation for alice, since his particle will be influenced by the magnetic field, right?

I guess what I'm asking is: If I introduce an external potential to 1 of 2 entangled particles such that it changes the quantum state of that particle without decohere-ing the entangled system, do the particles remain entangled when they finally do decohere?

Thks for your patience DrChinese!
 
Smacal1072 said:
I'm not sure I understand - the magnetic field is aligned on the same axis that bob and alice are making their observations. You would most likely never see a "down" observation for Bob, or a corresponding "up" observation for alice, since his particle will be influenced by the magnetic field, right?

I guess what I'm asking is: If I introduce an external potential to 1 of 2 entangled particles such that it changes the quantum state of that particle without decohere-ing the entangled system, do the particles remain entangled when they finally do decohere?

Thks for your patience DrChinese!

If you align Alice and Bob with the vertical, you will get UP/DOWN or DOWN/UP. The result will be random.

Technically, entangled particles are (potentially) entangled with respect to multiple degrees of freedom. It is possible to have collapse of one degree of freedom WITHOUT collapse of all degrees of freedom. So that is pretty weird if you think about it. The HUP can be respected a degree of freedom at a time, leaving other degrees of freedom entangled. Experiments have been performed to demonstrate that.

But once a degree of freedom collapses, entanglement ends for that particular portion.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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