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My layman's understanding of QFT is that once a system is observed, any future observation will always yield the same result. Would this hold true if the original-and-only observer, without any physical recording other than human memory, and not communicated to any other person, died before any subsequent observation?

If the plot of the film was, "the spy sent the vital, world-saving data back to headquarters, encoded with a quantum key, only to be intercepted and viewed by the antagonist of the story; but the dashing hero catches and kills him before he could reveal what he learned, and the data arrives still encrypted back at HQ", what would the physicists be muttering to each other as they left the theater?

Or put another way, If I flip a coin into a box, observe it, then bury the box until after my death, does the coin exist in a "superimposed" state of heads/tails/on the edge until observed again, or will any future observation always be what I saw; or would it be that "superimposed" state but always resolve to the same answer? (apologies if I'm using terms incorrectly).

My intuition says information is information, so even if lost to our access within decaying brain cells, HQ would know the data had been viewed, no matter the state of decomposition of the "person who did the measuring" but I've run into a road block in phrasing the questions to dive deeper down the rabbit hole.

What resources would one recommend --to a well-read layperson who has little trouble following most Royal Institute lectures, to give you an idea of my reading level --to further pursue this odd question?

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  • $\begingroup$ My choice of the analogy of a "quantum key" was because I was looking for an example where the outgoing effects that would need to be captured and reflected to preserve causality were on the quantum scale. $\endgroup$ – Harry Lerwill Apr 24 '15 at 20:11
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You are attaching too much significance to the role of the observer. Actually the term observe is an unfortunate one in this context as anything that interacts with the system observes it whether that something is conscious or not. Some related questions that you might find interesting are:

When you (or anything else) interact with a quantum system you perturb it and the interaction will force the system into (approximately) an eigenstate of whatever measurement you make. However once the interaction is over the system will start to evolve in time again.

For example you might measure a particle's position, which localises it at the position you measure, however the particle will immediately start to delocalise again. So a second measurement made immediately after the first will return (approximately) the same result, but if you wait a long time then make a third measurement the result will in general not be the same as the first two.

Back to your question, observer 2 will observe a different result to observer 1 but not due to observer 1's unfortunate demise. The difference is because the once observer 1 has finished their measurement the system will evolve with time. How big a difference observer 2 sees depends on how long they wait to do their measurement.

For completeness I should say that if the system is in an eigenstate of the Hamiltonian (an energy eigenstate) then it will not change with time. However the uncertainty principle tells us it's impossible to prepare a system in an energy eigenstate unless you take an infinite time to do it. So all real life systems will change with time.

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  • $\begingroup$ Thank you, John, that certainly helps my define my questions further. $\endgroup$ – Harry Lerwill Apr 24 '15 at 19:40

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