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Understanding superposition and entanglement 
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#1
Sep2910, 10:23 AM

P: 4

I am a newbie here, just an enthusiast with an above average understanding of math and physics, but exposed to QM after I left college.
I have looked through the posts and did not see a concise summary to the following. Please forgive me if I overlooked some threads. There are some assertions I have which I want to verify are generally accepted to be true. A. Measuring entangled particle A only impacts particle B by causing both to decohere where measured and both to go into superposition on a noncommuting measurement. B. The no communication theorem states that measuring one entangled particle cannot provide measurable information to the other entangled particle. C. There is no way to measure whether a particle is in superposition for a particular measurement. Do I understand all of this correctly? Many thanks, Marty 


#2
Sep2910, 10:43 AM

Sci Advisor
PF Gold
P: 5,378

Yes, I would say that you have it stated pretty well. I would comment on your A. that a measurement terminates the superposition and sends the particles into a mixed state, at least for noncommuting observables. Occasionally, you can observe entanglement continuing for commuting operators. 


#3
Sep2910, 11:28 AM

P: 4

Extending the above, if there were a way to determine whether or not an entangled particle was in superposition for a particular observable, then one could violate the nocommunication theorem? Thanks again, Marty 


#4
Sep2910, 12:27 PM

Sci Advisor
PF Gold
P: 5,378

Understanding superposition and entanglement



#5
Sep2910, 01:10 PM

P: 4

Suppose I had a device which produced two entangles particles A and B with noncommuting observables X and Y. In addition, suppose that X is not in superposition and is set to a predefined state (stay with me on this fantasy). Now if we measure observable Y on particle A, then observable X on both particles is in superposition. Once observable X on partible B is measured, it will either have the expected state, which tells us nothing, or it is in the unexpected state, which means that particle A has had observable Y measured. Where did I go wrong? Thanks, Marty 


#6
Sep2910, 01:43 PM

Sci Advisor
PF Gold
P: 5,378

Also: they cannot be in an entangled state with a known X or Y. X/Y must be unknown prior to the measurement. The result will therefore be random. You cannot send much of a signal using a random sequence. 


#7
Sep2910, 05:20 PM

P: 4

Ok.
Many thanks for the clarification. Cheers, Marty 


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