| Thread Closed |
Understanding superposition and entanglement |
Share Thread | Thread Tools |
| Sep29-10, 10:23 AM | #1 |
|
|
Understanding superposition and entanglement
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 |
| Sep29-10, 10:43 AM | #2 |
|
|
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 non-commuting observables. Occasionally, you can observe entanglement continuing for commuting operators. |
| Sep29-10, 11:28 AM | #3 |
|
|
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 no-communication theorem? Thanks again, Marty |
| Sep29-10, 12:27 PM | #4 |
|
|
Understanding superposition and entanglement |
| Sep29-10, 01:10 PM | #5 |
|
|
Suppose I had a device which produced two entangles particles A and B with non-commuting 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 |
| Sep29-10, 01:43 PM | #6 |
|
|
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. |
| Sep29-10, 05:20 PM | #7 |
|
|
Ok.
Many thanks for the clarification. Cheers, Marty |
| Thread Closed |
| Tags |
| entanglement, superposition |
| Thread Tools | |
Similar Threads for: Understanding superposition and entanglement
|
||||
| Thread | Forum | Replies | ||
| Superposition | Introductory Physics Homework | 3 | ||
| Am I understanding superposition correctly? Is it equivalent to "all-potential"? | Quantum Physics | 4 | ||
| superposition representation of particle state in 1-d infitne well (SUPERPOSITION?) | Advanced Physics Homework | 2 | ||
| Is understanding one branch of math conducive to understanding another? | General Math | 4 | ||
| Proving the superposition of initial conditions gives superposition of motion | Introductory Physics Homework | 2 | ||