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Is it possible to violate conservation of energy in quantum? |
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| Mar15-10, 09:34 PM | #1 |
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Is it possible to violate conservation of energy in quantum?
Me and my roommate have being talking about the conservation of energy and the uncertainty principle and we are wondering if its possible to violate conservation of energy for brief periods of time or not. We have seen various interpretations of the energy-time uncertainty relations but cannot come to a conclusion.
Can we get some feed back here on this idea. Thanks |
| Mar15-10, 10:28 PM | #2 |
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I'm no expert, but here is a link:
http://www.phys.ncku.edu.tw/mirrors/...particles.html (see conservation of energy) I think the energy-time uncertainty itself for normal particles does not violate energy conservation. The uncertainty relation is really only a different way to think about energy and time. You have to get into that QM maths to see the difference to classical concepts. Basically there isn't such a thing like a point particle with a velocity anymore. |
| Mar16-10, 04:50 AM | #3 |
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The Hamiltonian is an operator exactly conserved in time. Thus, if the initial state is an eigenstate of the Hamiltonian with the eigenvalue E, then, as far as evolution is governed by the Hamiltonian, the state at any other time is a Hamiltonian eigenstate with the same eigenvalue E.
However, the state at a given time can be expanded in any basis, including a basis which consists of states which are not Hamiltonian eigenstates. By doing an appropriate measurement at that time, the state may "collapse" to one of these states, which may cause apparent violation of energy conservation. However, the "collapse" is not a unitary evolution governed by the Hamiltonian. When the "collapse" is replaced by a more appropriate description of measurement in which a true collapse never occurs, it turns out that the total energy (i.e. the sum of energies of the measured system, measuring apparatus, and all the environment) is conserved again. |
| Mar16-10, 06:38 AM | #4 |
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Is it possible to violate conservation of energy in quantum? |
| Mar16-10, 06:50 AM | #6 |
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http://math.ucr.edu/home/baez/physic...particles.html He only says they are "an approximation to QM" |
| Mar16-10, 08:36 AM | #8 |
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| Mar16-10, 01:00 PM | #9 |
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I have been looking at Wikipedia and some other sources and we are still trying to figure this out :) and get settled on an interpretation. However I think there may be a conflict with your statement in your response and Wikipedia. What you want to do about it is up to you. Im not sure if its an issue. From you: And where does the time-energy uncertainty relationship come in ? It tells you esentially that *in order to perform an energy measurement with precision dE*, you will need to measure (to have your measurement apparatus interact with) the system for a time of at least dt. From Wiki: One false formulation of the energy-time uncertainty principle says that measuring the energy of a quantum system to an accuracy ΔE requires a time interval Δt > h / ΔE. This formulation is similar to the one alluded to in Landau's joke, and was explicitly invalidated by Y. Aharonov and D. Bohm in 1961. The time Δt in the uncertainty relation is the time during which the system exists unperturbed, not the time during which the experimental equipment is turned on. Thank you for your response! |
| Mar18-10, 01:42 PM | #10 |
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I want to to know this too. |
| Mar18-10, 03:51 PM | #11 |
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The issue of conservation of momentum/energy in quantum mechanics is best illustrated using the familiar single-slit setup. When an electron of definite momentum is directed at the plate with a single slit, it becomes a superposition of momentum states when it passes through the slit. With regard to conservation laws, this representes the uncontrollable exchange of momentum between the plate and the electron, since the plate is assumed to have effectively infinite mass, i.e. it is treated as a classical body. Because of the fact that the plate is treated as part of the experimental apparatus, i.e. the objects that define the possibility of space-time coordination of the quantum phenomenon under investigation, a control of energy-momentum exchange with it is excluded, which also happens to be the reason why in any given experimental set-up, the quantum formalism allows only probabilistic predictions. Here, one must keep in mind that the definition of energy-momentum is not the classical one, but the one connected with the relations (E, P) = (hf,hσ).
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| Mar19-10, 05:16 AM | #12 |
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| Mar19-10, 12:31 PM | #13 |
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| Mar19-10, 12:38 PM | #14 |
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I'm not a Bohmian, but even then I don't find your argument terribly convincing because I believe that the "indivisible quantum process" SHOULD be explicable with a complete "theory of eveything". Where a Bohmian argues for their Interpretation, and I won't put words into your mouth as to what you would do, I think recognizing that while QM formalism is incredibly useful it comes with more and more baggage. No interpretation based on such a blatantly incomplete theory with artifacts such as renormalization, can really be valid, given the foundations. |
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