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Does uncertainty principle imply nonconservation of energy? 
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#1
May906, 03:34 AM

P: 400

Hi,
I was wondering about the following question. Does Heisenberg's EnergyTime uncertainty inequality (ΔE.Δt=>h/2) imply nonconservation of energy? I mean, if the total energy of the system fluctuates, then how can the energy be conserved? Does the COnservation of Energy so fundamental to classcial and relativistic physics break down in quantum mechanics? Thanks. Molu 


#2
May906, 03:47 AM

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PF Gold
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It is possible to break the energy conservation law but if you look yourself at the equation you will see that this can only be achieved for very small periods of time. For example try working out how long a stationary electron can exist by plugging its rest mass in the equation and finding delta t.



#3
May906, 06:02 AM

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PF Gold
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The longer answer can be this: Given the probabilistic aspect of quantum theory, what do we mean now, by "conservation of energy" ? In quantum theory, it can be expressed in two different ways. The first way is this: A state with a precisely known energy will always keep this energy. The reason is that a state with a precisely known energy is an eigenstate of the Hamiltonian, and that's a stationary state under unitary evolution: so it remains (up to a phase factor) itself. The second way goes as follows: for a given state, look at its EXPECTATION VALUE of the energy <psi  H  psi>. This is the statistical average over many trials of measuring the energy. Well, this expectation value is to remain the same during time evolution. It simply means that the energy value was not a welldefined quantity (only its expectation value was), and hence we cannot talk about violation of its conservation, given that it wasn't fixed initially. And where does the timeenergy 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. So this means that when you are discussing about a system for a time less than dt, that there is no visible difference between a stationary state with precisely energy E, or with a superposition of stationary states of which the energy eigenvalues lie within dE of E. Indeed, below a time dt, the unitary evolution equation (Schroedinger equation integrated) will not have altered significantly the phases between these contributions as can easily be verified (each term taking a factor exp(i E t / hbar) ). So you're not able to find out the difference between the two situations, and hence you cannot know whether the system is in such a superposition, or in a precise energy eigenstate. As such, the uncertainty is a matter of uncertainty on the INITIAL condition (was the system in a pure energy state or not ?) or of the energy transfer during the measurement interaction between apparatus and system. It is not a question of "stealing energy from nowhere" or something of the kind. There are two typical cases: 1) the system was "created" in a time dt. This means that during its "creation interaction" one cannot be sure that it was in a pure energy eigenstate: it could be created in a superposition of eigenstates with eigenvalues spread over dE. So you measuring (precisely) the energy value just means you selected out one of the possible eigenstates of which the system was in a superposition: no violation of conservation of energy. This is often the case with "particle resonances" or other shortlived phenomena. 2) The system is *prepared* in a precise energy eigenstate, and you quickly measure, during time dt. In this case, it can be shown that the interaction between the system and the measurement apparatus can give rise to a transfer of energy of order of dE. So reading again another value (within dE) of the energy is then just part of the "perturbation" introduced by the energy measurement apparatus. Again no violation of energy conservation. Finally, in the long run, the *expectation value* will be recovered, as the average of a great many number of measurements. So there will never be a net gain or loss of energy. 


#4
May906, 07:57 AM

P: 4,006

Does uncertainty principle imply nonconservation of energy?
Nice explanation Vanesch. I am gonna steal this text and put my name under it when a future question like this pops up :)
marlon 


#5
May906, 11:27 AM

P: 2,954

Pete 


#6
May906, 01:12 PM

P: 400

Hmmm, the second half of that message went over my head, but I think I got my answer in the first part. As one can not assign a definite value to energy, there is not a definite value to be conserved. In a given time interval Δt, VonNeuman measured values of E will not stay the same but vary within ΔE.



#7
May906, 11:50 PM

P: 165

There exist many other formulations of the uncertainty principle for energy and time on which we shall only comment briefly. Some formulations are simply wrong, such as the statement that for a measurement of the energy with accuracy dE a time dt>hbar/dE is needed. This statement is wrong because it is an assumption of quantum mechanics that all observables can be measured with arbitrary accuracy in an arbitrarily short time and the energy is no exception to this. Indeed, consider a free particle; its energy is a simple function of its momentum and a measurement of the latter is, at the same time, a measurement of the former. Hence, if we assume that momentum can be accurately measured in an arbitrarily short time, so can energy...And right you are, Pete. 


#8
May1006, 04:31 AM

P: 400

So what DOES the ET UP mean? It seems to be a much more complicated concept than the xp one. We can measure the energy of a system in a particular instant of time, no time interval is needed for this. Then where does the inequality come in?



#9
May1106, 02:11 AM

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#10
May1206, 02:24 AM

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#11
May1206, 07:01 AM

P: 4,006

Also, the HUP does NOT imply a violation of total energyconservation. Suppose you have a process with initial and final state. When you compare the initial and final energies, there will not be a violation. The energyconservation law, by definition, only applies to initial and final states (ie the external Feynmann lines). It is in the intermediate states that energyconservation can be violated because of the energyuncertainty. However, the energyconservation law does not apply to such states (ie internal feynmann lines). Momentumconservation is respected always because it applies to both internal and external Feynmann lines. Virtual particles only arise in that specific period of time where the energy is uncertain (so in the internal Feynmann lines) and they are the QFT variant of the successive intermediate quantumstates that a system goes through when evolving from the initial to the final state in QM. Given this analogy, it always sounded strange to me that people bring up this apparent energy nonconservation within the context of virtual particles but not within QM context. regards marlon 


#12
May1206, 07:51 AM

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what *is* spe ial about internal lines is that they are offshell, that is their momenta and energies do not obet the dispersion relation E^2=c^2p^2 +m^2 c^4. But energy and momentum *is* conserved at the vertices. Patrick 


#13
May1206, 07:58 AM

P: 4,006

Total energy conservation only applies to final and initital states, NOT intermediate states. What i mean with this is the fact that during the intermediate states, the energy is uncertain. Now, indeed one can "pick (ie measure)" the "correct" energyvalue during these states, so that energyconservation is respected. But, other energyvalues are possible as well so in other cases energyconservation is NOT respected. To summarize, one can say the energy conservation CAN be respected in intermediate states but it does not have to be like that. This is also expressed by the fact that not all gauge bosons/ force carriers are virtual. regards marlon 


#14
May1206, 08:18 AM

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Since you may not be familiar with OFPT, let me put it this way. Take a oneloop Feynamn diagram, say. The "energy" integration can be carried trivially in the complex plane, the i epsilon prescription telling us how to do it. It picks discrete values for q_0, it does not integrate over a *continuum* of q_0. It really is different than the threemomentum integration. So one cannot say that *all* values of q_0 are included! What you are left with are two diagrams (corresponding to two OFPT diagrams). How would you define the energy of an intermediate state then? I am saying thatif you draw them as the two OFPT and use the values of q_0 that were picked by the poles, those values will correspond to the initial energy of the system! But I won't argue if this makes no sense to you. Regards Patrick 


#15
May1206, 08:27 AM

P: 4,006

The QM perturbationtheory that you learn in college (and i am pretty sure we are both talking about this one) the concept of virtual transistion states and the connection to energy conservation is very clear and is exactly as i have described it. No point in arguing that. marlon 


#16
May1206, 08:45 AM

P: 4,006

When talking about intermediate states and virtual particles, the fluctuation of energy during such states is a key concept. The influence of the HUP and the direct connection to virtual particles that, by definition, do NOT respect total energyconservation is very clear and straightforeward. Now, one can always start reciting old models "that nowbody knows about" and that somehow seem to be violating mainstream physics. While doing so, i ask you this : how do you bring in the above mentioned key concepts into your old model ? What is the connection between the two ? marlon 


#17
May1206, 10:06 AM

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