# Does uncertainty principle imply non-conservation of energy?

1. May 9, 2006

### loom91 Hi,

I was wondering about the following question. Does Heisenberg's Energy-Time uncertainty inequality (ΔE.Δt=>h/2) imply non-conservation 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. 3. May 9, 2006

### Kurdt Staff Emeritus
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.

4. May 9, 2006

### vanesch Staff Emeritus
The short answer is: no, there's no violation.

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 well-defined 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 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.

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 short-lived 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.

5. May 9, 2006

### marlon Nice explanation Vanesch. I am gonna steal this text and put my name under it when a future question like this pops up :)

marlon

6. May 9, 2006

### pmb_phy Note: ΔE.Δt=>h/2 isn't really a Heisenberg Uncertainty Relation because there t is not an observable, i.e. there is no hermetian operator corresponding to time. One must proceed with caution when interpreting the meaning of this expresssion.

Pete

7. May 9, 2006

### loom91 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, Von-Neuman measured values of E will not stay the same but vary within ΔE.

8. May 9, 2006

### koantum Huh? You can't know a given system will always keep a particular energy, can you? So how can you know its precise energy without waiting till the end of time?
The following is a quote from an excellent article by Jan Hilgevoord, The uncertainty principle for energy and time (American Journal of Physics 64 (12), pp 1451-6):
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.

9. May 10, 2006

### loom91 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?

10. May 11, 2006

### koantum As there are several wrong uses, so there are several correct uses, as Hilgevoord points out in the aforementioned article. Foremost among these is the relation between the energy width and the lifetime of a quantum state. Note that there is a completely analogous relation between the momentum width of a state and its spatial translation width, and that this is conceptually different from the limitation imposed on the simultaneous measurability of position and momentum by the commutation relation of the corresponding operators.

11. May 12, 2006

### loom91 You mean that a higher-energy quantum state will be short-lived? How can that be possible? A quantum state does not collapse until you measure it, so why would it be short-lived if it had higher-energy?

12. May 12, 2006

### marlon Again, you are not interpreting the HUP correctly. If the delta t is small, the delta E gets big. This does NOT imply that the energy of the state is bigger, it implies that the spread of energyvalues that you get after consecutive measurements is bigger. So we are talking about spread of energyvalues and not absolute energyvalues.

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 energy-uncertainty. 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 non-conservation within the context of virtual particles but not within QM context.

regards
marlon

13. May 12, 2006

### nrqed Actually, energy is conserved even by intermediate states (internal lines). The best way to see this is to use "old-fashioned perturbation theory" (where one uses those nifty time-ordered diagrams or "Z" diagrams) where one imposes explicitly conservation of both energy and momentum at all vertices. People don't use them much because for any loop process there are several OFPT diagrams and their individual expression is not covariant (for one thing, the integration is only over three-momenta). But when they are all summed up, the result *is* covariant (although it is not obvious). When they are summed up, one can actually introduce a 4th integration (over the energy), introduce an i epsilon prescription for handling of the poles and lo and behold, the total combination of all the OFPT diagrams sums up to a Feynman diagram with the usual Feynman rules (and it is obviously covariant)! So it is much faster and simpler to work with the Feynman rules. But they hide the fact that energy and momentum is conserved.

what *is* spe ial about internal lines is that they are off-shell, 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

14. May 12, 2006

### marlon I am sorry but i do not follow. I agree on the momentum conservation but the energy conservation is a totally different story. If total energy were conserved at all times (which is what you are saying because you also account for the intermediate states) , "virtuality" could not exist. What you are saying is therefore impossible.

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 energy-conservation 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

15. May 12, 2006

### nrqed What I am saying is that it is entirely possible to do the entire calculation is a consistent way by having enery conserved at all steps (using old fashioned perturbation theory which is, granted, almost completely unknown to most people). In that approach, one only integrates over three-momenta. Then the question is: what is the energy of an intermediate state? *If you use the mass-shell condition* for the intermediate states, then yes, energy will be violated. If they are off-shell, then you can always say that energy is conserved.

Since you may not be familiar with OFPT, let me put it this way. Take a one-loop 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 three-momentum 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

16. May 12, 2006

### marlon What exactly do you mean by "old fashioned perturbation theory" ?
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.

Indeed
I don't get this. The energy violation comes from the HUP (uncertainty in energy in the intermediate states) and as a consequence of that, particles are off mass shell. That is the story.

So but where is the spread in energy during the intermediate states. Indeed, if you pick the values by the poles, particles are on mass shell and all is ok. I agree, but what about the spread in energy?

marlon

17. May 12, 2006

### marlon Well, this is always the easy way out, so let me ask you this :

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

18. May 12, 2006

### nrqed you are right, I am just a crackpot.

19. May 12, 2006

### marlon Ok, what a mature answer. I asked you two specific questions that you do not seem to be able to answer. Good work my man...

regards

marlon

20. May 12, 2006

### George Jones Staff Emeritus
So, you're saying that 4-momentum, which has energy as one component, is not conserved at every vertex.

Regards,
George

21. May 12, 2006

### nrqed If you explain something and someone's first reaction is to say :"No, what you are saying is impossible. Case closed" then what is the point of trying to keep explaining? Your mind is already set. But I will give another try (even though it's probably a waste ot time since you already know that I am wrong)

The fact that you haven't heard about it does not imply that nobody has.
If you had read the original work of Feynman, Bethe, Stueckelberg, etc, you *would* be familiar with this (you obviously are not familiar with their original work). It is sometimes referred to as "Old fashioned Perturbation Theory" or "Non-covariant Perturbation theory" (which is a misnomer..it *is* covariant but not *explicityly* covariant) or "time ordered perturbation theory". ANY DIAGRAM in the "modern approach" can be written as a sum of time ordered diagrams, the two approaches are EQUIVALENT. (you just to do the contour integrals over the zeroth components and you end up with a sum of integrals over three-momenta). It is not a "model" or a discarded theory. It is of course more convenient to work directly with covariant diagrams because they contain all the time ordered ones and the expressions are manifestly covariant.

See for example (I did not have access to my books earlier today):
"Space-Time approach to Quantum Electrodynamics" by Feynman, Phys Rev vol 75, p.486. 1949. Look in particular at the two diagrams of figure 5 for Compton scattering. In the "modern" approach one would only draw on diagram.
(the article is reprinted in "Selected Papers on Quantum Electrodynamics", Edited by Julian Schwinger, Dover.)

Time ordered perturbation theory is covered briefly in "gauge Theories in Particle Physics" by Aitchison and Hey (sections 5.6 to 5.9..look in particular at Fig 5.15).

A quote from this book (p 155, second edition)"

"With this step we have made a fundamental reinterpretation,. In figures 5.7(a) and 5.7(b) the photon has zero mass but energy is not conserved at the vertices. This is, as repeatedly emphasized, perfectly normal in second-order quantum mechanical perturbation theory. One calls the states |n> "virtual" as well as "intermediate" for this reason. However, this is an intrinsically non-covariant statement, since energy is singled out. By contrast, we can interpret fig 5.8 quite differently. We can interpret q=p_a-p_a' as the *4-momentum* of the photon, assuming covariant 4-momentum conservation at each vertex. But then the mass of the photon is q^2 which is not zero! Here also we call the photon "virtual""

where figure 5.8 is simply a usual covariant Feynman diagram (figs 5.6 are time ordered diagrams). The term between ** is their emphasis. The bolded part is my emphasis.

The point is that four-momentum is conserved so energy as well as momentum is conserved at the vertices. It does not make sense to say that one quantity is conserved but not the other in a relativistically covariant theory!

I wil try to make it as simple as possible: consider any Feynman diagram vertex. The Feynman rule contains a four-dimensional delta function, right? This ensures that FOUR-MOMENTUM IS CONSERVED AT THE VERTICES! How can energy conservation be violated if four-momentum is conserved????? It does not matter if we are dealing with tree diagrams or loop diagrams...for *any* loop four-momentum the delta function insures that 4-momentum is conserved in all the intermediate states!! If this line has more 4-momentum, this other line will have less 4-momentum, in such a way that the total four-momentum of the intermediate state is equal to the total four momentum of the initial state. How can you have violation of energy in that case??

You seemed to think that I was naively going back to QM perturbation theory when it is actually you who was extrapolating a conclusion from QM to quantum field theory.