Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The way I learned it from practicing Fourier analysis and signal processing besides quantum mechanics, is that Energy conservation cannot be achieved in short time scales, and that limits energy conservation in Quantum mechanics.

In other words: Energy conservation is limited by the Heisenberg uncertainty principle in our universe.

$$\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$$

As I posted this answer somewhere, someone said this is wrong. So I posted this issue here, for maybe I don't understand it properly and you guys could tell me why this is wrong.


A beta decay produces a W-boson, which is ~85 times larger in mass than the initial particles (which decays eventually to neutrino and some beta particle), and this is possible due to the very short time scale, during which the W-boson is produced. I.e, energy is not conserved in very short time scales.

Do I understand this correctly? Thank you.

share|improve this question
I didn't say it was wrong, but it wasn't answering the question. –  Hunter Mar 16 '14 at 13:36
I would also be careful calling this 'energy is not conserved', even though I guess the term is defensible in this context. –  Danu Mar 16 '14 at 13:39
@Hunter not you, my friend. It was someone else. –  The Quantum Physicist Mar 16 '14 at 13:41
@Danu That's why I would say "energy conservation is limited by the Heisenberg uncertainty principle", and not energy isn't conserved. I was just expressing the situation for beta decay. –  The Quantum Physicist Mar 16 '14 at 13:43

2 Answers 2

There are several problems with the phrase in italics:

First of all, this is not "Heisenberg uncertainty". The Heisenberg uncertainty principle, as it is understood nowadays, involves the variances of two observables, which are not jointly measureable. Since time is no observable in quantum mechanics (neither in quantum field theory), this equation is not the "Heisenberg uncertainty principle". It is only often referred to as such, because it looks like the original formula. So much for the semantics.

Second, there is no problem with the conservation of energy. There also shouldn't be, because we have time translation invariance and hence should obtain conservation of energy by Noether (roughly speaking). So, energy is conserved within the framework of quantum mechanics. Can we see this? Yes, the unitary evolution of the state commutes with the Hamiltonian (as it is defined as exponential of the Hamiltonian) and thus the energy, which is the expectation value of the Hamiltonian, stays constant over all times.

Note also that you are talking about expectation values (since that's all we can do). The above energy-time uncertainty tells us something about the limits of measurements and preparations. Having a state with an energy E, if we measure this state, we will only be able to determine its energy up to some precision - which is limited by the amount of time we observe the particle. Losly speaking: If I only take a quick look, my measurement will likely be off. Similarly, a state living only a short time, will not have a well-defined energy.

Third, you mention a process in particle physics. It's true, your energy-time uncertainty gets mentioned a lot in quantum field theory and people like to interpret it as short time violation of energy conservation, but to my understanding, that's just not true. The problem is, that all these calculations (and the corresponding diagrams) come out of perturbation theory and if you have a look at nonperturbative exact calculations, the effects are gone - hence they are artifacts of perturbation theory. We just like to interpret them like this, because it gives a meaning to our calculations. In this vein, since all our "off-shell" particles are called "virtual particles", one should call the "borrowing" of energy a "virtual violation".

EDIT: Let me clarify a few of my points.

First of all, if we do agree that the laws of quantum mechanics are time-translation invariant, then we do agree that quantum mechanics has conservation of energy at all times. This is Noether's theorem and we can't get around it.

Now let's talk about two aspects of "violation of conservation of energy": On the one hand (and that is, what we should have in mind), we can have a look at what we measure. And here, as you will certainly agree, we never measure anything breaking the conservation of energy. Virtual particles can't be detected, a particle tunneling through a barrier can't be somewhere, where it would violate energy conservation and so forth. This leads us to conclude that there IS no breaking of energy conservation. See also e.g. here:


On the other hand, we can have a look at the formalism. Still, there should not be any violation, because for closed systems, the conservation of energy is built in as noted above. It all comes down to interpretation. Especially in quantum field theory, where we are even approximating our equations (e.g. perturbation theory), we must be extremely careful with interpreting, what is going on. We have "virtual particles" that seem to violate energy conservation. If however, we have a look at nonperturbative QFTs, lattice gauge theories are the only really interesting example, there are no virtual particles, which questions whether these "virtual particles" are in any way physically real (hint: they are not - we can't detect them), thereby also questioning our other interpretations of the diagrams.

With regard to the time-energy uncertainty relation (which, as time is no operator, does not simply follow from Fourier analysis in a rigorous sense), maybe this will be interesting for you:


Finally, let me remark that this is all very difficult on several levels, so in the end, we might want to focus on what we can actually analyze: the measurement outcomes of experiments, which is of course philosophically unsatisfying for many people.

share|improve this answer
Thank you for your response. Although I don't disagree with everything you said but I don't agree with everything either. And to save lots of unnecessary discussion, I would like to understand your claim: You're claiming that in QFT if we don't use perturbation theory, we won't have any energy violation and energy will be determined with infinite precision, is that what you're saying? If so, can you please provide proof for that in an example in QED? This is the first time I ever hear such a statement. Thank you. –  The Quantum Physicist Mar 16 '14 at 19:10
That's what I meant about Noether's theorem: because of time translation invariance of quantum mechanics, quantum mechanics obeys conservation of energy at all times. I can't give you mathematical proof other than that, because mostly, we don't know any way to do calculations without energy conservation. So instead, let me point you to another explication of the business (see subparagraph on the matter): math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html –  Martin Mar 16 '14 at 20:40
Thank you for the reference. Actually, the way I see it, some physicists are trying to escape the consequences of the wave nature of particles. What I see in that reference, is that they try to relate energy conservation problems to virtual particles, but this is very unjustified, because we definitely know the reason for uncertainty, which is the wave nature of the microscopic world (Fourier). However, then maybe we could go back to much simpler problems with no virtual particles, like Quantum Tunneling, which very clearly breaks energy conservation in the barrier. How can we understand that? –  The Quantum Physicist Mar 16 '14 at 22:34
I disagree, especially, the uncertainty principle does not follow from Fourier, because time is no observable - I try to clarify it with an edit. But in the end, it's interpretation... –  Martin Mar 17 '14 at 10:05
Thank you. I'll look into it and come back :) –  The Quantum Physicist Mar 17 '14 at 10:42

I think the following is a different way of saying what has been said above:

How can energy be conserved AND uncertain? Remember basic vector space theory: every state can be expressed as a linear combination of other states (which form a basis). So a state which is not a state of definite energy can be expressed as a linear combination of states which ARE of (different) definite energies. In each of those states, energy is conserved.

While I'm here, you may see this as related:

Pauli (or maybe Dirac) wrote in a book that there is a symmetry: energy-time is perfectly analogous to momentum-position, and one can think of energy as the momentum a thing has as it travels thru time.

Einstein tells us that one man's space is another man's time, so one man's momentum IS another man's energy.

share|improve this answer
No, energy and time are not perfectly analogous to position and momentum. Both are related by Fourier transform, but there is no operator for time. For the proper interpretation of the "energy-time uncertainty", see this post. –  ACuriousMind May 29 at 13:53
How can time not be perfectly analogous to position, when "proper time" is relativistically invariant (and thus physically real to all of us), and the way we divide that up into "time" and (relative) "position" varies according to how fast we are traveling past the events of interest? What I call (Dx, Dy, Dz, Dt) is what you call (Dx', Dy', Dz', Dt') so some of what I call "time" is -- IS -- part of what you call "position", so how can they not be perfectly analogous? The operator would be for energy, not for time –  Rod May 29 at 21:55
Yes, I am sure. Here's how the apparent contradicition is resolved: Ordinary quantum mechanics is not a relativistic theory, and it has position as an operator, but time as a parameter. Now, as you correctly say, relativity says time and space should be treated on equal footing, so a relativistic quantum theory must either have a time operator, or no position operators. QFT is that relativistic quantum theory, and it has, indeed, no position operators - spacetime coordinates are parameters for operators, not operators themselves. –  ACuriousMind May 29 at 21:56
So time is never an operator, and there is no naive uncertainty relation for time, while there is a non-relativistic position-momentum uncertainty relation. Thus, the pair energy-time is not analogous to the pair position-momentum in this sense. –  ACuriousMind May 29 at 21:56
And I do not disagree that relativity makes space-momentum and time-energy analogous, but this is misleading here, since position-momentum do have a non-relativistic relation of R-S type, while time-energy do not. As the question is about that difference, an answer stating there is essentially no difference is highly misleading, even if it would not be false in a classical relativistic context, and is also not false in a pure QFT context. –  ACuriousMind May 29 at 22:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.