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.


3 Answers 3


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.

  • $\begingroup$ +1 for pointing out the experimental errors in the measurement of $E$. $\endgroup$
    – Our
    Commented Feb 2, 2019 at 22:10
  • $\begingroup$ @martin how can you say virtual particles don't exist? what about the Casimir effect? $\endgroup$
    – user36446
    Commented Jun 14, 2019 at 11:43
  • $\begingroup$ @Pinhead: The Casimir effect is a side-effect of second quantization. If you have an electromagnetic field and promote the fields to operators with conjugation relations, you'll see that there is a difference in zero-point or vacuum energy (after renormalization). This will in turn generate the Casimir effect. No virtual particles required. See also physics.stackexchange.com/questions/109229/… and similar questions/answers. $\endgroup$
    – Martin
    Commented Jun 14, 2019 at 19:47
  • $\begingroup$ @Martin are you some kind of virtual-particle denier? In this video from Sixty Symbols (youtube.com/watch?v=IRcmqZkGOK4) the speaker talks about the Dynamical Casimir effect and he says it was proven experimentally not too long ago. Also, can you explain a little bit more about this electromagnetically induced Casimir effect? Don't ALL quantum fields result in so-called zero point energy after renormalization? $\endgroup$
    – user36446
    Commented Jun 14, 2019 at 20:34
  • $\begingroup$ @Pinhead As you can read in many many questions here, virtual particles are fine, but they are not measurable, hence no physical quantities. In fact, you can show (see the questions I linked) that they are an artifact of perturbation theory. I don't deny that the Casimir effect is real, I just say that you can explain and describe it without virtual particles - the Casimir effect results from a potential difference in the vacuum energy between the plates and around it. It's not interesting that you have a zero point energy after renormalization, it's that you have a difference. $\endgroup$
    – Martin
    Commented Jun 15, 2019 at 14:29

Energy might be conserved in very short time scales, even if it can not be measured, provided we complicate the analysis. Space-time containing no mass, waves, or particles (i.e. so called perfect vacuum or "free" space) exists at a cost of about 6.013x10^-10 J/m^3 (per Stephen Perrenod, Ph.D. astrophysics, Harvard). We might view space-time like a "sponge" that can be crushed or un-crushed as we store or release energy from it. For your example, a W-boson has an energy of about 1.2x10^-8 J. Perhaps space-time was temporarily expanded in the that particle's frame of reference during its formation? We also have to account for energy stored as a result of the presence of the particle's mass of about 1.3x10^-25 kg. I have not seen this derived anywhere but an approximation of energy storage in space-time as a result of a spherical mass M of radius r would be: $$\left.E=\frac{3GM^2}{20\pi r}\right.$$ This approximation does not include energy stored inside the W-boson, for which the radius r may only be about 2x10^-15 m. Using this, energy stored in space-time surrounding the W-boson is about 8.5x10^-47 J; negligible compared to the 1.2x10^-8 J of energy borrowed from space-time by the W-boson. No need to continue the calculation since the space surrounding the new W-boson may have contained even more sources besides space-time from which energy might have been borrowed. What was borrowed was paid back at particle decay time, less the 8.5x10^-17 J, and energy was conserved.

  • $\begingroup$ Can you include a specific reference for 6.013x10^-10 J/m^3? Thanks! $\endgroup$ Commented Nov 3, 2018 at 23:03

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.