The uncertainty principle is listed in most textbooks and articles as $$ \Delta E \Delta t \geq \frac{\hbar}{2}.$$ This can be derived in many ways in many different settings, most of them involving commutation relations with appropriate operators.

This is often interpreted such that $\Delta E$ is the amount of energy that can be "borrowed" and $\Delta t$ is the time for which it can be borrowed. The uncertainty principle is then used to argue that if $\Delta t$ is large (that is, if the energy is borrowed for long times), then $$ \Delta E \sim \hbar/2\Delta t$$ is small, making this effect negligible on large time scales.

However, the uncertainty relation is an inequality. If $\Delta E$ is actually the amount of energy we can borrow, and if $\Delta t$ is the time for which it can be borrowed, then if $\Delta t$ is large, $\hbar/2\Delta t$ is small, and $$\Delta E \geq \hbar/2\Delta t.$$

This puts no upper limit on $\Delta E$ at all, and in fact gives it a lower limit. $\Delta E$ must be at least $\hbar/2\Delta t$, but it could also be orders of magnitude larger and still satisfy the uncertainty principle. In fact, $\Delta E$ and $\Delta t$ can both be infinite and satisfy the uncertainty principle.

My question is, then, since this argument is so frequently cited in textbooks, what justifications are used for interpreting $\Delta E$ and $\Delta t$ as relating to energy nonconservation? Am I missing something? Is there a reason for the switch from $\geq$ to $\sim$ in the relation? Why don't we observe the infinite violations of conservation of energy (in time and energy) that are predicted by this interpretation of the uncertainty principle? Why are only the minimum-uncertainty cases used in the literature?

I'm assuming there is a reason, and trying to figure it out. Thanks in advance for clarification!

  • $\begingroup$ Related: physics.stackexchange.com/q/53802/2451 $\endgroup$
    – Qmechanic
    Commented Mar 20, 2014 at 1:44
  • $\begingroup$ Thanks! It's a good reference to have. I have seen that derivation before in lecture notes somewhere. That more or less gives a kind of relaxation time in the energy of the system to equilibrium. That is one way to derive the time-energy uncertainty principle, and illustrates that it takes a > and not a ~. $\endgroup$ Commented Mar 20, 2014 at 1:51

1 Answer 1


Why don't we observe the infinite violations of conservation of energy

The reason we don't see e.g. an atom spontaneously turning into a red giant for a fraction of a second is because of the extremely small timescales that such an energy difference would require - not even light could travel a tiny fraction of a proton radius in that time.

The heavy virtual particles which carry the "borrowed" energy (e.g. W bosons in $\beta$-decay) certainly can't get very far before decaying into more energy-cheap products.

I suppose, in fact, this kind of event would technically contribute a (negligible) cross-section to some processes... I can imagine a nice simple Feynman diagram of Compton scattering being augmented with $10^{60}$ $\tau^+ \tau^-$ pairs produced by low-energy photons at enormous energy cost (which would briefly outweigh the sun), but they wouldn't have time to travel any distance at all or interact with anything, let alone be detected by us. They'd also require a similarly huge number of vertices in the diagram and so the rate of such an interaction would be negligibly small (something like $10^{-120}$ per second and the universe is only around $10^{17}$ seconds old).

Why are only the minimum-uncertainty cases used in the literature

Switching from "≥" to "~" is not the same as switching to "=" so it's not using the minimum uncertainty case necessarily. The "~" just denotes a similar order of magnitude or that the uncertainty in energy is basically on the same scale as $\frac{\hbar}{2\Delta t}$.

I imagine it appears in many textbooks as a (sort of hand-wavy) explanation for how heavy virtual particles can be exchanged between lighter particles which lack sufficient invariant mass to produce such energies in the centre-of-mass frame.

  • 1
    $\begingroup$ You really should be studying for your exam on Wednesday! +1 though. $\endgroup$
    – PhotonBoom
    Commented May 3, 2015 at 1:44
  • 1
    $\begingroup$ Hahah yeah I should... at least I try to steer myself towards topics that might be loosely classed as revision. $\endgroup$
    – jamcowl
    Commented May 3, 2015 at 2:01

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