In some QFT books is said that a free electron can emit a virtual photon as long as it reabsorbs the photon and returns to its original state within a time:

$$\Delta t<\dfrac{\hbar}{2\Delta E}$$

That inequality DOES VIOLATE the Heisenberg Uncertainty Principle. Why is that POSSIBLE? If it were said in a time

$$\Delta t\geq \dfrac{\hbar}{2\Delta E}$$

I would not be so puzzled.

  • 1
    $\begingroup$ related: physics.stackexchange.com/a/103772/58382. In particular see the fourth paragraph $\endgroup$
    – glS
    Dec 14, 2014 at 12:00
  • $\begingroup$ Can you please give a link to a book with this formula? "some" does not give context. Could it be they are using "Order of magnitude " rather than less? in any case energy conservation would not allow an off mass shell photon to become real. $\endgroup$
    – anna v
    Dec 14, 2014 at 13:48
  • $\begingroup$ Anna, but the uncertainty principle refers to observable variables. What we measure here? Why should be a surprise if for variables that we can't measure, (in or case ΔE) this principle were violated? Can't it be that the above strange inequality tells us exactly that we can't observe ΔE? $\endgroup$
    – Sofia
    Dec 14, 2014 at 14:09
  • $\begingroup$ @Sofia you should quote with @ the user you are addressing otherwise they will never see the comment . Well, I can see it as a box volume delta(e)*delta(t) . virtual is within the box and there it is as you say. So it is a constraint on the box, I think. It makes the box an identity, but I would like to see the context ( if it is both larger than and equal and smaller than and equal, it is equal) $\endgroup$
    – anna v
    Dec 15, 2014 at 4:52
  • $\begingroup$ @annav, your explanations are always wise. Now, a small speculation. You told me that such a particle has a mass different (smaller?) than that of the real particle. Writing its energy as an uncertainty in the total energy involved in the experiment, $ΔE$, and assuming non-relativism, $v^2 << c^2$ we get $ΔE ~ Δm \ c^2$. So, the claimed inequality $ΔE∙Δt < ℏ/2$, becomes $Δt < ℏ/2c^2Δm$. That means, for the usual precision $Δm$ with which we can measure a particle mass, the life-time $Δt$ should be extremely small. (But, I begin to be quite convinced that such particles are only on the paper.) $\endgroup$
    – Sofia
    Dec 15, 2014 at 11:17

4 Answers 4


I think it is important to emphasize that the notion of 'virtual particles' is a very dangerous one, which seems to lead to countless (unnecessary) misconceptions. It appears to have originated from the diagrammatic technique that can be used to carry out perturbative quantum field theoretic computations (i.e. Feynman diagrams), but it is crucial to keep in mind that these pictures are nothing but a computational aid to simplify difficult calculations: They should not be assigned any ontological value!

In particular, the 'virtual photons' that people often mention in semi-popular expositions of quantum field theory don't really exist, and thus I think your fear that Heisenberg's principle may be violated here is unwarranted.


The uncertainty principle is still true in its usual form, but it refers to your knowledge of the state. Suppose your state is just one electron, you can confirm this if you observe the system for a time $\Delta t$ and you don't see additional particles. However, due to the uncertainty principle you can only measure particles that increase the energy of the system above the bound $\Delta E \geq \frac{\hbar}{2\Delta t}$. Therefore, virtual particles below this energy threshold can exist, you are just not able to observe them.


I have been thinking about this question. Tell me what you think:

1) The relation

$$\Delta \tau \Delta \Gamma \geq \dfrac{\hbar}{2}$$

is RIGHT, applied to the timelife AND width of RESONANT states in QFT. A completely stable particule would have "zero width" and infinite timelife.

2) The equation

$$\Delta t\Delta E<\dfrac{\hbar}{2}$$

is ALSO right, but applied not to REAL particles, but to VIRTUAL particles. Taking into account that the definition of virtual particle is one that is off-shell mass according to relativity, the opposite inequality in the Heisenberg Uncertainty Principle becomes possible. The propagator in QFT can not be ZERO, so the HUP vindicates the existence of particles popping out from vacuum in any moment of time and VIOLATING the relativistic dispersion relationship


Suppose that the transferred momentum is VERYLARGE, so we can neglect the rest mass term in the last equation. Thus we have TWO solutions, E=qc and E=-qc, where q is the transferred momentum. In particular, the transferred momentum can be negative...So the inverted inequality is possible. Let me know if I am making some terrible mistake...


The first relation you give is simply wrong (typo in the textbook?). Because for on-shell stable particle $\Delta E=0$, it leads to $\Delta t< \infty$, meaning that any value from 0 to infinity could be possible. It's a non sense, a stable particle cannot live 0 s. The second formula is the good one: the greater the mass (or energy) is shifted from its nominal value, the shorter the particle lives.

@riemannium: (I edit my answer because I don't have enough permission to post a comment about your second post!). You seem to make a distinction in nature between a real particle and a virtual particle. For me, there is no such distinction because all particles can actually be considered as virtual. For instance, consider a photon produced by a star that you detect on earth. I'm sure that you would qualify this photon as real. However, between its production and its detection which destroys it, a finite time has been spent. So, this particular photon has had a finite lifetime. Hence, because of Heisenberg uncertainty, this photon can be in principle slightly off-shell $E^2 - p^2c^2 \ne 0$. Thus this photon is finally virtual! If you think this way, you will see that all particles are actually (more or less) virtual.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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