I am self-studying QFT and I was learning about the Higgs field (lectures 7 - 8 in the link), and I learned how massless particles behave in the Higgs field. What I saw was that they follow this equation:

$$\Phi = fe^{i\alpha }$$

Where $f$ is the fixed magnitude of the field at the main circle of the potential (the bottom of it), and $\alpha$ is the rotation around this circle. Massless particles move around this main circle by slowly varying $\alpha$.

I learned that mass is the QF that oscillates back and forth when you displace it everywhere simultaneously homogeneously, and there is a restoring "force" that wants to bring it back in place that causes it to oscillate. Massless particles are the opposite, they have no restoring "force" that brings it back when you displace it, and therefore do not have mass.

The Question

A consequence of the Heisenberg uncertainty principle is that nothing can ever standstill or not oscillate. Everything has to oscillate.

With this consequence, then, would it make sense that massless fields would oscillate too like the mass fields? Would this give everything mass, or is there something critical that I am missing here, like misunderstanding the Heisenberg uncertainty principle?

  • 1
    $\begingroup$ Linked. Goldstone's sombrero is a shared metaphor, via a mechanical analog of classical macroscopic marbles (not particles!) rolling or oscillating at the bottom of the sombrero, to help you intuit the two degrees of freedom involved via the mass and massless pieces. You then apply them to the oscillations of the quantum fields in flavor space, not real space! You took the metaphor to a fantastical place. Write the quantum fields in terms of their constituent oscillators, if you wish, to see there is no quantum fluctuation at that level. $\endgroup$ – Cosmas Zachos Feb 19 at 15:53
  • $\begingroup$ That being said, radiative corrections to QF may "slightly" curl up flat directions, as Coleman and Weinberg demonstrated, but this far outranges the scope of your question... $\endgroup$ – Cosmas Zachos Feb 19 at 15:54
  • $\begingroup$ @CosmasZachos I know it is a metaphor and it is not a real space, but rather it is an abstract space that allows you to understand the idea. Now, I think that the natural question for me to ask is, does it really oscillate (the mass) in real space? I feel like I was lied to in the lectures when the professor said that the field oscillates and that is the mass. I am just trying to dig deeper and understand this. $\endgroup$ – Tachyon Feb 19 at 16:21
  • 1
    $\begingroup$ Fair enough; sloppy generic statements are not lies, but invitations for further study. Take a standard text, like Peskin & Schroeder, or whichever, and compute $\langle 0| m^2\int dx \phi (x) \phi(x) |0\rangle$ for a real field, to see how a mass term contributes to the vacuum energy, or not (if m vanishes). $\endgroup$ – Cosmas Zachos Feb 19 at 16:34
  • $\begingroup$ @CosmasZachos Thanks, I will continue my studies with this, and with time and experience, I'll understand it soon enough. $\endgroup$ – Tachyon Feb 19 at 16:55

A consequence of the Heisenberg uncertainty principle is that nothing can ever standstill or not oscillate. Everything has to oscillate.

I do not think this is a correct deduction. If one measure momentum there is a limit to the accuracy of measuring position , is all that the HUP says. It is an envelope where measurerements of momentum and position at the same time have to be. Nothing about oscillations there.

Also, quantum mechanics is a probabilistic theory, many measurements under the same boundary conditions must be carried out to get a probability distribution that can be checked against calculations.



defines a probability for the variables to fluctuate, the fluctuations due to the uncertainty are in the probability distributions, a probable state to exist, but it is not oscillating in space time,(or energy-momentum) as it is the wavefunction that has the wave behavior. In space time there is just a probability of measurement that can be calculated. ( example: orbitals not orbits).

  • $\begingroup$ Then why did I hear many people (even physicists) saying that a consequence of the Heisenberg uncertainty principle is that nothing stands still and because of this, there are quantum fluctuations in the vacuum which gives rise to virtual particles coming in and out of existence? I am quite confused now as I see these contradictions. But I do see what you are trying to say. $\endgroup$ – Tachyon Feb 19 at 16:27
  • $\begingroup$ the fluctuations due to the uncertainty are in the probability distributions, the HUP allows for a Delta(p) given a Delta(x) for probable states to exist, or delta(E) versus delta(t)., but it is not oscillations $\endgroup$ – anna v Feb 19 at 16:32
  • $\begingroup$ I see what you are saying. I think that clears up my confusion, thanks. If you could include what you just said in your answer, I will checkmark it. $\endgroup$ – Tachyon Feb 19 at 16:52
  • $\begingroup$ just edited . I am not addressing your main question, "whether an elementary particle can have a variable mass", because I am bad in calculations. One would have to see at what delta(t) and delta(E) the HUP would be violated. As h_bar is tiny and the mass of the electron is measured to 30 decimal places, I suspect the experiment cannot be done. $\endgroup$ – anna v Feb 20 at 5:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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