# Do particles with exactly zero energy exist?

In my understanding, in Newtonian mechanics if something has no mass it cannot be said to "exist" since it cannot possibly have energy or momentum and thus cannot participate in interactions or be detected.

I believed that this is also the case in relativistic physics with energy in place of mass. The complete absence of energy is only possible for a massless particle of zero momentum. The question is whether such particles "exist", i.e. affect physical processes in any way?

I always assumed that the answer is negative. On the other hand, consider the massless scalar field with creation operator $a^\dagger(\vec{p})$. Then the state $$a^\dagger(\vec{0})|0\rangle:=a^\dagger(\vec{p})|0\rangle\Big|_{\vec{p}=0}$$ does not look to me as flawed in any respect compared with the states of non-vanishing $\vec{p}$.

To summarize: is the concept of a massless particle with vanishing momentum meaningful experimentally or theoretically?

Edit: making it harder to ignore let's assume that the particle we've created above posseses an electric charge. As far as I am aware there is no principle prohibiting massless particle to carry a charge (even if it has zero energy).

• I would disagree. For instance, properties of massless and massive (although arbitrarily light) particles are quite different. The question whether there are particles of exactly zero mass makes sense. It seems to me that a massless particle having vanishing momentum is also quite special (for example, this property is frame-independent). It is not obvious for me that the question of their existence (or at least theoretical importance) does not have a direct meaning. Commented Jul 14, 2015 at 10:00
• I agree there is a discontinuity between massive and massless particles. But this is because the symmetry of the system is enhanced in the limit where the masses of gauge bosons are zero (you get a gauge symmetry). There is no enhanced symmetry when you go between zero and non-zero energy particles... Furthermore, energy isn't a fundamental property of a particle. Its not even Lorentz invariant. Commented Jul 14, 2015 at 10:06
• In special relativity all zero mass particles move with velocity c, and E^2-p^2 =0, (the mass). special relativity is validated innumerable times in the lab. Commented Jul 14, 2015 at 11:10
• Massless particles cannot have an electric charge physics.stackexchange.com/a/7919/23473
– Jim
Commented Jul 14, 2015 at 13:31
• My comment means that data trumps theory (physics), philosophy expects theory to form the data. As I tried to show in my example with the expansion in a series fitting an object, the mathematics does not form the object , it just fits it as a tool to be able to enter it in a computer, for example, or another calculation. Those series components are virtual, they have no physics status. Commented Jul 14, 2015 at 13:58

The concept of a particle with exactly zero energy is rigorously meaningless.

The issue is that the quantum field is not an operator, it is an operator-valued distribution. Therefore, strictly speaking, you can't apply $\phi(x)$, $a(p)$ or $a^\dagger(p)$ to anything, but you have to smear these things out. Strictly speaking, $\phi(x)$ doesn't even mean anything, as distributions live on the space of test functions, not on spacetime itself. Therefore, you can't actually speak of the state $a^\dagger(p)\lvert \Omega \rangle$, but should speak about something like $\int a^\dagger(\vec p) f(\vec p) \mathrm{d}^3p \lvert \Omega \rangle$ for some "profile" $f\in C_c^\infty(\mathrm{R}^3)$, which does not possess a definite energy, in particular not zero.

This is analogous to saying that the QM momentum eigen"states" $\lvert p \rangle$ for a free Hamiltonian do not lie in the Hilbert space of states, but only the wavepackets of uncertain momentum constructed from them.

• This sounds quite reasonable, hm. Initially, my question arose in connection with the symmetry breaking. In case of a broken symmetry it said to be realized non-linearly: degenerate are states with different number of the Goldstone bosons (which are massless) of zero energy. I wondered if such states make sense and your answer seems to imply that they do not. However, this may be simply an abuse of terminology and the statement could be properly cast in terms of distributions. I'm not sure whether it is a reasonable or a purely pedantry question. Commented Jul 14, 2015 at 15:05
• Is it the case then that the concept of a particle with exactly $E$ energy is rigorously meaningless? Does this argument apply generally? I cannot see where it is limited to $E=0$? Commented Jul 14, 2015 at 15:22
• @innisfree: Yes. We use it nevertheless because smearing everything with functions narrowly peaked around the momentum/energy we wish to speak of is annoying, and doesn't change the results (or at least the form of the reasoning). Commented Jul 14, 2015 at 15:24
• @ACuriousMind Another question. Is the "vacuum" a state in a Hilbert space? Usually, we assume it to be both i) normalizible $\langle0|0\rangle$ and ii) possesing definite energy (say zero) $H|0\rangle=0$. Your answer imply that these two properties are in contradiction for a generic state. Is this also true for the vacuum or we have an exception here? Commented Jul 14, 2015 at 16:51
• @WeatherReport: The vacuum $\lvert \Omega \rangle$ is not of the form $a^\dagger(p)\lvert \Omega \rangle$, so my argument doesn't apply to it. It is usually by assumption an eigenstate of $H$ (which is a proper operator). Commented Jul 14, 2015 at 17:06

The emission of massless particles (e.g. photons) with zero momentum (or momentum tending towards zero) in the rest frame of a charged particle is called collinear emission.

Collinear emission is somewhat problematic for massless particles, because it results in a so-called IR divergence that cannot be removed by renormalization (cf. UV divergences). The resolution to this problem is resolution: the collinear emission is experimentally indistinguishable from the case in which there was no emission, as one cannot detect arbitrarily low-energy photons. When making a prediction, one must sum the differential cross-sections for collinear emission and no emission, which are both divergent. The sum results in a cancellation of the divergent terms.

So, does the zero energy particle exist? This really depends on what you mean by exist. I would say that the particle didn't exist, because the of arbitrarily low-energy photons cannot be distinguished from no emission at all. On the other hand, though, without them, the IR singularities wouldn't cancel, so the inclusion of real, zero-energy emission is important.

• Ha! "The resolution to the problem is resolution" Great line. Though, you could say "...is the resolution of the problem" and still have it carry the meaning you intend. Then it's both a tautology and a meaningful explanation. :)
– Jim
Commented Jul 14, 2015 at 15:29
• You seem to be talking about photons with arbitrarily small (but non-zero) energies. There is no doubt that they exist since an arbitrarily low energy could be obtained/observed simply by going to the appropriate reference frame. In my view, your answer does not address the issue of exactly zero-energy photons. Commented Jul 14, 2015 at 15:29
• @WeatherReport A photon is a self-propagating disturbance/wave through the background EM fields. A photon with zero energy would have an infinite wavelength. Having an infinite wavelength would mean it is no longer a disturbance in the background fields, but rather is a "DC" signal that would then be the background fields. Thus, it wouldn't actually be a photon. It would be nothing.
– Jim
Commented Jul 14, 2015 at 15:33
• @JimsBond Would a DC electric potential in the universe really be nothing? Perhaps the answer should address that. Commented Jul 14, 2015 at 15:38
• @WeatherReport I am talking about photons of all energies $E<\Delta E$, where $\Delta E$ is the smallest detectable energy. Commented Jul 14, 2015 at 15:41

Do particles with exactly zero energy exist?

No.

if something has no mass it cannot be said to "exist" since it cannot possibly have energy or momentum and thus cannot participate in interactions or be detected.

A photon has no mass, but it does have energy-momentum, it does participate in interactions, and it can be detected. It exists. Perhaps the word "mass" is the issue here. When we say mass without qualification it's assumed to mean "rest mass". The photon has no rest mass, but it does have a non-zero "active gravitational mass" and a non-zero "inertial mass".

I believed that this is also the case in relativistic physics with energy in place of mass. The complete absence of energy is only possible for a massless particle of zero momentum. The question is whether such particles "exist", i.e. affect physical processes in any way?

They don't exist. Nor do any zero-inch rulers.

I always assumed that the answer is negative. On the other hand, consider the massless scalar field with creation operator...

The problem here is that the creation operator is an abstract mathematical "construct" that substitutes for a clear physics understanding of how say gamma-gamma pair production actually works. The gamma photons do not pop out of existence courtesy of an annihilation operator, and the electron and positron do not pop into existence courtesy of a creation operator. Have you ever read the given explanation for this? "A photon can, within the bounds of the uncertainty principle, fluctuate into a charged fermion–antifermion pair, to either of which the other photon can couple". Pair production occurs because pair production occurs. Spontaneously, like worms from mud. As if a 511keV photon is forever fluttering along turning into a 511keV electron and a 511keV positron in defiance of conservation of energy, which obligingly turns back into a single 511keV photon in defiance of conservation of momentum, which nevertheless manages to propagate at c. It's tautological garbage I'm afraid.

To summarize: is the concept of a massless particle with vanishing momentum meaningful experimentally or theoretically?

No.

Making it harder to ignore let's assume that the particle we've created above possesses an electric charge. As far as I am aware there is no principle prohibiting massless particle to carry a charge (even if it has zero energy).

There is. You can't have charge without mass. Think of photon momentum as resistance to change-in-motion for wave propagating linearly at c. Then remember your pair production, and the wave nature of matter, and that in atomic orbitals electrons "exist as standing waves". And think of magnetic moment and electron spin and the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Then think of electron mass as resistance to change-in-motion for a chiral "spinor" wave going round and round at c, whereupon the electromagnetic field-variation now looks like a standing field. Standing wave, standing field. The label we apply to this standing field, is charge.

• I think that your answer only backs up the natural intuition. However the question was how is this intuition related to the formalism of QFT. Maybe I did not stress that enough in the body of the starting post. I think that the ACuriousMind's answer does the job of explaining the apparent tension. Commented Jul 14, 2015 at 20:38