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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).

  • This sounds more like a philosophical question then a physical one. Nothing in this world has exactly a given value of a continuous quantity. Its like saying how many people way exactly 100 kg. If you keep enough decimal places the answer will be zero... A physical question would be how many particles have energies within $\epsilon$ of zero. – JeffDror Jul 14 '15 at 09:54
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    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. – Weather Report Jul 14 '15 at 10:00
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    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. – JeffDror Jul 14 '15 at 10:06
  • Well, it is Lorentz-invariant. If the four-momentum of a particle is zero in one frame it is zero in any inertial frame. – Weather Report Jul 14 '15 at 10:32
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    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. – anna v Jul 14 '15 at 11:10
  • this is similar to asking whether a massless particle such as a photon could have $\lambda\to\infty$, and thus have zero energy. I think the answer is YES, as can be seen in IR divergences associated with massless particles in QFT. – innisfree Jul 14 '15 at 11:23
  • @anna v However, this is not obvious to me that having velocity must be the same as having momentum. Even if was so, what would be wrong with the theory where a photon of zero speed may exist? The famous relation you've quoted is also satisfied by $p^\mu\equiv0$ (both energy and momentum zero). – Weather Report Jul 14 '15 at 11:23
  • both energy and momentum zero leaves no object for measurement. Just space time points. (0,0,0,0) is an origin not a particle – anna v Jul 14 '15 at 11:26
  • @annav Does this mean that the one-particle state of zero energy and momentum $a^\dagger(\vec{0})|0\rangle$ (mentioned in the body of the question) is the same as the vacuum state $a^\dagger(\vec{0})|0\rangle=|0\rangle$? – Weather Report Jul 14 '15 at 11:29
  • @annav won't it contribute to collinear emission/IR divergences? in the end, you can't measure them because of resolution, but if they weren't there, you'd get a divergence that didn't cancel with the one from a renormalized vertex... – innisfree Jul 14 '15 at 11:30
  • @JeffDror, what do you make of my comments re IR divergences? surely zero energy modes for massless particles in QFT are IR divergences? – innisfree Jul 14 '15 at 11:32
  • Those are mathematical constructs, virtual, very useful in calculating measurable cross sections etc. but not physical/measurable. They are a necessary part of a mathematical model . Take a complete set of functions and fit a three dimensional shape to high accuracy, a real ball as an example. Does it mean that the ball is built up by the fourier( or what not) components that go to infinity? – anna v Jul 14 '15 at 12:50
  • @annav What is the exact sense in which sense the state I mention in the post is not physical? To make the state more respectfull, I will empower it with an electric charge. – Weather Report Jul 14 '15 at 13:05
  • @JimsBond Firstly, I did not defined a "non-existing particle" as the particle of zero energy. I only said that this is my natural intuition which I am questioning here. Secondly, In my opinion your argument is just a rephrasing of this natural intuition and hence I do not find it sufficient. I would like to see some exact statements about the triviality of the quantum state defined in the question. – Weather Report Jul 14 '15 at 13:11
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    Massless particles cannot have an electric charge http://physics.stackexchange.com/a/7919/23473 – Jim Jul 14 '15 at 13:31
  • You are thinking on platonic lines, i.e. mathematics defines nature , (philosophy) and not nature is modeled with mathematical functions/theories ( physics) – anna v Jul 14 '15 at 13:38
  • @annav Your comment feels a bit like a permission to ignore some predictions of the theories that we do not like (say, zero-energy particles). I think that in the case at hand should either have a way to explain why the corresponding state is unphysical or to prove the contrary. – Weather Report Jul 14 '15 at 13:47
  • @JimsBond Thank you for the link. However, the "no-go" result you've quoted is not a self-consistency requirement of the QFT, but rather a specific detail of a particular model (we simply haven't added such a particle to our Lagrangian). Though this "particular model" is quite an important one:), I do not see how this invalidates the question in general. – Weather Report Jul 14 '15 at 13:53
  • It doesn't, it simply points out that you can't increase the complexity by making it an electrically charged particle – Jim Jul 14 '15 at 13:55
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    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. – anna v Jul 14 '15 at 13:58
  • @JimsBond I would insist that I can. Not in the framework of the Standard Model (which simply does not contain such a particle), of course. Alternatively if we really want to stick with the SM, let $a^\dagger$ (with appropriate indices added) be the creation operator for a gluon, which is color-charged. – Weather Report Jul 14 '15 at 14:00
  • In the SM mathematics a zero mass zero energy gluon is a singularity and on the lines of Lubos' answer linked above in a comment, it would have been observed if it existed !!!. – anna v Jul 14 '15 at 14:03
  • I have no issues with using the colour charge – Jim Jul 14 '15 at 14:08
  • @annav Nevertheless, any state in the Hilbert space of a theory has a well-defined physical status. If we assume that a state with a zero-energy particle is equivalent to the vacuum state, our theory should be able to prove that. That's what I am asking for. Or, if they are not equivalent (which would surprise almost everyone who commented here, including me) that's an interesting question what differs them and what are the properties of this zero-energy state. – Weather Report Jul 14 '15 at 14:14
  • No, the Hilbert space is a mathematical space extremely useful for describing nature. It does not define nature. If a theory based on this hilber space included that, i.e. the lagrangian whose solutions give the standard model, it would be a falsified theory because the data falsifies it. – anna v Jul 14 '15 at 14:23
  • @annav Could you please elaborate on the claim

    " zero mass zero energy gluon is a singularity...it would have been observed if it existed".

    I do not quite understand it. If existed, how exactly would them affect the observations?

    – Weather Report Jul 14 '15 at 14:40
  • The strong force gets stronger the smaller the energy, is more attractive.zero mass glue glue would glue to each other with numerous bound states approaching zero energy and I do not know what would happen at the exact zero. these glue balls have not been seen except with jets at high energies. – anna v Jul 14 '15 at 16:21
  • @annav Although being a novice here, I always believed that this is the exchanged energy in the process of interest that affects the strength of the interaction, not the energies of the individual participant particles. – Weather Report Jul 14 '15 at 16:29
  • The gluons are gauge bosons for strong interactions, i.e. the exchange particle.http://hyperphysics.phy-astr.gsu.edu/hbase/forces/feyns.html – anna v Jul 14 '15 at 16:47
  • See also infrared divergence and infrared cutoff. https://en.wikipedia.org/wiki/IR_divergence Prior to renormalization, such particles exist but in unbounded numbers. Renormalization helps keep these particle numbers under control. (I've been known to ask students if it's possible for something to have a wavelength greater than the diameter of the universe, which is clearly gibberish, but helps get the idea motivated.) – Eric Towers Jul 14 '15 at 20:50
  • @innisfree: I agree that IR divergences exist and hence we should have photons with very large wavelength. But I think the question is more trivial then that this it tries to ask about an exact value of a continuous quantity (which has no enhanced symmetry at that value). – JeffDror Jul 15 '15 at 02:38

3 Answers3

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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.

ACuriousMind
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  • 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. – Weather Report Jul 14 '15 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 $E=0$? – innisfree Jul 14 '15 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). – ACuriousMind Jul 14 '15 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? – Weather Report Jul 14 '15 at 16:51
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    @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). – ACuriousMind Jul 14 '15 at 17:06
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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.

innisfree
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    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 Jul 14 '15 at 15:29
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    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. – Weather Report Jul 14 '15 at 15:29
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    @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 Jul 14 '15 at 15:33
  • @JimsBond Would a DC electric potential in the universe really be nothing? Perhaps the answer should address that. – Blackbody Blacklight Jul 14 '15 at 15:38
  • @WeatherReport I am talking about photons of all energies $E<\Delta E$, where $\Delta E$ is the smallest detectable energy. – innisfree Jul 14 '15 at 15:41
  • @BlackbodyBlacklight I didn't literally mean a DC potential. I used DC as a metaphor for a signal with a frequency of $0,Hz$ – Jim Jul 14 '15 at 15:42
  • @innisfree However, I believe that the contribution of the exactly "energyless" photons to the effect you describe is absent. The reason is that the point $p^\mu=0$ of zero measure and does not contribute to the any finite-range (energy-wise) integral. – Weather Report Jul 14 '15 at 15:46
  • @WeatherReport that couldn't be further from the truth! The divergences in the amplitudes occur at $E=0$. – innisfree Jul 14 '15 at 15:48
  • @innisfree but this is the behaviour of the integrand near $E=0$ that causes the divergence, the very value at $E=0$ is irrelevant unless there are delta-functions in the integrand, which is not the case (or is it?). – Weather Report Jul 14 '15 at 15:51
  • @WeatherReport not sure what you're getting at. the value is irrelevant because it disappears once real and virtual emission is summed. – innisfree Jul 14 '15 at 15:53
  • @innisfree I'm just saying that the integral $\int_0^1\frac{dE}{E}$ is divergent not because function $1/E$ is infinite at $E=0$, but because it grows too fast near $E=0$. So it's the contribution of arbitrarily soft (but non-zero) photons that gives rise to a divergence, not the value at $E=0$. – Weather Report Jul 14 '15 at 16:02
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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.

John Duffield
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    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. – Weather Report Jul 14 '15 at 20:38