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I learned today in class that photons and light are quantized. I also remember that electric charge is quantized as well. I was thinking about these implications, and I was wondering if (rest) mass was similarly quantized.

That is, if we describe certain finite irreducible masses $x$, $y$, $z$, etc., then all masses are integer multiples of these irreducible masses.

Or do masses exist along a continuum, as charge and light were thought to exist on before the discovery of photons and electrons?

(I'm only referring to invariant/rest mass.)

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There are a couple different meanings of the word that you should be aware of:

  • In popular usage, "quantized" means that something only ever occurs in integer multiples of a certain unit, or a sum of integer multiples of a few units, usually because you have an integer number of objects each of which carries that unit. This is the sense in which charge is quantized.
  • In technical usage, "quantized" means being limited to certain discrete values, namely the eigenvalues of an operator, although those discrete values will not necessarily be multiples of a certain unit.

As far as we know, mass is not quantized in either of these ways... mostly. But let's leave that aside for a moment.

For fundamental particles (those which are not known to be composite), we have tabulated the masses, and they are clearly not multiples of a single unit. So that rules out the first meaning of quantization. As for the second, there is no known operator whose eigenvalues correspond to (or even are proportional to) the masses of the fundamental particles. Many physicists suspect that such an operator exists and that we will find it someday, but so far there is no evidence for it, and in fact there is basically no concrete evidence that the masses of the fundamental particles have any particular significance. This is why I would not say that mass is quantized.

When you consider composite particles, though, things get a little trickier. Much of their mass comes from the kinetic energy and binding energy of the constituents, not from the masses of the constituents themselves. For instance, only a small part of the mass of the proton comes from the masses of its quarks. Most of the proton's mass is actually the kinetic energy of the quarks and gluons. These particles are moving around inside the proton even when the proton itself is at rest, so their energy of motion contributes to the rest mass of the proton. There is also a contribution from the potential energy that all the constituents of the proton have by virtue of being subject to the strong force. This contribution, the binding energy, is actually negative.

When you put together the mass energy of the quarks, the kinetic energy, and the binding energy, you get the total energy of what we call a "bound system of $\text{uud}$ quarks." Why not just call it a proton? Well, there is actually a particle exactly like the proton but with a higher mass, the delta baryon $\Delta^+$. Technically, a $\text{uud}$ bound system could be either a proton or a delta baryon. But we've observed that when you put these three quarks together, you only ever get $\mathrm{p}^+$ (with a mass of $938\ \mathrm{MeV/c^2}$) or $\Delta^+$ (with a mass of $1232\ \mathrm{MeV/c^2}$). You can't get any old mass you want. This is a very strong indication that the mass of a $\text{uud}$ bound state is quantized in the second sense. Now, the calculations involved are very complicated, so I'm not sure if the operator which produces these two masses as eigenvalues can be derived in detail, but there's basically no doubt that it does exist.

You can take other combinations of quarks, or even include leptons and other particles, and do the same thing with them - that is, given any particular combination of fundamental particles, you can make some number of composite particles a.k.a. bound states, and the masses of those particles will be quantized given what you're starting from. But in general, if you start without assuming the masses of the fundamental particles, we don't know that mass is quantized at all.

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Shouldn't mass be as quantized as energy? I believe E = mc^2 holds even in QT!? – Sklivvz Nov 3 '10 at 7:30
Energy is actually not always quantized in the sense of being discrete ;-) True, quantum theory does not invalidate E = mc^2, but you can only conclude that mass is quantized if you do the math for a particular particle at rest and find that it has discrete bound states. (By the way, it's E^2 = m^2c^4 + p^2c^2 in general) – David Z Nov 3 '10 at 7:40
I understand the relation between bound states and quantization (i.e. discrete vs. continuous spectra). However, the concept of "quanta" of energy applies in both cases, since you have photons anyways, and as such - should we entertain the concept of corresponding "quanta" of mass? – Sklivvz Nov 3 '10 at 11:24
Sure, there's nothing wrong with entertaining the concept, and some theories do predict quantization of mass, although it may be more complicated than just multiples of a certain value. But there's no experimental evidence that mass is actually quantized, at least not for fundamental particles. – David Z Nov 3 '10 at 20:24
@Gordon or more precisely, we haven't reached the stage to determine the quantization of mass, a quick calculation shows that corresponding to $h$ one get $ m = 7.36229778 * 10^{-53} kg $ – Vineet Menon Dec 8 '11 at 16:17

I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a consistent quantum field theory - is inherently related to the properties of the symmetry it is the "charge" of: Poincaré invariance.

All symmetry groups in quantum physics must be represented by a projective unitary representation on the Hilbert space of states (cf. Wigner's theorem). It is this fact alone that forces discretization of many quantities.

The archetypical example of a discrete quantity in quantum physics is spin, coming in integral multiples of $\frac{1}{2}$. Spin is discrete because it is the value of the quadratic Casimir operator $S^2$ of the rotation group $\mathrm{SO}(3)$, which is constant on its irreducible representations, and the rotation group has only countably many irreducible unitary representations, since as a compact group, all its irreducible representations are finite-dimensional, and there are only countably many finite-dimensional vector spaces. Furthermore, it turns out that, purely representation-theoretic, the only existent irreducible representations are those where $S^2$ takes values as $\frac{n}{2},n\in\mathbb{N}$.

Likewise, mass is the (square root of the value of) the Casimir operator $P^2$ of the Poincaré group. The Poincaré group is non-compact, which means it does not have finite-dimensional unitary representations. Therefore there is no reason to expect there to be only countably many of them, and, in fact, there are not. By Wigner's classification, there is a unitary representation (several in fact, labeled by the spin of the massive particle) of the Poincaré group for every possible positive real value $P^2 = m^2\in\mathbb{R}_{>0}$. Therefore, there is no reason that the usual procedure of quantization (which is not "associating discrete values to things" but rather something like "representing all physical observables and symmetries as operators on a Hilbert space") should confine masses to a discrete spectrum, let alone one where all masses are integral multiples of a fundamental mass unit. Physically only finitely many of the representations will be realized because we only have a finite number of distinct particle species, but, in contrast to spin, there is no a priori constraint of the masses in a quantum theory.

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This is a very mathy answer, but it gets to the heart of the matter. I like it. – Paul T. Apr 27 at 15:19

I think just the opposite of David Zaslavski, and assert:

The rest mass of particles is quantized, [edit] being the spectrum of the component P_0 of 4-momentum in the Hilbert space of states where the particle is at rest. (For example, quarks and neutrinos have a 3-dimensional mass matrix, each eignevalue being infinitely degenerate.)

This doesn't conflict with David's supporting facts but with his use of the terminology. For:

(i) A quantity is conventionally called quantized if its spectrum (the set of possible values it can attain) is discrete. This is the case for mass, as mass is defined in quantum field theory as values of the energy where the S-matrix in the rest frame becomes singular (''poles of the S-matrix''). Such poles must be discrete in each instance, for purely mathematical reasons. More specifically, the masses of the known elementary (and less elementary) particles are tabulated and can be seen to take fixed, discrete values.

(ii) Being quantized has nothing to do with being a parameter. Indeed, electromagnetic carge is quantized, although the value of the electric charge is a free parameter of the standard model.

(iii) Being quantized has nothing to do with having a (simply or not) understandable pattern. Most spectra of chemical substances have know pattern, but they are all expained by the discreteness of the spectrum of the corresponding Hamiltonian - the most conspicuous case of quantization.

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I see your point - of course the rest masses of the fundamental particles are specific discrete values, but what I meant was that there is no known mechanism that restricts the masses to those values specifically. In other words, mass does not correspond to any known operator that has a discrete spectrum. (Also, I was using "pattern" rather broadly, in particular the spectrum of eigenvalues of an operator counts as a pattern for my purposes.) – David Z Mar 9 '12 at 19:49
I understood, but in the sense ''what I meant'', charge would also be not quantized, while in the sense ''in other words'', mass is quantized, see the addition to my answer. – Arnold Neumaier Mar 9 '12 at 19:56
Well, yes, charge is not quantized in the same sense as something like electron binding energy. I would agree that you could consider it quantized in the same sense as rest masses of fundamental particles (we only ever observe certain discrete values), but I didn't think that's what is being addressed here. Perhaps it will be worthwhile clarifying that in my answer. – David Z Mar 9 '12 at 20:04
The inventor of the question mentioned that charge is quantized, so this constrains the interpretation of his question. – Arnold Neumaier Mar 9 '12 at 20:30
That is true, when I wrote my original answer I suppose I was focusing more on the second paragraph. Anyway I've rewritten it in a way that I think might be more precise. – David Z Mar 9 '12 at 21:43

masses are on a continuum, as they must be, because you can make a box, put photons in it, and make the photons have arbitrarily low momentum, so that the energy in the box is quantized in as small a unit as you like. The mass of an object is the energy it has when at rest, and this energy can change by arbitrarily small amounts by adding a few bound photons. This is for macroscopic objects.

For elementary particles, the rest masses have no sensible quanization rule because they are energies derived from complicated interactions. Even if you have some condition at high energy that determines the mass, it gets corrections from the energy due to interactions with low lying fields, and you get a new mass which is effectively like binding photons to the particle (although it isn't photons, but it's not quantized either). It makes no sense to quantize mass in units, because energy is not discrete in a Lorentz invariant theory, and the mass is the energy bound into a particle.

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Could it be, that at least the masses of black holes are quantized ? – jjcale Jul 14 '12 at 15:12
@jjcale: Black holes don't have well defined masses--- they decay. If you are asking about extremal black holes, they don't decay and have well defined masses, and there the mass is quantized in the semi-classical limit. It is equal to a combination of charge and angular momentum, both of which are quantized quantities, although if you have both charge and angular momentum, it isn't a linearly spaced tower of masses (and the spacing is different for L and Q). This property persists in strings, it is a much more interesting question, and you should post it separately. – Ron Maimon Jul 14 '12 at 16:46

Rest mass of elementary particles is not quantized: in the standard model, the masses are free parameters of the theory; they must be measured and introduced in the model experimentally.

However, the mass of, say the Hydrogen atom is given by the mass of its constituent (proton and electron whose mass are given) minus the binding energy which is quantized.

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Since we r talking of quantization of mass,

First of all its shouldnt be just mass, because the level of which we r talking off ,it makes it impossible for a particle to stay at rest, (The Uncertainity Principle)


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The momentum of a particle does not affect its rest mass by definition. Momentum itself is usually quantised in bound states, like electrons in an atom. – Phil H Mar 8 '12 at 12:27
"mass and momentum for sub atomic particles cannot be differentiated from one another" IS this statement correct ? – Tomarinator Mar 8 '12 at 13:06
For example for photon, its rest mass is 0 but it can have definite momentum, – Tomarinator Mar 8 '12 at 13:06
Relativity shows us that there is a rest state of a particle; it has innate properties when no force is acting on it. In the frame of a particle moving at a constant speed, it is at rest, and thus no property of it can depend on its speed. To an observer in a different frame, however, it can have momentum and other kinetic properties. – Phil H Mar 8 '12 at 13:33
In electrons angular momentum is quantised, does that mean its linear momentum is also quantised? – Tomarinator Mar 8 '12 at 13:38

It may be, but in a different way than the electric charge. If the Compton wavelengths hbar/mc of particles are quantized, then 1/m is quantized, preferably as n/m_P where n is a positive integer and m_P is the Planck mass, but here n would be much larger than 1 for all the elementary fermions in SM.

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Just guessing, or have you a proposal for how this is supposed to work? In the former case this is probably in violation of the FAQs prohibition on "pitches for your personal theories" and needs to be deleted. In the latter the answer is woefully incomplete without a explanation and will almost certainly attract more downvotes. – dmckee Nov 7 '12 at 14:09

consider the smallest unit of energy as h.. plancks constant.. and considering all mass is interconvertible with energy, according to e=mc^2... therefore, the smallest mass possible must be the mass that can produce h amount of energy... i.e e=mc^2...... 6.6 * 10^-34 = m * 9 * 10^16... therefore the smallest mass possible is 7.3 * 10^51.. hence mass is quantised....

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First, I don't have a clue as to what units you're using. Second, Planck's constant is not in units of mass. Third, we know that there are lots of things smaller than the Planck mass. There are even insects whose mass is smaller than the Planck mass. – Peter Shor Mar 9 '12 at 20:02
The Planck mass usually defined as sqrt(hbar * c / G) or about 22 micro-grams. It is not the smallest mass possible. – Paul T. Apr 27 at 15:15

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