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Why is the mass of the proton such a precise value?

A proton is composed of 3 net valence quarks and what is often described as "binding energy" or "a zillion gluons and quarks and anti-quarks self annihilating and popping into existence". The quarks are only about 1% of the mass but the mystery to me is understanding why all of this 99% "amorphous" dynamic binding energy in a proton (or neutron) amounts to the consistent and very exact mass or energy that it is.

I would have maybe guessed that an individual proton's mass would just have more of a range, that the amount of binding energy would be more variable

Qmechanic
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ParityViolator
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    Do you wonder whether the ground state energy of a hydrogen atom is variable? – G. Smith Sep 01 '20 at 19:21
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    Just because your metaphorical QCD picture of the proton appears fuzzy and "anything goes", it doesn't mean that it does not represent a precise and well-defined theory (QCD): it's just hard to simulate, let alone calculate analytically. – Cosmas Zachos Sep 01 '20 at 19:25
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    As long as a proton is a stable particle, whatever it is composed of, you can measure its mass with arbitrary precision.You may need to measure over a long enough time interval. The error you see is based on the measurement device, not on the proton. – Ross Millikan Sep 02 '20 at 03:35
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    Measuring mass doesn't happen in a (proverbial) vacuum. When you design an experiment to measure the proton mass you will have to relate your measurement to other experiments by theoretical considerations. On the lowest level you need a procedure to relate the MeV$/c^2$ that the mass is usually given in to kg. On a higher level, you will compare the proton's mass to another object (a positron, say), and you will have to account for how they behave differently in ways unrelated to mass. This to say that what you describe is a small part of the complexities of extracting a meaningful number. – tobi_s Sep 02 '20 at 05:58
  • Short version of my comment: the value is a theoretical value which attempts to be as free as possible of prejudice. The recent confusion about the proton radius after new measurements using muons is a good lesson into how much modelling goes into these averaged numbers that end up being tabulated in references. ps I just noticed that there is a Wikipedia article on the proton radius puzzle, but I have yet to read it: https://en.wikipedia.org/wiki/Proton_radius_puzzle – tobi_s Sep 02 '20 at 06:12
  • Hi! I've read once it took 200 hours for a supercomputer to calculate the proton's spin to be one halve. I know this ain't the same as mass, but nevertheless. The proton's spin is "easily" measured to be one halve though (not precisely but quite near, due to QED, just as famous calculation of the electron's magnetic moment which corresponded up to 10(?) digits after the point). – Deschele Schilder Sep 02 '20 at 12:51
  • Also this article may help to you Nuclear masses calculated from scratch – Agnius Vasiliauskas Sep 02 '20 at 13:37
  • Related: https://physics.stackexchange.com/q/207644/226902 https://physics.stackexchange.com/q/474084/226902 – Quillo Mar 23 '23 at 10:45

5 Answers5

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You say:

a zillion gluons and quarks and anti-quarks self annihilating and popping into existence

and while this is a very common way to describe the interior of a hadron like a proton it is actually rather misleading. Nothing is popping into existence then disappearing again. But explaining what is actually happening is a little involved.

Our current best theory for describing particle is quantum field theory. In this theory the fundamental objects are quantum fields that exist everywhere in the universe. Particles like quarks are not fundamental objects. Instead they are just states of the quantum field. This nicely explains how particles can be created and annihilated at colliders like the LHC, because we can start with a zero particle state of the quantum field and add energy to it to excite it to states that correspond to non-zero numbers of particles. Likewise a state of the field that corresponds to particles can decay to a state with fewer or no particles.

But while there are states of the field that do correspond to what we call particles, this is actually a rather special case. Specifically this is only the case when we have an isolated particle that isn't interacting with any other particles. These are called the Fock states of the field. But the field has an infinite number of other states that aren't Fock states so they don't correspond to particles. The problem is that we don't know how to solve the equations of the field to get these states. Instead we have to use approximate methods to calculate properties like their mass.

And this is the case for the bound states we call hadrons. A proton is a state of the quantum field but it isn't a Fock state. In principle we could write down the equation for the field and solve it to get the state corresponding to a proton, but in practice we simply don't know how to do this so we have to approximate it. We do this by approximating the state as a collection of virtual particles, and this is why popular science descriptions talk about particles popping into existence and disappearing again. Where the popular science articles go wrong is that these virtual particles are a computational device and they do not exist. I cannot emphasise this enough: the virtual particles are just a way of calculating the properties of field states that are not Fock states and therefore do not correspond to particles.

This has taken us a long way from your question, but we can now understand why the mass of a proton is well defined. It is because it is a well defined state of quantum fields and as such has a well defined mass. It just doesn't correspond to a well defined number of particles, which is why it isn't just three quarks or $n$ quarks and $m$ gluons or any other collection of particles.

If you're interested in finding out more about this you might want to look at my answer to Are vacuum fluctuations really happening all the time? where I use a similar argument to explain why the vacuum isn't actually fluctuating either.

John Rennie
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    This is a good discussion, but the actual answer "because ... a well defined state [has] a well defined mass" should go at the top. – Xerxes Sep 02 '20 at 13:31
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    A hadron is not a Fock state but it can still be expressed as a linear combination of Fock states because the Fock states form a basis for the Hilbert space (up to UV issues). And if so, the question is a good one -- how can some linear combination of Fock states be an energy eigenstate? – Eric David Kramer Sep 02 '20 at 20:26
  • Even though the sea quarks don't exist, we can bounce neutrinos off of them. (Look for the description of the CDHS experiment.) – Will Orrick Sep 03 '20 at 08:00
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    @EricDavidKramer the free particle states do not form a basis for the Hilbert state of an interacting field. This is Haag's theorem. So you cannot write a hadron as a linear combination of Fock states. – John Rennie Sep 03 '20 at 09:00
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    @JohnRennie Haag's theorem is a UV issue, which is why I wrote "up to UV issues". If you take a cutoff and renormalize then the Fock states do in some sense form a basis for the regulated Hilbert space. And in any case, we should be able to talk about a proton without worrying about what's happening in the UV. – Eric David Kramer Sep 03 '20 at 10:10
  • Thank you. The answer clarifying on the nature of what is inside the protons basically rejecting the pop sci notion of "a sea of particles", gives me a better way of looking at my assumptions which led me to ask this question. The "binding energy" cannot be considered variable, its quantum states are precisely defined in the proton as precisely defined as the notion of their being 3 valence quarks...I suppose the short answer is that the set of states of quantum fields that define a proton is very precise, added together, they yields the property of mass to a value that is also quite precise – ParityViolator Sep 03 '20 at 16:12
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The mass of the proton has been measured to be $938.27208816(29) MeV/c^2$, the value in parenthesis the error in the measurement.

The job of a theoretical model in modeling the proton is to attempt to explain the measurement.

what is often described as "binding energy"

It is not a good description, as if you imagine an atomic type model just more complicated because of more particles. The strong interaction is involved in modeling hadrons, and the simple solutions and even the tool of quantum field theory is useless due to the large coupling constant of strong interactions.

The theory of QCD on the lattice

Lattice QCD is a well-established non-perturbative approach to solving the quantum chromodynamics (QCD) theory of quarks and gluons. It is a lattice gauge theory formulated on a grid or lattice of points in space and time. When the size of the lattice is taken infinitely large and its sites infinitesimally close to each other, the continuum QCD is recovered

It has managed to model the hadronic spectra given some inputs. Here is a presentation of the status of the theoretical model.

The hadronic spectrum is research in progress both experimentally and theoretically.

To find out the origin of the proton mass experimentally is one the primary goals of the upcoming electron-ion collider. This work has opened the door for more numerical calculations and theoretical understanding of nucleon structure, including the spin decomposition of the proton in terms of quark spin, quark orbital angular and glue angular momentum, whose measurement is also a leading goal of the electron-ion collider.

anna v
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The origin of proton mass is rather a subtle issue, it is really created by a complicated dynamics of quark and gluon fields. As far I understand this issue, the known value for the proton mass is an energy of a proton ground state - which is some eigenstate of corresponding Hamiltonian, excitation in the QCD theory. There are excited states with higher energy, mass correspondingly. However, in case of not very high temperature and density - the ground state will strongly dominate.

spiridon_the_sun_rotator
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    But don't we need a reference energy level to describe the ground state energy level? Is this reference the vaccum energy? – NiRVANA Sep 01 '20 at 23:49
  • @Astik - well, I would say yes. However the issue is a bit subtle, because the structure of QCD vacuum is highly nontrivial, and creation of proton particle is possible in the presence of other nucleons or particles. It is usually assumed, that the influence of the surroundings is smaller significantly than of the QCD scale – spiridon_the_sun_rotator Sep 02 '20 at 04:45
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In quantum mechanics, and energy eigenstate is a stationary state: that is, unchanging with time (other than a phase). This picture doesn't work with the idea of virtual particles popping in and out of existence, and that is a problem with overly-classical analogies.

JEB
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A particle, be definition, is an energy eigenstate. That means that the state is constant in time.

The "vacuum", i.e. the state with no gluons and quarks, is not an energy eigenstate when interactions are included, because, as you say, quark and gluons will pop in and out of existence. But a bound state of quarks and gluons is an energy eigenstate. That means that there is a way to combine quarks and gluons in such a way that when the interactions are turned on, the stuff annihilating is exactly replaced by what is being created and the state is constant. One of those states is called a proton. And the energy of that state that stays constant is called the proton mass.

Eric David Kramer
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    In an interacting theory a particle is not an energy eigenstate because the energy eigenstates are not also eigenstates of the number operator. The particles are only energy eigenstates in the free field theory. By contrast the vacuum state in interacting theories is an energy eigenstate (though also not an eigenstate of the number operator). So it is exactly the opposite of what you have written. – John Rennie Sep 03 '20 at 06:20
  • But a free proton is an energy eigenstate, isn't it? – Anton Tykhyy Sep 03 '20 at 06:55
  • @AntonTykhyy By particle we mean a fundamental particle like a quark or gluon i.e. a Fock state, not a composite state like a proton. The proton is indeed an energy eigenstate of the quantum fields but it is not a fundamental particle. – John Rennie Sep 03 '20 at 08:03
  • @JohnRennie The reason I said that the vacuum is not an energy eigenstate is that I am calling the vacuum the state with no particles: $a|0\rangle$. If a particle is not an eigenstate, then neither can this vacuum be one. You must be calling the vacuum the state such that $H|\psi\rangle = 0$. I'm not sure what the physical significance of that state is. – Eric David Kramer Sep 03 '20 at 10:04
  • @JohnRennie And the reason I said the proton is an eigenstate is that I am claiming that if you take a state that is already an eigenstate (such as your $H\psi\rangle=0) and add a proton, the new state will be an eigenstate of the "QCD Hamiltonian" (which of course doesn't really exist), that depends only on quarks and gluons, but has no field operator for protons. But I am still confused about this and I am open to your thoughts. – Eric David Kramer Sep 03 '20 at 10:08
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    The vacuum state of an interacting field is the lowest energy state i.e. the condensate. This is an energy eigenstate. However it is not an eigenstate of the number operator so it is not $|0\rangle$ nor can it be written as a sum of the Fock states, though we usually wave our arms around and assume it can be approximated in this way. – John Rennie Sep 03 '20 at 10:20
  • I agree that the lowest energy state is also called the vacuum, but in any physical process we assume the interaction vanishes in the asymptotic past and future (although this should be technically illegal according to Haag's theorem) and so the vacuum we use to define the S-matrix is the $a|0\rangle=0$ vacuum, and this is the "vacuum" we use to calculate the propagator, so it's also OK to call it the vacuum. And I think your condensate vacuum can be approximated as a sum of Fock states, once again up to UV issues. – Eric David Kramer Sep 03 '20 at 10:28