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This is a short and simple question...

I have been reading my book on particle physics and quantum physics when I had thought of a question that it failed to answer: "Does a particle enter/interact with the Higgs Field when created, or at some other time? And if the answer is neither of these two, does it constantly go through/react with the Higgs Field?"

I have been looking for this everywhere (here, Wikipedia and other sites) and haven't found an answer yet. Does anyone know that answer (or possible answers) to this?

Graviton
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    Your question does not make sense to me. What do you mean when you say that a particle "enters" or "goes through" the Higgs field? – ACuriousMind Jul 12 '14 at 14:45
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    In our current understanding of quantum field theory, the fields it describes exist throughout the entire universe. As such, the Higgs field is everywhere and interaction with it is unavoidable. Does this answer your question? – Wouter Jul 12 '14 at 15:01
  • The way in which the Higgs mechanism was initially described to me is that particles are taken through the Higgs field like pearls on a string going through a soft medium (and gaining mass in the process). I looked up on it and it was described to me through the "champagne bottle/Mexican hat" analogy. And I know that the Higgs field is said to be omnipresent, but that does not tell me when in a particle's life it interacts with the field. I've changed the question to reflect that. – Graviton Jul 12 '14 at 15:13

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I don't think you understand QFT. To be fair, I'm no expert myself, but I can certainly point out where you're going wrong here. A particle does not enter the Higgs field. However, the particle field that gets mass from the Higgs field does interact with the Higgs field. What this means is that in the Lagrangian of your model, there exists a term that will combine both your particle's field (denoted $\psi$) and the Higgs field (denoted $\phi$). Roughly speaking, interaction with the Higgs field adds a term of the form $G_\psi\bar{\psi} \phi \psi$, which is a scalar term with non zero expectation value i.e. a mass term. The coupling constant $G_\psi$ changes with each $\psi$, hence why each fermion has a different mass.

Incidentally it makes no sense to talk about a particle going through the Higgs field or any other particle field. In a sense individual particles are mnemonic tools meant to gain an intuitive vision of the solutions provided by QFT since these indeed provide a quantum number that represents the "number of particles", but always remember that the fundamental physical object is the field, not the particles. In that sense, given that "number of particles" is a quantum number, an individual particle should be interpreted somewhat the same way we interpret individual energy levels : they are not by themselves physical objects, rather they describe the current state of a physical object. In this case, what this means is that it makes no sense to talk about individual particles going through some other field. All a particle is is a description of the state of its underlying field. Keeping with the analogy, an energy level is a description of an electron in a hydrogen atom, it makes no sense to talk about the interaction of an energy level with the electron of some other atom. What you can talk about, however, is the interaction of the 2 electrons, and then use energy levels to describe the impact of that interaction. Likewise, you can't talk about a particle interacting with a field. You can talk about 2 fields interacting, and then use the concept of particles to describe the impact of that interaction.

ticster
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  • Okay, so I understand that the Higgs mechanism is understood to be an interaction (in which case those who have described it to me as a "string of pearls" are considered to be wrong), but if this is true, then does this interaction occur during the whole of the particle's life? My confusion comes from this: the "champagne bottle" analogy; it states that a particle slips into a state where Higgs bosons can be released (similar to VEV dynamics), but this implies that the process only occurs once. Is this the case? – Graviton Jul 12 '14 at 16:03
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    You're lumping together many things here. I won't give you a complete overview of all of those things, that would take a book. Instead I'll just try to disentangle those ideas :

    1 Completely forget the "string of pearls" analogy (it's just nonsense).

    2 The interaction is always in the Lagrangian, think of it as an intrinsic property of the field. It doesn't just turn on and off.

    3 The "champagne bottle" likely refers to the spontaneous symmetry breaking of the Higgs potential, whereby it acquires a VEV. This happened earlier in the universe's history, and permanently changes the Lagragian.

    – ticster Jul 12 '14 at 16:14
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Does a particle enter/interact with the Higgs Field when created, or at some other time?

After reading your question a couple of times as well as your comments, it occurs to me that you're picturing something like this: a massless particle is created, interacts once with the Higgs field to acquire a permanent classical like mass which it then 'keeps'.

But, this isn't a valid picture at all. A much better picture is to make an analogy with how photons become 'massive' within a superconductor.

Essentially, in this picture, space is an electroweak superconductor with the apparent mass of particles due to continuous interaction with the superconducting 'fluid'. From the Wikipedia article "Higgs Mechanism":

The Higgs mechanism is a type of superconductivity which occurs in the vacuum. It occurs when all of space is filled with a sea of particles which are charged, or, in field language, when a charged field has a nonzero vacuum expectation value. Interaction with the quantum fluid filling the space prevents certain forces from propagating over long distances (as it does in a superconducting medium; e.g., in the Ginzburg–Landau theory).

There's much more to this but hopefully this gets you pointed in the right direction.

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If I'm understanding, you're asking: How does the Higgs field "give" particles mass? This is an explanation given by Henry of MinutePhysics

(To be clear, we're talking about the Higgs field and NOT the Higgs Boson, which is merely an excitation leftover after the process we're about to explain. But I digress…back to mass!)

To begin, we need to know what we even mean by "mass" - so we'll head the other direction and talk about what it means to be massless: This may sound crazy, but the defining feature of any particle without mass is that it travels at the speed of light. In fact, if we're honest it should really be called the "speed of massless particles," but since the first massless particles we knew about were photons of light, the name has stuck.

Anyway, the point is that all massless particles travel 300 million meters every second. The details of this are explained by special relativity, but simply put, it's physically impossible for a massless particle to NOT travel at 300 million meters per second. They can travel in a straight line or bounce off of things and change direction, but the speed of a massless particle never changes.

And so mass is just the property of NOT HAVING TO always travel at the speed of light. As a side effect, this also means not being ABLE to travel at the speed of light, but the key is that particles with mass are lucky enough that they get to travel at ANY speed they want – as long as it's slower than light. The amount of mass something has just tells how hard it is for it to change from one of these speeds to another. Now, in Part I we mentioned that if there were no Higgs field in the Standard Model, ALL particles should be massless and thus travel at the speed of light. But you and I and swiss cheese clearly have mass, because we have the beautiful luxury of being able to sit still.

So how does the Higgs field help us do that? Well, while massless particles can only travel at the speed of light, they ARE allowed to bounce off of things. Things like particles, which are really just excitations in a quantum field. For example, the electron field is more concentrated at certain places called "electrons" - and everywhere else is "empty space". But the Higgs field is unusual in that it has a high value EVERYWHERE – and to be clear, this high value is NOT the famous Higgs Boson - that's an extra excitation in addition to this already elevated field. But because the Higgs field has this everywhere non-zero value, any particle that CAN interact with it is pretty much bouncing off of it all the time. And if a massless particle bounces back and forth and back and forth (or, since it's quantum mechanics, does both at the same time), then even though in between bounces it travels at the speed of light, when you add everything up it LOOKS like the particle is going slower than light. Even... like it's not moving! And since only things with mass are allowed to not move, our massless particle now looks and acts like it has mass. Well done, Higgs!

What's more, the Higgs field can even interact with its own excitations, which is to say, it can give mass to the Higgs Boson, too. Actually, the Higgs field likes to interact with itself so much more than with the lowly electrons and protons that make us up, that the Higgs Boson has a great deal more mass – and that's part of why it's been so hard to find. But we shouldn't complain, because even though the Higgs has given us a lot of trouble and only a little bit of mass, at least we have mass, which allows us the simple pleasure of not moving."

Source

  • I understand (this is also vey interesting, thanks for answering), but I know what the Higgs boson and the Higgs Mechanism are. My question was: "Do particles always interact with the Higgs field, or does this happen once, and if so, when in a given particle's life?" – Graviton Jul 13 '14 at 13:53
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    You said : "in between bounces it travels at the speed of light". So according to your analogy all particles actually travels at C and just average velocity is under C. So I can't imagine why time dilation must happens for massless particles and not for all particles. It was a interesting analogy but make more question for me. Can anyone explain it without mass. I know if you consider E2=m2c4+p2c2 and put some mass in m then v must be below C but according to this analogy relativity is not as fundamental as quantum mechanic. Because in quantum mechanic all particles moves at ---> – Alberto Jul 03 '16 at 11:23
  • <--- speed of light and relativity formula just considers the average velocity. – Alberto Jul 03 '16 at 11:23
  • There are exponential decay continuous operator matrix multiplication series rather than the simple E^2 = m^2 c^4 + p^2 c^2. Also, Higgs interacts with itself a lot. – Brian Cannard Aug 10 '23 at 17:07