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How can a particle be a combination of other particles?

  1. Sep 24, 2015 #1
    Moderator's note: this is a spin off from a previous thread here.

    I confess to a very limited understanding of QM. I understand (somewhat) that a state of a particle can be a combination of possible states. I do not understand how a particle can be a combination of two particles. In particular, a particle has measurable properties such as: energy, momentum, position, and time of measurement. Of course there are uncertainty constraints on the precision of such measurements, and predictions about measurements are probabilistic. Can you explain the physics of how the energy, momentum, position, or time of measurement of a single particle can be a combination of such measurements of two particles.
     
    Last edited by a moderator: Sep 24, 2015
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  3. Sep 24, 2015 #2

    PeterDonis

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    That's not what is being described when a particle is expressed as a "combination of two particles".

    The example in the previous thread was the electroweak bosons. At very high energy (such as in the very early universe, before the electroweak phase transition), the electroweak bosons are ##W^1##, ##W^2##, ##W^3##, and ##B##. At low energy (after the electroweak phase transition), they are ##W^+##, ##W^-##, ##Z##, and ##A##, where ##A## is the photon. We can write the second set in terms of the first as follows:

    $$
    W^+ = \frac{1}{\sqrt{2}} \left( W^1 + i W^2 \right)
    $$

    $$
    W^- = \frac{1}{\sqrt{2}} \left( W^1 - i W^2 \right)
    $$

    $$
    Z = \cos \theta W^3 - \sin \theta B
    $$

    $$
    A = \sin \theta W^3 + \cos \theta B
    $$

    What do these equations mean? They do not mean that the ##Z## boson, for example, is measured by measuring a ##W^3## and measuring a ##B## and combining the measurements in some way. What these equations mean is this: what we call the ##Z## boson, for example, is really a quantum field state which is a combination of the ##W^3## state and the ##B## state, in a certain ratio. That state happens to be the one we can detect in particle physics experiments; we can't detect ##W^3## or ##B## states at the energies accessible to us. But all of them are just quantum field states, they are not "different particles". They are just different possible states of the field, which can combine in different ways to form new states.

    Another way to look at this, which was touched on in the earlier thread, is to think of the two groups of bosons, ##W^1##, ##W^2##, ##W^3##, and ##B##, vs. ##W^+##, ##W^-##, ##Z##, and ##A##, as two different sets of basis vectors in the "space" of electroweak quantum field states. On this view, the equations above just express a transformation in electroweak field space from one basis to another. Any possible electroweak field state can be expressed in either basis, and you get the same predictions for all observables no matter which basis you use. It's just that the second basis is more convenient at the energies we can currently probe experimentally, because those basis states are the ones we can detect.
     
  4. Sep 24, 2015 #3

    ohwilleke

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    I don't really disagree with this which basically explains the answer. But, I do feel like there are one or two unstated assumptions lurking in the electroweak unification story that are not articulated which makes this reasoning less obvious than it should be, although I can't quite put my finger on exactly what those assumption are or should be.

    Something similar comes up with the lighter pseudo-scalar mesons like the neutral pion (which is a "combination" of u/u bar and d/d bar) and neutral kaon (which is a "combination" of long and short versions) (I recognize that I may not be using precisely the right terminology here, but bear with me). It is all good and well to say that this is how the reality works, but there are unstated assumptions about what that means and about why it can happen in some circumstances and not others that aren't well articulated in the standard simplified quark model explanation of hadrons.

    For the neophyte, the practical solution is to simply treat them like irregular verbs in French that you memorize and don't ask why. But, it would be useful to have an intermediate level of explanation between "it's that way because Nature says so" and a full blown understanding of QCD at a level why some hadrons are mixes, and other are not which probably goes along the lines of "all of the quark content models are oversimplifications are what are really combinations of all possibilities that can produce the same quantum numbers for the composite particle, but if its 98%+ one state, it is convenient to treat them as if they are just one state", which would be O.K. is there was some explanation of why we see the mixes that we do. Hell maybe we really don't know (in which case this should be made more clear as an unsolved problem of physics), but I think that we know more than we do a good job of explaining to the interested educated layman.
     
  5. Sep 24, 2015 #4

    PeterDonis

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    It's going to be hard to discuss this if you can't be specific. Certainly my previous post was not intended to be a complete or even close to complete discussion of electroweak unification. I was only explaining the particular point that the OP asked about.

    I'm not sure what you mean by "the standard simplified quark model of hadrons". Again, it's hard to discuss if you can't be specific. Do you have a particular reference you can give that illustrates what you're talking about?

    I'm not sure what this means either. What are "mixes", and how do they differ from things that are not "mixes"? Again, a reference would really be helpful because I'm not understanding the issue.

    This is probably true, but I'm not sure "doing a good job" involves quite what you think it does. Many people seem to think that there should always be a way of explaining these things that makes them simple and intuitive, without realizing that often, the reason it seems like experts in the field see them as simple and intuitive is that they have painstakingly retrained their intuitions. A layman's intuitions are simply not well matched to physical phenomena that are outside of normal everyday experience.
     
  6. Sep 25, 2015 #5
    You are incorrect about the measurement being probabilistic, nature itself is. It's still one particle, it just exists in every location in it's path at the same time, that's not a fault of the measurement, that's how nature actually seems to be. You say it has a measurable properties, that's true, but position is not one of them. It only has a position upon being measured. It's not that we didn't know it's location because we didn't measure it, it's that it literally doesn't have position until it's measured.
     
  7. Sep 25, 2015 #6

    PeterDonis

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    This is not really a good description of what is going on; it implies that the particle has multiple positions at one time, rather than that the particle doesn't have a well-defined position at all until it's measured (which is a better description, and the one you give later in your post).

    (Also, even if we admit "having multiple positions at one time" as a candidate for a sloppy heuristic description of what's going on, based on the path integral formalism, it's still not right as you state it. The multiple "positions" that the particle has at one time are not multiple positions along one path; they're single positions on multiple possible paths that the particle could take.)
     
  8. Sep 25, 2015 #7

    Orodruin

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    This is exactly the point. Below electroweak symmetry breaking, the fields with definite masses are the ##Z##, ##A##, and ##W^\pm##, not the ##W^{1,2,3}## and ##B##. Above electroweak symmetry breaking the gauge bosons all become massless and it becomes more meaningful to talk about the fields in terms of their interactions. The ##W^{1,2,3}## are the gauge bosons of the non-Abelian SU(2) gauge group with coupling constant g', while the hypercharge gauge boson ##B## belongs to the Abelian hypercharge U(1) with coupling constant g.

    Similar concepts are behind all particle mixing, the mass eigenstates are generally not the states in which the interactions are diagonal. This leads to phenomena as K-Kbar or neutrino oscillations as well as to weak decays of heavy quark generations to lighter.
     
    Last edited: Sep 25, 2015
  9. Sep 25, 2015 #8

    PeterDonis

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    Do you mean "massless"? If electroweak symmetry is unbroken, the ##W^1##, ##W^2##, ##W^3##, and ##B## bosons are all massless (like all of the Standard Model fields), correct?
     
  10. Sep 25, 2015 #9

    Orodruin

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    Yes, thanks. Updating the post.
     
  11. Sep 25, 2015 #10
    Hi @Orodruin:

    I had the imprssion that the emerging of the Higgs boson and the resulting particle masses took place when the strong force separated from the electro-weak, not when the EM force separated from the weak? If I am incorrect about this, it seems to me there are two main possible scenarios (with some additional possible minor variations):

    1. Step 1: Higgs boson emerges and EM separates from weak. Strong remains unified with weak and also with EM. Step 2. Strong separates from weak or EM in some order. Step 3. Strong separates from the last still unified force.

    2. Higgs boson emerges and EM, weak and strong all de-unify concurrently. I confess this possibiity seems most unlikely to me; its too much of a coincidence that all of this distinct separations would happen concurrently.
    Which of these is right, or is there another explnation? If (2) is right, can you explian tha apparent coincidences?

    Regards,
    Buzz




     
  12. Sep 25, 2015 #11

    Orodruin

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    Neither is correct. Unification is assumed to be broken at an even higher scale (which is why you talk about three forces and not one).

    The electroweak symmetry breaking breaks the electroweak SU(2)xU(1) symmetry of the Standard Model down to a different U(1), which is the electromagnetic interaction. the SU(3) of strong interactions remains unbroken.
     
  13. Sep 25, 2015 #12
    Hi newjerseyrunner:

    I apologize for my sloppy phraseology. How about:
    A particle has properties about which the distribution of probabilities of possible measured values can be calculated.​
    Is this OK?

    Regards,
    Buzz


     
  14. Sep 25, 2015 #13
    Hi @Urodruin:

    Sometimes I get a strong impression that my failure to understand physics is not so much due to the difficulty of the concepts, but to the odd dialect of english that physisists use to explain these concepts. Perhaps it's just the physicsese. ;)

    Can you explain in simple language what is different in the two statements:
    (1) What I quoted above.
    (2) What I quote below.

    You seem to saying that the breaking of symmetry is not all concurrent, and that electo-weak breaks first. Is that correct?

    Regards,
    Buzz
     
    Last edited: Sep 25, 2015
  15. Sep 25, 2015 #14
    Hi @Peter:

    I think a lot of my confusion relates to a misunderstanding of how (1) QM theory about particles and their states relate to (2) QFT and quantum fields whose states are particles. I have had the confused idea that a quantum field was similar in some physical way to a classical field, e.g., an EM field. Is there a QFT about photons, and if so, is there a theorectical or mathematical connection between such a QFT and the Maxwell equations that define the behavior of the EM fields?

    Regards,
    Buzz
     
  16. Sep 25, 2015 #15

    PeterDonis

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    No. Here's the order of symmetry breaking (with reference to the Standard Model interactions):

    First, the grand unification symmetry (which unifies all three of the interactions--strong, weak, and electromagnetic) is broken. After this symmetry breaking, there are two interactions: strong and electroweak. In terms of symmetry groups, this symmetry breaking splits up the complete symmetry group of the Standard Model, which is SU(3) x SU(2) x U(1), into two parts: the SU(3) of the strong interaction, and the SU(2) x U(1) of the electroweak interaction.

    Then, the electroweak symmetry (which unifies the weak and electromagnetic interactions) is broken. This involves the Higgs field acquiring a vacuum expectation value, which causes the Standard Model fermions and the weak interaction gauge bosons to acquire mass, and the Higgs boson to emerge. After this symmetry breaking, there are three interactions: strong, weak, and electromagnetic. In terms of symmetry groups, this symmetry breaking splits up the SU(2) x U(1) symmetry of the electroweak interaction into two parts: the SU(2) symmetry of the weak interaction, and the U(1) symmetry of the electromagnetic interaction. (There is a twist here in that this U(1) is a different U(1) from the one that appears in the unbroken SU(2) x U(1) symmetry of the electroweak interaction.) The SU(3) of the strong interaction remains unchanged by electroweak symmetry breaking.
     
  17. Sep 25, 2015 #16
    Does this imply that at some energy scales, before electroweak symmetry breaking, that fermions are massless?
     
  18. Sep 26, 2015 #17

    Orodruin

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    Yes.
     
  19. Sep 26, 2015 #18
    Is this region accessible by current methods? It sounds very interesting.
     
  20. Sep 26, 2015 #19
    Hi Peter:

    Thank you very much for your post. Your post has the most clear statements I have ever seen about the sequence of symmetry breaking events and the role of the Higgs field/particle. I finally feel that I now understand this, althought the Lie group stuff is still over my head.

    I think the main reason for all my confusion was accepting the following as correct from https://en.wikipedia.org/wiki/Grand_Unified_Theory :
    The novel particles predicted by GUT models are expected to have masses around the GUT scale. . .​
    This seems to me to be clearly inconsistent with the quotes above by implying that the Higgs field caused particles to acquire mass before or during the GUT symmetry breaking, rather than later during the electroweak symmetry breaking. I hope you don't mind my suggesting that you would help many future interested users of Wikipedia regarding this topic by your correcting this Wikipedia error.

    Regards,
    Buzz
     
    Last edited: Sep 26, 2015
  21. Sep 26, 2015 #20

    PeterDonis

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    Unfortunately, no, not by many orders of magnitude.
     
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