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Aidan Randle-Conde | Université Libre de Bruxelles | Belgium

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Spinning out of control!

We’ve all heard the big news from CERN by now (if not then you might want to catch up on the latest gossip!) Right now most of the focus at ATLAS and CMS is on measuring the properties of the new boson we’ve found. The numbers of events are small, so studies are very difficult. One of the most important properties that we need to study is the particle’s spin, and luckily we can say something about that right now!

A typical Higgs boson candidate in the "golden mode" (ATLAS Collaboration)

The big news: One of many Higgs boson candidates in the "golden mode" (ATLAS Collaboration)

There are two ways to study the spin of this boson, the hard way and the easy way. The hard way involved looking at angles between the final state particles and that’s tricky, but it can be done with the existing data. This method is hard because we have to model both signal and background to get it right. The easy way is to look at the decays of the boson and see which ones happen and which ones don’t. We need a little more data to do this, but we can perform this study by the end the data taking for the year. Richard has already discussed the “hard” method, so I’m going to show the “easy” method. It comes with nice pictures, but there are a few subtleties.

I want to consider four decays: a decay to two photons, a decay to two \(Z\) bosons (the same applies to two \(W\) bosons), a decay to two \(\tau\) leptons, and a decay to two \(b\) quarks. All of these decay modes should be seen by both experiments if what we have seen is the Standard Model Higgs boson.

We need to label our particles properly and describe them a little before we begin. We can never measure the spin of a particle exactly, and the best we can do is measure its total spin, and its projection along a certain axis. The spin along the other two axes remains a mystery, because as soon as we measure its spin along one axis, the other two components of spin become indeterminate. That’s quantum mechanics for you! A component of spin can be increased or decreased with “raising” and “lowering” operators, and the change is always in natural units of 1. (This is just a result of the universe having three spatial dimensions, so if the answer was any different then the universe would look very different!)

Let’s take the electron and work out what spin states it can have. The electron’s total spin has been measured to be 1/2, so we need to project this spin onto an axis and find out the allowed values. A little thought shows that there are only two states that can exist: spin +1/2 and spin -1/2 (which we call “spin up” and “spin down”.) The \(J/\psi\) meson has spin 1, so it’s allowed states are +1, 0, -1. When the \(J/\psi\) is in state spin 0 what really mean is that it has “hidden” its spin at 90 degrees to the axis, so it’s total spin is still 1 and its projection along our chosen axis is 0.

So let’s get on with the job of considering the spins of all these other particles. The photon is a massless boson with spin 1, and it can only arrange its spin transversely (for obscure reasons that Flip explains very well), so it can’t hide its spin when it projects along an axis. That means that it can only have spin of +1 and -1. (There’s one more particle we’re going to use in these arguments, and that’s the gluon. The gluon is the same as the photon, except it interacts with a different field, so like the photon it can only have spin states +1 and -1):

The spin projections of the photon

The spin projections of the photon

The \(Z\) and \(W\) bosons are similar, except they have mass, so they have the luxury of hiding their spin. This means that they can have spin -1, 0, and 1, just like the \(J/\psi\) did:

Spin projections of the massive boson

Spin projections of the massive boson

Both the \(b\) quark and \(\tau\) lepton are fermions, which means that they have spin 1/2. We already know what spin states are allowed for fermions, spin up and spin down:

Spin projections of fermions

Spin projections of fermions

Now that we know the spin states of all these particles we can just add them up and confirm or refute which spin our new boson has. Let’s see how we can get spin 0:

Possible decays of a spin 0 particle

Possible decays of a spin 0 particle

It looks like we can a spin 0 particle by combining any of our particles.

Let’s try spin 1:

Possible decays of a spin 1 particle

Possible decays of a spin 1 particle

Uh-oh, it looks like we can’t make a spin 1 particle from photons! To align the spins correctly the photons must be in an antisymmetric state, which is absolutely forbidden by Bose-Einstein statistics. (Incidentally the term “boson” comes from the name Bose.) That means that this new boson is definitely not spin 1, because we see it decay to two photons.

So that means we have to do things the hard way to measure the spin of this new particle. For those who are interested, one of the main challenges presented here comes from the “acceptance” of the detectors- the kinematics of the final states we observe are significantly biased by the geometry of the detector. Even for a spin-0 boson, which decays isotropically, the distributions of the final decay products in the detector will not be isotropic, because the detectors do not have completely hermetic coverage. Fortunately since this post was first written we’ve gathered more data, and detailed studies have been performed eliminating all but the spin 0 hypothesis with a positive parity, indicating that what we have seen is most likely the long sought Standard Model Higgs boson after all.

Errata: In the original post I incorrectly made an argument stating that the decay of a spin 2 boson to a pair of quarks would be significantly more probable than the decay to a pair of leprons. Following discussions with Frank Close and Bob Cousins it was pointed out that well established graviton models would give a tensor interaction that would decay to leptons roughly 2% of the time per lepton flavour, making these final states accessible to the LHC experiments, and likely before the dijet final states would be accessible. My thanks go out to Close and Cousins for their correction.

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  • Sebastian

    Very nice article!

    But shouldn’t it read “We can exclude a spin 2 boson if we see bbˉ AS WELL AS ττ, and that’s why these two modes are so important to us!”?

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Oops, yes, good catch. Thanks Sebastian!

  • Mushfiq

    Very nicely written! Enjoyed!

  • Xezlec

    I love your articles. I wish I knew the difference between a Z and a Z* particle though.

  • Gil

    Nice analysis for us schmucks, since I think the CMS/Atlas did a poor job of explaining what was known of the particle’s spin.

    No mention of the of a spin -1 or spin -2 particle though.
    Are these a real possibility ?

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Xezlec, thanks for your comment! The difference between a Z and a Z* is a bit subtle. Whenever there’s an interaction that turns one set of particles into another then we have to consider all possible processes and sum them in order to get the rate correct. The probability that a particle will mediate a process depends on how far away its mass from the “nominal” mass. For the Z boson this mass is 91.2GeV, with a natural width of about 2GeV.

    When we have a spectrum of events, there will be several components, and one of them will be described by the natural lineshape of the Z boson (mean=91.2GeV, width=2GeV), and these are the “real” Z bosons. Another one will be “virtual” Z bosons, where the mass can take any value and the probability depends on the mass. There are two competing factors that suppress the production of virtual Z bosons, where higher masses are kinematically suppressed and masses further from the nominal mass are suppressed due to quantum mechanical effects.

    In the case of the new boson decay, with a mass of about 126GeV, and a Z boson with a mass of about 91GeV, we only have 31GeV left over for second Z boson. Since this isn’t enough to make a second real Z boson, it must me a virtual Z boson, and the mass spectrum can be predicted using quantum mechanics. A lot of people just call the decay H->ZZ and ignore all the mess about what is on shell and what is off shell, but then the pedants (like me!) will point out that ZZ is not the same as ZZ*. The difference may be small, but it’s important.

    (In fact, to make things even more complicated the virtual Z boson and virtual photon mix with each other, so the spectrum is even harder to compute than we’d like.)

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Gil, thanks for the comment! If you’re referring to the spin itself, then there can be no negative values, since we always take the magnitude of the spin. If you mean the charge conjugation+parity (CP) of the boson then finding that value is very complicated and will have to wait quite a long time! To get the CP of a particle we need to consider what happens when we change from matter to antimatter, and see if there’s any difference in the rate of production/decay of the particle. We can have CP even (+1) and CP odd (-1). When we perform the CP transformation, if the wavefunction changes sign then the particle is CP odd, and if it stays the same it’s CP even.

    Recently one of my colleagues gave a talk (over beer and fries!) about the measurement of the CP of the new boson, and to be honest I found the whole thing a little confusing. It seems that the easiest way to measure the CP of the new boson is to look at the tau tau and WW* final states, which are difficult because neutrinos take away momentum, and where the boson is produced in association with two sets of jets at small angles to the bean axis, which is difficult because of the acceptance of the detector. I doubt we’ll have that measurement until after the Phase I upgrade.

    There may be an easier way to measure the CP of this boson, but if there is I can’t think of it!

  • Chris Lester

    Hey Aiden,

    This is one of the coolest articles ive seen.

    I had a question about this: angular momentum sum rules allow for 1+1 = 0, 1, 2. The photon is a spin one particle with only two polarizations because of its masslessness, which seems to suggest the new higgsy particle isnt spin 1. but couldnt the two photons couple c to the |1 0> state of the Higgs. |1 0> = 1/sqrt(2) |1 +1> – 1/sqrt(2) |1 -1>. I think this is allowed but it would mean that the Higgs has the wrong CP value. This can be reconciled if the photons are emitted in a p-wave of orbital angular momentum.

    Chris

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Chris, I’m glad you liked this article! If this boson has spin 1 then as you say there would be a |1 0> state, but then as you also say the CP would be wrong. I think that if the boson was spin 1 and CP odd then it would be almost impossible for it to decay to two photons, even in a p-wave state. (For an example of a limit of a 1- state decaying to two photons, look at the J/psi and Z boson limits for diphoton decays- they’re very stringent.) I’ll take a closer look at the math and get back to you if I got it wrong.

  • Seth

    Hi,

    Thanks for the article. I’m wondering why this new boson can’t be spin 1? After all, a spin 1 boson can decay to two photons through a triangle diagram of fermions. Is it because the decay rate of a triangle diagram won’t match the current observation?

    Thanks a lot!

  • Pablo

    Hi,

    Quite nice the article, I was looking for this kind of “nice” explanation.
    But still have the doubt, even if the gluons appear with the b quarks, accounting for the conservation of momenta, the higgs-b-bbar vertex wouldn’t be correct if the higgs appear to have s=2, isn’t it? Otherwise
    angular momenta wouldn’t be conserved …

    Any idea about how the hypothetical s=2 higgs boson may couple to fermions?

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Seth, thanks for the question! The problem is that a spin 1 boson would have the wrong CP (charge-parity) to decay to two photons. The parity of a particle describes how it arranges its wavefunction (it’s probability field) in space, and for a pointlike object this is either odd or even. Odd means that if you flip it through the origin it inverts, and even means if you flip it through the origin it stays the same. When calculating the probability of transition from one state to another essentially what we do find the overlap of the wavefunctions in all space and integrate to get the total probability. Each photon has CP odd, so a pair of photons (with no net angular momentum between them) is CP even. The spin 1 boson is CP odd, therefore the overlap of the wavefunctions is zero, and the only way to see the transition is if the spin 1 boson it’s really fundamental, but has some structure to it.

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Pablo, thanks for the question! As it happens, there is no spin 2 Higgs boson, and if you read the post carefully you’ll notice I only talk about a “boson”. The point of the post is that we have definitely seen a boson, but we don’t know if it’s the Higgs boson or not. If it turns out to be spin 2, then it cannot be a Higgs boson at all, and we can tell its spin by studying its decays. Sorry if there was some confusion!

  • frankclose

    spin2 can decay into two taus in a P-wave. there is plenty of momentum and no general principle that forbids it. also the forbidden decay of spin 1 to two vectors applies only to IDENTICAL vectors, hence two real photons cannot but a real and a virtual (or ZZ* for example) can. or am i misunderstanding something.

  • Chris Spencer

    Hi, I’m a 16 year old high school student from Australia, who is definitely interested in a career in particle physics. This may be a little off topic, but from my understanding, if the particle is spin 0, then it conforms with the current understanding of the Higgs, but if it’s spin 2, then it must be a graviton. I could be completely wrong here, as I have limited understanding of this field (although that’s what I love about researching these things!), but based on your understanding, what would the implications of a graviton have on our understanding of the standard model (since it doesn’t fit in the current model), and if so, does it start supporting ideas like the requirement of a string theory or quantum gravitation to tie in particle physics with relativity? Also, do you have any advice for a student who would love to have the opportunity to work at CERN?

  • Greg Myers

    Hi Aiden…
    Great blog. Are there any updates as to the decay limits to tau-tau?? Is is still suspicially low, or has enough data come in yet to tell? Thanks, Greg.

  • Andreas

    Perhaps getting back to Pablo’s real question above, i.e. semantics about Higgs or Higgs-like bosons aside, I’d appreciate a little further explanation regarding the gluon argument in the spin-2 b-quark case. Are the referenced gluons meant to be those involved in hadronization? Also, if the newly discovered state is composite and spin-2, then that surely can afford to produce two taus in a P-wave. But, as I think Pablo was trying to point out, if the new state is point-like then one must be careful to conserve angular momentum at the interaction vertex with the new state.

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Greg! Thanks for the comment. More and more data has been coming in since July 4th, but it’s still being analyzed, and it’s only when we are sure we understand it that we “unblind” the analysis and look at the results. We can expect an update at the Hadron Physics Conference which takes place in Japan from November 12th. Until then we have to wait. I suspect we’ll see the tau tau channel from ATLAS at that point, and possible see something from CMS as well. But if we don’t, then that will be very exciting!

  • frankclose

    J=2 can decay into two taus whether the J=2 interaction is pointlike or not. There is no need to invoke compositeness. Likewise, there is no need to invoke a gluon in the b case: a J=2 can decay into bb(bar). The decay goes in P=wave. There is a large amount of momentum in the decay and there is no suppression of the P-wave.This is basic quantum; nothing subtle.If this blog site is to be useful and not mislead, please correct the table! If you are in doubt, send me an email at f.close@physics.ox.ac.uk or ask some neighbourhood theorist.

  • frankclose

    And the explanation of the spin 1 being forbidden to decay to two photons: this is not forbidden by CP if the initial J=1 is 1++ (axial). Bose symmetry + gauge invariance (“Yang’s theorem”) forbids the decay into two IDENTICAl photons, such as two real photons. But nothing forbids the decay into one real and one virtual.

  • frankclose

    even for pointlike there is no forbidding of P-wave. Possible confusion is one’s familiarity with Fermi beta decay where the pointlike is in S-wave. But this is because it is strictly non-relativistic with effectively no momentum transfer. For a decay with large momentum released (as for a 125 GeV decay into fermion-antifermion) P-wave is allowed (in fact very large angular momenta could be allowed, so almost any value of initial J could decay into two taus, or into quark-antiquark without suppression. There is no need to require a gluon to accompany the quarks (not a photon to accompany the tatus!)

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Chris, thanks for the questions! As far as models go, I have only come across one model which has a fundamental spin 2 particle, and that is the graviton. The graviton has spin 2 because it’s a tensorial force and carries two indices around with it. (There’s probably some additional subtlety to the argument somewhere.) If this is the graviton then this means a few things. First of all it means there is probably no Higgs boson (since most reasonable regions that are not around 126GeV have already been ruled out.) Second of all the graviton has been excluded from most models at this mass by direct searches, so it would need to be a very strange model to account for lack of observation before now. I’m afraid I don’t know enough about how to incorporate gravity into the rest of the Standard Model, especially with respect to things like string theory. If it is a graviton then some theorists will certainly come up with some models and tell the experimentalists (like me) where to look and what signatures to expect, and this will help us to form a new model of the forces.

    If you want to work at CERN then the Summer Student program is a great place to start! You get the chance to spend a summer at CERN working on a project that fits into a timeline of several weeks. You’d get to meet all kinds of other students, enjoy the local area and the social life is great fun. Take a look at https://hr-recruit.web.cern.ch/hr-recruit/summies/default%20page/default.asp

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Hi Frank, thanks for the comments! To clarify something before I begin, as you point out the decay of a spin 2 object to two taus is not absolutely forbidden (nearly no decay is absolutely forbidden, even μ->eγ) but the purposes of working on a hadron collider, severely suppressed is no different to absolutely forbidden. If a decay is suppressed enough that we’ll never see it then it may as well not exist for us, especially if we have a well defined expectation of when a given decay mode should be seen.

    As for the rate of a spin 2 object decaying to two fermions, I found no clear literature on the topic, so I worked from analogy. The Clebsch-Gordon coefficients forbid the transition of 2→1/2 1/2 so in the non relativistic limit this decay does not happen. To double check this I compared the decays of heavy mesons to fermions and observed that in the ccbar system a spin 0 object rarely decays to two fermions, a spin 1 object decays this way quite frequently, and for a spin 2 objects it hasn’t been seen at all. Now this is, of course, a different system, and it has different constraints, so we shouldn’t be taking the analogy too seriously. Even so, it suggests that the decay of a spin 2 object to two fermions is suppressed to some extent, even in the relativistic limit. Furthermore most of the graviton searches (and other fundamental spin 2 object searches) at hadron colliders tend to look for a diphoton state, in favor of a difermion state, even though two leptons would be a much cleaner signal. Presumably this is because of the suppression to the two fermion final state. If you can point me in the direction of a reputable paper that outlines non-radiative decays of a spin 2 object to two fermions I’ll be happy to read it and revise this blog post based on this information.

    For the decay of the spin 1 object to two photons surely the argument of identical bosons is incorrect, since the spins of the photons must be opposed to each other and this can (in principle) be measured.

  • frankclose

    Aidan

    Please give me an email address where I can contact you. The arguments in your blog are seriously wrong and the analogies you give in answer to me even more so.

    Simple example: the quark model. Spin 2 meson is qqbar in P wave. Ergo a spin 2 could decay to two fermions. Your clebsch gordans completely neglect relative orbital angular momentum.

    The decay of spin 1 to two photons being forbidden is subtle. It is combintaion of Bose and gauge invariance (known as Yangs theorem) It forbids tow photons of same q^2. Two photons of different photons are allowed (e.g. the f1 1++ meson decays into gamma + gamma* experimentally).

    Look at my textbook Introduction to Quarks and Partons if it hasnt been stolen from the CERN library or look at any standard textbook.

  • frankclose

    Aidan a 0^+ state decays to tau-antitau in P wave just like a J=2. So any “suppressions” would be the same. A decay in S-wave would come from 0^- not 0^+. I am sending an email to aidan@cern.ch to try and explain these basic physics [For anyone reading this wondering why 0^+ is Pwave and 0^- is S-wave: it is because fermion and antifermion have opposite intrinsic parity.

  • Ed Williams

    What do you mean “if this is the graviton”. Gravity has infinite range like electromagnetism and graviton has to be massless like the photon. So whatever this particle is it aint the graviton.

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  • Greg Myers

    So, at the Hardon Collider Symposium, what kinds of things may we find out about the Higgs? :)

  • Tin

    Why didn’t you correct the table after Prof. Close corrections? If I hadn’t read all the comments, or if I hadn’t read Prof. Close books, I would have perhaps believed in what you wrote. If this blog is taken with so little care by the authors themselves, than I am afraid I won’t read more of it.

  • frank close

    reply by Aidan is false and does not represent what i emailed him.
    a spin 2 decays to fermion antifermion in P wave just like a spin 0 does.
    there is no “penalty” for spin 2 decay any more than for spin 0+
    aidan – and annyone who really cares – a 0- decays in S wave but 0+ (higgs) decays in P wave EXACTLY the same wave as a spin 2

    post my emails to you and not your misunderstood summary of them.

  • http://aidanatcern.wordpress.com Aidan Randle-Conde

    Prof Close and I disagree on some of these points and I think that he may have misunderstood some of my blog post. He followed up on this privately but we continued to disagree, but I do not think that the disagreements invalidate the overall message of the blog post.

    In a comment reply I misrepresented Prof Close, and I apologise for this. It’s now been amended.