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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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