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The title says it all really, I searched this website and came across a post with a question titled Why is the Higgs boson spin 0?. But it doesn't really answer my question in the title.

But this next post I found on this site is much more strongly related to my question.

In fact, it is one of the answers to the question in that post which I am essentially questioning. Here is a direct quote for the answer given by @Chiral Anomaly:

The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.

The problem with this quote is that it just seems to be just concerned with conserving the spin projection quantum number - which I denote here by $s_z$ and $s$ for the spin quantum number. The $z$-direction in $s_z$ is arbitrary and could be any direction.

As I understand it, photons always have spin, $s=1$. Just like electrons always have a spin of $s=1/2$, but can have $2s+1 (= 2)$ spin projections, namely, $s_z=\pm 1/2$.

So the only way to conserve spin projection would be to have one photon with $-1$ spin projection and the other photon with $+1$ spin projection. Or, have both photons with zero spin projection. That is the only way to conserve spin projection in the $H \to \gamma\gamma$ decay mode. But, the problem is we are not trying to conserve spin projection, but rather spin, $s$, itself.

I tried searching the internet for explanations of why that decay shows the spin of the Higgs boson is different from one, but I cannot get a straight answer. It must be something quite trivial as the original discovery paper in July 2012 simply stated in its abstract that "The decay to two photons indicates that the new particle is a boson with spin different from one" which can be found here after selecting the appropriate title paper which looks like:

Higgs original discovery paper


So, just to summarize, why does the Higgs diphoton decay prove that the spin of the Higgs is different from one?

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In simple terms :

Spin had to be invented and assigned studying the interactions of elementary particles because of conservation of angular momentum. There are three laws which go through all physics theories/models as axioms, conservation of energy, conservation of momentum, and conservation of angular momentum.

Conservation of energy led to special relativity for the study of nuclear and elementary particle interactions , special relativity reduces to Galilean mathematically smoothly . Conservation of momentum at the center of mass of a particle forces it to decay to two particles.

For conservation of angular momentum:

The pi0 decays to two photons and the decay was seen long before a quark model was proposed for the theory of elementary particles.

The discovery of particles is a process a century long, and started with cosmic rays, where the pi+ and pi- were seen , and the pi0 inferred.

The existence of the neutral pion was inferred from observing its decay products from cosmic rays, a so-called "soft component" of slow electrons with photons. The π0 was identified definitively at the University of California's cyclotron in 1950 by observing its decay into two photons. Later in the same year, they were also observed in cosmic-ray balloon experiments at Bristol University.

The symmetry models developed at the time gave the pion spin zero, and this has not been falsified. I use the pi0 decay into two photons to give an example of entanglement, that if the spin projection of one photon is identified as +1, the other has -1, because of conservation of angular momentum, spin is part of the angular momentum of a system of particles.

The studies led to the eightfold way of symmetries for the particles discovered,

octet

where the pi0 has to have spin zero from the symmetry.

The eightfold way eventually led to the symmetries of the present standard model

So the decay of Higgs to two photons, within our current model of particle physics with its symmetries , by conservation of angular momentum algebraically leads to the conclusion that it either has spin 0 or spin 2.

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  • $\begingroup$ Thanks for your answer, I just have a related question to this post but not directly related. As I understand it spin-parity is represented by the notation $J^P$, where $P=\pm$, and $J=L+S$, the sum of the orbital angular momentum quantum no. ($L$) for the particle and $S$, the spin quantum no. for the particle. But for photons $L$ must be zero right? As the photon is not like an electron bound to an atom, which 'orbits' the nucleus. If this is true all of the angular momentum for the photon ($\gamma$) comes from its' spin. So we might as well write $S^P$. Do we have $L=0$ for photons? $\endgroup$ May 6, 2022 at 9:36
  • $\begingroup$ @N.Ginlabs the notation of J is only for bound states, AFAIK, and the photon is never bound as you note so it does not need the notation. $\endgroup$
    – anna v
    May 6, 2022 at 10:27
  • $\begingroup$ Thanks for your reply, I'm sorry I don't know what "AFAIK" means, but you are saying that $L=0$, right? (for unbound photons) $\endgroup$ May 6, 2022 at 12:23
  • $\begingroup$ @N.Ginlabs As Far As I Know . No, L has not meaning/value for aphoton .J is a quantum number. S is the quantum number for the photon spin. see my answer here physics.stackexchange.com/questions/706931/… $\endgroup$
    – anna v
    May 6, 2022 at 13:59
  • $\begingroup$ Really nice answer. $\endgroup$ May 9, 2022 at 20:58
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You surely learned at the beginning of your elementary particles' course that the two photons are bosons, so the γγ state is symmetric in them.

Addition of two spin 1 states yields a spin s=1 only when they are in an antisymmetric state, $$ |1, 0\rangle= \frac{1}{\sqrt{2}}\bigl (|\uparrow\downarrow\rangle -|\downarrow\uparrow\rangle \bigr ), $$ but in which they cannot be, by boson symmetry, above. (Recall the cross product of two 3-vectors is a 3-pseudovector, antisymmetric in the two constituent vectors!) Spin projection and spin are preserved in this expression, but the antisymmetry of the wavefunction kills it for symmetric photons.

So the spin of the Higgs must be 0 or 2. Spin 2 is firmly excluded from production estimates, cf Fig 2 in ATLAS Aad, Georges, Tatevik Abajyan, Brad Abbott, Jalal Abdallah, S. Abdel Khalek, Rosemarie Aben, B. Abi et al. "Evidence for the spin-0 nature of the Higgs boson using ATLAS data" Physics Letters B 726 (2013) 120-144 .

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  • $\begingroup$ Thanks for your answer. "You surely learned at the beginning of your elementary particles' course that the two photons are bosons, so the γγ state is symmetric in them. Addition of two spin 1 states yields a spin 1 when they are in an antisymmetric state, which they cannot be, by above." Yes, I knew the part about photons being bosons which have integer spin, but the rest I didn't know. This business about symmetic/antisymmetric states is not something I knew about. Do you have a link to some resource that explains what this means please? $\endgroup$ May 4, 2022 at 16:59
  • $\begingroup$ Most group theory courses for physics more or less start with this review of angular momentum, and there should be questions on this site on 1⊗1. You already know of this fact by the manifest antisymmetry of the J =1 column in the Clebsch Matrix. $\endgroup$ May 4, 2022 at 17:14
  • $\begingroup$ "You already know of this fact by the manifest antisymmetry of the J =1 in the Clebsch Matrix". No, I did not know about this and have never even heard of a Clebsch Matrix. Is it really necessary to turn to group theory to explain this, I mean you started off by saying that I should already know about this from my "elementary particles' course", and now you just brought group theory into it. I'm sorry, I don't mean to sound ungrateful but I really think there is a simpler way to explain this. I do remember learning that the overall wavefunction must be antisymmetric ..... $\endgroup$ May 4, 2022 at 17:26
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    $\begingroup$ As I already indicated above, for three bosonic 3-vectors A, B and C, the cross product $C= A\times B$ is manifestly antisymmetric in A and B... $\endgroup$ May 4, 2022 at 17:53
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    $\begingroup$ Thanks for your reply, I'm sorry but I have literally no idea what are talking about. There must be a simpler way to explain this and I have updated my post to address this concern. Thanks all the same :) $\endgroup$ May 5, 2022 at 3:28

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