On the 2012 Independence Day, the world of physics has made a transition. The question "Is there a Higgs-like particle with mass near \(126\GeV\)?" has been answered.
Weinberg's toilet, i.e. Glashow's model of the SM Higgs boson. Explanations below.
The ATLAS Collaboration gave a 5-sigma "Yes" answer; via the rules of the normal distribution, this translates to the risk "1 in a million" that the finding is noise i.e. that it is a false positive.
Even CMS, despite its inability to use the correct Comic Sans fonts, has been able to do the same thing: 5 sigma discovery, "1 in a million" certainty. If you combine the two independent experiments, the risk of a false positive multiplies: the LHC says that there is a 7-sigma (via the Pythagorean theorem), or "1 in a trillion" (via a product), risk of a false positive. Compare it with 1-in-10 or 1-in-3 or 9-in-10 risk of a false positive that is tolerable in the climate "science" or various medical "sciences".
Once this question is answered, the new question is: What is the right theory that describes the precise behavior of the new particle?
There is one popular candidate: the so-called Standard Model. Steven Weinberg gave it its name and he also constructed much of it. Besides old and well-known things, it also includes one Higgs doublet which gives rise to one physical Higgs polarization.
Various trained and active physicists have worshiped the God particle by rituals that only physicists have mastered so well. Stephen Wolfram called it a hack and this word was actually used by some currently active physicists, too. High school and Nobel trip pal of Weinberg, Shelly Glashow, called it the Weinberg toilet. It's something your apartment needs but you're not necessarily too proud about it.
The Standard Model is popular because it's minimal in a certain sense that is easily comprehensible for humans: it contains a minimum number of such toilets. Whether it's so great for Nature or for a landlord or for a crowded aircraft serving strange combinations of food for lunch to have the minimum possible (consistent) number of toilets remains to be seen; minimizing the number or the sophistication of toilets actually isn't necessarily Nature's #1 priority. Now, however, we may already give an answer to the question:
How much are the 2011-2012 LHC data compatible with the minimum solution, the Standard Model?
To answer the question, we must look at somewhat more detailed properties of the new particle and apply a statistical procedure. As an undergraduate freshman, I didn't consider fancy statistics to be a full-fledged part of physics. Can't you measure things accurately and leave the discussion of errors to the sloppy people who make errors? However, the good enough experimenters in Prague forced me to learn some basic statistics and I learned much more of it later. Statistics is needed because even physicists who love accuracy and precision inevitably get results that are incomplete or inaccurate for various reasons.
The text will be getting increasing technical as we continue. Please feel free to stop and escape this blog with a horrified face. Of course, I will try to make it as comprehensible as possible.
Collecting the data
We could consider very detailed properties of the new particle but instead, we will only look at five "channels" – five possible recipes for final products into which the Higgs particle may decay (not caring about some detailed properties of the final products except their belonging to one of the five types). We will simply count how many times the LHC has seen the final products of one of the five types that look just like if they come from the Higgs.
Karel Gott (*1939 Pilsen), a perennial star of the Czech pop-music, "Forever Young" (in German; Czech). The video is from Lu-CERN, Switzerland.
In each of the five groups, some of them really come from the Higgs; others come from processes without any Higgs that are just able to fully emulate the Higgs. The latter collisions – called the "background" – are unavoidable and inseparable and they make the Higgs search or any other search in particle physics harder.
The LHC has recorded over a quadrillion (1,000 trillions or million billions) collisions. We will only look at the relevant ones. For each of the five groups below, the Standard Model predicts a certain number, plus minus a certain error, and the LHC (ATLAS plus CMS which we will combine, erasing the artificial separation of the collisions into two groups) has measured another number. The five groups are these processes:\[
\eq{
pp&\to h\to b\bar b \\
pp&\to h\to \gamma\gamma \\
pp&\to h\to \tau^+\tau^- \\
pp&\to h\to W^+ W^- \\
pp&\to h\to ZZ
}
\] You see that in all of them, two protons collide and create a Higgs particle, among other things. This particle quickly decays – either to a bottom quark-antiquark pair; or two photons; or two tau-leptons; or two W-bosons; or two Z-bosons.
Now open the freshly updated Phil Gibbs' viXra Higgs combo java applet (TRF) and play with the "Channel" buttons only, so that you don't damage this sensitive device. ;-)
If you're brave enough, switch the button from "Official" to "Custom" and make five readings. In each of them, you just check one of the boxes corresponding to the five channels above; \(Z\gamma\) yields no data and we omitted this "sixth channel". Then you look at the chart you get at the top.
Oh, I forgot, you must also switch "Plot Type" (the very first option) from "Exclusion" to "Signal" to make things clear. The graphs then show you whether you're close to the red, no-Higgs prediction or the green, yes-Higgs prediction.
In each of the five exercises, you move your mouse above the Higgs mass \(126\GeV\) or, if not possible, \(125\GeV\) which is close to the \(126\GeV\) that the LHC currently predicts (the average of ATLAS and CMS). A small rectangle shows up and you read the last entry, "xsigm". It tells you how much the combined LHC data agree with the yes-Higgs prediction of the Standard Model. "xsigm=0" means a perfect agreement, "xsigm=-1" means that the measured number of events is one standard deviation below the predictions, and so on. You got the point. Of course, you could read "xsigm" from the graph itself.
For the five processes, you get the following values of "xsigm":\[
\begin{array}{r|l}
\text{Decay products}& \text{xsigm}\\
\hline
b\bar b & +0.35 \\
\gamma\gamma& +2.52 \\
\tau^+\tau^-& -1.96 \\
W^+ W^- & -1.75 \\
Z^0 Z^0 & +0.32
\end{array}
\] I added the sign to indicate whether the observations showed an excess or a deficit; I am not sure why Phil decided to hide the signs. You see that only the \(\gamma\gamma\) i.e. diphoton channel has a deviation greater than two standard deviations. But there are two others that are close to two standard deviations, too: a deficit of ditau events and the W-boson pairs.
On the other hand, there are five channels which is a rather high number: it matches the number of fingers on a hand of a vanilla human. With a high number of entries, you expect some entries to deviate by more than one sigma by chance, don't you? How do we statistically determine how good the overall agreement is?
We will perform the so-called chi-squared test
Because both signs indicate a discrepancy between the Standard Model and the observations, we will square each value of "xsigm" and sum these squares. In other words, we will define\[
Q = \sum_{i=1}^5 {\rm xsigm}_i^2.
\] The squaring is natural for general mathematical reasons; it's even more natural if you realize that \({\rm xsigm}\) are actually quantities distributed along the normal distribution with the standard deviation of one and the normal distribution is Gaussian. It has \(-{\rm xsigm}^2/2\) in the exponent.
Fine, so how much do we get with our numbers?\[
\eq{
Q&=0.35^2+2.52^2+1.96^2+1.75^2+0.32^2\\
Q&= 13.48
}
\] The result is approximate. Yes, the signs cancelled here, anyway.
At this moment, we would like to know whether the \(Q\) score is too high or too low or just normal. First, it is clearly too high. It's because \(Q\) was a sum of five terms and each of them was expected to equal one in average; this "expectation value of \({\rm xsigm}^2\)" is what defines the standard deviation. So the average value of \(Q\) we should get is \(5\). We got \(13.48\).
How bad the disagreement is? What is the chance that we get \(13.48\) even though the expected average is just five?
Looking at the \(\chi^2\) distribution
To answer this question, we must know something about the statistical distribution for \(Q\). We know that the average is \(\langle Q\rangle=5\) but we need to know the probability that \(Q\) belongs to any interval. By definition of the \(\chi^2\)-distribution, the quantity \(Q\) defined as the sum of squares of (thanks, JP) five normally distributed quantities (with a vanishing mean and with the unit standard deviation) is distributed according to the chi-squared distribution. If you prefer mathematical symbols,\[
Q\sim \chi_5^2.
\] The subscript "five" is the number of normally distributed quantities we're summing. This distribution is no longer normal. After all, the negative values are strictly prohibited, something that is unheard of in the world of normal distributions.
The probability density function (PDF) for this distribution – and the PDF is what we normally call "the distribution function" – may be explicitly calculated by composing the five normal distributions in a certain way and computing the resulting integral in spherical coordinates. The result is\[
\eq{
d\,Prob(Q\in(Q,Q+dQ))&= dQ\cdot \rho(Q)\\
\rho(Q) &= \frac{Q^{k/2-1}\exp(-Q/2)}{2^{k/2}\Gamma(k/2)}
}
\] The exponent in the exponential is just \(-Q/2\) without squaring because it comes from the Gaussians but \(Q\) already includes terms like \({\rm xsigm}^2\). The additional power of \(Q\) in the numerator arises from the integral over the four angular spherical coordinates in the 5-dimensional space. The remaining factors including the Gamma-function are needed to normalize \(\int\rho(Q)\dd Q=1\).
Now, the cumulative distribution function (CDF) – the probability that \(Q\) is found in the interval \((-\infty,Q)\) for some value of \(Q\) (hope you won't confuse the \(Q\)'s: there's no real potential for confusion here) is an integral of the PDF and it may be expressed in terms of some special function. Then you just substitute the numbers to get the probability.
However, such integrals may be calculated directly if you have Wolfram Mathematica 8. For example, if you ask the system to compute
Probability[q > 5., Distributed[q, ChiSquareDistribution[5]]]you will be told that the result is 41.588%, not far from the naive "50%" you would expect. The deviation arises because 5 is really the average and that's something else than the "median" which is the value that divides equally likely (50%) intervals of smaller and larger values.
Now, you just modify one number above and calculate
Probability[q > 13.48, Distributed[q, ChiSquareDistribution[5]]]The result is \(0.019\) i.e. \(1.9\%\). The probability is about 1-in-50 that the \(Q\) score distributed according to this \(\chi^2\) distribution is equal to \(13.48\) or it is larger. How should you think about it?
Counting civilizations
Imagine that the Standard Model is exactly correct. There are millions of planets in the Universe. Each of them has an LHC that has accumulated – if we count the combination from \(7\TeV\) as well as \(8\TeV\) collisions and from the year 2011 as well as 2012 after their own Jesus Christ – about 20 inverse femtobarns of high-energy proton-proton collisions.
There is an Internet on each planet, invented by their own Al Gore. If I have horrified you by the idea of a million of Al Gores, let me mention that there is also The Reference Frame on each planet that has processed the local LHC results and calculated \(Q\). However, the collisions were random so they also got a different number of events of various kinds and a different value of \(Q\).
The probability \(1.9\%\) means that \(1.9\%\) of the planets found \(Q\) that was at least \(13.48\). So if you believe that the Standard Model is correct, you must also believe that by chance, we belong among the top \(1.9\%\) of the planets that were just "unlucky" enough because their LHC measured a pretty large deviation from the Standard Model.
Is that shocking that by chance, we belong among the "top \(1.9\%\) offenders" within the planetary physics communities? This value of \(1.9\%\) is slightly greater than a 2-sigma signal. So you may say that the group of five channels we have looked at provides us with a 2-sigma evidence or somewhat stronger evidence that the Standard Model isn't quite right.
This is too weak a signal to make definitive conclusions, of course. It may strengthen in the future; it may weaken, too. After the number of collisions that is \(N\) times larger than those that have been used so far, the number "2" in this overall "2 sigma" will get multiplied roughly by \(\sqrt{N}\) if the deviation boils down to Nature's real disrespect for the exact laws of the Standard Model.
However, it will be mostly moving between 0 and 1.5 sigma in the future (according to the normal distribution: a higher spurious confidence level in sigmas would drop there following the function \(CL/\sqrt{N}\)) if the deviation from the Standard Model we observe and we quantified is due to noise and chance. It's too early to say but the answer will be getting increasingly clear in the near future. For example, if the signal is real, those overall 2 sigma may grow to up to 4 sigma after the end of the 2012 run.
Stay tuned.
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