Difference: PolarizationAtlasPartonLevel (r91 vs. r90)

Analysis of Single Top Polarization using Parton Level Atlas samples

1. Abstract

I looked at polarization distributions of Atlas parton level samples in the Optimal/Spectator basis for the s-channel and the t-channel. Trying to do this I ran into missing information and had to reconstruct quarks.

2. Parton Level Samples

The used parton samples are generated by AcerMC for the Atlas Single Top analysis. The parton level samples can be found in the Truth tree. This truth tree doesn't contain the initial particles, nor the spectator quark for the t-channel. This complicates things, as explained in the following.

3. Degree of Polarization

The degree of polarization is defined as D=(N--N+)/(N-+N+) where N- is the number of events with a negative polarized top and N+ is the number of events with a positive polarized top.

There are two ways to get the N- and the N+ from the histogram:


The first method means you just take the number of events at the left and right corner of the histogram (cos(?)=-1 and cos(?)=1 respectively). In the second method you actually count events and plug the total number of events with negative/positive polarized tops. This requires a fit.

The first method works as D is a ratio and using the actual number of N-/N+ we can simplify it by cancelling, so that we end up with the values at cos(?)=-1 and cos(?)=1.

Although half of the time we use the wrong direction of the antiquark, it is interesting to look at the s-channel and see the degree of polarization as a function of the number of jets in our event.


4. Optimal basis/ Spectator basis

For the Spectator basis in the t-channel use spectator quark and the antilepton in the CM frame of the top quark.

For the Optimal basis (s-channel) you have to take the incoming antiquark and the antilepton in the CM-frame of the top quark.
The Optimal Basis in the s-channel corresponds to the Spectator basis in the t-channel.

1. The t-channel

For the t-channel we need the spectator quark, which isn't given in the Truth tree. So we have to use its jets to get information about the original direction of the spectator quark. The jets have a pdgid of 0 or 5, where 5 stands for the b-tagged jets. We use only events that contain two jets and use the one with a pdgid of 0. If we condider consider the fact that electrons can be misinterpreted as jets, we should use events with three jets and use the first jet that fullfills our requirements.

Here is the result for the t-channel, spectator basis.


2. The s-channel

For the Optimal basis, s-channel, we need the direction of the incoming antiquark (the d quark). At the Tevatron we collide protons and antiprotons. Therefore the guess that the antiquark has the same direction as the antiproton is right in most cases, but not in all as there is a certain percentage of events in which the antiquark has its origins in the quark sea of the proton.

At the LHC protons are collided with protons, which means that there is no preferred direction for the antiquark. The guess that the antiquark comes, let's say from the right, should lead to the correct result in 50% of all events.


To get an idea why the polarization distribution looks the way it does let us take a look at the Tevatron first (ONETOP sample) and what happens when I guess the direction of the incoming d quark wrong.


In the left graph we assumed the antiquark would have the same direction as the incoming antiproton, the middle graph assumes an antiquark coming in the proton direction. The histogram on the right hand side is the sum of the two other histograms.

If I consider the fact that we expect a 100% polarized top quark in the s-channel, the offset in the first histogram must be due to a wrong assumption of the direction of the incoming antiquark. It makes sense that we see this offset, as a certain percentage of the antiquarks do not come from the antiproton but from the quark sea of the incoming proton.

The question is now: Does guessing the wrong side just give me the opposite polarization for the top quark? When we look at the second graph, the answer is no, as in that case we would see exactly the same offset in both the first and the second graph.
This can be explained by the fact that the polarization is measured not in the restframe of the incoming quarks (which would mean it should be symmetric in exchanging the directions of the quark and the antiquark) but in the top quark restframe.

If we assume the top to be produced at rest (which can be achieved by putting a cut onto the top PT), we therefore expect that guessing the wrong side just means opposite polarization for the top quark. As a consequence our offset would give us directly the number of antiquark whose direction we guessed wrong and therefore directly the percentage of antiquarks coming from the quark sea of the proton. Furthermore we would expect the two left graphs to have the same offset and the sum to be a horizontal line.
On the other hand by applying a minimum cut on the top PT we expect the assymetry between the two left graphs to be bigger and therefore the sum to show a steeper curve, a larger "Degree of Polarization".


Here we put a PT cut on the top quark, Px>100 GeV. If we compare the "Degree of Polarization" of the sum without and the sum with the cut, we get:

D(without cut)=0.102
D(with cut)=0.55

So how does the momentum of the topquark effect the polarization measurement? For the Optimal basis the angle between the incoming antiquark and the decay lepton is measured in the top quark restframe. Guessing the wrong direction for the antiquark doesn't affect our distribution as long as the topquark is close to rest, as then we have a totally symmetric situation and simply get the opposite polarization for the top quark. If our top quark has PT momentum and we guess the wrong direction, the situation is not symmetric any more but the chosen reference axis (which is the direction of the antiquark) is off by a certain angle compared to the direction of the optimal reference axis which is the direction of the quark. topmomentum1.jpg
Thus our basis is not optimal any more and we cannot expect to see full polarization.


What does all that mean for the LHC?
At the LHC half the time we guess the wrong direction for the antiquark. Therefore the polarization distribution for the s-channel should look like the sum of the two graphs, assuming the antiquark comes from the right, respectively from the left. That is it should look like the right histogram in the upper graph.

The polarization dsitribution s-channel (AcerMC) for Atlas with the beamline chosen as basis:

The "Degree of Polarization" is:
D=0.63 +/- 0.02

The high "degree of polarization" means we have a big assymetry between guessing the direction of the incoming antiquark right/ wrong. This means we have a high top PT, which makes sense, as the Atlas samples in contrast to the ONETOP samples have cuts on the minimum PT.
-- SarahHeim - 10 Mar 2008

2jettchannelfirstjet.jpgjpg2jettchannelfirstjet.jpgmanage 26.0 K 26 Mar 2008 - 20:17SarahHeim  
degreejets.jpgjpgdegreejets.jpgmanage 23.2 K 26 Mar 2008 - 20:41SarahHeim  
degreepolarization.jpgjpgdegreepolarization.jpgmanage 28.0 K 21 Mar 2008 - 18:38SarahHeim  
schanneloptimal2.jpgjpgschanneloptimal2.jpgmanage 22.9 K 26 Mar 2008 - 20:16SarahHeim  
schanneloptimalonetop2008.jpgjpgschanneloptimalonetop2008.jpgmanage 19.3 K 26 Mar 2008 - 20:47SarahHeim  
tevatronsum.jpgjpgtevatronsum.jpgmanage 24.8 K 10 Apr 2008 - 13:29SarahHeim  
tevatronsum1.jpgjpgtevatronsum1.jpgmanage 23.9 K 14 Apr 2008 - 16:14SarahHeim  
tevatronsum100.jpgjpgtevatronsum100.jpgmanage 24.9 K 14 Apr 2008 - 16:17SarahHeim  
topmomentum.jpgjpgtopmomentum.jpgmanage 8.7 K 10 Apr 2008 - 14:24SarahHeim  
topmomentum1.jpgjpgtopmomentum1.jpgmanage 6.9 K 10 Apr 2008 - 14:29SarahHeim  

This site is powered by FoswikiCopyright © by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
Ideas, requests, problems regarding Foswiki? Send feedback