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. This effects the polarization distributions, in a way that is described below.

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. Optimal basis/ Spectator basis

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.


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

1. The t-channel

For the t-channel we need the spectator quark, which isn't given in the Truth tree, either. 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 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.

2jettchannelfirstjet.jpg

2. The s-channel

schanneloptimal2.jpg

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.

tevatronsum1.jpg

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. %Br% 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 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 the two left graphs to have the same offset and the sum to be a horizontal line.

4. 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:

degreepolarization.jpg

The first method means you just take the number of events at the left and right corner of the histogram (cos(θ)=-1 and cos(θ)=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(θ)=-1 and cos(θ)=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.

degreejets.jpg

-- SarahHeim - 10 Mar 2008
Topic attachments
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schanneloptimalonetop2008.jpgjpg schanneloptimalonetop2008.jpg manage 19.3 K 26 Mar 2008 - 20:47 SarahHeim  
schanneloptimal2.jpgjpg schanneloptimal2.jpg manage 22.9 K 26 Mar 2008 - 20:16 SarahHeim  
degreejets.jpgjpg degreejets.jpg manage 23.2 K 26 Mar 2008 - 20:41 SarahHeim  
tevatronsum.jpgjpg tevatronsum.jpg manage 24.8 K 10 Apr 2008 - 13:29 SarahHeim  
2jettchannelfirstjet.jpgjpg 2jettchannelfirstjet.jpg manage 26.0 K 26 Mar 2008 - 20:17 SarahHeim  
degreepolarization.jpgjpg degreepolarization.jpg manage 28.0 K 21 Mar 2008 - 18:38 SarahHeim  
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Topic revision: r55 - 10 Apr 2008, SarahHeim
 

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