To correct the cross-feed, we solve equations N0 = R0 + R1*f1, N1 = R1 + R0*f0.
N is the number of "signal" obtained from fit. R is the number that with cross-feed
extracted. f is the probability to feed up or feed down relative to the reconstruction
efficience which is obtained from MC. Please note that in MC we fixed the width of
gaussian to that of real data.
f0 = 0.0908 ± 0.0068
f1 = 0.840 ± 0.042
R0 = 154.8 ± 22.5
R1 = 41.2 ± 11.2
We also fit the Ds* sideband subtracted spectrum for Ds*pi0. As the feed up amount
in the Ds* signal region is the same as in the sideband, this is an independent
way to extract feed up.
N = 45.7 ± 11.6, M = 351.2 ± 1.7, sigma = 5.5 ± 1.1
Selina has independent measurement of DsPi0 signal. The numbers of DsPi0 "signal" are different between two. We did check and found that the spectra are very similar. The difference is mainly due to fluctuation that causes different fit result. The difference is about 1 sigma. I think we should not worry too much about that.
List of plots and brief explanations:
2d_1011_b.eps and 2d_1011_c.eps M(DsGam)-M(Ds) vs M(DsGamPi0) - M(DsGam). This was suggested by John. We should see two bands vertical one corresponds to Ds* signal, horizontal one corresponds to 2460 signal and feed up of 2320.
2d_1012_b.eps and 2d_1012_c.eps M(DsGam) vs M(DsGamPi0). Similar as first one except that now the feedup spreads out.
2d_1013_b.eps and 2d_1013_c.eps M(DsPi0) - M(Ds) vs M(DsGamPi0) - M(DsGam). Because we use differences, the two are really correlated. It is difficult to see things except that we know the cross-feed is easy.
2d_1014_b.eps and 2d_1014_c.eps M(DsPi0) vs M(DsGamPi0). Samilar as group 3. Now it spreads along Y direction. We see the more entries in (2460) signal region, and less at its side band. This indicate that the signal is not all come from feed up of (2317).
All CLEO II and CLOE II.V datum are used except for 4sR and 4sS due to problem on computer.
The D0's are reconstructed from K-pi+ and K-pi+pi0. The D0 is further required to form
a D*+ with a soft pi+ and M(D*)-M(D0)-M(expectation) within 2 MeV (about ~2.5 sigma).
I also require P(D0)>1.0 GeV which I need to use to suppress bkg. The pi0's are
reconstructed same way as for Ds(*)pi0 studies including E gamma > 0.1. The mass spectra
of D0 are fit with 1-gaussian + 2nd order polynomial function. We get
M(D0->Kpi) = 1864.0 ± 0.5 (with sigma ~ 14.3, N~2850)
M(D0->Kpipi0) = 1863.6 ± 0.8 (with sigma ~ 19.1, N~4890)
There exists one problem that the momentum of pi0 is different from that of Ds(*)pi0
studies. So we select compare pi0 momentum spectra in DsPi0 signal region and D0 signal
region. The sample from D0 is then reweighted so that the spectra are the same.
With all these, we obtain
M(D0->Kpipi0) = 1864.2 ± 0.9 (with sigma ~17.5)
That only D0->Kpipi0 is reweighted makes me unhappy. So I try to reweight event based on D0 momentum by checking weight profile vs P(D0) from above method. Unfortunately the smearing is too big that if I do that it does not produce correct pi0 momentum at all. In fact the profile is very close to be flat. At this stage I give up further work.
So the difference is 0.4 ± 0.9 without reweighting and 0.2 ± 1.0 with reweighting.
d0_ppi0.eps The momentum spectra of pi0 candidates. The blue histogram is for D0 signal sample, red for Dspi0 signal sample, and black for reweighted D0 signal sample.
d0_kpipi0_1.eps
Kpipi0 invariant mass spectrum before reweighting.
d0_kpipi0_2.eps
Kpipi0 invariant mass spectrum after reweighting.
d0_kpi_1.eps Kpi invariant mass spectrum with fit, single gaussian for signal, par1 for area, par2 for mass, par3 for width.
d0_kpi_2.eps Kpi invariant mass spectrum with fit, two gaussians for signal. par1 TOTAL area, par2 mass, par3 width, par4 area of wide one, par5 mass, par6 width. The widths are 7.6+/-1.7, and 23.3+/-10.5. The wide one contribute 2/3. Please note that D0 from continuum events also included. So the width is wider than "expected".
Track candidates are required to pass a serie of cuts: Track must pass TNG approval (trackman $\ge$ 0); It is classified as good primary track (kincd = 0); It is consistent with primary vertex: For tracks above 250 MeV we require |dbcd| $<$0.005 and $|$z0cd-zvptx$|<$0.03, and |dbcd| $<$0.01 and $|$z0cd-zvptx$|<$0.05 for softer tracks; If dE/dX is available (iqaldi $>0$), we require that dE/dX information is consistent with the track hypothesis ($\pi/K$) within $2.5\sigma$.
We use XBAL package to reconstruct photon. Photon candidates are required to be within the ``good barrel region''($\cos\theta < 0.707$); The E9/E25 distribution must look like a photon (e925u $>$ c92501); The candidate must not be shower fragments (ibstop = 0); Finally, the angle with closest charged track that does not match with other bump must be $>20^{\circ}$.
We reconstruct $\pi^0$ in two-photon mode. The bump energy of each of the two photons is required to be $>$ 100 MeV. If both bumps belong to a multi-bump region, they must come from the same region. The invariant mass of two photon candidates is consistent with that of $\pi^0$ within -3.0$\sigma$ and +2.5$\sigma$, where momentum dependent $\sigma \approx$ 5.46 MeV. For candidates that pass above criteria we perform kinetic fit with two-photon invariant mass constraint to $\pi^0$ mass.
$\phi$ is reconstructed using oppositely charged Kaon pair. The invariant mass of Kaon pair is consistent with $\phi$ mass within 10 MeV ($~2.5\sigma$).
$D_s^+$ is reconstructed with a $\phi$ candidate and a charged pion candidate. The invariant mass is consistent with $M_{D_s}$ within 12 MeV ($~2.5\sigma$). To further reject background, we take advantage of the polarization of $\phi$ as it is spin-1 while other particles are spin-0. We require that the $\phi$ decay helicity angle be $|cos \theta_h| > 0.3$, where $\theta_h$ is the open angle between $D_s^+$ and one of the changed Kaons in $\phi$ rest frame.
We combine a $D_s$ candidate with a photon candidate to form $D_s^{*+}$. The photon energy is required to be $> 50$ MeV. The invariant mass is required to be consistent with $D_s^{*+}$ within 13 MeV ($~2.5\sigma$).
Finally we combine $D_s^+$ or $D_S^(*+)$ with a $\pi^0$ candidate. The momentum of the combination is required to be greater than 3.5 GeV/c to reject large amount of combinatorial background at lower momentum. The numbers of ``signal'' is obtained by fitting the invariant mass spectra of the combination with a gaussian function for signal and 2nd order polynomial for background.
In the report to the collabration and in previous paper, although we tried to unify selection criteria in two studies, and we did use same kinetic cuts, there are still quite significant differences: In study A we use CCFC package for photon reconstruction and study B we use XBAL package instead. In study A one of the two photons from $\pi^0$ could come from endcap calorimeter, while study B only uses good barrel part. By checking the spectra of $M(D_s\pi^0) - M(D_s)$ from study A (fig.xx) and study B (fig.yy) we noticed that there is a structure at about 0.2 GeV from study A, while the structure is absent in study B.
In the process to find the source of the structure, we note some interesting features. Please note the plots shown in this section is not with final selection criteria. It is just for sake of discussion of these features. In order to compare the two studies, we limit photon to good barrel region for study A. The spectra from two studies are shown in Figures compare1.eps and compare2.eps We first check the multiple entries per event with $M(D_s\pi^0) - M(D_s) < 0.6$. As photon is used in the reconstuction, multiple entries in signal event could create struture like this, as an experience. However, this is not the case. This was also proved by continuum MC which simulate the structure very well (ref Selina's plot). Shown in shadowed histogram are contibution from multiple entry events. In study A, 2935 entries with $M(D_s\pi^0) - M(D_s) < 0.6$ coming from 2388 events, and 968 entries are from multi-entry event. In study B, 1793 total entries comes from 1609 events with 353 entries from multi-entry event.
Figure compare3.eps shows the difference of the two spectra from the two studies as shown in Fig.~\ref{multi_spect}. Study A has more signal around 0.35 GeV, about 20-30 more, and also structure at 0.2 GeV. Apparently spectrum A has more entries and may include all entries from spectrum B. So checking events not appeared in spectrum B may shed light on the source of the structure. Actually the situation is more complicated. 1387 out of 1607 events in spectrum B are included in spectrum A. And multiple entries further complicate the study. In stead of finding out the source with this method, we check the mass difference between two studies as shown in Figure compare4.eps. Please note that the total number of entries is more that 1387 due to multiple entries in some events. The sigma of mass difference is about 3 MeV. With a cut of -20 MeV to +20 MeV we have 1351 entries. This gives an idea that the two studies are actually very close except for that structure.
We change photon reconstruction package in study A from CCFC to XBAL with selection of photon same as in study B. The spectrum shown as dot in Figure compare5.eps is from study A with this change. It agrees well with spectrum from study B. The difference is due to slightly different track selection. This proves that the two studies are consistent.
we further noticed that the EM shower shape requirement was skipped by mistake in selection of photons from $\pi^0$ decay, as there is pi0 list created with in CCFC. With this requirement implimented, we obtain the spectrum in study A as shown Fig.(zzz). The structure is gone. The two studies with CCFC or XBAL are consistent..
I used all basic selection criteria in (DsGamma)Pi0 but leave M(DsGamma) untouched. Further the single photon (not from Pi0) is required to have energy more than 150 MeV. At this stage the efficiency for two modes are the same ( ~4.2% with P>3.5). In following discussion the efficiency is calculated after all these cuts.
The difference due to lorentz boost is shown in mc_dspi0_dm.eps and mc_dsgam_dm.eps.
The first plot shows spectra of M(DsPi0)-M(Ds). In the mode 1, it contains real resonance so the width is 6.7 ± 0.2. For mode 2, it is merely reflection and the width is 13.9 ± 0.3. We can select sample consistent with ~0.350 within 10 MeV. With this cut, the efficiency of mode 1 drops to 94%, while mode 2 drops to 62%. It is too high for mode 2.
The second plot shows spectra of M(DsGam)-M(Ds). In mode 2, there are real Ds* so the width is 4.4 ± 0.1. And in mode 1, the width is 13.2±0.3 MeV. I made a very tough cut by requires that |DM-center|>10MeV, that is, most of real Ds*'s are rejected. With only this cut, the efficiency of mode 1 drops to 54% And mode 2 drops to 6.9%. It is great. But there is a catch. The M(DsPi0Gam)-M(DsPi0) spectrum has a dip in the signal region for background as shown in cut_ds2320gam_4.eps and cut_dsspi0_4.eps, where the green solid for background with signal add on top (Please note that there is photon involved here thus MC tagging is not very right which cause a background peak at the later plot). This dip will be reduced by applying M(DsPi0)-M(Ds) selection. But It will real create some prolem in counting number of events.
Maybe if I choose something else associated with DsGamma mass but not as direct
as this may help. Shown in mc_pi0_gam_2d.eps are
2D plot for P_gam and P_pi0, where the momentum is calculated in the rest frame
of (DsPi0Gam). The left for mode 1 and right for mode 2. The upper ones are with
all tagged DsPi0Gam and bottom ones for all sample. Projections of tagged signal
on Ppi0 axis are shown in mc_ppi0.eps. Similarly I can
reject a band with Ppi0 between (0.288, 0.308). Now the mass spectra are shown in
cut_ds2320gam_5.eps and
cut_dsspi0_5.eps and with this cut alone.
This time it is a hump. Again with M(DsPi0)-M(Ds) selection, the effect will be reduced.
Combination of all three selection criteria works. I adjust the cuts to be
Now we can fit the mass spectra shown in
mc_ds2320gam_mass.eps and
mc_dsspi0_mass.eps.
I use single gaussian plus 2nd order polynomial fitting from 0.05 GeV
to 0.5 GeV. For the signal MC, the width is 5.0 ± 0.2 MeV.
And for the main background signal (mode 2) the width is 23.0 ± 2.4 MeV.
To extract the number of signals in data(data_dspi0gam.eps),
we will need to fix the width to that of signal MC.
So in estimating the contribution from mode 2 we also need to fix the width.
Here are the number of signals:
Calculations here:
From a toy MC simultion, we get N_real = 3.084 ± 3.567,
and ratio upper limit = 0.52 at 90% CL.
In the simulation I assume all efficiencies and numbers have Gaussian fluctuation.
The upper limit of ratio at 90% CL is found in
ratio_limit.eps as:
Num (0 < ratio < "limit") / Num (0 < ratio) = 0.9.
The plot that can be included in paper is data_dspi0gam.eps.
I use samples in mc_right_dsspi0.eps for signal MC.
From 0 - 0.6 GeV, there are 1549 candidates, from fit we have 1000.9 signals, or with fixed
width we have 936.6 signals. I randomly pickup 68 out of 1549 candidates, including signal
and background. The number corresponds to 43.9 signal with floating width,
or 41.1 with fixed width.
I use samples in mc_wrong_dsspi0.eps for feed up MC.
From 0 - 0.6 GeV, there are 817 candidates, from fit we have 464.2 signals, or with fixed
width we have 223.8 signals. I randomly pickup 51 out of 817 candidates, corresponds to
28.9 "signal" with floating width, or 14.0 with fixed width.
Sum the two sources, there will be about 72 "signal" with floating width or 55 signal with
fixed width. There is also fluctuation on these number apparently since we are not just throwing
"signals" but also background from the two spectra.
In the data spectrum there are 605 candidates. So I randomly throw 486 events between 0.15 to 0.6 GeV
for more background.
I made 1000 independent spectra and fit with floating width and mass. Following is the is of plots.
First of all, in data we get a width of 6.0 ± 1.0 MeV, which is smaller than either the real
signal or the feedup signal. The plot demonstrates that there is sizable possibility that
the width is smaller than 6.0 MeV. We are just that "lucky" to have a narrow width.
The total number of "signal" with floating fit can be 50 or smaller although we throw about 74
"signal" in the spectrum. This is due to two reasons: the fluctuation, and the width.
We can check the second one by looking at 2D plot of number of signal
vs width (Thanks for Roy for providing better explanation) as shown in
simu_2d_nw_float.eps. With narrow width we count less number
of signals. Signals in the tail of the broad DsPi0 feed up are not counted. That is the case
we have with real data. So to estimate the feed up contribution to the data signal peak
(not the spectrum), the right thing is to fix the width to that of data.
To further prove that we count less with narrow width, I also generate the mass spectra from
the feed up, and real signal separaterly. Since I don't have exactly number of "signal", a
fit to the spectra is a good way. The plots are shown as follows.
Due to limited amount of events as shown in y_sideband_sample.eps,
I do not cut narrow slice. Instead I select samples with |M(DsPi0)-M(Ds)-M(expect)| (lets call it DM)
between 0.05 and 0.1 GeV in magenta, and shift M(DsPi0Gam)-M(DsGam) spectrum by DM. This is an
reasonable approximation to that DsPi0 from 2317 with width=0.
With these magenta sample, we obtain the spectrum of DsPi0Gamma as shown in
y_sideband_feedup1.eps.
The width is 13.0 ± 0.7 MeV. Please note that the contribution from DsPi0 width (6-7 MeV)
is removed here.
We can also simulate the "signal" plus "background". This time I include the sample in green also.
For samples with 0.05 < | DM | < 0.10, I shift M(DsPi0Gam)-M(DsGam) by DM as in last plot.
And for samples with |DM| > 0.10, I shift the mass by ± 0.1, that is, to fill the gap between
± 0.1 GeV. The final spectrum is shown in y_sideband_feedup2.eps.
And the width is 15.1 ± 1.7.
Well, it is an interesting test.
And the mass spectra of MC are shown in
cut_ds2320gam_9.eps and
cut_dsspi0_9.eps.
The total efficiency over number of generated signals with P(>3.5)
are ~2.5% and ~0.24% respectively from tagging. Please note that due to
anti Ds* cut. Most of the candidates left in mode 2 are with lousy photon
which is difficult to tag. That is the reason that the "background" peaking
at signal. The proper efficiency will be derived from fit not tagging.
For data, we reconstructed (DsGam)Pi0 number to be N4 = 41.2 ± 11.2 after crossfeed
removal. The efficiency corresponding is eff4 = (5.70 ± 0.19)%.
N_real = N3 - eff2 * (N4/eff4) = (N3*eff4 - N4*eff2) / eff4
Ratio = (N_real/eff1) / (N4/eff4) = (N3*eff4 - N4*eff2) / (N4*eff1)
Toy MC for Ds*Pi0 spectrum
If we combine MC Ds*Pi0 signal and feed-up from DsPi0 MC and random background, what will happen to the spectrum? Here is the detail on what I did.
If I fit with fixed width and center of peak, the plots are as follows.
What do the plots tell us? I think it solves the following puzzles if they are puzzles to you.
DsPi0 side band to test feedup
We know that there is feedup from Ds(2317)&to;DsPi0 signal to DsPi0Gamma peak at 2460, since the
extra gamma is soft. The width of the contribution is width due to Lorentz boost.
This was proved in MC simulation.
To satisfy the curiosity, we also test with real data. The idea is that, fake Ds(2317)'s should
behave the same in creating the peak. So a narrow slice of DsPi0 sideband sample should do the trick.
Miscellaneous
mc_momentum.eps The momentum spectra of Ds(2320) and Ds(2460).
dave.eps David Miller suggested this test.
I reject Ds*pi0 events with DsPi0 mass consistent with 2317 within 2.5 MeV.
As the width of DsPi0 from DsPi0 MC is 7.0, this will reduce the contribution from DsPi0
to Ds*Pi0 by 0.357 * 14.1 = 5.03.
As the width of DsPi0 from Ds*Pi0 MC is 14.9, this will reduce the real Ds*Pi0 by
0.133 * 41.2 = 5.48.
There is another subtraction, the combinatorial bkg in DsPi0 spectrum under the peak.
If you don't agree, think it in this way. If you add a huge peak at DsPi0 peak, will it contribute
to Ds*Pi0 peak just like a real DsPi0? of course. The effect is much small. 17 * 0.09 = 1.55
So for a perfect test, one expect the number = 55.3 - 5.03 - 5.48 -1.55 = 43.2.
The fit gives 37.8 ± 8.9. Not perfect, but acceptable I consider.