New version page

# Polarization Correlation in the Gamma

## This preview shows page 1-2-3 out of 9 pages.

View Full Document
Do you want full access? Go Premium and unlock all 9 pages.
Do you want full access? Go Premium and unlock all 9 pages.
Do you want full access? Go Premium and unlock all 9 pages.

Unformatted text preview:

Bin LIIntroductionMethodExperiment Set-Up and ProceduresExperimental DataPolarization Correlation in the Gamma-Gamma Decay of Positronium Bin LI Department of Physics & Astronomy, University of Pittsburgh, PA 15260, U.S.A April 25, 2001 Introduction Positronium is an unstable bound state of a positron and an electron. It is very like hydrogen atom, except that a positron replaces the proton. The ground state of positronium can be either a singlet or triplet state , where the angular momentum number L=0, and these two states are corresponding to spin zero and spin one respectively. Due to the conservation principles of linear momentum, angular momentum, parity and charge conjugation, we know the singlet decays through annihilation into a pair of gamma particles, and the triplet decays into three gamma particles. Theoretical calculation shows that the decay rate of singlet is three orders of magnitude fasters than that of triplet state. So most of the decaying gamma rays come from singlet state annihilation and they will exist as two back-back gamma particles with same magnitude but vice-visa momentums. 01S13S Now, we begin to discuss about the polarizations of the two gamma particles. In order to satisfy the conservation laws mentioned previously, both of the two back-back gamma particles must be RHC (right hand circulated) or LHC (left hand circulated). Meanwhile, we know the parity of singlet state of positronium is (-1) L+1=(-1) 0+1=-1, so the two-photon state should also be negative parity, we can write it as following: >−>=2121|| LLRRψ, where is the uncoupled states of and ; same with . >21| RR >1| R >2| R>21| LL We are supposed to set up a counter that only accepts the lights with x-polarization on one side of our source ----- , and another counter that only accepts the lights with y-polarization on the other side ----- . So we can calculation the amplitude that >1| X>2|Yψwill be in the state of since we know the RHC gamma is =>21| YX>1| R2||11>+> YiX or =>2| R2||22>+> YiX, and the LHC gamma is =|1>L2||11>−> YiX or =|2>L2||22>−> YiX.So from ><ψ|21YX=><−><21212121|| LLYXRRYX=)2(2ii−−=i, we get the possibility to obtain one photon with x-polarization on one side and another photon with y-polarization on the other side is unit (=1). Similarly, we can also get the possibility to obtain both of these two photons with x-polarizations ( or both with y-polarizations) is . Since the directions of x-axis and y-axis are arbitrary, the only constraint is that they are perpendicular, so more generally, we get that the polarizations of the two detectable correlated back-back gamma particles should be orthogonal. 2|| i0|||<221=>ψXX Method In our experiment, we use as irradiation source. As decays into , a positron will be emitted. And this positron forms positronium when it becomes bound to an electron. Most of positroniums are in singlet state, and they will decay into two back-back gamma rays. Now we use Compton Scattering to measure the relative polarizations of the two photons since the signal rate is proportional to the scattering cross section, which is strongly dependent on the polarizations of incident gamma particle and out-going gamma particle. From reference, we have the formula of deferential cross section for Compton Scattering: 22Na22Na22Ne 220222||)'()(→∗→⋅=Ωεεσkkmcedd, If we consider , and are the momentum and polarization of the incident gamma ray, and are the momentum and polarization of the scattering gamma ray. From the point of classical electrodynamics, we can estimate the differential scattering cross section as the following: (omit) 1=h0k→ε'k∗→ε In this diagram, and are the propagation directions of the incident wave and scattering wave respectively, so these two vectors build up a plane called scattering plane. and are two special choices for the possible polarization of the two outgoing wave after the scattering, where is in the plane containing and (scattering plane), is perpendicular to it. Since both cases and are perpendicular to , in term of unit vectors parallel to the coordinate axes, we have: 0→k→n1→ε2→ε1→ε0→k→n2→ε1→ε2→ε→n θφφθεsin)sincos(cos1 zyxeee→→→→−+=φφεcossin2 yxee→→→+−= If we assume that the out-going wave is unpolarized (this might be a reasonable assumption for circular polarized wave), we will get: (1) For an incident linearly polarized wave with polarization parallel to x-axis, the angular distribution summed over final polarizations is: . φφθ222sincoscos +(2) For the incident wave with polarization on the y-direction, the over all angular distribution for final polarization is: φφθ222cossincos + In order to see the apparent difference due to the variation of the polarizations of the gamma particles, we choose θ= . That is to say, the propagation direction of incident momentum and that of the scattering momentum are perpendicular. So we will put our detector for the out-going gamma particle at some place perpendicular to the incident direction with some azimuthal orientation. The following figure shows how our detecting system works. We use the combination of two detectors. °90 (1) Parallel Configuration (2)Diagonal ConfigurationEverything is the same with the previous configure for parallel case except that the relative orientations of the two detectors are not parallel but perpendicular and they are still kept orthogonal to the incident directions respectively. Now we know, for both of these two cases, θ= . Using Compton Scattering formula:°90)cos1(1110θ−=−cmkke and the energy conservation for Positronium annihilated to two back-back gamma particles with same magnitude momentum: , we can easily get 20cmcke=kk0=2. So we have 2222sin)2(φσmcedd=Ω for the incident wave with x-polarization; while 2222cos)2(φσmcedd=Ωfor the incident wave with y-polarization. If assuming that spherical angle of the surface area of detector due to the scattering center is small, we can calculate Ωddσ at an approximate constant °≈ 90θ. So we obtain the scattering cross section at that orientation is 222220sin)2(φφσπmced∫=, or 222220cos)2(φφσπmced∫=. So the signal rate (due to cross section) is totally depended on the polarizations of the incident gamma rays. (Since here these Unlocking...