At the tree level of QED there are two diagrams related by interchanging of the two photon s in the 64257nal state 1 The net amplitude due to these diagrams is 57559M ie ie ie ie 2 where and Note the opposite orders of the and vertices in the ID: 35782 Download Pdf

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At the tree level of QED there are two diagrams related by interchanging of the two photon s in the 64257nal state 1 The net amplitude due to these diagrams is 57559M ie ie ie ie 2 where and Note the opposite orders of the and vertices in the

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ANNIHILATION In these notes I explain the annihilation process. At the tree level of QED, there are two diagrams related by interchanging of the two photon s in the ﬁnal state: (1) The net amplitude due to these diagrams is , , M = )( ie ie = )( ie ie (2) where and . Note the opposite orders of the and vertices in the and the amplitudes. We may re-write these amplitudes without matrix denominators using and (3) Consequently, v u, v u. (4)

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Ward Identity Before we go any further, lets check the Ward identities for the an nihilation amplitude: for

the ﬁrst photon we should have = 0, and for the second photon = 0. Let’s start with the ﬁrst photon and the ﬁrst diagram. Multiplying th e second factor in the ﬁrst eq. (4) by , we have v = v −6 = v because = 0 = v 2( − 6 = 2( v because ( = 0 = ( v (5) and consequently = + v u. (6) Note the non-zero right hand side — the ﬁrst diagram does not satis fy the Ward identity all by itself. As for the second diagram, we have v = −6 = −6 because = 0 = 2( ) + ( 2( v because ) = 0 v (7) and consequently v u. (8) Againwehaveanon-zeroresult

—theseconddiagramalsodoesnots atisfytheWardidentity all by itself. However, the right hand sides of eqs. (6) and (8) canc el each other, so together

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the two diagrams do satisfy the Ward identity: = 0 (9) This is an example of a general rule: The Ward Identity does not work diagram by diagram but only for entire amplitudes, or for partial sums of all diagrams re lated by permutations of photonic vertices on the same fermionic line. The Ward identity = 0 for the second photon works similarly to the ﬁrst, and I see no point in repeating the argument. Indeed, it would be an

exac tly similar argument because the net annihilation amplitude is symmetric with respect to th e two photons. Summing over the Spins and Polarizations Earlier in class I explained how to use Ward identities to sum |M| over polarizations of the two photons: , |M| = + (10) Combining the two diagrams, we have , |M| = + + 2Re (11) Note that this formula works despite the fact that and do not satisfy the Ward identities by themselves — it’s enough that the sum satisﬁes the identities. Thus, in light of eqs. (4), , |M| v u v u )( Re v u (12) This takes care of the photon polarizations. The

next step is to ave rage over spins of

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the initial electron and positron. Proceeding is usual, we have |M| ,s , |M| 11 22 )( Re 12 (13) where 11 Tr 22 Tr 12 Tr (14) Traceology 1 Our next task is to evaluate the traces (14). Let’s start with the 11 Back in homework set #5, you saw that = 4 and p . Applying these formulae to the expression inside the trace in 11 , we have 2( , 2( +2 (15) and consequently 11 = Tr +2 )( )( )( (16) Next, we expand the parentheses inside this trace and throw away terms with odd numbers of momenta or . This gives us 11 = Tr( ) + Tr( Tr( + 2 Tr( Tr(

Tr(1) = 8( 4( + 4 16 + 16 16 16 = 8( )( 4( 16 ) + ( ) + (17) We may further simplify this formula by expressing all the momenta pr oducts in terms

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of the Mandelstam’s variables , and . Using and = 0, we have = ( t, qp = ( ) = + qp = ( (18) Consequently, on the last line of eq. (17), the last term vanishes ) + ( ) + ) + = 0 (19) — while the remaining terms add up to 11 = 8( )( 4( 2( 2( tm + 2 + 2 tu tm um = 2 tu tm um = 2( )( (20) This completes our evaluation of the ﬁrst trace. Now consider the second trace 22 . Instead of working through the calculation, we may use the

photon exchange / crossing symmetry between the two dia grams (1). This symmetry exchanges and also 11 22 , thus 22 = 2( )( (21) Traceology 2 Now we need to evaluate the third trace 12 which accounts for the interference between the two diagrams (1). This trace is more complicated, so let’s start b y simplifying the

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part. Back in homework #5, we had a a, b = 4( ab , c a, (22) which now gives us + 4 q. (23) Plugging this formula into eq. (14) for the 12 and remembering that we need an even number of slash-momentum or factors inside the trace, we obtain 12 Tr = Tr h

− 6 q q i Tr − 6 q Tr q Tr q q q = 4 Tr 4( + 2 = 4 ) + ( qp ( qp ) + ( 8( )( ) + 4 ( qp ) + 8 (24) To simplify this rather messy formula, we need to work out the kinema tics. Besides eqs. (18), we have qp = ( ) = + qp = ( qq = ( )( ) = ) + ) = (25)

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Consequently, 12 = 4 ) + ) + 8 = 4 (2 +2 +2 (26) Annihilation Summary Having worked out the traces, let’s plug them into eq. (13): |M| 2( )( 2( )( )( = 2 = 2 (27) or more compactly |M| = 2 + 1 1 + (28) This is our ﬁnal result; the rest is kinematics.

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Annihilation Kinematics In the center of mass

frame, = ( E, ) where = + , and = ( ω, where . Consequently, = 4 + 2 cos θ, cos θ, −| cos cos (29) and therefore + 1 = cos −| cos −| cos cos + 1 cos cos (3+cos sin −| cos cos cos sin 1 + sin sin (30) Thus |M| = 2 (3+cos sin sin sin (31) and ﬁnally the partial cross section of annihilation d |M| 64 (3+cos sin sin sin (32)

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For the non-relativistic electron and positron with | , the expression in the square brackets becomes 3 1) = 2, hence isotropic partial cross section d (slow (33) And the total cross section in this limit is tot

(slow ) = (34) where total solid angle is 4 π/ 2 because of 2 identical photons in the ﬁnal state. In the opposite limit of ultra-relativistic and with | , we have 3+cos sin 1 = 2(1+cos sin (35) and hence highly un-isotropic cross section d (fast 1+cos sin (36) Note how this cross-section is strongly peaked in the forward direc tion = 0 where one photon continues the electron’s motion while the other continues th e positron’s motion. According to eq. (36), the total annihilation cross-section tot (fast ) = 2 π/ d sin d cm (37) divergesatsmallangles,

butthat’sanartefactoftheapproximat ion(35)becominginaccurate

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at small angles where sin . Instead, for small angles we have (1) (38) and consequently d (fast (39) This cross-section is strongly peaked in the forward direction, but it does not diverge. In- stead, tot (fast ) = log (40) Compton Scattering Compton scattering of an electron and a photon is related by crossing symmetry to the annihilation. Indeed, at the tree level there are two diagrams −0 −0 (41) which are obviously related by crossing to the annihilation diagrams (1). Hence, given eq. (28) for the

annihilation, we may immediately write down a simila r formula for the Compton scattering without doing any work. All we need is to exchan ge in eq. (28) and change the overall sign because we cross one fermion, thus |M Compton = 2 1 + 1 + (42) This is it; all we need to do now is kinematics. 10

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Compton scattering is usually studied in the lab frame where the initial electron is at rest, = ( m, ). In this frame, the initial and the ﬁnal photon energies and are related to photon’s scattering angle via Compton’s formula cos (43) originally written by Arthur Compton in

terms of the photon’s wavelen gths as (1 cos (44) According to this formula, there is an upper limit on the energy of the ﬁnal photon for any ﬁxed = 0: regardless of the initial energy , the ﬁnal energy can never exceed (1 cos ). The Compton’s formula follows from the energy–momentum conserv ation and (45) which imply = ( cos (46) while = ( + 2 ωm (47) Subtracting these two formulae and canceling similar terms gives us ωm = 2 + 2 (1 cos ) (48) and hence eq. (43). 11

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The Mandelstam variables and in the lab frame are = ( = 2 ωm = ( (49) and

hence = +2 mω, u m (50) Plugging these values into eq. (42), we have 1 = + 1 + = 1 + cos (51) where the last equality follows from eq. (43), and therefore |M Compton = 2 1 + cos (52) Finally, we need the phase space factor for the lab frame. For a gen eric 2 2 scattering process, d |M| where (2 (2 (2 (4) 64 64 (53) Specializing to the Compton scattering and the lab frame for initial ele ctron, we immediately obtain 64 ωm 1 + dE d (54) The only non-trivial issue here is the derivative in the parentheses. This derivative should be taken for a ﬁxed photon angle and before

applying the energy conservation rule 12

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. Instead, we use the momentum conservation , hence eq. (46) for the and consequently cos (55) For ﬁxed and dE = 2 | = 2( cos d (56) and hence dE d cos (57) Once we have taken this derivative, me may now use energy conserv ation, thus 1 + dE d cos cos ωm (58) where the last equality follows from the Compton formula (43). Plugg ing the derivative (58) into eq. (54), we arrive at 64 (59) and hence the Klein–Nishina formula for the partial cross-section: d Compton lab sin (60) where is given by eq. (43). For low photon

energies , the Compton’s formula gives , and the Klein Nishina cross-section (60) becomes the good old Thompson cross- section d Compton lab d Thompson lab (2 sin = 1+cos (61) 13

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and the total cross-section is Thompson total 663barn (62) On the other hand, for very high photon energies and 6 0, we have sin (63) and the Klein–Nishina formula becomes d Compton lab cos (64) This approximation is not accurate at small angles / for which 6 , so the cross section does not really diverge for 0. Instead, at small angles we have large but ﬁnite partial cross-section d Compton

lab (2 / )+2(2 / +(2 / )) 6 (65) and hence ﬁnite total cross-section Compton total log (66) 14

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