Srednicki Chapter 20

Transcription

1 Srednicki Chaper QFT Problems & Solions. George Ocober 4, Srednicki.. Verify eqaion.7. Using eqaion.7,., and he fac ha m = in his limi, or ask is o evalae his inegral:! x x x dx dx dx x sx + x + x + x x + x These are all dmmy indices, so le s swap x x in he inegrand:! I is rivial ha: x x x dx dx So we solve his inegral o find:! dx x sx + x + x + x x + x dx x + = x + x [ dx dx x x sx + x + x + x x + x ] [ x + x x + x ] Using properies of he logarihm o simplify his, we find ha:! If we le = x sx, we have:! dx x x = x dx d x = dx s + x x x sx sx +sx Le s noice ha he denominaors in boh logarihms cancel sing properies of he logarihm. Then,! x = x sx dx d + sx x =

2 This firs inegral is easy o evalae:! + s dx sx s sx x = x x = d + sx.. The second inegral needs o be evalaed on Mahemaica and simplified by hand; we find ha: d + sx = sx Polylog sx Now le s go back o x so we can apply he endpoins: d + sx = x sx sx Polylog x sx sx nd applying he endpoins: x = x x = d + sx = x sx sx Polylog x sx sx sx sx + Polylog sx sx Simplifying, and sing he definiion of he polylogarihm in he las erm, we find: x = x d + sx x sx = sx Polylog sx x sx + π sx x = which is: x = x x = d + sx Using his in eqaion..: + s = s sx + s sx Polylog s sx + π! dx Polylog + s s π sx.. Now le s redefine some hings for simpliciy. We ll ake x x, = +, and =. s s We ll also sppress he!/, hogh i will be very imporan o add i in laer. Finally, le s ignore he endpoins of he inegral for now, defining he indefinie inegral o be a fncion insead. Then, fx = dxpolylog x This can be evalaed on Mahemaica: fx = xpolylog + + x x x x + + x x +

3 + x x x x + x x Polylog + Polylog Using eqaion.., or inegral is eqal o: Firs le s evalae f:! [f f π ].. f = Polylog Polylog + Polylog I is easy o see sing properies of logs ha here are no infiniies in he hird-o-las erm, so ses he enire erm o zero: f = Polylog Polylog + Polylog Le s disribe he erms on he second line: f = Polylog Polylog + Polylog Nex, noe ha if we spli he nmeraor from he denominaor on he firs erm of he second line, here is a cancellaion. Hence: f = Polylog Polylog + Polylog Now, noe ha =. This les s simplify many erms: f = Polylog + + +

4 Polylog + Polylog This las erm vanishes: f = Polylog Polylog Using he definiion of he Polylog, we find he vale of his firs erm: f = π Polylog Now le s go back o he definiion of fx o calclae f: fx = xpolylog + + x x x x + + x x + + x x x x + x x Polylog + Polylog The firs erm vanishes as x. The easies way o verify his is o ry a few vales on Mahemaica, or check he graph. I also makes sense concepally, since he polylog is in some sense a heavy-dy logarihm, which will grow mch slower han x shrinks. Le s plg in where possible on he oher erms, leaving he erms ha appear o diverge as x for he momen: f = x + x Polylog + Polylog We know ha Polylog =. We can also combine he wo added logarihms on he firs line. This gives: f = + + x + x + Polylog The hird and firs erms can now be combined, sing he properies of logarihms: f = x x Polylog Seing he las remaining xs o, we have: + f = 4 + Polylog

5 Now we re ready o combine hese: f f = π Polylog Polylog Le s separae he denominaor in he second erm inside he braces: f f = π Polylog Polylog Combining he firs and hird erms: f f = π Polylog Polylog Combining he firs and fifh erms: f f = π Polylog Polylog Combining he firs and forh erms: f f = π Polylog =, so he firs erm becomes: f f = π + Polylog + + Polylog Polylog 5

6 The second erm has wo componens mliplied ogeher: he firs forces he fncion oward, he oher oward. Using L hôpial s rle or anoher means, we see ha his vanishes. Then, f f = π + Polylog Polylog Using his in.., he vale of he inegral is: Polylog / = /s +, so: which implies: s + s + Polylog s + s + + Polylog s s + Polylog s s s s + s + iπ + Polylog + Polylog s Now we need o se an ideniy, which holds when x <. π = iπ x + PolyLog + /x + PolyLog + x + x I can find a saemen of his ideniy anywhere. Mahemaically proving his ideniy orselves wold be very difficl, b fornaely we re no mahemaicians. We re convinced if we js look a a graph of he righ hand side: where he ble is he real par and he prple is he imaginary par. Indeed, he ideniy appears o be re for x <. Using his in or expression, wih x = s/, we obain eqaion.7: df 4 D 4 s, = π + [s/] s +

7 Noe: Srednicki s solion presens an easier way o derive his. Criosly enogh, by comparing his answer wih my answer, we are able o mahemaically prove he ideniy I saed above since or answers are eqivalen if and only if he ideniy is re. I presened his solion insead, becase I hink i is nlikely ha he rick Srednicki sed wold be obvios o a sden wiho frher help. Of corse, I have had o pay dearly for his recide my solion is mch more complicaed han Srednicki s! Noe : gain, I m nsre wha he poin of his problem was. Calcls pracice? I is sefl o pracice Feynman s Formla and he inrodcion o polylogarihms was cerainly ineresing b his seems like a very difficl problem wih very lile prpose. Perhaps some leading hins cold have a leas allowed s o spend mch less ime on his problem. Srednicki.. Compe he Oα correcion o he wo-paricle scaering amplide a hreshold, ha is, for s = 4m and = =, corresponding o zero hree-momenm for boh he incoming and ogoing paricles. ll we have o do is evalae some npleasan inegrals. Le s sar wih V : V = g gα! We can rewrie his as: V = g gα x dx dx [ x + x x x ] x dx dx [ x + x x + x + x ] This inegral is an absole mess, b afer doing a lo of simplificaion, and sing Mahemaica, we find ha: [ V = g gα dx + x + π x an x ] x + x Ping his on Mahemaica, we find: V = g + gα π Nex we ll do V 4m. Noice ha he only difference is an exra erm of 4x x m. Then, V 4m = g gα x dx dx [ x + x x + x + x ] gain, afer inegraing and doing a lo of simplificaion, we re lef wih: [ V 4m = g gα dx + x + + x 4x 8x an an 8x 8x 7

8 Ping his on Mahemaica, we find + 4x 4x + 4x + + x x + x ] V 4m = g + gα 8 π The for-poin verex fncion is given by: V 4 4m,, = g α x x x dx m dx dx 4x x + x + x x x + x + x x x + 4x x x x + x + x x x I is bes o do his on Mahemaica; he resl is: V 4 4m,, = g α m 5π 8 Noe: In an earlier version of his solion, I had a Mahemaica workbook ha replicaed Srednicki s solion hogh he answer did no seem iniive. year laer, I received an from a reader who had yped in exacly he same hing, and goen a differen b more iniive resl o he firs inegral, resling in a slighly differen answer. I ended p reprodcing his resl, hogh I see no reason for he discrepency. Maybe Mahemaica was pdaed, and he bg was fixed? In any case, please see he aached workbook, pdaed, for a derivaion of his answer noe ha i does slighly disagree wih Srednicki s solion. Thanks o gso Medeiros Washingon Universiy in S Lois for bringing his o my aenion. The self-energy a zero is given by: Using Mahemaica: which gives: Π = αm dx x + x + Π = αm = m αm [ π ] [ π ] Simplifying, and expanding he denominaor o firs order in α, we have: = m [ + α π ] 8

9 Finally, he self-energy a 4m is given by: Π 4m = α dx4x 4x + Doing he inegral and simplifying, we find ha: which gives: Expanding o firs order in α: αm π 9 4x 4x + α x x + m 4m = [ m + α ] π 4 [ 4m = + α m 4 ] π Now we re ready o combine hese sing eqaion.: g τ loop = m + g α 8m 4 π 5 + g m + g α 5 π m + g α m 5π 8 Simplifying: τ loop = [ ] g 8m 8 + α489 4 π s a fn fac, we can wrie his as: [ ] τ loop = 8g + α489 4 π 8m 8 8g 8m.59g showing ha hese one-order correcions have an absolely iny impac on he overall scaering amplide. In ligh of his, or decision o neglec all erms higher han Oα seems exremely reasonable. Noe: David Griffihs once wroe a problem prefaced by he commen for masochiss only. Srednicki wold do well o add sch a disclaimer o his problem. 9

10 In[]:= Inegrae 4 x x x x x x x x x x 4 x x x x x x x x, x,, x x O[]= CondiionalExpression x x x x x x x x x x x x x rctan xx xx x x, x x x x x x Im Im Re x x x x x x && x x x x x x x x x x Reals Re x x x x x x Re && x x x x x x x x Reals x Re x x x x x x Re x x x x x In[4]:= O[4]= Inegrae x x x x x x x x x^ x x x x rctan x x Sqrx x Sqrx x, x rctan rctan x x x x x x rctan 8 x 8 x x x x x rctan x x Log x x x x x x 4 x rctan 8 x x x 8 x In[]:= Fx_, x_ : Sqr rctansqr x x rctan x x Sqr Sqr rctan x x Sqr 8 x Sqr 8 x 4 x rctan x x Sqr 8 x Sqr 8 x Sqrx x rctan x x Sqrx x Log x x^ x x x x^

11 Sred.nb In[7]:= Fx, x Fx, rctan O[7]= rctan x x rctan x x rctan x x x rctan x x rctan x 8 x 8 x 4 x rctan 8 x x 8 x x x rctan 8 x 8 x 4 x rctan 8 x Log x x Log x x x x x x 8 x In[8]:= O[8]= In[9]:= O[9]= FllSimplify 8 8 x 5 Π 9 4 x 8 x 9 4 x rcco 8 x x x rctan x 8 4 x rctan x 8 x rctan 4 x 8 x 8 x x rctan Log x Log x x Inegrae 8 Sqr 8 x 5 Π Sqr9 4 x 8 Sqr9 4 x rcco x Sqr Sqr 8 x Sqrx rctan x Sqrx 8 4 x rctan x Sqr 8 x rctan 4 x Sqr 8 x Sqr 8 x Sqr rctan x Sqr Log x^ Log x x, x,, 5 Π 8