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by: Malvina Orn


Malvina Orn
GPA 3.77


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Class Notes
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This 11 page Class Notes was uploaded by Malvina Orn on Monday September 7, 2015. The Class Notes belongs to PHY 396K at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 60 views. For similar materials see /class/181826/phy-396k-university-of-texas-at-austin in Physics 2 at University of Texas at Austin.




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Date Created: 09/07/15
PHY7396 K Solutions for problem set 9 Textbook Problem 42 We begin by developing Feynman rules for the theory at hand The Hamiltonian clearly decomposes into H H0 V where no nge H366 s1 and V Md3xltigtx 32x 32 In perturbation theory Feynman propagators are contractions of the free elds which follow from the free Hamiltonian 81 Thus 7 ymassM7 F 83 39 G iymass7n7 m and there are no mixed b ltIgt contractions In the momentum basis this gives us two distinct propagators 239 1 d 84 qLM2 0 an q27m2 0 The Feynman vertices follow from the perturbation Hamiltonian 82 which involves one power of the ltIgt eld and two powers of the b eld hence the vertices involve one double line and two single lines net valence 3 I 72m 85 b where the factor 2 comes from the interchangeability of the two identical 5 elds in the vertex Now consider the decay process ltIgt a b b To the lowest order of the perturbation theory we have a single diagram 5 with one vertex one incoming double line two outgoing single lines and no internal lines of either kind Thus 1 2T llt1gtgt E if X 2W464p 71937195 72m X 2W454p 71917 195 37 239e1ltIgt a 451 5 72 It remains to calculate the decay rate of the ltIgt particle in its rest frame Using eq 486 of the textbook we have P L A amp M2X2W463 6EE7M 2M 22Ei 27732E P1 pg 1 2 1 d3p 7 mWXWQW 1E2 M 88 2 2 M W dlpil d9 6 E E 7M 871727 17L 2 Now fdQ 471392 because of two identical bosons in the nal state while the rest of the phase space integral evaluates to 2 lpllgdlpll lpll 1 2m 6E1E27M7 7x 17 7 39 EiE EiE E 1E 2M 2 M Therefore 2 2 M 2m 1 7 810 87TM X M l Textbook Problem 43a Similar to the previous problem7 the propagators are contractions of the free elds7 thus for N distinct elds ltIgti of the same mass m we have Hi I 6 G iymass7n7 or in momentum space7 Wk j o q27m2 0 312 The vertices follow from the perturbation operator V d3xltlt1gtlt1gt2 EZMW Zgltigti2ltigtk27 313 j jltk hence two vertex types 1 a vertex involving 4 lines of the same eld species lt137 with the vertex factor 72 gtlt 4 76M and 2 a vertex involving 2 lines of one eld species ltIgtj and 2 lines of a different species Pk7 with the vertex factor 72 gtlt 22 722 The combinatoric factors arise from the interchanges of the identical elds in the same vertex7 thus 4 for the rst vertex type and 22 for the second type Equivalently7 we may use a single vertex type involving 4 elds of whatever species7 with the species dependent vertex factor ltIgti ltIgtZ 72 A6jk6m WW WW 314 ltIgtk ltIgtm Now consider the scattering process ltIgtj ltIgtk a ltIgtZ lt13 At the lowest order of the perturbation theory7 there is just one Feynman diagram for this process it has one vertex7 4 external legs and no internal lines Consequently at the lowest order Mltigti ltigtk a lt1gtZ W 72A6jk6m WNW WW 315 independent of the particles7 momenta Speci cally Mltigt1 ltIgt2 a ltIgt1 lt19 72A Mltigt1 ltigt1 ltIgt2 lt19 72A 816 Mltigt1 ltigt1 ltIgt1 ltIgt1 76A and consequently using eq 485 of the textbook dalt1gt1 ltIgt2 a ltIgt1 lt19 7 A dalt1gt1 ltigt1 ltIgt2 lt19 7 A S 17 dam 167T2Eg m 7 39 dalt1gt1 ltigt1 ltIgt1 ltIgt1 7 9A These are partial cross sections To calculate the total cross sections we integrate over d9 which gives the factor of 47139 when the two nal particles are of distinct species but for the same species we only get 27139 because of Bose statistics Hence A2 UtotltIgt1ltIgt2 a lt1gt1lt1gt2 W 211 ltIgt1 ltIgt1 ltIgt2 ltIgt2 A72 S18 Utotlt 87TE2 211 1 1 1 1 9A2 Ut0t ltIgt q 87TE2 211 Textbook Problem 43b The linear sigma model was discussed earlier in class The classical potential Vlt1gt2 nger 319 319 with a negative mass term m2 in lt 0 has a minimum or rather a spherical shell of minima for 2 i 2 i 2 ltigtlt1gtlt1gtu 7 A gt0 s20 Semi classically7 we expect a non zero vacuum expectation value of the scalar elds7 lt1gt 31 0 with lt132 1127 or equivalently7 lt1gtj MWN modulo the ON symmetry of the problem Shifting the elds according to ltIgtNltzgt v altzgt7 ltIgtjltzgt M j ltNgt7 s21 and re writing the Lagrangian in terms of the shifted elds7 we obtain 602 7 p202 612 7 AUUUQ 2 7 A02122 const 822 where 1 stands for the N 7 17plet of the 7Tj elds7 thus 12 2AM The free part of the Lagrangian 822 the rst 3 terms describe one massive real scalar eld U of mass ma n and N 7 1 massless real scalars 7Tjx which are the Goldstone particles of the ON symmetry spontaneously broken down to ON 7 1 thus N 7 1 broken symmetry generators7 forming a vector multiplet of the unbroken ON 7 1 symmetry Consequently7 the non zero contractions of the free 039 and 7139 elds are l I HQ 09 GFW ymassmo 7 S 23 7T Wky 6jk 7 ymass07 which give us two distinct Feynman propagators in the momentum basis7 239 039 039 q2 7 2M2 20 7 824 26376 7139 Wk q2 20 The last two terms in the Lagrangian 822 give rise to the interaction Hamiltonian of the linear sigma model7 namely V d3xltwf73 mag 3amp4 gm 209 325 The ve terms in this interaction Hamiltonian give rise to ve types of Feynman vertices Proceeding exactly as in part a of the problem7 we obtain 7Tj HZ 72 A5ik5m WNW WW 826 Wk Wm and similarly 7Tj 039 039 039 gtlt immik and 762 327 k U U 0 7139 The remaining two vertices have valence 3 and follow from the cubic terms in the inter action Hamiltonian S25 The analysis proceeds exactly as in the previous problem and yields 039 71439 Ult 7229116376 and 039 76211 828 7Tk This completes the Feynman rules of the linear sigma model Textbook Problem 43c In this part of the problem7 we use the Feynman rules we have just derived to calculate the tree level 7m 7 7m scattering amplitudes As explained in class7 a tree diagram L 0 with E 4 external legs has either one valence 4 vertex and hence no propagators or two valence 3 vertices and hence one propagator Altogether7 there are four such diagrarns contributing to the tree level 2M Wj p1 Wkp2 7 ap1 Wmp392 7 they are shown in the textbook The diagrams evaluate to 7T4101 W223 X 72 A6jk6m6 6km6jm6kl7 77k102 WmP392 7T101 W39ZCD39D 7 Z39U39k 7iyzm gtlt 72A639101p22 2226 39 WWW WmP392 Wm W3 53929 7 7 2 2 j 7 2 2 km I 72A639p7pl22ug26 7 Wklt102gt WmWQ 7T 101 W24 7 7 2 2 39m 7 2 2 M Elt 72A639p7p 22ug267 102 WmP392 which gives the net scattering amplitude 2A1 J k l m 7 k in MW 101 W 102 w W 101 W 102 2A5 5 1 p1p22 2M2 2A1 E mallow 1 330 19171932 7 2M2 2 E mama 1 1917192 72M Now7 according to eq 820 v2 27 which makes for 2 2 1 2M 2 p1 52 2 831 101102 E 2M 101102 E 2M and ditto for the other two terms in the amplitude 830 Altogether7 we now have 1022 39l k 101E1902 M swam gtlt m 6 6 m gtlt 191 p22 7 2M2 1917 pa2 7 2M2 jm kl X 10171952 7 1917 1922 7 2M2 332 which vanishes in the zero momentum limit for any one of the four pions7 initial or nal lndeed7 since the pions are massless7 pl2 pg2 go12 go22 0 and hence df 8 e 191p22 E ID1P2 2p1p2 7 2plp27 df t 2 ID171912 E ID271922 72plp1 7 72p2p27 533 df u 2 ID171922 E ID271932 72p1p 7 72plp27 this whenever any one of the four momenta becomes small7 all three numerators in the amplitude 832 become small as well7 thus M Osmall p Please note that although eq 832 gives only the tree level approximation to the ac tual scattering amplitude7 its behavior in the small pion momentum limit is correct and completely general According to the Goldstone theorem7 not only the Goldstone particles such as pions7 in this linear sigma model are exactly massless7 but also any scattering amplitude involving any Goldstone particle vanishes as 01 when the Goldstone particle s momentum pr goes to zero To complete this part of the problem7 let us now assume that all four pion7s momenta are small compared to the U particle7s mass ma n In this limit7 all three denominators in eq 832 are dominated by the 72112 term7 hence M jkWpi 122 kwpi 7 pa Wka 712 0 834 For generic species of the four pions7 this amplitude is of the order 0102197 but there is a cancellation when all for pions belong to the same species this is unavoidable for N 2 lndeed7 191 p22 p17 pl2 p17 1922 2p1p2 2p1p1 2p1p 2 835 2p1p2 7191 7192 7191 72191 0 and hence 1 1 1 1 1 104 M7r 7r a7r7r 00 mfg 836 QED Finally7 let us translate the amplitude 834 into the low energy scattering cross sections da7r1 7T2 a 7T1 7139 i Egm X Sin4 9cm dQC m i 647TQU4 7 1 2 1 2 E2 Ut0t7f 7T aw 7T 482147 dU7T1 7T1 a 7T2 7139 Egm m 7 837 dQC m 647139 1 2 Ut0t7f1 711 7T2 7139 322147 1 1 1 1 7 om 07T 7T aw 7T i Oltv4mg Textbook Problem 43d The linear term A1 7altlgtN in the classical potential for the N scalar elds ewplz39cz39tly breaks the 0N symmetry of the theory Consequently7 as we saw in homework set 4 problem 17 the potential VltIgt p319 7 amp 7 altigtltNgt 338 now has a non degenerate minimum at 2 ltlgtjgt MVN where 1 m E 1 0 ag S39 2M M W Shifting the elds according to eq 821 for the new value of 1 now gives us 602 7 m302 aag 7 m3rE27 AUUUQ 2 7 A02122 840 plus an irrelevant constant where m3 2M2 3mr and mi E gt 0 841 1 Thus7 the pions are no longer exactly massless Goldstone bosons but rather pseud07Goldst0ne bosons with small but non zero masses Comparing the Lagrangians 840 and 822 we immediately see identical interaction terms7 hence the Feynman vertices of the modi ed sigma model are exactly as in eqs 8267 827 and 8287 without any modi cation except for the new value of 1 On the other hand7 the Feynman propagators need adjustment to accommodate the new masses 8417 thus 239 U 7 q27m3 07 342 j k7 Wk 7T 7T 7 q27m3r 0 The tree level 7139 7r 7 7139 7139 scattering amplitude is governed by the same four Feynman 10 diagrams as before7 thus 2A1 M j k I m 72A6Jk lm 7T 101 W 102 w W 101 W 102 p1p22 7mg 2M gm gkm 1 843 lt 19171932 77713 2M 7 mama 1 7 191 7 pg2 7 m3 exactly as in eq 8307 except for the new 1 and new The exact equation for the minimum 839 is M2 7 M Z W 344 hence W 7 m3 7 345 and 2 2 2 1 1012 7 m3 E2 232 846 and ditto for the other two terms in the amplitude 843 Therefore7 instead of eq S32 we now have 2 2 2 2 7 m 7 7 m M 72Alt63k65m X 101 192 r km X 101 1012 r 191 192 7 mg 1917 pl 7 ma 2 2 347 jm ki 101 7 102 m7r Xltp 7220277712 7 2 a which in the low energy limit Ec m ltlt ma simpli es to 2A 1 M 72 m 72 mm 221 122 7 m3 mm 121 723 7 mi ma 1 4 848 6J7n6kp17p22 7 m0 ln particular7 near the threshold 10115022 Egm 4mr while 101 71012 m p2 71012 m 0 11


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